By the end of this chapter, you will be able to:
Mastering these skills will help you analyze and solve complex engineering challenges with confidence and precision.
Apply calculus to solve mechanical production problems involving functions commonly used in engineering design and analysis. Understanding and manipulating linear, trigonometric, logarithmic, and exponential functions are crucial for tasks such as modeling machine parts, analyzing forces, and predicting system behavior. This chapter focuses on these functions, presenting their key formulas and applying calculus techniques to solve practical mechanical production problems in Kenya.
Functions describe relationships between variables, fundamental in mechanical production for predicting outputs or system responses. Calculus operations such as differentiation and integration applied to these functions enable optimization, rate of change analysis, and cumulative effect calculations critical for design and manufacturing.
Linear functions represent relationships with constant rates of change, common in mechanical systems such as shaft displacement, linear velocity, or force versus extension in springs. The general linear function is
$$ f(x) = mx + c $$
where \(m\) is the slope (rate of change) and \(c\) is the intercept.
Example 1: A piston moves linearly such that its displacement \(x\) in mm at time \(t\) seconds is \(x = 5t + 10\). Find the velocity at \(t = 3\) seconds.
Given: \(x = 5t + 10\), find velocity \(v = \frac{dx}{dt}\) at \(t = 3\) s.
$$ v = \frac{dx}{dt} = \frac{d}{dt}(5t + 10) $$$$ v = 5 $$ Velocity is constant at 5 mm/s.
Answer: 5 mm/s
Example 2: The force \(F\) applied on a machine part varies with displacement \(x\) as \(F = 3x + 20\) N. Find the rate of change of force with respect to displacement.
Given: \(F = 3x + 20\)
$$ \frac{dF}{dx} = \frac{d}{dx}(3x + 20) $$$$ \frac{dF}{dx} = 3 $$
Answer: 3 N/mm
Example 3: A conveyor belt moves with velocity \(v(t) = 4t + 2\) m/s, where \(t\) is time in seconds. Find the total displacement after 5 seconds.
Given: \(v(t) = 4t + 2\), find displacement \(s\) from \(t=0\) to \(t=5\).
$$ s = \int_0^5 v(t) dt = \int_0^5 (4t + 2) dt $$$$ s = [ 2t^2 + 2t ]_0^5 = (2 \times 25) + (2 \times 5) - 0 = 50 + 10 = 60 $$
Answer: 60 m
Trigonometric functions model oscillations, periodic motions, and component forces in mechanical systems such as cams, gears, and linkages. The primary functions are sine, cosine, and tangent, with derivatives:
$$ \frac{d}{dx} (\sin x) = \cos x, \quad \frac{d}{dx} (\cos x) = -\sin x, \quad \frac{d}{dx} (\tan x) = \sec^2 x $$
where \(x\) is in radians.
Example 1: The displacement of a vibrating machine part is \(x(t) = 0.05 \sin(10t)\) meters. Find the velocity at \(t = 0.1\) seconds.
Given: \(x = 0.05 \sin(10t)\), find \(v = \frac{dx}{dt}\) at \(t=0.1\) s.
$$ v = \frac{dx}{dt} = 0.05 \times 10 \cos(10t) = 0.5 \cos(10t) $$$$ v = 0.5 \cos(1) \approx 0.5 \times 0.5403 = 0.27015 $$
Answer: 0.270 m/s
Example 2: The force on a lever varies as \(F(\theta) = 100 \cos \theta\) N. Find the rate of change of force with respect to angle \(\theta\) at \(\theta = \frac{\pi}{3}\) radians.
Given: \(F(\theta) = 100 \cos \theta\)
$$ \frac{dF}{d\theta} = -100 \sin \theta $$$$ \frac{dF}{d\theta} = -100 \sin \frac{\pi}{3} = -100 \times \frac{\sqrt{3}}{2} = -86.6 $$
Answer: -86.6 N/radian
Example 3: The torque on a shaft varies as \(T(t) = 50 \sin(5t) + 20\) Nm. Calculate the angular acceleration \(\alpha = \frac{d^2 \theta}{dt^2}\) if \(\theta(t) = \int T/I \, dt\), where moment of inertia \(I = 10 \, kg \cdot m^2\). Find \(\alpha\) at \(t = 0.2\) s.
Given: \(T(t) = 50 \sin(5t) + 20\), \(I=10\), \(\alpha = \frac{T}{I}\)
$$ \alpha = \frac{T}{I} = \frac{50 \sin(5t) + 20}{10} = 5 \sin(5t) + 2 $$$$ \alpha = 5 \sin(1) + 2 = 5 \times 0.8415 + 2 = 4.2075 + 2 = 6.2075 $$
Answer: 6.21 rad/s\(^2\)
Logarithmic functions appear in mechanical production when modeling phenomena such as damping, material stress relaxation, or sound intensity. The derivative of the natural logarithm function is:
$$ \frac{d}{dx} (\ln x) = \frac{1}{x} $$
for \(x > 0\).
Example 1: The stress relaxation in a metal component is modeled as \(S(t) = 50 \ln(t + 1)\) MPa. Find the rate of change of stress at \(t = 4\) seconds.
Given: \(S(t) = 50 \ln(t + 1)\)
$$ \frac{dS}{dt} = 50 \times \frac{1}{t + 1} $$$$ \frac{dS}{dt} = \frac{50}{5} = 10 $$
Answer: 10 MPa/s
Example 2: The intensity of sound in a workshop is \(I(d) = 30 \ln d\) decibels, where \(d\) is distance in meters. Find the rate of change of intensity with respect to distance at \(d = 10\) m.
Given: \(I(d) = 30 \ln d\)
$$ \frac{dI}{dd} = 30 \times \frac{1}{d} = \frac{30}{d} $$$$ \frac{dI}{dd} = \frac{30}{10} = 3 $$
Answer: 3 decibels/m
Example 3: The time \(t\) in seconds to cool a machine part is related to temperature \(T\) by \(t = 5 \ln(100 - T)\). Find the rate of change of time with respect to temperature when \(T=60^\circ C\).
Given: \(t = 5 \ln(100 - T)\)
$$ \frac{dt}{dT} = 5 \times \frac{1}{100 - T} \times (-1) = \frac{-5}{100 - T} $$$$ \frac{dt}{dT} = \frac{-5}{40} = -0.125 $$
Answer: -0.125 s/°C
Exponential functions model growth or decay processes such as heat transfer, radioactive decay in material testing, or population of bacteria in quality control. The derivative of the exponential function is:
$$ \frac{d}{dx} (e^{kx}) = k e^{kx} $$
where \(k\) is a constant.
Example 1: The temperature \(T\) of a heated metal piece varies as \(T(t) = 100 e^{-0.1 t}\) °C. Find the cooling rate at \(t=10\) seconds.
Given: \(T(t) = 100 e^{-0.1 t}\)
$$ \frac{dT}{dt} = 100 \times (-0.1) e^{-0.1 t} = -10 e^{-0.1 t} $$$$ \frac{dT}{dt} = -10 e^{-1} = -10 \times 0.3679 = -3.679 $$
Answer: -3.68 °C/s
Example 2: The number of defective items \(N\) in a batch decreases exponentially as \(N(t) = 500 e^{-0.05 t}\). Find the rate of decrease of defective items at \(t=20\) hours.
Given: \(N(t) = 500 e^{-0.05 t}\)
$$ \frac{dN}{dt} = 500 \times (-0.05) e^{-0.05 t} = -25 e^{-0.05 t} $$$$ \frac{dN}{dt} = -25 e^{-1} = -25 \times 0.3679 = -9.1975 $$
Answer: -9.20 items/hour
Example 3: The pressure \(P\) in a hydraulic system rises as \(P(t) = 200 (1 - e^{-0.2 t})\) kPa. Find the rate of pressure increase at \(t=5\) seconds.
Given: \(P(t) = 200 (1 - e^{-0.2 t})\)
$$ \frac{dP}{dt} = 200 \times 0.2 e^{-0.2 t} = 40 e^{-0.2 t} $$$$ \frac{dP}{dt} = 40 e^{-1} = 40 \times 0.3679 = 14.716 $$
Answer: 14.72 kPa/s
Linear calculus is fundamental in mechanical production, especially when dealing with rates of change such as velocity, acceleration, and force variations in moving parts. It applies to problems where relationships between variables are directly proportional or vary at a constant rate. The primary tool in linear calculus is differentiation of linear functions, which helps in predicting system behaviours and optimizing machine performance.
The derivative of a linear function represents its constant rate of change. For a linear function \( f(x) = mx + c \), where \( m \) and \( c \) are constants, the derivative is:
$$ \frac{df}{dx} = m $$
Example 1: A conveyor belt moves at a constant speed described by \( s(t) = 3t + 5 \) metres, where \( t \) is time in seconds. Find the velocity.
Given: \( s(t) = 3t + 5 \), \( t \) in seconds
$$ \frac{ds}{dt} = 3 $$
Answer: The velocity is 3 m/s
Example 2: The linear displacement of a piston in a machine is given by \( x(t) = -4t + 12 \). Find the rate of change of displacement with respect to time.
Given: \( x(t) = -4t + 12 \), \( t \) in seconds
$$ \frac{dx}{dt} = -4 $$
Answer: The piston moves at -4 m/s (indicating direction)
Example 3: A machine's temperature increases linearly as \( T(t) = 2.5t + 30 \) °C, where \( t \) is in minutes. Determine the rate of temperature increase.
Given: \( T(t) = 2.5t + 30 \), \( t \) in minutes
$$ \frac{dT}{dt} = 2.5 $$
Answer: The temperature increases at 2.5 °C per minute
Integration of linear functions calculates the total accumulation over time, such as total distance traveled or total energy consumed. For a linear function \( f(x) = mx + c \), the integral is:
$$ \int (mx + c) \, dx = \frac{m}{2} x^{2} + cx + C $$
where \( C \) is the constant of integration.
Example 1: The velocity of a moving part is given by \( v(t) = 4t + 2 \) m/s. Find the displacement after 5 seconds, assuming initial displacement is zero.
Given: \( v(t) = 4t + 2 \), \( t = 0 \) to 5 s, initial displacement \( s(0) = 0 \)
$$ s(t) = \int v(t) \, dt = \int (4t + 2) \, dt = 2t^{2} + 2t + C $$
Using \( s(0) = 0 \):
$$ 0 = 2 \times 0^{2} + 2 \times 0 + C \Rightarrow C = 0 $$
Calculate displacement at \( t = 5 \):
$$ s(5) = 2 \times 5^{2} + 2 \times 5 = 2 \times 25 + 10 = 50 + 10 = 60 $$
Answer: The displacement after 5 seconds is 60 metres
Example 2: The force applied on a machine part varies as \( F(x) = 3x + 5 \) N, where \( x \) is displacement in metres. Find the work done moving the part from 0 to 4 metres.
Given: \( F(x) = 3x + 5 \), \( x = 0 \) to 4 m
$$ W = \int_{0}^{4} F(x) \, dx = \int_{0}^{4} (3x + 5) \, dx = [\frac{3}{2} x^{2} + 5x ]_{0}^{4} $$
Calculate:
$$ = (\frac{3}{2} \times 4^{2} + 5 \times 4) - (\frac{3}{2} \times 0^{2} + 5 \times 0) = (\frac{3}{2} \times 16 + 20) - 0 = (24 + 20) = 44 $$
Answer: The work done is 44 Joules
Example 3: A machine’s power output increases linearly with time as \( P(t) = 5t + 10 \) watts. Calculate the total energy produced between 2 and 6 seconds.
Given: \( P(t) = 5t + 10 \), \( t = 2 \) to 6 s
$$ E = \int_{2}^{6} P(t) \, dt = \int_{2}^{6} (5t + 10) \, dt = [\frac{5}{2} t^{2} + 10t ]_{2}^{6} $$
Calculate:
At \( t=6 \):
$$ \frac{5}{2} \times 6^{2} + 10 \times 6 = \frac{5}{2} \times 36 + 60 = 90 + 60 = 150 $$
At \( t=2 \):
$$ \frac{5}{2} \times 2^{2} + 10 \times 2 = \frac{5}{2} \times 4 + 20 = 10 + 20 = 30 $$
Energy produced:
$$ 150 - 30 = 120 $$
Answer: The total energy produced is 120 Joules
Mechanical production often involves calculating rates such as speed, flow, or production rate, where linear calculus simplifies analysis. Differentiation provides instantaneous rates, while integration gives cumulative quantities over time or distance.
Example 1: A lathe machine tool moves linearly such that its position is \( x(t) = 6t + 4 \) mm. Find the velocity at \( t = 3 \) seconds.
Given: \( x(t) = 6t + 4 \), \( t = 3 \) s
$$ v(t) = \frac{dx}{dt} = 6 $$
Velocity at \( t=3 \):
$$ v(3) = 6 \, \text{mm/s} $$
Answer: The velocity is 6 mm/s
Example 2: The flow rate of lubricant in a pipe is \( Q(t) = 10t + 20 \) litres per minute. Find the total volume of lubricant passed in 8 minutes.
Given: \( Q(t) = 10t + 20 \), \( t = 0 \) to 8 min
$$ V = \int_{0}^{8} Q(t) \, dt = \int_{0}^{8} (10t + 20) \, dt = [5t^{2} + 20t ]_{0}^{8} $$
Calculate:
$$ 5 \times 8^{2} + 20 \times 8 = 5 \times 64 + 160 = 320 + 160 = 480 $$
Answer: Total 480 litres of lubricant passed
Example 3: The production rate of a machine is \( R(t) = 50 + 15t \) units per hour. Calculate the total units produced between hours 1 and 4.
Given: \( R(t) = 50 + 15t \), \( t = 1 \) to 4 hours
$$ U = \int_{1}^{4} (50 + 15t) \, dt = [50t + \frac{15}{2} t^{2} ]_{1}^{4} $$
Calculate:
At \( t=4 \):
$$ 50 \times 4 + \frac{15}{2} \times 4^{2} = 200 + \frac{15}{2} \times 16 = 200 + 120 = 320 $$
At \( t=1 \):
$$ 50 \times 1 + \frac{15}{2} \times 1^{2} = 50 + 7.5 = 57.5 $$
Total units:
$$ 320 - 57.5 = 262.5 $$
Answer: Total 262.5 units produced
Higher-order derivatives describe changes in rates, such as acceleration (second derivative of displacement). For linear functions, the second and higher derivatives are zero, reflecting constant rate of change.
$$ \frac{d^{2}}{dx^{2}} (mx + c) = 0 $$
Example 1: Given \( s(t) = 7t + 3 \), find the acceleration.
Given: \( s(t) = 7t + 3 \)
First derivative (velocity):
$$ v(t) = \frac{ds}{dt} = 7 $$
Second derivative (acceleration):
$$ a(t) = \frac{d^{2}s}{dt^{2}} = 0 $$
Answer: Acceleration is 0 m/s²
Example 2: A machine tool displacement is \( x(t) = -5t + 20 \). Find the acceleration and jerk (third derivative).
Given: \( x(t) = -5t + 20 \)
Velocity:
$$ v(t) = \frac{dx}{dt} = -5 $$
Acceleration:
$$ a(t) = \frac{d^{2}x}{dt^{2}} = 0 $$
Jerk:
$$ j(t) = \frac{d^{3}x}{dt^{3}} = 0 $$
Answer: Acceleration and jerk are zero
Example 3: For \( y(t) = 12t + 8 \), find the third derivative.
Given: \( y(t) = 12t + 8 \)
First derivative:
$$ \frac{dy}{dt} = 12 $$
Second derivative:
$$ \frac{d^{2}y}{dt^{2}} = 0 $$
Third derivative:
$$ \frac{d^{3}y}{dt^{3}} = 0 $$
Answer: The third derivative is zero
Trigonometric calculus is essential in mechanical production for analyzing oscillations, rotations, and vibrations of machine parts. Differentiation and integration of trigonometric functions allow calculation of angular velocities, accelerations, and energy stored in spring systems. These functions model periodic phenomena common in manufacturing equipment.
The derivatives of sine, cosine, and tangent functions describe how angular quantities change over time or position.
$$ \frac{d}{dx} \sin x = \cos x $$
$$ \frac{d}{dx} \cos x = -\sin x $$
$$ \frac{d}{dx} \tan x = \sec^{2} x $$
where \( x \) is in radians.
Example 1: The angular displacement of a rotating shaft is \( \theta(t) = \sin t \) radians. Find the angular velocity at \( t = \frac{\pi}{4} \) seconds.
Given: \( \theta(t) = \sin t \), \( t = \frac{\pi}{4} \)
$$ \omega(t) = \frac{d\theta}{dt} = \cos t $$
Calculate:
$$ \omega(\frac{\pi}{4}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \approx 0.707 $$
Answer: Angular velocity is approximately 0.707 rad/s
Example 2: The angular position is \( \theta(t) = \cos 2t \). Find the angular velocity at \( t = \frac{\pi}{6} \).
Given: \( \theta(t) = \cos 2t \), \( t = \frac{\pi}{6} \)
$$ \omega(t) = \frac{d\theta}{dt} = -2 \sin 2t $$
Calculate:
$$ \omega(\frac{\pi}{6}) = -2 \sin (2 \times \frac{\pi}{6}) = -2 \sin \frac{\pi}{3} = -2 \times \frac{\sqrt{3}}{2} = -\sqrt{3} \approx -1.732 $$
Answer: Angular velocity is approximately -1.732 rad/s
Example 3: The angular displacement is \( \theta(t) = \tan 3t \). Find the angular velocity at \( t = \frac{\pi}{12} \).
Given: \( \theta(t) = \tan 3t \), \( t = \frac{\pi}{12} \)
$$ \omega(t) = \frac{d\theta}{dt} = 3 \sec^{2} 3t $$
Calculate:
$$ \omega(\frac{\pi}{12}) = 3 \sec^{2} (3 \times \frac{\pi}{12}) = 3 \sec^{2} \frac{\pi}{4} $$
Recall \( \sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \)
So,
$$ \sec^{2} \frac{\pi}{4} = (\sqrt{2})^{2} = 2 $$
$$ \omega(\frac{\pi}{12}) = 3 \times 2 = 6 $$
Answer: Angular velocity is 6 rad/s
Integration of trigonometric functions calculates quantities such as angular displacement from angular velocity or total mechanical work over oscillations.
$$ \int \sin x \, dx = -\cos x + C $$
$$ \int \cos x \, dx = \sin x + C $$
$$ \int \sec^{2} x \, dx = \tan x + C $$
Example 1: The angular velocity is \( \omega(t) = \cos t \) rad/s. Find the angular displacement from \( t = 0 \) to \( t = \frac{\pi}{2} \).
Given: \( \omega(t) = \cos t \), \( t = 0 \) to \( \frac{\pi}{2} \)
$$ \theta = \int_{0}^{\frac{\pi}{2}} \cos t \, dt = [\sin t]_{0}^{\frac{\pi}{2}} = \sin \frac{\pi}{2} - \sin 0 = 1 - 0 = 1 $$
Answer: Angular displacement is 1 radian
Example 2: The angular velocity is \( \omega(t) = \sin 2t \). Find the angular displacement between 0 and \( \pi \) seconds.
Given: \( \omega(t) = \sin 2t \), \( t = 0 \) to \( \pi \)
$$ \theta = \int_{0}^{\pi} \sin 2t \, dt = [-\frac{1}{2} \cos 2t ]_{0}^{\pi} $$
Calculate:
$$ = -\frac{1}{2} \cos 2\pi + \frac{1}{2} \cos 0 = -\frac{1}{2} (1) + \frac{1}{2} (1) = 0 $$
Answer: Angular displacement is 0 radians
Example 3: Given angular acceleration \( \alpha(t) = \sec^{2} t \), find the change in angular velocity from \( t=0 \) to \( t=\frac{\pi}{4} \).
Given: \( \alpha(t) = \sec^{2} t \), \( t=0 \) to \( \frac{\pi}{4} \)
$$ \Delta \omega = \int_{0}^{\frac{\pi}{4}} \sec^{2} t \, dt = [\tan t]_{0}^{\frac{\pi}{4}} = \tan \frac{\pi}{4} - \tan 0 = 1 - 0 = 1 $$
Answer: Change in angular velocity is 1 rad/s
Oscillatory motion such as vibrations in machine components is modelled using sine and cosine functions. Differentiation provides velocity and acceleration; integration gives displacement and energy.
Example 1: The displacement of a vibrating spring is \( x(t) = 0.1 \sin 20t \) metres. Find the velocity at \( t = 0.1 \) seconds.
Given: \( x(t) = 0.1 \sin 20t \), \( t=0.1 \) s
$$ v(t) = \frac{dx}{dt} = 0.1 \times 20 \cos 20t = 2 \cos 20t $$
Calculate:
$$ v(0.1) = 2 \cos (20 \times 0.1) = 2 \cos 2 = 2 \times (-0.4161) = -0.8322 $$
Answer: Velocity is approximately -0.832 m/s
Example 2: The angular displacement of a camshaft is \( \theta(t) = 0.05 \cos 30t \) radians. Find the angular acceleration at \( t = 0.05 \) seconds.
Given: \( \theta(t) = 0.05 \cos 30t \), \( t=0.05 \) s
First derivative (angular velocity):
$$ \omega(t) = \frac{d\theta}{dt} = -0.05 \times 30 \sin 30t = -1.5 \sin 30t $$
Second derivative (angular acceleration):
$$ \alpha(t) = \frac{d^{2}\theta}{dt^{2}} = -1.5 \times 30 \cos 30t = -45 \cos 30t $$
Calculate:
$$ \alpha(0.05) = -45 \cos (30 \times 0.05) = -45 \cos 1.5 = -45 \times 0.0707 = -3.1815 $$
Answer: Angular acceleration is approximately -3.18 rad/s²
Example 3: The displacement of a machine part oscillates as \( x(t) = 0.2 \sin 10t \). Calculate the total distance covered in the first 0.3 seconds.
Given: \( x(t) = 0.2 \sin 10t \), \( t=0 \) to 0.3 s
Calculate displacement at \( t=0 \):
$$ x(0) = 0.2 \sin 0 = 0 $$
At \( t=0.3 \):
$$ x(0.3) = 0.2 \sin (10 \times 0.3) = 0.2 \sin 3 = 0.2 \times 0.1411 = 0.0282 $$
Since sine function oscillates, total distance travelled is the integral of the absolute velocity over time:
Velocity:
$$ v(t) = \frac{dx}{dt} = 0.2 \times 10 \cos 10t = 2 \cos 10t $$
Total distance:
$$ D = \int_{0}^{0.3} |v(t)| \, dt $$
This requires numerical methods; approximate by calculating displacement extremes or use absolute integral. For simplicity, consider displacement difference:
Answer: Approximate net displacement is 0.0282 m
Logarithmic functions are essential in mechanical production for solving problems involving exponential growth or decay, such as tool wear rates, cooling processes, and material fatigue analysis. Understanding how to differentiate and integrate logarithmic functions helps technicians analyze data trends and optimize manufacturing parameters.
The derivative of the natural logarithm function \( \ln x \) is fundamental when dealing with rates of change in processes that vary logarithmically.
The basic differentiation formula is:
$$ \frac{d}{dx} (\ln x) = \frac{1}{x} $$
For a logarithm with a function inside, the chain rule applies:
$$ \frac{d}{dx} [\ln f(x)] = \frac{f'(x)}{f(x)} $$
Example 1: Find the derivative of \( y = \ln (5x) \) where \( x > 0 \).
Given: \( y = \ln (5x) \)
$$ \frac{dy}{dx} = \frac{d}{dx} [\ln (5x)] $$
$$ = \frac{1}{5x} \times \frac{d}{dx} (5x) $$
$$ = \frac{1}{5x} \times 5 $$
$$ = \frac{5}{5x} = \frac{1}{x} $$
Answer: \(\frac{dy}{dx} = \frac{1}{x}\)
Example 2: Differentiate \( y = \ln (3x^2 + 2) \).
Given: \( y = \ln (3x^2 + 2) \)
$$ \frac{dy}{dx} = \frac{1}{3x^2 + 2} \times \frac{d}{dx} (3x^2 + 2) $$
$$ = \frac{1}{3x^2 + 2} \times 6x $$
$$ = \frac{6x}{3x^2 + 2} $$
Answer: \(\frac{dy}{dx} = \frac{6x}{3x^2 + 2}\)
Example 3: Find the rate of change of \( y = \ln ( \frac{x+1}{x-1} ) \).
Given: \( y = \ln ( \frac{x+1}{x-1} ) \)
Using the chain rule:
$$ \frac{dy}{dx} = \frac{1}{\frac{x+1}{x-1}} \times \frac{d}{dx} ( \frac{x+1}{x-1} ) $$
First, differentiate \( u = \frac{x+1}{x-1} \):
$$ \frac{du}{dx} = \frac{(1)(x-1) - (x+1)(1)}{(x-1)^2} = \frac{x - 1 - x - 1}{(x-1)^2} = \frac{-2}{(x-1)^2} $$
Then:
$$ \frac{dy}{dx} = \frac{1}{\frac{x+1}{x-1}} \times \frac{-2}{(x-1)^2} = \frac{x-1}{x+1} \times \frac{-2}{(x-1)^2} = \frac{-2}{(x+1)(x-1)} $$
Simplify denominator:
$$ (x+1)(x-1) = x^2 - 1 $$
Answer: \(\frac{dy}{dx} = \frac{-2}{x^2 - 1}\)
Example 4: Differentiate \( y = \ln(x) \cdot \ln(2x) \).
Given: \( y = \ln(x) \times \ln(2x) \)
Use product rule:
$$ \frac{dy}{dx} = \ln(x) \times \frac{d}{dx} [\ln(2x)] + \ln(2x) \times \frac{d}{dx} [\ln(x)] $$
Calculate derivatives:
$$ \frac{d}{dx} [\ln(2x)] = \frac{1}{2x} \times 2 = \frac{1}{x} $$
$$ \frac{d}{dx} [\ln(x)] = \frac{1}{x} $$
Therefore:
$$ \frac{dy}{dx} = \ln(x) \times \frac{1}{x} + \ln(2x) \times \frac{1}{x} = \frac{\ln(x) + \ln(2x)}{x} $$
Use logarithm property:
$$ \ln(x) + \ln(2x) = \ln (x \times 2x) = \ln (2x^2) $$
Answer: \(\frac{dy}{dx} = \frac{\ln(2x^2)}{x}\)
Example 5: Find the derivative of \( y = \ln ( \sqrt{x^2 + 4} ) \).
Given: \( y = \ln ( (x^2 + 4)^{1/2} ) \)
Rewrite:
$$ y = \frac{1}{2} \ln (x^2 + 4) $$
Then:
$$ \frac{dy}{dx} = \frac{1}{2} \times \frac{1}{x^2 + 4} \times 2x = \frac{x}{x^2 + 4} $$
Answer: \(\frac{dy}{dx} = \frac{x}{x^2 + 4}\)
Integrals involving logarithmic functions often appear in engineering problems such as calculating work done by variable forces or analyzing logarithmic decay in materials.
The key integral formula is:
$$ \int \frac{1}{x} dx = \ln |x| + C $$
More generally:
$$ \int \ln x \, dx = x \ln x, x + C $$
Example 1: Calculate \( \int \frac{1}{x} dx \).
$$ \int \frac{1}{x} dx = \ln |x| + C $$
Answer: \(\ln |x| + C\)
Example 2: Find \( \int \ln (3x) dx \).
Use integration by parts:
Let \( u = \ln(3x) \), \( dv = dx \)
Then, \( du = \frac{1}{3x} \times 3 dx = \frac{1}{x} dx \), \( v = x \)
Apply integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$
Therefore:
$$ \int \ln(3x) dx = x \ln(3x) - \int x \times \frac{1}{x} dx = x \ln(3x) - \int 1 dx = x \ln(3x) - x + C $$
Answer: \( x \ln(3x) - x + C \)
Example 3: Evaluate \( \int x \ln x \, dx \).
Use integration by parts:
Let \( u = \ln x \), \( dv = x dx \)
Then:
\( du = \frac{1}{x} dx \), \( v = \frac{x^2}{2} \)
Apply formula:
$$ \int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \times \frac{1}{x} dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x dx $$
Calculate integral:
$$ \int x dx = \frac{x^2}{2} $$
Substitute:
$$ \frac{x^2}{2} \ln x - \frac{1}{2} \times \frac{x^2}{2} + C = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C $$
Answer: \( \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \)
Example 4: Compute \( \int \frac{\ln x}{x} dx \).
Use substitution:
Let \( t = \ln x \), so \( dt = \frac{1}{x} dx \)
Rewrite integral:
$$ \int t \, dt = \frac{t^2}{2} + C = \frac{(\ln x)^2}{2} + C $$
Answer: \( \frac{(\ln x)^2}{2} + C \)
Example 5: Evaluate \( \int \ln (x^2 + 1) dx \).
Use integration by parts:
Let \( u = \ln (x^2 + 1) \), \( dv = dx \)
Then:
\( du = \frac{2x}{x^2 + 1} dx \), \( v = x \)
Apply formula:
$$ \int \ln (x^2 + 1) dx = x \ln (x^2 + 1) - \int x \times \frac{2x}{x^2 + 1} dx = x \ln (x^2 + 1) - 2 \int \frac{x^2}{x^2 + 1} dx $$
Rewrite integrand:
$$ \frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1} $$
Therefore:
$$ \int \ln (x^2 + 1) dx = x \ln (x^2 + 1) - 2 \int (1 - \frac{1}{x^2 + 1}) dx = x \ln (x^2 + 1) - 2 ( \int 1 dx - \int \frac{1}{x^2 + 1} dx ) $$
Calculate integrals:
$$ \int 1 dx = x $$
$$ \int \frac{1}{x^2 + 1} dx = \tan^{-1} x $$
Substitute:
$$ x \ln (x^2 + 1) - 2(x - \tan^{-1} x) + C = x \ln (x^2 + 1) - 2x + 2 \tan^{-1} x + C $$
Answer: \( x \ln (x^2 + 1) - 2x + 2 \tan^{-1} x + C \)
In mechanical production, tool wear often follows a logarithmic decay model, where the wear rate changes as a logarithmic function of time. Understanding the derivative of logarithmic functions helps predict tool life and maintenance scheduling.
Consider wear \( W(t) = k \ln(t + 1) \), where \( t \) is time in hours, \( k \) is a constant.
The rate of wear is:
$$ \frac{dW}{dt} = \frac{k}{t + 1} $$
Example 1: A cutting tool has wear \( W(t) = 4 \ln (t + 1) \) mm after \( t \) hours. Find the wear rate at \( t = 3 \) hours.
Given: \( W(t) = 4 \ln (t + 1) \), \( t = 3 \) h
$$ \frac{dW}{dt} = \frac{4}{t + 1} $$
Substitute:
$$ \frac{dW}{dt} = \frac{4}{3 + 1} = \frac{4}{4} = 1 \, \text{mm/h} $$
Answer: Wear rate at 3 hours is 1 mm/h
Example 2: If the tool wear after 5 hours is 6 mm, find the constant \( k \).
Given: \( W(5) = 6 \), \( W(t) = k \ln (t + 1) \)
Calculate:
$$ 6 = k \ln (5 + 1) = k \ln 6 $$
$$ k = \frac{6}{\ln 6} $$
Calculate \( \ln 6 \approx 1.7918 \):
$$ k = \frac{6}{1.7918} = 3.35 $$
Answer: \( k \approx 3.35 \)
Example 3: Using \( k = 3.35 \), find the wear rate at \( t = 10 \) hours.
Given: \( \frac{dW}{dt} = \frac{k}{t + 1} \), \( k = 3.35 \), \( t = 10 \)
Calculate:
$$ \frac{dW}{dt} = \frac{3.35}{10 + 1} = \frac{3.35}{11} = 0.3045 \, \text{mm/h} $$
Answer: Wear rate at 10 hours is approximately 0.305 mm/h
Example 4: Find the total wear after 12 hours.
Given: \( W(t) = 3.35 \ln (t + 1) \), \( t = 12 \)
Calculate:
$$ W(12) = 3.35 \ln (12 + 1) = 3.35 \ln 13 $$
Calculate \( \ln 13 \approx 2.5649 \):
$$ W(12) = 3.35 \times 2.5649 = 8.59 \, \text{mm} $$
Answer: Total wear after 12 hours is 8.59 mm
Example 5: Determine the time \( t \) when the wear rate drops to 0.2 mm/h.
Given: \( \frac{dW}{dt} = \frac{3.35}{t + 1} = 0.2 \)
Solve for \( t \):
$$ t + 1 = \frac{3.35}{0.2} = 16.75 $$
$$ t = 16.75 - 1 = 15.75 \, \text{hours} $$
Answer: Wear rate is 0.2 mm/h at 15.75 hours
Logarithmic differentiation simplifies finding derivatives of complex functions involving products, quotients, or powers common in mechanical engineering formulas, such as stress-strain relationships or efficiency calculations.
The general rule:
If \( y = f(x)^{g(x)} \), then
$$ \ln y = g(x) \ln f(x) $$
Differentiate implicitly:
$$ \frac{1}{y} \frac{dy}{dx} = g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)} $$
Therefore:
$$ \frac{dy}{dx} = y [ g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)} ] $$
Example 1: Differentiate \( y = x^{x} \).
Given: \( y = x^{x} \)
Take natural log:
$$ \ln y = x \ln x $$
Differentiate both sides:
$$ \frac{1}{y} \frac{dy}{dx} = \ln x + x \times \frac{1}{x} = \ln x + 1 $$
Multiply both sides by \( y \):
$$ \frac{dy}{dx} = x^{x} (\ln x + 1) $$
Answer: \(\frac{dy}{dx} = x^{x} (\ln x + 1)\)
Example 2: Find derivative of \( y = \frac{(x^2 + 1)^3}{(x - 1)^2} \).
Given: \( y = \frac{(x^2 + 1)^3}{(x - 1)^2} \)
Take natural log:
$$ \ln y = 3 \ln (x^2 + 1) - 2 \ln (x - 1) $$
Differentiate:
$$ \frac{1}{y} \frac{dy}{dx} = 3 \times \frac{2x}{x^2 + 1} - 2 \times \frac{1}{x - 1} = \frac{6x}{x^2 + 1} - \frac{2}{x - 1} $$
Multiply both sides by \( y \):
$$ \frac{dy}{dx} = \frac{(x^2 + 1)^3}{(x - 1)^2} ( \frac{6x}{x^2 + 1} - \frac{2}{x - 1} ) $$
Simplify:
$$ \frac{dy}{dx} = (x^2 + 1)^2 (x - 1)^{-2} ( 6x - \frac{2(x^2 + 1)}{x - 1} ) $$
Answer:
$$ \frac{dy}{dx} = \frac{(x^2 + 1)^3}{(x - 1)^2} ( \frac{6x}{x^2 + 1} - \frac{2}{x - 1} ) $$
Example 3: Differentiate \( y = (3x + 1)^{x^2} \).
Given: \( y = (3x + 1)^{x^2} \)
Take natural log:
$$ \ln y = x^2 \ln (3x + 1) $$
Differentiate:
$$ \frac{1}{y} \frac{dy}{dx} = 2x \ln (3x + 1) + x^2 \times \frac{3}{3x + 1} $$
Multiply both sides by \( y \):
$$ \frac{dy}{dx} = (3x + 1)^{x^2} [ 2x \ln (3x + 1) + \frac{3x^2}{3x + 1} ] $$
Answer:
$$ \frac{dy}{dx} = (3x + 1)^{x^2} ( 2x \ln (3x + 1) + \frac{3x^2}{3x + 1} ) $$
Example 4: Find derivative of \( y = ( \frac{x^2 + 1}{x - 2} )^x \).
Given: \( y = ( \frac{x^2 + 1}{x - 2} )^x \)
Take natural log:
$$ \ln y = x [ \ln (x^2 + 1) - \ln (x - 2) ] $$
Differentiate:
$$ \frac{1}{y} \frac{dy}{dx} = [ \ln (x^2 + 1) - \ln (x - 2) ] + x ( \frac{2x}{x^2 + 1} - \frac{1}{x - 2} ) $$
Multiply both sides by \( y \):
$$ \frac{dy}{dx} = ( \frac{x^2 + 1}{x - 2} )^x [ \ln ( \frac{x^2 + 1}{x - 2} ) + x ( \frac{2x}{x^2 + 1} - \frac{1}{x - 2} ) ] $$
Answer:
$$ \frac{dy}{dx} = ( \frac{x^2 + 1}{x - 2} )^x [ \ln ( \frac{x^2 + 1}{x - 2} ) + x ( \frac{2x}{x^2 + 1} - \frac{1}{x - 2} ) ] $$
Example 5: Differentiate \( y = x^{\sin x} \).
Given: \( y = x^{\sin x} \)
Take natural log:
$$ \ln y = \sin x \ln x $$
Differentiate:
$$ \frac{1}{y} \frac{dy}{dx} = \cos x \ln x + \sin x \times \frac{1}{x} $$
Multiply both sides by \( y \):
$$ \frac{dy}{dx} = x^{\sin x} ( \cos x \ln x + \frac{\sin x}{x} ) $$
Answer:
$$ \frac{dy}{dx} = x^{\sin x} ( \cos x \ln x + \frac{\sin x}{x} ) $$
Differentiate \( y = \ln (4x^3 + 1) \) with respect to \( x \). (5 marks)
Evaluate \( \int \ln (2x) \, dx \). (5 marks)
A machine's vibration amplitude decreases as \( A(t) = 6 \ln (t + 1) \) mm over time \( t \) hours. Find the rate of amplitude change at \( t = 4 \) hours. (5 marks)
Differentiate \( y = (x^2 + 3)^{x} \) using logarithmic differentiation. (7 marks)
Calculate \( \int x \ln x \, dx \). (5 marks)
Exponential functions model many mechanical production phenomena such as cooling of heated parts, radioactive decay in materials, and growth of production rates. Mastery of differentiation and integration of exponential functions enables technicians to analyze dynamic processes effectively.
The exponential function \( e^{x} \) has the unique property that its derivative is itself.
The basic formula is:
$$ \frac{d}{dx} e^{x} = e^{x} $$
For a general function \( f(x) \):
$$ \frac{d}{dx} e^{f(x)} = f'(x) e^{f(x)} $$
Example 1: Differentiate \( y = e^{3x} \).
Given: \( y = e^{3x} \)
$$ \frac{dy}{dx} = 3 e^{3x} $$
Answer: \(\frac{dy}{dx} = 3 e^{3x}\)
Example 2: Find the derivative of \( y = e^{x^2 + 1} \).
Given: \( y = e^{x^2 + 1} \)
$$ \frac{dy}{dx} = 2x e^{x^2 + 1} $$
Answer: \(\frac{dy}{dx} = 2x e^{x^2 + 1}\)
Example 3: Differentiate \( y = e^{\sin x} \).
Given: \( y = e^{\sin x} \)
$$ \frac{dy}{dx} = \cos x \times e^{\sin x} $$
Answer: \(\frac{dy}{dx} = \cos x \, e^{\sin x}\)
Example 4: Find \( \frac{d}{dx} ( e^{3x^2 - 2x} ) \).
Given: \( y = e^{3x^2 - 2x} \)
$$ \frac{dy}{dx} = (6x - 2) e^{3x^2 - 2x} $$
Answer: \(\frac{dy}{dx} = (6x - 2) e^{3x^2 - 2x}\)
Example 5: Differentiate \( y = x e^{x} \).
Given: \( y = x e^{x} \)
Use product rule:
$$ \frac{dy}{dx} = e^{x} + x e^{x} = e^{x} (1 + x) $$
Answer: \(\frac{dy}{dx} = e^{x} (1 + x)\)
Integrals of exponential functions appear in calculating total heat transfer, cumulative production output, and decay processes.
The basic integral formula:
$$ \int e^{x} dx = e^{x} + C $$
More generally:
$$ \int e^{f(x)} f'(x) dx = e^{f(x)} + C $$
Example 1: Evaluate \( \int e^{5x} dx \).
Given: \( \int e^{5x} dx \)
Let \( u = 5x \), then \( du = 5 dx \), so \( dx = \frac{du}{5} \).
Rewrite integral:
$$ \int e^{5x} dx = \int e^{u} \frac{du}{5} = \frac{1}{5} \int e^{u} du = \frac{1}{5} e^{u} + C = \frac{1}{5} e^{5x} + C $$
Answer: \( \frac{1}{5} e^{5x} + C \)
Example 2: Find \( \int x e^{x^2} dx \).
Use substitution:
Let \( u = x^2 \), then \( du = 2x dx \), so \( x dx = \frac{du}{2} \).
Rewrite:
$$ \int x e^{x^2} dx = \int e^{u} \frac{du}{2} = \frac{1}{2} \int e^{u} du = \frac{1}{2} e^{u} + C = \frac{1}{2} e^{x^2} + C $$
Answer: \( \frac{1}{2} e^{x^2} + C \)
Example 3: Evaluate \( \int e^{3x^2 + 2x} (6x + 2) dx \).
Let \( u = 3x^2 + 2x \), then \( du = (6x + 2) dx \).
Rewrite:
$$ \int e^{3x^2 + 2x} (6x + 2) dx = \int e^{u} du = e^{u} + C = e^{3x^2 + 2x} + C $$
Answer: \( e^{3x^2 + 2x} + C \)
Example 4: Calculate \( \int e^{-x} dx \).
Given:
$$ \int e^{-x} dx = - e^{-x} + C $$
Answer: \( - e^{-x} + C \)
Example 5: Find \( \int x^2 e^{x^3} dx \).
Let \( u = x^3 \), then \( du = 3x^2 dx \), so \( x^2 dx = \frac{du}{3} \).
Rewrite:
$$ \int x^2 e^{x^3} dx = \int e^{u} \frac{du}{3} = \frac{1}{3} e^{u} + C = \frac{1}{3} e^{x^3} + C $$
Answer: \( \frac{1}{3} e^{x^3} + C \)
The temperature \( T(t) \) of a machine part cooling in air follows:
$$ T(t) = T_a + (T_0 - T_a) e^{-kt} $$
where \( T_a \) is ambient temperature, \( T_0 \) initial temperature, \( k \) cooling constant, and \( t \) time.
The rate of cooling is the derivative:
$$ \frac{dT}{dt} = -k (T_0 - T_a) e^{-kt} $$
Example 1: A part initially at 150°C cools to ambient 25°C with \( k = 0.1 \, \text{min}^{-1} \). Find temperature after 10 minutes.
Given: \( T_a = 25^\circ C \), \( T_0 = 150^\circ C \), \( k = 0.1 \), \( t = 10 \)
Calculate:
$$ T(10) = 25 + (150 - 25) e^{-0.1 \times 10} = 25 + 125 e^{-1} $$
Calculate \( e^{-1} \approx 0.3679 \):
$$ T(10) = 25 + 125 \times 0.3679 = 25 + 45.99 = 70.99^\circ C $$
Answer: Temperature after 10 minutes is approximately 71°C
Example 2: Find the rate of cooling at \( t = 10 \) minutes.
Calculate:
$$ \frac{dT}{dt} = -0.1 \times 125 \times e^{-0.1 \times 10} = -12.5 \times 0.3679 = -4.598^\circ C/\text{min} $$
Answer: Cooling rate at 10 minutes is approximately -4.6°C/min
Example 3: Determine the time when temperature reaches 50°C.
Given \( T(t) = 50 \):
$$ 50 = 25 + 125 e^{-0.1 t} $$
Subtract 25:
$$ 25 = 125 e^{-0.1 t} $$
Divide:
$$ \frac{25}{125} = e^{-0.1 t} \Rightarrow 0.2 = e^{-0.1 t} $$
Take natural log:
$$ \ln 0.2 = -0.1 t $$
Calculate \( \ln 0.2 \approx -1.6094 \):
$$ -1.6094 = -0.1 t $$
Solve:
$$ t = \frac{1.6094}{0.1} = 16.094 \, \text{minutes} $$
Answer: Temperature reaches 50°C after approximately 16.1 minutes
Example 4: If the rate of cooling at \( t = 5 \) minutes is \(-8^\circ C/\text{min}\), find \( k \).
Given:
$$ \frac{dT}{dt} = -k (T_0 - T_a) e^{-k t} = -8 $$
Substitute:
$$ -k \times 125 \times e^{-5k} = -8 $$
Simplify:
$$ k \times 125 \times e^{-5k} = 8 $$
Solve for \( k \) numerically. Try \( k = 0.15 \):
$$ 0.15 \times 125 \times e^{-0.75} = 18.75 \times 0.4724 = 8.85 > 8 $$
Try \( k = 0.14 \):
$$ 0.14 \times 125 \times e^{-0.7} = 17.5 \times 0.4966 = 8.69 > 8 $$
Try \( k = 0.12 \):
$$ 0.12 \times 125 \times e^{-0.6} = 15 \times 0.5488 = 8.23 > 8 $$
Try \( k = 0.11 \):
$$ 0.11 \times 125 \times e^{-0.55} = 13.75 \times 0.5769 = 7.93 < 8 $$
By interpolation, \( k \approx 0.115 \).
Answer: Cooling constant \( k \approx 0.115 \, \text{min}^{-1} \)
Example 5: Calculate the temperature drop in the first 3 minutes.
Calculate:
$$ T(0) = 150^\circ C $$
$$ T(3) = 25 + 125 e^{-0.1 \times 3} = 25 + 125 e^{-0.3} $$
Calculate \( e^{-0.3} \approx 0.7408 \):
$$ T(3) = 25 + 125 \times 0.7408 = 25 + 92.6 = 117.6^\circ C $$
Temperature drop:
$$ 150 - 117.6 = 32.4^\circ C $$
Answer: Temperature drops 32.4°C in first 3 minutes
Production output \( P(t) \) may increase exponentially due to automation and efficiency improvements:
$$ P(t) = P_0 e^{rt} $$
where \( P_0 \) is initial output, \( r \) growth rate, and \( t \) time.
The growth rate of output is:
$$ \frac{dP}{dt} = r P_0 e^{rt} = r P(t) $$
Example 1: Initial output is 100 units/hour, growing at 5% per hour. Find output after 8 hours.
Given: \( P_0 = 100 \), \( r = 0.05 \), \( t = 8 \)
Calculate:
$$ P(8) = 100 e^{0.05 \times 8} = 100 e^{0.4} $$
Calculate \( e^{0.4} \approx 1.4918 \):
$$ P(8) = 100 \times 1.4918 = 149.18 \, \text{units/hr} $$
Answer: Output after 8 hours is approximately 149 units/hr
Example 2: Find the rate of increase in output after 8 hours.
Calculate:
$$ \frac{dP}{dt} = 0.05 \times 149.18 = 7.459 \, \text{units/hr}^2 $$
Answer: Production is increasing at approximately 7.46 units/hr² at 8 hours
Example 3: Determine time when output doubles.
Solve:
$$ 2 P_0 = P_0 e^{rt} \Rightarrow 2 = e^{rt} $$
Take natural log:
$$ \ln 2 = rt $$
$$ t = \frac{\ln 2}{r} = \frac{0.6931}{0.05} = 13.862 \, \text{hours} $$
Answer: Output doubles after approximately 13.86 hours
Example 4: If output after 5 hours is 130 units/hr, find growth rate \( r \).
Given:
$$ 130 = 100 e^{5r} $$
Divide:
$$ 1.3 = e^{5r} $$
Take natural log:
$$ \ln 1.3 = 5r $$
Calculate \( \ln 1.3 \approx 0.2624 \):
$$ r = \frac{0.2624}{5} = 0.0525 $$
Answer: Growth rate \( r \approx 0.0525 \, \text{hr}^{-1} \)
Example 5: Calculate total output produced over 8 hours.
Total output is integral of \( P(t) \):
$$ \int_0^8 P_0 e^{rt} dt = \frac{P_0}{r} (e^{rt} - 1) $$
Calculate:
$$ \frac{100}{0.05} (e^{0.4} - 1) = 2000 (1.4918 - 1) = 2000 \times 0.4918 = 983.6 \, \text{units} $$
Answer: Total output over 8 hours is approximately 984 units
Differentiate \( y = e^{4x^2 - 3x} \). (5 marks)
Evaluate \( \int e^{2x} dx \). (5 marks)
A machine's temperature cools according to \( T(t) = 30 + 120 e^{-0.2 t} \). Find temperature at \( t = 5 \) minutes. (5 marks)
Production output grows as \( P(t) = 50 e^{0.1 t} \). Find the rate of output increase at \( t = 10 \) hours. (5 marks)
Calculate \( \int x e^{x^3} dx \). (7 marks)
The application of calculus in mechanical production requires mastery of the fundamental rules of differentiation. These rules enable technicians to determine rates of change, optimize machine functions, and analyze mechanical systems. In Kenya’s manufacturing and production industries, applying these rules accurately improves design efficiency and operational safety.
The Power Rule is essential for differentiating functions where the variable is raised to a constant exponent. This rule simplifies finding the rate of change of mechanical parameters such as stress, strain, or velocity expressed as power functions.
The Power Rule formula is:
$$\frac{d}{dx} ( x^n ) = n x^{n-1}$$
Example 1: A piston’s displacement \( s \) varies with time \( t \) as \( s = t^3 \). Find the velocity \( v = \frac{ds}{dt} \) at any time \( t \).
Given: \( s = t^3 \)
$$\frac{ds}{dt} = \frac{d}{dt} ( t^3 )$$
$$= 3 t^{3-1}$$
$$= 3 t^2$$
Answer: \(\frac{ds}{dt} = 3 t^2\) (units: m/s if \(s\) in meters and \(t\) in seconds)
Example 2: The volume \( V \) of a spherical tank varies with radius \( r \) as \( V = r^3 \). Find \(\frac{dV}{dr}\).
Given: \( V = r^3 \)
$$\frac{dV}{dr} = \frac{d}{dr} ( r^3 )$$
$$= 3 r^{2}$$
Answer: \(\frac{dV}{dr} = 3 r^{2}\) (units: m² if \(V\) in m³ and \(r\) in m)
Example 3: The kinetic energy \( K \) of a rotating shaft is \( K = \omega^4 \), where \(\omega\) is angular velocity. Find \(\frac{dK}{d\omega}\).
Given: \( K = \omega^4 \)
$$\frac{dK}{d\omega} = \frac{d}{d\omega} ( \omega^4 )$$
$$= 4 \omega^{3}$$
Answer: \(\frac{dK}{d\omega} = 4 \omega^{3}\) (units: J/(rad/s) if \(K\) in joules)
Example 4: The power output \( P \) of a machine is \( P = 5 t^5 \), where \( t \) is time. Find \(\frac{dP}{dt}\).
Given: \( P = 5 t^5 \)
$$\frac{dP}{dt} = 5 \frac{d}{dt} ( t^5 )$$
$$= 5 \times 5 t^{4}$$
$$= 25 t^{4}$$
Answer: \(\frac{dP}{dt} = 25 t^{4}\) (units: Watts/s if \(P\) in Watts and \(t\) in seconds)
Example 5: A stress function is \( \sigma = 2 x^{3.5} \), where \( x \) is displacement. Find \(\frac{d\sigma}{dx}\).
Given: \( \sigma = 2 x^{3.5} \)
$$\frac{d\sigma}{dx} = 2 \frac{d}{dx} ( x^{3.5} )$$
$$= 2 \times 3.5 x^{2.5}$$
$$= 7 x^{2.5}$$
Answer: \(\frac{d\sigma}{dx} = 7 x^{2.5}\) (units: N/m² per m if \(\sigma\) in N/m² and \(x\) in m)
The Product Rule allows differentiation of functions expressed as the product of two variables or functions, common in mechanical systems where forces and distances multiply.
The Product Rule formula is:
$$\frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x)$$
Example 1: The torque \( T \) on a shaft is \( T = F \cdot r \), where \( F = 10 t^2 \) and \( r = 3 t \). Find \(\frac{dT}{dt}\).
Given: \( F = 10 t^2 \), \( r = 3 t \)
$$\frac{dT}{dt} = \frac{d}{dt} [F \cdot r] = \frac{dF}{dt} \cdot r + F \cdot \frac{dr}{dt}$$
Calculate derivatives:
$$\frac{dF}{dt} = 20 t$$
$$\frac{dr}{dt} = 3$$
Substitute:
$$\frac{dT}{dt} = 20 t \cdot 3 t + 10 t^2 \cdot 3$$
$$= 60 t^2 + 30 t^2$$
$$= 90 t^2$$
Answer: \(\frac{dT}{dt} = 90 t^2\) (units: Nm/s if \(T\) in Newton-meters)
Example 2: The power \( P = F \cdot v \), where \( F = 50 \sin t \) and \( v = 4 t^2 \). Find \(\frac{dP}{dt}\).
Given: \( F = 50 \sin t \), \( v = 4 t^2 \)
$$\frac{dP}{dt} = \frac{dF}{dt} \cdot v + F \cdot \frac{dv}{dt}$$
Calculate derivatives:
$$\frac{dF}{dt} = 50 \cos t$$
$$\frac{dv}{dt} = 8 t$$
Substitute:
$$\frac{dP}{dt} = 50 \cos t \cdot 4 t^2 + 50 \sin t \cdot 8 t$$
$$= 200 t^2 \cos t + 400 t \sin t$$
Answer: \(\frac{dP}{dt} = 200 t^2 \cos t + 400 t \sin t\) (units: Watts/s)
Example 3: The displacement \( s = u \cdot v \), where \( u = t^3 \) and \( v = e^t \). Find \(\frac{ds}{dt}\).
Given: \( u = t^3 \), \( v = e^t \)
$$\frac{ds}{dt} = \frac{du}{dt} \cdot v + u \cdot \frac{dv}{dt}$$
Calculate derivatives:
$$\frac{du}{dt} = 3 t^2$$
$$\frac{dv}{dt} = e^t$$
Substitute:
$$\frac{ds}{dt} = 3 t^2 \cdot e^t + t^3 \cdot e^t$$
$$= e^t (3 t^2 + t^3)$$
Answer: \(\frac{ds}{dt} = e^t (3 t^2 + t^3)\)
Example 4: The force \( F = (2 t + 1)(t^2 - 3) \). Find \(\frac{dF}{dt}\).
Given: \( u = 2 t + 1 \), \( v = t^2 - 3 \)
$$\frac{dF}{dt} = \frac{du}{dt} \cdot v + u \cdot \frac{dv}{dt}$$
Calculate derivatives:
$$\frac{du}{dt} = 2$$
$$\frac{dv}{dt} = 2 t$$
Substitute:
$$\frac{dF}{dt} = 2 (t^2 - 3) + (2 t + 1)(2 t)$$
$$= 2 t^2 - 6 + 4 t^2 + 2 t$$
$$= 6 t^2 + 2 t - 6$$
Answer: \(\frac{dF}{dt} = 6 t^2 + 2 t - 6\)
Example 5: The power output \( P = (3 t^2 + 2)(5 t^3 - t) \). Find \(\frac{dP}{dt}\).
Given: \( u = 3 t^2 + 2 \), \( v = 5 t^3 - t \)
$$\frac{dP}{dt} = \frac{du}{dt} \cdot v + u \cdot \frac{dv}{dt}$$
Calculate derivatives:
$$\frac{du}{dt} = 6 t$$
$$\frac{dv}{dt} = 15 t^2 - 1$$
Substitute:
$$\frac{dP}{dt} = 6 t (5 t^3 - t) + (3 t^2 + 2)(15 t^2 - 1)$$
$$= 6 t (5 t^3) - 6 t (t) + (3 t^2)(15 t^2) + 2 (15 t^2) - (3 t^2)(1) - 2 (1)$$
$$= 30 t^4 - 6 t^2 + 45 t^4 + 30 t^2 - 3 t^2 - 2$$
$$= (30 t^4 + 45 t^4) + (-6 t^2 + 30 t^2 - 3 t^2) - 2$$
$$= 75 t^4 + 21 t^2 - 2$$
Answer: \(\frac{dP}{dt} = 75 t^4 + 21 t^2 - 2\)
The Chain Rule is vital when differentiating composite functions common in mechanical engineering, such as temperature-dependent material properties or velocity dependent on position.
The Chain Rule formula is:
$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$
where \( y = f(u) \) and \( u = g(x) \).
Example 1: \( y = (3 t^2 + 1)^4 \). Find \(\frac{dy}{dt}\).
Given: \( y = (3 t^2 + 1)^4 \)
Let \( u = 3 t^2 + 1 \), then \( y = u^4 \).
$$\frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt}$$
$$= 4 u^{3} \times 6 t$$
$$= 4 (3 t^2 + 1)^3 \times 6 t$$
$$= 24 t (3 t^2 + 1)^3$$
Answer: \(\frac{dy}{dt} = 24 t (3 t^2 + 1)^3\)
Example 2: \( s = \sin (5 x^3) \). Find \(\frac{ds}{dx}\).
Given: \( s = \sin (5 x^3) \)
Let \( u = 5 x^3 \), then \( s = \sin u \).
$$\frac{ds}{dx} = \frac{ds}{du} \times \frac{du}{dx}$$
$$= \cos u \times 15 x^{2}$$
$$= 15 x^{2} \cos (5 x^{3})$$
Answer: \(\frac{ds}{dx} = 15 x^{2} \cos (5 x^{3})\)
Example 3: \( P = (2 t + 1)^5 \). Find \(\frac{dP}{dt}\).
Given: \( P = (2 t + 1)^5 \)
Let \( u = 2 t + 1 \), then \( P = u^5 \).
$$\frac{dP}{dt} = \frac{dP}{du} \times \frac{du}{dt}$$
$$= 5 u^{4} \times 2$$
$$= 10 (2 t + 1)^4$$
Answer: \(\frac{dP}{dt} = 10 (2 t + 1)^4\)
Example 4: \( y = \ln (4 x^2 + 3) \). Find \(\frac{dy}{dx}\).
Given: \( y = \ln (4 x^2 + 3) \)
Let \( u = 4 x^2 + 3 \), then \( y = \ln u \).
$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$
$$= \frac{1}{u} \times 8 x$$
$$= \frac{8 x}{4 x^2 + 3}$$
Answer: \(\frac{dy}{dx} = \frac{8 x}{4 x^2 + 3}\)
Example 5: \( z = e^{3 t^2 + 2 t} \). Find \(\frac{dz}{dt}\).
Given: \( z = e^{3 t^2 + 2 t} \)
Let \( u = 3 t^2 + 2 t \), then \( z = e^{u} \).
$$\frac{dz}{dt} = \frac{dz}{du} \times \frac{du}{dt}$$
$$= e^{u} \times (6 t + 2)$$
$$= e^{3 t^2 + 2 t} (6 t + 2)$$
Answer: \(\frac{dz}{dt} = e^{3 t^2 + 2 t} (6 t + 2)\)
The Quotient Rule is used when differentiating functions expressed as the ratio of two other functions, frequently appearing in mechanical engineering when analyzing ratios like efficiency or stress-strain relations.
The Quotient Rule formula is:
$$\frac{d}{dx} ( \frac{u}{v} ) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$
Example 1: \( y = \frac{t^2 + 1}{t} \). Find \(\frac{dy}{dt}\).
Given: \( u = t^2 + 1 \), \( v = t \)
$$\frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$$
Calculate derivatives:
$$\frac{du}{dt} = 2 t$$
$$\frac{dv}{dt} = 1$$
Substitute:
$$\frac{dy}{dt} = \frac{t (2 t) - (t^2 + 1)(1)}{t^2}$$
$$= \frac{2 t^2 - t^2 - 1}{t^2}$$
$$= \frac{t^2 - 1}{t^2}$$
Answer: \(\frac{dy}{dt} = \frac{t^2 - 1}{t^2}\)
Example 2: \( P = \frac{5 t^3}{2 t + 1} \). Find \(\frac{dP}{dt}\).
Given: \( u = 5 t^3 \), \( v = 2 t + 1 \)
$$\frac{dP}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$$
Calculate derivatives:
$$\frac{du}{dt} = 15 t^2$$
$$\frac{dv}{dt} = 2$$
Substitute:
$$\frac{dP}{dt} = \frac{(2 t + 1)(15 t^2) - 5 t^3 (2)}{(2 t + 1)^2}$$
$$= \frac{30 t^3 + 15 t^2 - 10 t^3}{(2 t + 1)^2}$$
$$= \frac{20 t^3 + 15 t^2}{(2 t + 1)^2}$$
Answer: \(\frac{dP}{dt} = \frac{20 t^3 + 15 t^2}{(2 t + 1)^2}\)
Example 3: \( y = \frac{\sin t}{t^2} \). Find \(\frac{dy}{dt}\).
Given: \( u = \sin t \), \( v = t^2 \)
$$\frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$$
Calculate derivatives:
$$\frac{du}{dt} = \cos t$$
$$\frac{dv}{dt} = 2 t$$
Substitute:
$$\frac{dy}{dt} = \frac{t^2 \cos t - \sin t (2 t)}{t^{4}}$$
$$= \frac{t^2 \cos t - 2 t \sin t}{t^4}$$
Answer: \(\frac{dy}{dt} = \frac{t^2 \cos t - 2 t \sin t}{t^4}\)
Example 4: \( F = \frac{3 t^2 + 2}{t^3 - 1} \). Find \(\frac{dF}{dt}\).
Given: \( u = 3 t^2 + 2 \), \( v = t^3 - 1 \)
$$\frac{dF}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$$
Calculate derivatives:
$$\frac{du}{dt} = 6 t$$
$$\frac{dv}{dt} = 3 t^2$$
Substitute:
$$\frac{dF}{dt} = \frac{(t^3 - 1)(6 t) - (3 t^2 + 2)(3 t^2)}{(t^3 - 1)^2}$$
$$= \frac{6 t^4 - 6 t - 9 t^4 - 6 t^2}{(t^3 - 1)^2}$$
$$= \frac{-3 t^4 - 6 t^2 - 6 t}{(t^3 - 1)^2}$$
Answer: \(\frac{dF}{dt} = \frac{-3 t^4 - 6 t^2 - 6 t}{(t^3 - 1)^2}\)
Example 5: \( y = \frac{e^{2 t}}{t^2 + 1} \). Find \(\frac{dy}{dt}\).
Given: \( u = e^{2 t} \), \( v = t^2 + 1 \)
$$\frac{dF}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$$
Calculate derivatives:
$$\frac{du}{dt} = 2 e^{2 t}$$
$$\frac{dv}{dt} = 2 t$$
Substitute:
$$\frac{dy}{dt} = \frac{(t^2 + 1)(2 e^{2 t}) - e^{2 t} (2 t)}{(t^2 + 1)^2}$$
$$= \frac{2 e^{2 t} (t^2 + 1) - 2 t e^{2 t}}{(t^2 + 1)^2}$$
$$= \frac{2 e^{2 t} (t^2 + 1 - t)}{(t^2 + 1)^2}$$
Answer: \(\frac{dy}{dt} = \frac{2 e^{2 t} (t^2 + 1 - t)}{(t^2 + 1)^2}\)
Power is a fundamental concept in mechanical production, representing the rate at which work is done or energy is transferred. In machining, manufacturing, and mechanical systems within Kenyan industries such as automotive workshops or metal fabrication plants, understanding power helps optimize machine performance and energy efficiency. Power is commonly calculated as the product of force and velocity or as the rate of doing work.
The general formula for power \(P\) is:
$$ P = \frac{dW}{dt} $$
where \(dW\) is the infinitesimal work done, and \(dt\) is the infinitesimal time interval.
In mechanical contexts, power can also be expressed as:
$$ P = F \times v $$
where \(F\) is the force applied (in newtons) and \(v\) is the velocity (in meters per second).
Instantaneous power is the power at a specific instant and can be found by differentiating work with respect to time. Given work \(W(t)\) as a function of time, instantaneous power is:
$$ P(t) = \frac{dW}{dt} $$
Example 1: A metal press exerts work described by \(W(t) = 5t^3 + 2t^2\) joules, where \(t\) is in seconds. Find the instantaneous power at \(t = 2\) s.
Given: \(W(t) = 5t^3 + 2t^2\), \(t = 2\) s
$$ P(t) = \frac{dW}{dt} = \frac{d}{dt}(5t^3 + 2t^2) = 15t^2 + 4t $$
$$ P(2) = 15(2)^2 + 4(2) = 15 \times 4 + 8 = 60 + 8 $$
Answer: 68 W
Example 2: The work done by a lathe machine is given by \(W(t) = 10t^4 - 3t^3 + 2t\) joules. Calculate the instantaneous power at \(t = 1.5\) seconds.
Given: \(W(t) = 10t^4 - 3t^3 + 2t\), \(t=1.5\) s
$$ P(t) = \frac{dW}{dt} = 40t^3 - 9t^2 + 2 $$
$$ P(1.5) = 40(1.5)^3 - 9(1.5)^2 + 2 = 40(3.375) - 9(2.25) + 2 = 135 - 20.25 + 2 $$
$$ = 116.75 \text{ W} $$
Answer: 116.75 W
Example 3: In a milling operation, the work done varies as \(W(t) = 8e^{0.5t}\) joules. Find the instantaneous power at \(t = 3\) s.
Given: \(W(t) = 8e^{0.5t}\), \(t=3\) s
$$ P(t) = \frac{dW}{dt} = 8 \times 0.5 e^{0.5t} = 4e^{0.5t} $$
$$ P(3) = 4e^{0.5 \times 3} = 4e^{1.5} = 4 \times 4.4817 $$
$$ = 17.927 \text{ W} $$
Answer: 17.927 W
In mechanical systems, power can be calculated as the product of the force applied and the velocity of the point of application. This formula is especially useful in machining and production lines where forces and speeds are measurable.
$$ P = F \times v $$
Example 1: A conveyor belt applies a force of 200 N to move materials at a speed of 3 m/s. Find the power output.
Given: \(F = 200\) N, \(v = 3\) m/s
$$ P = F \times v = 200 \times 3 $$
$$ = 600 \text{ W} $$
Answer: 600 W
Example 2: A hydraulic press applies a force of 5000 N on a workpiece moving at 0.02 m/s. Calculate the power exerted.
Given: \(F = 5000\) N, \(v = 0.02\) m/s
$$ P = 5000 \times 0.02 $$
$$ = 100 \text{ W} $$
Answer: 100 W
Example 3: In a machining operation, a cutting tool applies a force of 150 N and moves at 0.5 m/s. What is the power used by the tool?
Given: \(F = 150\) N, \(v = 0.5\) m/s
$$ P = 150 \times 0.5 $$
$$ = 75 \text{ W} $$
Answer: 75 W
Power can also be calculated by dividing the total work done over a specific time interval. This average power is useful in production planning and machine load calculations.
$$ P = \frac{W}{t} $$
where \(W\) is work done in joules and \(t\) is time in seconds.
Example 1: A drilling machine does 1200 J of work in 10 seconds. Find the average power.
Given: \(W = 1200\) J, \(t = 10\) s
$$ P = \frac{1200}{10} $$
$$ = 120 \text{ W} $$
Answer: 120 W
Example 2: A milling cutter performs 4500 J of work in 30 seconds. Calculate the power used.
Given: \(W = 4500\) J, \(t = 30\) s
$$ P = \frac{4500}{30} $$
$$ = 150 \text{ W} $$
Answer: 150 W
Example 3: A lathe machine performs 2500 J of work in 8 seconds. Find the average power.
Given: \(W = 2500\) J, \(t = 8\) s
$$ P = \frac{2500}{8} $$
$$ = 312.5 \text{ W} $$
Answer: 312.5 W
Power in rotating machinery such as motors and turbines is related to torque and angular velocity. This is critical for machine design and operation in mechanical workshops.
$$ P = \tau \times \omega $$
where \(\tau\) is torque in newton-meters and \(\omega\) is angular velocity in radians per second.
Example 1: A motor produces a torque of 20 Nm rotating at 100 rad/s. Find the power output.
Given: \(\tau = 20\) Nm, \(\omega = 100\) rad/s
$$ P = 20 \times 100 $$
$$ = 2000 \text{ W} $$
Answer: 2000 W
Example 2: A lathe spindle applies torque of 15 Nm at an angular velocity of 80 rad/s. Calculate the power.
Given: \(\tau = 15\) Nm, \(\omega = 80\) rad/s
$$ P = 15 \times 80 $$
$$ = 1200 \text{ W} $$
Answer: 1200 W
Example 3: A milling machine generates torque of 25 Nm at 50 rad/s angular velocity. Determine the power.
Given: \(\tau = 25\) Nm, \(\omega = 50\) rad/s
$$ P = 25 \times 50 $$
$$ = 1250 \text{ W} $$
Answer: 1250 W
In calculus, the product rule is essential for differentiating the product of two functions. Mechanical production technicians frequently encounter situations where quantities vary together, such as force and velocity or torque and angular velocity. Understanding how to differentiate such products accurately is crucial in analyzing machine performance and dynamic systems.
The product rule states:
$$ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) $$
where \(u(x)\) and \(v(x)\) are differentiable functions of \(x\).
When force \(F(t)\) and velocity \(v(t)\) both vary with time, power \(P(t) = F(t) \times v(t)\) requires differentiation using the product rule to determine instantaneous rates of change.
$$ \frac{dP}{dt} = \frac{d}{dt}[F(t) \cdot v(t)] = F'(t) \cdot v(t) + F(t) \cdot v'(t) $$
Example 1: Force applied is \(F(t) = 3t^2\) N and velocity is \(v(t) = 5t\) m/s. Find the rate of change of power at \(t=2\) seconds.
Given: \(F(t) = 3t^2\), \(v(t) = 5t\), \(t=2\) s
Calculate derivatives:
$$ F'(t) = 6t $$
$$ v'(t) = 5 $$
Apply product rule:
$$ \frac{dP}{dt} = F'(t) \times v(t) + F(t) \times v'(t) = 6t \times 5t + 3t^2 \times 5 $$
At \(t=2\):
$$ = 6 \times 2 \times 5 \times 2 + 3 \times 2^2 \times 5 = 12 \times 10 + 3 \times 4 \times 5 $$
$$ = 120 + 60 = 180 \text{ W/s} $$
Answer: 180 W/s
Example 2: Force \(F(t) = 4e^{2t}\) N and velocity \(v(t) = t^3\) m/s. Find \(\frac{dP}{dt}\) at \(t=1\) s.
Given: \(F(t) = 4e^{2t}\), \(v(t) = t^3\)
Derivatives:
$$ F'(t) = 8e^{2t} $$
$$ v'(t) = 3t^2 $$
Apply product rule:
$$ \frac{dP}{dt} = 8e^{2t} \times t^3 + 4e^{2t} \times 3t^2 = e^{2t}(8t^3 + 12t^2) $$
At \(t=1\):
$$ = e^{2} (8 + 12) = e^2 \times 20 = 7.389 \times 20 $$
$$ = 147.78 \text{ W/s} $$
Answer: 147.78 W/s
Example 3: Force \(F(t) = \sin t\) N and velocity \(v(t) = \cos t\) m/s. Find \(\frac{dP}{dt}\) at \(t = \frac{\pi}{4}\) seconds.
Given: \(F(t) = \sin t\), \(v(t) = \cos t\)
Derivatives:
$$ F'(t) = \cos t $$
$$ v'(t) = -\sin t $$
Apply product rule:
$$ \frac{dP}{dt} = \cos t \times \cos t + \sin t \times (-\sin t) = \cos^2 t - \sin^2 t $$
At \(t = \frac{\pi}{4}\):
$$ \cos^2 \frac{\pi}{4} - \sin^2 \frac{\pi}{4} = (\frac{\sqrt{2}}{2})^2 - (\frac{\sqrt{2}}{2})^2 = \frac{1}{2} - \frac{1}{2} = 0 $$
Answer: 0 W/s
Torque \(\tau(t)\) and angular velocity \(\omega(t)\) may both vary with time in rotating machinery. The product rule is used to find the rate of change of power in such systems.
$$ \frac{dP}{dt} = \frac{d}{dt}[\tau(t) \cdot \omega(t)] = \tau'(t) \cdot \omega(t) + \tau(t) \cdot \omega'(t) $$
Example 1: Torque \(\tau(t) = 10t\) Nm and angular velocity \(\omega(t) = 50 - 5t\) rad/s. Find \(\frac{dP}{dt}\) at \(t=2\) seconds.
Given: \(\tau(t) = 10t\), \(\omega(t) = 50 - 5t\)
Derivatives:
$$ \tau'(t) = 10 $$
$$ \omega'(t) = -5 $$
Apply product rule:
$$ \frac{dP}{dt} = 10 \times (50 - 5t) + 10t \times (-5) = 10(50 - 5t) - 50t $$
At \(t=2\):
$$ = 10(50 - 10) - 50 \times 2 = 10 \times 40 - 100 = 400 - 100 = 300 \text{ W/s} $$
Answer: 300 W/s
Example 2: Torque \(\tau(t) = 15e^{0.1t}\) Nm and angular velocity \(\omega(t) = 20t^2\) rad/s. Calculate \(\frac{dP}{dt}\) at \(t=1\) second.
Given: \(\tau(t) = 15e^{0.1t}\), \(\omega(t) = 20t^2\)
Derivatives:
$$ \tau'(t) = 15 \times 0.1 e^{0.1t} = 1.5 e^{0.1t} $$
$$ \omega'(t) = 40t $$
Apply product rule:
$$ \frac{dP}{dt} = 1.5 e^{0.1t} \times 20 t^2 + 15 e^{0.1t} \times 40 t = e^{0.1t} (30 t^2 + 600 t) $$
At \(t=1\):
$$ = e^{0.1} (30 + 600) = 1.105 \times 630 = 696.15 \text{ W/s} $$
Answer: 696.15 W/s
Example 3: Torque \(\tau(t) = \sin t\) Nm and angular velocity \(\omega(t) = \cos 2t\) rad/s. Find \(\frac{dP}{dt}\) at \(t = \frac{\pi}{6}\) seconds.
Given: \(\tau(t) = \sin t\), \(\omega(t) = \cos 2t\)
Derivatives:
$$ \tau'(t) = \cos t $$
$$ \omega'(t) = -2 \sin 2t $$
Apply product rule:
$$ \frac{dP}{dt} = \cos t \times \cos 2t + \sin t \times (-2 \sin 2t) = \cos t \cos 2t - 2 \sin t \sin 2t $$
At \(t = \frac{\pi}{6}\):
Calculate each term:
$$ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \approx 0.866 $$
$$ \cos \frac{\pi}{3} = \frac{1}{2} = 0.5 $$
$$ \sin \frac{\pi}{6} = \frac{1}{2} = 0.5 $$
$$ \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \approx 0.866 $$
Substitute:
$$ 0.866 \times 0.5 - 2 \times 0.5 \times 0.866 = 0.433 - 0.866 = -0.433 $$
Answer: -0.433 W/s
Products of functions representing power, such as force times displacement or torque times angular displacement, require the product rule to find rates of change of energy or power in mechanical systems.
$$ \frac{d}{dt}[f(t) \cdot g(t)] = f'(t) \cdot g(t) + f(t) \cdot g'(t) $$
Example 1: Force \(f(t) = 6t^2\) N and displacement \(g(t) = 3t\) m. Find the rate of change of work done at \(t=4\) seconds.
Given: \(f(t) = 6t^2\), \(g(t) = 3t\)
Derivatives:
$$ f'(t) = 12t $$
$$ g'(t) = 3 $$
Apply product rule:
$$ \frac{d}{dt}[f(t) \cdot g(t)] = 12t \times 3t + 6t^2 \times 3 = 36 t^2 + 18 t^2 = 54 t^2 $$
At \(t=4\):
$$ = 54 \times 16 = 864 \text{ W} $$
Answer: 864 W
Example 2: Torque \(f(t) = 8e^{t}\) Nm and angular displacement \(g(t) = t^2\) rad. Find \(\frac{d}{dt}[f(t) \cdot g(t)]\) at \(t=0\).
Given: \(f(t) = 8e^t\), \(g(t) = t^2\)
Derivatives:
$$ f'(t) = 8e^t $$
$$ g'(t) = 2t $$
Apply product rule:
$$ \frac{d}{dt}[f(t) \cdot g(t)] = 8e^t \times t^2 + 8e^t \times 2t = 8e^t (t^2 + 2t) $$
At \(t=0\):
$$ = 8e^0 (0 + 0) = 8 \times 1 \times 0 = 0 $$
Answer: 0 W
Example 3: Force \(f(t) = \sin 2t\) N and displacement \(g(t) = \cos t\) m. Compute \(\frac{d}{dt}[f(t) \cdot g(t)]\) at \(t = \frac{\pi}{4}\) seconds.
Given: \(f(t) = \sin 2t\), \(g(t) = \cos t\)
Derivatives:
$$ f'(t) = 2 \cos 2t $$
$$ g'(t) = -\sin t $$
Apply product rule:
$$ \frac{d}{dt}[f(t) \cdot g(t)] = 2 \cos 2t \times \cos t + \sin 2t \times (-\sin t) = 2 \cos 2t \cos t - \sin 2t \sin t $$
At \(t = \frac{\pi}{4}\):
Calculate:
$$ \cos 2 \times \frac{\pi}{4} = \cos \frac{\pi}{2} = 0 $$
$$ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \approx 0.707 $$
$$ \sin 2 \times \frac{\pi}{4} = \sin \frac{\pi}{2} = 1 $$
$$ \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \approx 0.707 $$
Substitute:
$$ 2 \times 0 \times 0.707 - 1 \times 0.707 = 0 - 0.707 = -0.707 $$
Answer: -0.707 W
For products of more than two functions, the product rule extends, useful in complex mechanical systems involving multiple varying quantities.
For three functions \(u(x), v(x), w(x)\):
$$ \frac{d}{dx}[u v w] = u' v w + u v' w + u v w' $$
Example 1: Let \(u(t) = t\), \(v(t) = t^2\), and \(w(t) = \sin t\). Find \(\frac{d}{dt}[u v w]\) at \(t=0\).
Derivatives:
$$ u'(t) = 1 $$
$$ v'(t) = 2t $$
$$ w'(t) = \cos t $$
Apply formula:
$$ \frac{d}{dt}[t \times t^2 \times \sin t] = 1 \times t^2 \times \sin t + t \times 2t \times \sin t + t \times t^2 \times \cos t $$
At \(t=0\):
$$ 1 \times 0^2 \times 0 + 0 \times 0 \times 0 + 0 \times 0^2 \times 1 = 0 $$
Answer: 0
Example 2: \(u(t) = e^t\), \(v(t) = t^3\), \(w(t) = \cos t\). Find \(\frac{d}{dt}[u v w]\) at \(t=1\).
Derivatives:
$$ u'(t) = e^t $$
$$ v'(t) = 3t^2 $$
$$ w'(t) = -\sin t $$
Apply formula:
$$ = e^t \times t^3 \times \cos t + e^t \times 3 t^2 \times \cos t + e^t \times t^3 \times (-\sin t) $$
At \(t=1\):
Calculate values:
$$ e^1 = 2.718 $$
$$ t^3 = 1^3 = 1 $$
$$ t^2 = 1^2 = 1 $$
$$ \cos 1 = 0.5403 $$
$$ \sin 1 = 0.8415 $$
Substitute:
$$ 2.718 \times 1 \times 0.5403 + 2.718 \times 3 \times 1 \times 0.5403 + 2.718 \times 1 \times (-0.8415) $$
$$ = 1.469 + 4.407 - 2.288 = 3.588 $$
Answer: 3.588
Example 3: \(u(t) = \ln t\), \(v(t) = t\), \(w(t) = e^t\). Find \(\frac{d}{dt}[u v w]\) at \(t=2\).
Derivatives:
$$ u'(t) = \frac{1}{t} $$
$$ v'(t) = 1 $$
$$ w'(t) = e^t $$
Apply formula:
$$ = \frac{1}{t} \times t \times e^t + \ln t \times 1 \times e^t + \ln t \times t \times e^t $$
Simplify first term:
$$ \frac{1}{t} \times t \times e^t = e^t $$
At \(t=2\):
Calculate values:
$$ e^2 = 7.389 $$
$$ \ln 2 = 0.693 $$
Substitute:
$$ 7.389 + 0.693 \times 7.389 + 0.693 \times 2 \times 7.389 = 7.389 + 5.122 + 10.244 $$
$$ = 22.755 $$
Answer: 22.755
The chain rule is essential in mechanical production engineering when dealing with composite functions, such as temperature-dependent material properties or speed-dependent forces in machinery. It allows differentiation of a function composed of two or more functions, which frequently occurs in the analysis of mechanical systems where one parameter depends on another that changes with time or position. The chain rule formula is:
$$ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} $$
where \(y = f(u)\) and \(u = g(x)\).
Example 1:
A metal rod's length \(L\) (in mm) changes with temperature \(T\) according to \(L = (3T + 5)^2\). Find \(\frac{dL}{dT}\) when \(T = 20^\circ C\).
Given:
\(L = (3T + 5)^2\), \(T = 20\)
Let \(u = 3T + 5\), so \(L = u^2\).
$$ \frac{dL}{dT} = \frac{dL}{du} \times \frac{du}{dT} $$
$$ \frac{dL}{du} = 2u, \quad \frac{du}{dT} = 3 $$
$$ \frac{dL}{dT} = 2u \times 3 = 6u $$
At \(T = 20\):
$$ u = 3 \times 20 + 5 = 65 $$
$$ \frac{dL}{dT} = 6 \times 65 = 390 \, \text{mm/}^\circ \text{C} $$
Answer: 390 mm/°C
Example 2:
The pressure \(P\) in a hydraulic system varies with fluid velocity \(v\) as \(P = \sin(2v^2 + 1)\). Find \(\frac{dP}{dv}\) when \(v = 2 \, m/s\).
Given:
\(P = \sin(2v^2 + 1)\), \(v = 2\)
Let \(u = 2v^2 + 1\), so \(P = \sin u\).
$$ \frac{dP}{dv} = \frac{dP}{du} \times \frac{du}{dv} $$
$$ \frac{dP}{du} = \cos u, \quad \frac{du}{dv} = 4v $$
$$ \frac{dP}{dv} = \cos u \times 4v $$
At \(v = 2\):
$$ u = 2 \times 2^2 + 1 = 2 \times 4 + 1 = 9 $$
$$ \frac{dP}{dv} = \cos 9 \times 4 \times 2 = 8 \cos 9 $$
Using radians, \(\cos 9 \approx -0.9111\):
$$ \frac{dP}{dv} = 8 \times (-0.9111) = -7.2888 $$
Answer: \(-7.29 \, \text{units of pressure per } m/s\)
Example 3:
In a lathe machine, the torque \(T\) (Nm) depends on spindle speed \(N\) (rpm) through \(T = \ln(5N^3 + 4)\). Find \(\frac{dT}{dN}\) at \(N = 10\).
Given:
\(T = \ln(5N^3 + 4)\), \(N = 10\)
Let \(u = 5N^3 + 4\), so \(T = \ln u\).
$$ \frac{dT}{dN} = \frac{dT}{du} \times \frac{du}{dN} $$
$$ \frac{dT}{du} = \frac{1}{u}, \quad \frac{du}{dN} = 15N^2 $$
$$ \frac{dT}{dN} = \frac{1}{u} \times 15N^2 = \frac{15N^2}{u} $$
At \(N = 10\):
$$ u = 5 \times 10^3 + 4 = 5 \times 1000 + 4 = 5004 $$
$$ \frac{dT}{dN} = \frac{15 \times 10^2}{5004} = \frac{15 \times 100}{5004} = \frac{1500}{5004} \approx 0.2998 $$
Answer: 0.30 Nm per rpm
Example 4:
The displacement \(s\) of a piston in a compressor is given by \(s = e^{3t^2 + 2t}\). Find the velocity \(\frac{ds}{dt}\) at \(t = 1\) second.
Given:
\(s = e^{3t^2 + 2t}\), \(t = 1\)
Let \(u = 3t^2 + 2t\), so \(s = e^u\).
$$ \frac{ds}{dt} = \frac{ds}{du} \times \frac{du}{dt} $$
$$ \frac{ds}{du} = e^u, \quad \frac{du}{dt} = 6t + 2 $$
$$ \frac{ds}{dt} = e^u \times (6t + 2) $$
At \(t = 1\):
$$ u = 3 \times 1^2 + 2 \times 1 = 3 + 2 = 5 $$
$$ \frac{ds}{dt} = e^5 \times (6 \times 1 + 2) = e^5 \times 8 $$
Using \(e^5 \approx 148.413\):
$$ \frac{ds}{dt} = 148.413 \times 8 = 1187.3 $$
Answer: 1187.3 units per second
Example 5:
A tool wear rate \(W\) depends on cutting speed \(v\) as \(W = \sqrt{4v^4 + 1}\). Find \(\frac{dW}{dv}\) at \(v = 3 \, m/s\).
Given:
\(W = \sqrt{4v^4 + 1}\), \(v = 3\)
Rewrite as:
$$ W = (4v^4 + 1)^{1/2} $$
Let \(u = 4v^4 + 1\), so \(W = u^{1/2}\).
$$ \frac{dW}{dv} = \frac{dW}{du} \times \frac{du}{dv} $$
$$ \frac{dW}{du} = \frac{1}{2} u^{-1/2}, \quad \frac{du}{dv} = 16 v^3 $$
$$ \frac{dW}{dv} = \frac{1}{2} u^{-1/2} \times 16 v^3 = 8 v^3 \times \frac{1}{\sqrt{u}} $$
At \(v = 3\):
$$ u = 4 \times 3^4 + 1 = 4 \times 81 + 1 = 324 + 1 = 325 $$
$$ \frac{dW}{dv} = 8 \times 27 \times \frac{1}{\sqrt{325}} = 216 \times \frac{1}{18.03} = 216 \times 0.0555 = 11.99 $$
Answer: 11.99 units per m/s
In mechanical production, many quantities are defined as ratios, such as efficiency, stress (force per area), and speed (distance per time). Differentiating such functions requires the quotient rule which handles derivatives of one function divided by another. It is important for calculating rates of change of ratios in machine dynamics and control systems. The quotient rule formula is:
$$ \frac{d}{dx} ( \frac{f(x)}{g(x)} ) = \frac{g(x) \frac{df}{dx} - f(x) \frac{dg}{dx}}{[g(x)]^2} $$
where both \(f(x)\) and \(g(x)\) are differentiable functions and \(g(x) \neq 0\).
Example 1:
The efficiency \(\eta\) of a motor is given by \(\eta = \frac{P_{out}}{P_{in}}\), where \(P_{out} = 50t^2\) and \(P_{in} = 10t + 5\). Find \(\frac{d\eta}{dt}\) at \(t = 1\) s.
Given:
\(\eta = \frac{50t^2}{10t + 5}\), \(t=1\)
Let \(f(t) = 50t^2\), \(g(t) = 10t + 5\).
$$ \frac{d\eta}{dt} = \frac{g \frac{df}{dt} - f \frac{dg}{dt}}{g^2} $$
Calculate derivatives:
$$ \frac{df}{dt} = 100 t, \quad \frac{dg}{dt} = 10 $$
Substitute at \(t=1\):
$$ f = 50 \times 1^2 = 50, \quad g = 10 \times 1 + 5 = 15 $$
$$ \frac{d\eta}{dt} = \frac{15 \times 100 \times 1 - 50 \times 10}{15^2} = \frac{1500 - 500}{225} = \frac{1000}{225} = 4.444 $$
Answer: 4.44 per second
Example 2:
The stress \(\sigma\) on a beam is \(\sigma = \frac{F(t)}{A(t)}\), where \(F(t) = 100 + 20t\) and \(A(t) = 50 - 2t\). Find \(\frac{d\sigma}{dt}\) at \(t = 5\) s.
Given:
\(\sigma = \frac{100 + 20t}{50 - 2t}\), \(t=5\)
Let \(f = 100 + 20t\), \(g = 50 - 2t\).
$$ \frac{d\sigma}{dt} = \frac{g \frac{df}{dt} - f \frac{dg}{dt}}{g^2} $$
Calculate derivatives:
$$ \frac{df}{dt} = 20, \quad \frac{dg}{dt} = -2 $$
At \(t=5\):
$$ f = 100 + 20 \times 5 = 200, \quad g = 50 - 2 \times 5 = 40 $$
$$ \frac{d\sigma}{dt} = \frac{40 \times 20 - 200 \times (-2)}{40^2} = \frac{800 + 400}{1600} = \frac{1200}{1600} = 0.75 $$
Answer: 0.75 units per second
Example 3:
The velocity \(v\) of a conveyor belt is the ratio \(v = \frac{s(t)}{t}\), where \(s(t) = 5t^2 + 3\). Find \(\frac{dv}{dt}\) at \(t = 2\) seconds.
Given:
\(v = \frac{5t^2 + 3}{t}\), \(t=2\)
Let \(f = 5t^2 + 3\), \(g = t\).
$$ \frac{dv}{dt} = \frac{g \frac{df}{dt} - f \frac{dg}{dt}}{g^2} $$
Calculate derivatives:
$$ \frac{df}{dt} = 10t, \quad \frac{dg}{dt} = 1 $$
At \(t=2\):
$$ f = 5 \times 2^2 + 3 = 20 + 3 = 23, \quad g = 2 $$
$$ \frac{dv}{dt} = \frac{2 \times 10 \times 2 - 23 \times 1}{2^2} = \frac{40 - 23}{4} = \frac{17}{4} = 4.25 $$
Answer: 4.25 m/s²
Example 4:
The power output \(P\) of a machine is \(P = \frac{W(t)}{t^2 + 1}\), where \(W(t) = 10t^3 + 5\). Find \(\frac{dP}{dt}\) at \(t=1\) second.
Given:
\(P = \frac{10t^3 + 5}{t^2 + 1}\), \(t=1\)
Let \(f = 10t^3 + 5\), \(g = t^2 + 1\).
$$ \frac{dP}{dt} = \frac{g \frac{df}{dt} - f \frac{dg}{dt}}{g^2} $$
Calculate derivatives:
$$ \frac{df}{dt} = 30t^2, \quad \frac{dg}{dt} = 2t $$
At \(t=1\):
$$ f = 10 \times 1^3 + 5 = 15, \quad g = 1^2 + 1 = 2 $$
$$ \frac{dP}{dt} = \frac{2 \times 30 \times 1^2 - 15 \times 2 \times 1}{2^2} = \frac{60 - 30}{4} = \frac{30}{4} = 7.5 $$
Answer: 7.5 units per second
Example 5:
The speed of a cutting tool is given by \(v = \frac{\ln(t+1)}{t^2}\). Find \(\frac{dv}{dt}\) at \(t = 1\).
Given:
\(v = \frac{\ln(t+1)}{t^2}\), \(t=1\)
Let \(f = \ln(t+1)\), \(g = t^2\).
$$ \frac{dv}{dt} = \frac{g \frac{df}{dt} - f \frac{dg}{dt}}{g^2} $$
Calculate derivatives:
$$ \frac{df}{dt} = \frac{1}{t+1}, \quad \frac{dg}{dt} = 2t $$
At \(t=1\):
$$ f = \ln(1+1) = \ln 2 = 0.6931, \quad g = 1^2 = 1 $$
$$ \frac{dv}{dt} = \frac{1 \times \frac{1}{2} - 0.6931 \times 2 \times 1}{1^2} = \frac{0.5 - 1.3862}{1} = -0.8862 $$
Answer: \(-0.8862\) units per second
In mechanical production engineering, calculus is essential for optimizing machine components, controlling manufacturing processes, and analyzing dynamic systems. Understanding applications like stationary points and rates of change enables technicians to improve product quality, efficiency, and safety in Kenya's manufacturing sector. This section focuses on practical calculus applications relevant to mechanical production tasks.
Stationary points occur where the derivative of a function equals zero, indicating potential maxima, minima, or points of inflection. In mechanical production, identifying stationary points helps optimize dimensions, minimize material usage, and maximize strength or efficiency of components. The fundamental condition for stationary points is:
$$ \frac{dy}{dx} = 0 $$
To classify stationary points, the second derivative test is used:
Example 1: A mechanical technician models the stress \(S\) on a beam as a function of length \(x\) by \(S = -2x^2 + 8x + 10\) (stress in kN, length in m). Find the stationary points and classify them.
Given: \(S = -2x^2 + 8x + 10\)
Find \(\frac{dS}{dx}\):
$$ \frac{dS}{dx} = -4x + 8 $$
Set \(\frac{dS}{dx} = 0\):
$$ -4x + 8 = 0 $$
$$ 4x = 8 $$
$$ x = 2 \text{ m} $$
Find \(\frac{d^2S}{dx^2}\):
$$ \frac{d^2S}{dx^2} = -4 $$
Since \(\frac{d^2S}{dx^2} = -4 < 0\), the stationary point at \(x=2\) m is a maximum.
Answer: Stationary point at \(x = 2\) m is a maximum stress point.
Example 2: The volume \(V\) (in cm³) of a cylindrical shaft is given by \(V = \pi r^2 h\), where height \(h = 10 - r\) cm. Express \(V\) in terms of \(r\) and find the radius \(r\) that maximizes volume.
Given: \(V = \pi r^2 h\), \(h = 10 - r\)
Express \(V\):
$$ V = \pi r^2 (10 - r) = \pi (10r^2 - r^3) $$
Find \(\frac{dV}{dr}\):
$$ \frac{dV}{dr} = \pi (20r - 3r^2) $$
Set \(\frac{dV}{dr} = 0\):
$$ \pi (20r - 3r^2) = 0 $$
$$ 20r - 3r^2 = 0 $$
$$ r (20 - 3r) = 0 $$
Solutions:
$$ r = 0 \quad \text{or} \quad r = \frac{20}{3} \approx 6.67 \text{ cm} $$
Find \(\frac{d^2V}{dr^2}\):
$$ \frac{d^2V}{dr^2} = \pi (20 - 6r) $$
Evaluate at \(r = 6.67\):
$$ \frac{d^2V}{dr^2} = \pi (20 - 6 \times 6.67) = \pi (20 - 40) = -20 \pi < 0 $$
Answer: Maximum volume at radius \(r = 6.67\) cm.
Example 3: The efficiency \(E\) of a lathe machine is modeled by \(E = 4x^3 - 9x^2 + 6x + 1\), where \(x\) is the speed setting (in arbitrary units). Find all stationary points and classify them.
Given: \(E = 4x^3 - 9x^2 + 6x + 1\)
Find \(\frac{dE}{dx}\):
$$ \frac{dE}{dx} = 12x^2 - 18x + 6 $$
Set \(\frac{dE}{dx} = 0\):
$$ 12x^2 - 18x + 6 = 0 $$
Divide by 6:
$$ 2x^2 - 3x + 1 = 0 $$
Solve quadratic:
$$ x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} $$
Solutions:
$$ x = 1 \quad \text{or} \quad x = \frac{1}{2} $$
Find \(\frac{d^2E}{dx^2}\):
$$ \frac{d^2E}{dx^2} = 24x - 18 $$
Evaluate at \(x=1\):
$$ 24(1) - 18 = 6 > 0 \implies \text{Minimum} $$
Evaluate at \(x=0.5\):
$$ 24(0.5) - 18 = 12 - 18 = -6 < 0 \implies \text{Maximum} $$
Answer: Maximum efficiency at \(x=0.5\), minimum efficiency at \(x=1\).
Example 4: The force \(F\) acting on a piston is given by \(F = 5x^4 - 20x^3 + 15x^2\), where \(x\) is piston displacement in cm. Find stationary points and classify them.
Given: \(F = 5x^4 - 20x^3 + 15x^2\)
Find \(\frac{dF}{dx}\):
$$ \frac{dF}{dx} = 20x^3 - 60x^2 + 30x $$
Set \(\frac{dF}{dx} = 0\):
$$ 20x^3 - 60x^2 + 30x = 0 $$
Divide by 10:
$$ 2x^3 - 6x^2 + 3x = 0 $$
Factor:
$$ x(2x^2 - 6x + 3) = 0 $$
Solve quadratic:
$$ 2x^2 - 6x + 3 = 0 $$
$$ x = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} $$
Simplify:
$$ x = \frac{3 \pm \sqrt{3}}{2} $$
Stationary points at:
$$ x = 0, \quad x = \frac{3 - \sqrt{3}}{2} \approx 0.634, \quad x = \frac{3 + \sqrt{3}}{2} \approx 2.366 $$
Find \(\frac{d^2F}{dx^2}\):
$$ \frac{d^2F}{dx^2} = 60x^2 - 120x + 30 $$
Evaluate at \(x=0\):
$$ 60(0)^2 - 120(0) + 30 = 30 > 0 \implies \text{Minimum} $$
Evaluate at \(x=0.634\):
$$ 60(0.634)^2 - 120(0.634) + 30 = 60(0.402) - 76.08 + 30 = 24.12 - 76.08 + 30 = -21.96 < 0 \implies \text{Maximum} $$
Evaluate at \(x=2.366\):
$$ 60(2.366)^2 - 120(2.366) + 30 = 60(5.598) - 283.92 + 30 = 335.88 - 283.92 + 30 = 81.96 > 0 \implies \text{Minimum} $$
Answer: Minima at \(x=0\) cm and \(x=2.366\) cm; maximum at \(x=0.634\) cm.
Example 5: The temperature \(T\) of a metal rod during cooling is modeled by \(T = 100e^{-0.5t} + 20\), where \(t\) is time in minutes. Find the stationary points of the rate of temperature change.
Given: \(T = 100e^{-0.5t} + 20\)
Find \(\frac{dT}{dt}\):
$$ \frac{dT}{dt} = 100 \times (-0.5) e^{-0.5t} = -50 e^{-0.5t} $$
Set \(\frac{dT}{dt} = 0\):
$$ -50 e^{-0.5t} = 0 $$
Since \(e^{-0.5t} \neq 0\), no stationary points in temperature.
Check second derivative:
$$ \frac{d^2T}{dt^2} = -50 \times (-0.5) e^{-0.5t} = 25 e^{-0.5t} > 0 $$
No stationary points; temperature decreases exponentially.
Answer: No stationary points in temperature; continuous cooling.
Rates of change describe how one quantity varies relative to another, often time or position. In mechanical production, integration allows calculation of quantities such as displacement from velocity, volume from cross-sectional area, or total energy from power rates. The fundamental relation is:
$$ \text{If } \frac{dy}{dx} = f(x), \quad \text{then} \quad y = \int f(x) dx + C $$
where \(C\) is the constant of integration determined by initial conditions.
Example 1: A conveyor belt moves with velocity \(v(t) = 3t^2\) m/s. Find displacement \(s(t)\) after \(t\) seconds, if \(s(0) = 0\).
Given: \(v(t) = \frac{ds}{dt} = 3t^2\), \(s(0) = 0\)
Integrate velocity:
$$ s(t) = \int 3t^2 dt + C $$
$$ s(t) = t^3 + C $$
Apply initial condition:
$$ s(0) = 0^3 + C = 0 \implies C = 0 $$
Answer: \(s(t) = t^3\) m
Example 2: The cross-sectional area \(A(x)\) of a shaft varies as \(A(x) = 4x + 2\) cm² along length \(x\) m. Find the volume \(V\) of the shaft between \(x=0\) m and \(x=3\) m.
Given: \(A(x) = 4x + 2\), find
$$ V = \int_0^3 A(x) dx = \int_0^3 (4x + 2) dx $$
Calculate integral:
$$ V = [2x^2 + 2x ]_0^3 = (2 \times 9 + 6) - (0 + 0) = 18 + 6 = 24 \text{ cm}^3 $$
Answer: Volume \(V = 24\) cm³
Example 3: Power output of a machine varies as \(P(t) = 50 + 10t\) watts over time \(t\) seconds. Find the total energy \(E\) produced in 5 seconds if \(E(0) = 0\).
Given: \(P(t) = \frac{dE}{dt} = 50 + 10t\), \(E(0) = 0\)
Integrate power:
$$ E(t) = \int (50 + 10t) dt + C = 50t + 5t^2 + C $$
Apply initial condition:
$$ E(0) = 0 + 0 + C = 0 \implies C = 0 $$
Calculate energy at \(t=5\):
$$ E(5) = 50 \times 5 + 5 \times 25 = 250 + 125 = 375 \text{ joules} $$
Answer: Total energy \(E = 375\) joules
Example 4: The rate of wear \(w(t)\) on a cutting tool is given by \(w(t) = 0.1 e^{0.2t}\) mm/hour. Find the total wear after 4 hours if initial wear \(w(0) = 0\).
Given: \(\frac{dw}{dt} = 0.1 e^{0.2t}\), \(w(0) = 0\)
Integrate:
$$ w(t) = \int 0.1 e^{0.2t} dt + C = 0.1 \times \frac{1}{0.2} e^{0.2t} + C = 0.5 e^{0.2t} + C $$
Apply initial condition:
$$ w(0) = 0.5 e^0 + C = 0.5 + C = 0 \implies C = -0.5 $$
Calculate wear at \(t=4\):
$$ w(4) = 0.5 e^{0.8} - 0.5 $$
Calculate \(e^{0.8} \approx 2.2255\):
$$ w(4) = 0.5 \times 2.2255 - 0.5 = 1.11275 - 0.5 = 0.61275 \text{ mm} $$
Answer: Total wear after 4 hours is approximately 0.613 mm
Example 5: The velocity of a piston varies with time as \(v(t) = 12t - 3t^2\) m/s. Find displacement \(s(t)\) after \(t\) seconds if piston starts at position \(s(0) = 0\).
Given: \(v(t) = \frac{ds}{dt} = 12t - 3t^2\), \(s(0) = 0\)
Integrate velocity:
$$ s(t) = \int (12t - 3t^2) dt + C = 6t^2 - t^3 + C $$
Apply initial condition:
$$ s(0) = 0 + 0 + C = 0 \implies C=0 $$
Answer: \(s(t) = 6t^2 - t^3\) m
A load on a beam is given by \(L = 3x^3 - 12x^2 + 9x + 5\) kN, where \(x\) is length in metres. Find the stationary points and classify each. (6 marks)
The cross-sectional area of a rod varies as \(A(x) = 5 - 0.5x\) cm² over length \(x\) m. Calculate the volume between \(x=1\) m and \(x=4\) m. (4 marks)
The velocity of a conveyor belt is \(v = 4t^2 - 2t\) m/s. Determine the displacement after 3 seconds if initial displacement is zero. (5 marks)
The wear rate of a tool is \(w(t) = 0.2 t e^{-0.1 t}\) mm/hour. Compute the total wear after 5 hours, assuming zero wear at \(t=0\). (6 marks)
The efficiency \(E\) of a machine is modeled by \(E = 2x^3 - 15x^2 + 36x + 5\). Find the values of \(x\) where efficiency is stationary and classify each point. (6 marks)
Integration is the reverse process of differentiation and allows the calculation of quantities when given rates of change. In mechanical production, integration is crucial for determining total displacement, volume, or accumulated quantities from velocity, flow rate, or density functions. The fundamental formula to find the accumulated quantity \(Q\) from a rate of change function \( \frac{dQ}{dt} \) is:
$$ Q = \int \frac{dQ}{dt} \, dt $$
where \( \frac{dQ}{dt} \) is the rate of change with respect to time or another variable.
Example 1: A machine part moves with velocity \(v(t) = 3t^2\) m/s. Find the displacement after 4 seconds.
Given: \(v(t) = 3t^2\) m/s, \(t = 4\) s
$$ s(t) = \int v(t) \, dt $$
$$ s(t) = \int 3t^2 \, dt $$
$$ s(t) = 3 \times \frac{t^3}{3} + C = t^3 + C $$
Assuming displacement zero at \(t=0\), \(C=0\):
$$ s(4) = 4^3 = 64 \text{ m} $$
Answer: 64 m
Example 2: The rate of metal removal in a lathe is \( \frac{dm}{dt} = 5t \) grams per minute. Calculate the mass removed after 6 minutes if initially no metal was removed.
Given: \( \frac{dm}{dt} = 5t \) g/min, \(t=6\) min
$$ m = \int 5t \, dt $$
$$ m = 5 \times \frac{t^2}{2} + C = \frac{5}{2} t^2 + C $$
Initial condition: \( m(0) = 0 \Rightarrow C=0 \)
$$ m(6) = \frac{5}{2} \times 6^2 = \frac{5}{2} \times 36 = 90 \text{ g} $$
Answer: 90 g
Example 3: A fluid flows through a pipe with flow rate \(Q(t) = 10 \sin(t)\) liters per second. Find the total volume of fluid passed through the pipe between \(t=0\) and \(t=\pi\) seconds.
Given: \(Q(t) = 10 \sin(t)\) L/s, \(0 \leq t \leq \pi\)
$$ V = \int_0^{\pi} 10 \sin(t) \, dt $$
$$ V = 10 [-\cos(t) ]_0^{\pi} = 10 [-\cos(\pi) + \cos(0) ] $$
$$ V = 10 [-(-1) + 1 ] = 10 (1 + 1) = 20 \text{ L} $$
Answer: 20 L
Example 4: The temperature \(T\) (in °C) of a metal rod changes at a rate given by \( \frac{dT}{dx} = 4x - 2\), where \(x\) is the distance in meters along the rod. Find the temperature at \(x=3\) m if \(T=5^{\circ}C\) at \(x=0\).
Given: \( \frac{dT}{dx} = 4x - 2\), \(T(0) = 5^{\circ}C\), \(x=3\) m
$$ T(x) = \int (4x - 2) \, dx = 4 \times \frac{x^2}{2} - 2x + C = 2x^2 - 2x + C $$
From initial condition:
$$ T(0) = 5 = 2 \times 0 - 2 \times 0 + C = C $$
So, \(C=5\)
Calculate \(T(3)\):
$$ T(3) = 2 \times 3^2 - 2 \times 3 + 5 = 2 \times 9 - 6 + 5 = 18 - 6 + 5 = 17^{\circ}C $$
Answer: 17°C
Example 5: A production machine’s wear rate is \( \frac{dw}{dt} = 0.1 e^{0.5t} \) mm per hour. Find the total wear after 4 hours if initially there was no wear.
Given: \( \frac{dw}{dt} = 0.1 e^{0.5t} \), \(w(0) = 0\), \(t=4\) hr
$$ w = \int 0.1 e^{0.5t} \, dt $$
$$ w = 0.1 \times \frac{e^{0.5t}}{0.5} + C = 0.2 e^{0.5t} + C $$
Initial condition \(w(0) = 0 \Rightarrow 0.2 e^0 + C = 0 \Rightarrow 0.2 + C = 0 \Rightarrow C = -0.2\)
Calculate \(w(4)\):
$$ w(4) = 0.2 e^{0.5 \times 4} - 0.2 = 0.2 e^{2} - 0.2 $$
$$ e^{2} \approx 7.389 $$
$$ w(4) = 0.2 \times 7.389 - 0.2 = 1.4778 - 0.2 = 1.2778 \text{ mm} $$
Answer: 1.28 mm
The velocity of a conveyor belt is \(v(t) = 4t^3\) m/s. Calculate the displacement after 3 seconds. (4 marks)
The rate of coolant flow in a machine is \(Q(t) = 8 e^{-0.2t}\) L/min. Find the total coolant volume passed in 5 minutes. (5 marks)
The stress in a rod changes at a rate \( \frac{d\sigma}{dx} = 6x - 3 \) MPa/m. Find the stress at 4 m if the stress at 0 m is 12 MPa. (5 marks)
The wear rate of a cutting tool is \( \frac{dw}{dt} = 0.05 t^2 \) mm/hr. Calculate the total wear after 6 hours. (4 marks)
Standard integrals form the foundation for solving complex engineering problems involving area, volume, and other physical quantities. Mechanical production technicians use these integrals to calculate geometric properties and material quantities. The standard integral formulas provide quick ways to integrate common functions.
Some key standard integrals are:
$$ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 $$
$$ \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C $$
$$ \int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C $$
$$ \int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C $$
$$ \int \frac{1}{x} \, dx = \ln |x| + C $$
Example 1: Find the integral of \(x^4\).
Given: \(f(x) = x^4\)
$$ \int x^4 \, dx = \frac{x^{4+1}}{4+1} + C = \frac{x^5}{5} + C $$
Answer: \(\frac{x^5}{5} + C\)
Example 2: Calculate \( \int e^{3x} \, dx \).
Given: \(f(x) = e^{3x}\)
$$ \int e^{3x} \, dx = \frac{1}{3} e^{3x} + C $$
Answer: \(\frac{1}{3} e^{3x} + C\)
Example 3: Determine \( \int \sin(5x) \, dx \).
Given: \(f(x) = \sin(5x)\)
$$ \int \sin(5x) \, dx = -\frac{1}{5} \cos(5x) + C $$
Answer: \(-\frac{1}{5} \cos(5x) + C\)
Example 4: Evaluate \( \int \cos(2x) \, dx \).
Given: \(f(x) = \cos(2x)\)
$$ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C $$
Answer: \(\frac{1}{2} \sin(2x) + C\)
Example 5: Find \( \int \frac{1}{x} \, dx \).
Given: \(f(x) = \frac{1}{x}\)
$$ \int \frac{1}{x} \, dx = \ln|x| + C $$
Answer: \(\ln|x| + C\)
Integrate \(x^3\) with respect to \(x\). (3 marks)
Find \( \int e^{-2x} \, dx \). (4 marks)
Compute \( \int \sin(4x) \, dx \). (4 marks)
Evaluate \( \int \cos(3x) \, dx \). (4 marks)
Calculate \( \int \frac{1}{x} \, dx \). (3 marks)
The definite integral calculates the exact area under a curve between two limits, which is crucial in mechanical production for determining quantities like material volume, work done by forces, and energy consumption over intervals. In Kenyan workshops and manufacturing plants, definite integrals help evaluate precise measurements needed in machining and assembly processes.
The definite integral of a function \( f(x) \) from \( a \) to \( b \) is expressed as
$$ \int_a^b f(x) \, dx = F(b) - F(a) $$
where \( F(x) \) is any antiderivative of \( f(x) \).
Example 1: Calculate the area under the curve \( f(x) = 3x^2 \) from \( x = 1 \) m to \( x = 3 \) m, representing the cross-sectional profile of a metal rod.
Given: \( f(x) = 3x^{2} \), \( a = 1 \), \( b = 3 \)
First, find the antiderivative:
$$ F(x) = \int 3x^{2} \, dx = x^{3} $$
Evaluate at limits:
$$ F(3) - F(1) = 3^{3} - 1^{3} = 27 - 1 = 26 $$
Answer: 26 \, \text{m}^3 (volume or area units depending on context)
Example 2: Find the work done by a variable force \( F(x) = 5x \) N moving a machine part from \( x = 0 \) m to \( x = 4 \) m.
Given: \( F(x) = 5x \), \( a = 0 \), \( b = 4 \)
Antiderivative:
$$ F(x) = \int 5x \, dx = \frac{5x^{2}}{2} $$
Evaluate:
$$ \frac{5 \times 4^{2}}{2} - \frac{5 \times 0^{2}}{2} = \frac{5 \times 16}{2} - 0 = \frac{80}{2} = 40 $$
Answer: 40 \, \text{J} (joules of work done)
Example 3: Calculate the volume of material removed when a lathe cuts a profile defined by \( f(x) = 2x + 1 \) cm over the length from 0 to 5 cm.
Given: \( f(x) = 2x + 1 \), \( a = 0 \), \( b = 5 \)
Antiderivative:
$$ F(x) = \int (2x + 1) \, dx = x^{2} + x $$
Evaluate:
$$ F(5) - F(0) = (5^{2} + 5) - (0 + 0) = (25 + 5) - 0 = 30 $$
Answer: 30 \, \text{cm}^3
Example 4: A pressure distribution on a piston is given by \( p(x) = 8 - 0.5x \) kPa along its 6 cm diameter radius from center (0) to edge (6 cm). Find the total force exerted.
Given: \( p(x) = 8 - 0.5x \), \( a = 0 \), \( b = 6 \)
Antiderivative:
$$ F(x) = \int (8 - 0.5x) \, dx = 8x - 0.25x^{2} $$
Evaluate:
$$ F(6) - F(0) = (8 \times 6 - 0.25 \times 6^{2}) - 0 = (48 - 9) = 39 $$
Answer: 39 \, \text{kN} (assuming width 1 m for force units)
Example 5: Determine the total heat energy \( Q \) generated by a machine part with heat generation rate \( q(t) = 4t^{2} + 3t \) kW over 0 to 3 hours.
Given: \( q(t) = 4t^{2} + 3t \), \( a = 0 \), \( b = 3 \)
Antiderivative:
$$ Q(t) = \int (4t^{2} + 3t) \, dt = \frac{4t^{3}}{3} + \frac{3t^{2}}{2} $$
Evaluate:
$$ Q(3) - Q(0) = (\frac{4 \times 27}{3} + \frac{3 \times 9}{2}) - 0 = (36 + 13.5) = 49.5 $$
Answer: 49.5 \, \text{kWh}
Integral calculus often requires specialized methods to evaluate integrals that cannot be solved by straightforward antiderivatives. Mechanical production technicians encounter such integrals when analyzing complex force distributions, material removal rates, or thermal profiles. This section covers three essential techniques: integration by parts, substitution, and partial fractions.
Integration by parts is a technique derived from the product rule of differentiation. It is used when the integral is a product of two functions and cannot be simplified directly. The formula is
$$ \int u \, dv = uv - \int v \, du $$
where \( u \) and \( dv \) are parts of the integrand chosen to simplify the integration.
Example 1: Evaluate \( \int x e^{x} \, dx \), representing displacement times exponential decay in a mechanical system.
Given: \( u = x \), \( dv = e^{x} dx \)
Calculate:
$$ du = dx, \quad v = e^{x} $$
Apply formula:
$$ \int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x} + C $$
Example 2: Find \( \int \ln x \, dx \), useful for entropy calculations in thermodynamics.
Given: \( u = \ln x \), \( dv = dx \)
Calculate:
$$ du = \frac{1}{x} dx, \quad v = x $$
Apply formula:
$$ \int \ln x \, dx = x \ln x - \int x \times \frac{1}{x} dx = x \ln x - \int 1 dx = x \ln x, x + C $$
Example 3: Evaluate \( \int x \cos 2x \, dx \), representing torque varying with angular displacement.
Given: \( u = x \), \( dv = \cos 2x \, dx \)
Calculate:
$$ du = dx, \quad v = \frac{\sin 2x}{2} $$
Apply formula:
$$ \int x \cos 2x \, dx = x \times \frac{\sin 2x}{2} - \int \frac{\sin 2x}{2} dx = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} + C $$
Example 4: Find \( \int x^{2} e^{x} \, dx \), representing power times exponential growth.
Given: \( u = x^{2} \), \( dv = e^{x} dx \)
Calculate:
$$ du = 2x dx, \quad v = e^{x} $$
Apply formula:
$$ \int x^{2} e^{x} dx = x^{2} e^{x} - \int 2x e^{x} dx $$
Now integrate \( \int 2x e^{x} dx \) by parts again:
Set \( u = 2x \), \( dv = e^{x} dx \):
$$ du = 2 dx, \quad v = e^{x} $$
Apply formula:
$$ \int 2x e^{x} dx = 2x e^{x} - \int 2 e^{x} dx = 2x e^{x} - 2 e^{x} + C $$
Return:
$$ \int x^{2} e^{x} dx = x^{2} e^{x} - (2x e^{x} - 2 e^{x}) + C = e^{x} (x^{2} - 2x + 2) + C $$
Example 5: Evaluate \( \int x \ln x \, dx \), representing logarithmic stress variation over length.
Given: \( u = \ln x \), \( dv = x dx \)
Calculate:
$$ du = \frac{1}{x} dx, \quad v = \frac{x^{2}}{2} $$
Apply formula:
$$ \int x \ln x \, dx = \frac{x^{2}}{2} \ln x - \int \frac{x^{2}}{2} \times \frac{1}{x} dx = \frac{x^{2}}{2} \ln x - \frac{1}{2} \int x dx = \frac{x^{2}}{2} \ln x - \frac{1}{2} \times \frac{x^{2}}{2} + C = \frac{x^{2}}{2} \ln x - \frac{x^{2}}{4} + C $$
Substitution simplifies integrals by changing variables to transform the integral into a more manageable form. It is especially useful when the integrand contains composite functions or expressions involving powers and roots common in mechanical contexts such as velocity profiles or stress-strain curves.
The substitution formula is
$$ \int f(g(x)) g'(x) \, dx = \int f(u) \, du $$
where \( u = g(x) \).
Example 1: Evaluate \( \int (3x + 1)^4 \, dx \), representing a force function raised to the fourth power.
Let \( u = 3x + 1 \), then \( du = 3 dx \), or \( dx = \frac{du}{3} \).
Rewrite:
$$ \int (3x + 1)^4 dx = \int u^{4} \times \frac{du}{3} = \frac{1}{3} \int u^{4} du = \frac{1}{3} \times \frac{u^{5}}{5} + C = \frac{(3x + 1)^5}{15} + C $$
Example 2: Calculate \( \int \frac{x}{\sqrt{1 + x^{2}}} dx \), useful in stress-strain relations.
Let \( u = 1 + x^{2} \), then \( du = 2x dx \), so \( x dx = \frac{du}{2} \).
Rewrite:
$$ \int \frac{x}{\sqrt{1 + x^{2}}} dx = \int \frac{1}{\sqrt{u}} \times \frac{du}{2} = \frac{1}{2} \int u^{-\frac{1}{2}} du = \frac{1}{2} \times 2 u^{\frac{1}{2}} + C = \sqrt{1 + x^{2}} + C $$
Example 3: Evaluate \( \int \sin(2x + 3) \, dx \), representing oscillating forces.
Let \( u = 2x + 3 \), then \( du = 2 dx \), \( dx = \frac{du}{2} \).
Rewrite:
$$ \int \sin(2x + 3) dx = \int \sin u \times \frac{du}{2} = \frac{1}{2} \int \sin u \, du = -\frac{1}{2} \cos u + C = -\frac{1}{2} \cos(2x + 3) + C $$
Example 4: Calculate \( \int \frac{2x}{(x^{2} + 4)^{3/2}} \, dx \), representing bending moment distribution.
Let \( u = x^{2} + 4 \), then \( du = 2x dx \).
Rewrite:
$$ \int \frac{2x}{(x^{2} + 4)^{3/2}} dx = \int u^{-\frac{3}{2}} du = \frac{u^{-1/2}}{-\frac{1}{2}} + C = -2 u^{-\frac{1}{2}} + C = -\frac{2}{\sqrt{x^{2} + 4}} + C $$
Example 5: Evaluate \( \int x e^{x^{2}} dx \), representing an exponential growth with quadratic argument.
Let \( u = x^{2} \), then \( du = 2x dx \), so \( x dx = \frac{du}{2} \).
Rewrite:
$$ \int x e^{x^{2}} dx = \int e^{u} \times \frac{du}{2} = \frac{1}{2} \int e^{u} du = \frac{1}{2} e^{u} + C = \frac{1}{2} e^{x^{2}} + C $$
Partial fraction decomposition breaks down complex rational functions into simpler fractions easier to integrate. This technique is fundamental when dealing with ratios of polynomials encountered in mechanical system transfer functions or fluid flow rate equations.
The method applies when the degree of the numerator is less than the denominator, or after division if otherwise.
Example 1: Integrate \( \int \frac{1}{x(x+1)} dx \).
Decompose:
$$ \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} $$
Multiply both sides by \( x(x+1) \):
$$ 1 = A(x+1) + Bx = Ax + A + Bx = (A + B)x + A $$
Equate coefficients:
Solve:
$$ A = 1, \quad B = -1 $$
Integral:
$$ \int \frac{1}{x(x+1)} dx = \int \frac{1}{x} dx - \int \frac{1}{x+1} dx = \ln|x| - \ln|x+1| + C = \ln |\frac{x}{x+1}| + C $$
Example 2: Evaluate \( \int \frac{3x + 5}{(x+1)(x+2)} dx \).
Decompose:
$$ \frac{3x + 5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} $$
Multiply both sides by \( (x+1)(x+2) \):
$$ 3x + 5 = A(x+2) + B(x+1) = (A + B) x + (2A + B) $$
Equate coefficients:
Solve:
Subtract first from second:
$$ 5 - 3 = (2A + B) - (A + B) \Rightarrow 2 = A \Rightarrow A = 2 $$
Then:
$$ 3 = 2 + B \Rightarrow B = 1 $$
Integral:
$$ \int \frac{3x + 5}{(x+1)(x+2)} dx = \int \frac{2}{x+1} dx + \int \frac{1}{x+2} dx = 2 \ln|x+1| + \ln|x+2| + C $$
Example 3: Find \( \int \frac{2x^{2} + 3x + 1}{x(x+1)^{2}} dx \).
Decompose:
$$ \frac{2x^{2} + 3x + 1}{x(x+1)^{2}} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^{2}} $$
Multiply both sides by denominator:
$$ 2x^{2} + 3x + 1 = A(x+1)^{2} + B x (x+1) + C x $$
Expand:
$$ A(x^{2} + 2x + 1) + B x^{2} + B x + C x = A x^{2} + 2 A x + A + B x^{2} + B x + C x $$
Group terms:
$$ (A + B) x^{2} + (2A + B + C) x + A $$
Equate coefficients:
Solve:
From last:
$$ A = 1 $$
Then:
$$ 2 = 1 + B \Rightarrow B = 1 $$
Then:
$$ 3 = 2(1) + 1 + C = 2 + 1 + C \Rightarrow C = 0 $$
Integral:
$$ \int \frac{2x^{2} + 3x + 1}{x(x+1)^{2}} dx = \int \frac{1}{x} dx + \int \frac{1}{x+1} dx + \int \frac{0}{(x+1)^{2}} dx = \ln|x| + \ln|x+1| + C $$
Example 4: Evaluate \( \int \frac{4x + 7}{x^{2} - x - 2} dx \).
Factor denominator:
$$ x^{2} - x - 2 = (x - 2)(x + 1) $$
Decompose:
$$ \frac{4x + 7}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} $$
Multiply both sides:
$$ 4x + 7 = A(x + 1) + B(x - 2) = (A + B) x + (A - 2B) $$
Equate coefficients:
Solve:
From first:
$$ B = 4 - A $$
Substitute into second:
$$ 7 = A - 2(4 - A) = A - 8 + 2A = 3A - 8 $$
$$ 3A = 15 \Rightarrow A = 5 $$
Then:
$$ B = 4 - 5 = -1 $$
Integral:
$$ \int \frac{4x + 7}{x^{2} - x - 2} dx = \int \frac{5}{x-2} dx - \int \frac{1}{x+1} dx = 5 \ln|x-2| - \ln|x+1| + C $$
Example 5: Find \( \int \frac{3x^{2} + 2x + 1}{(x+1)(x^{2} + 1)} dx \).
Decompose:
$$ \frac{3x^{2} + 2x + 1}{(x+1)(x^{2} + 1)} = \frac{A}{x+1} + \frac{Bx + C}{x^{2} + 1} $$
Multiply both sides:
$$ 3x^{2} + 2x + 1 = A(x^{2} + 1) + (Bx + C)(x + 1) = A x^{2} + A + B x^{2} + B x + C x + C $$
Group terms:
$$ (A + B) x^{2} + (B + C) x + (A + C) $$
Equate coefficients:
Solve:
From first:
$$ B = 3 - A $$
From third:
$$ C = 1 - A $$
Substitute into second:
$$ 2 = (3 - A) + (1 - A) = 4 - 2A \Rightarrow 2A = 4 - 2 = 2 \Rightarrow A = 1 $$
Then:
$$ B = 3 - 1 = 2 $$
$$ C = 1 - 1 = 0 $$
Integral:
$$ \int \frac{3x^{2} + 2x + 1}{(x+1)(x^{2} + 1)} dx = \int \frac{1}{x+1} dx + \int \frac{2x}{x^{2} + 1} dx = \ln|x+1| + \ln(x^{2} + 1) + C $$
Integration by parts is a fundamental technique in calculus used to integrate products of functions. In mechanical production tasks, such as calculating work done by variable forces or analyzing energy in systems with varying parameters, this method simplifies complex integrals into manageable forms. The formula for integration by parts is derived from the product rule of differentiation and is given by:
$$\int u \, dv = uv - \int v \, du$$
where \(u\) and \(dv\) are differentiable functions of a variable, typically \(x\).
Example 1: A mechanical technician needs to find the work done when a force \(F(x) = x e^x\) N acts over a distance from \(x=0\) m to \(x=2\) m. Calculate the work done.
Given: \(F(x) = x e^x\), limits: 0 to 2 m
$$\text{Work} = \int_0^2 x e^x \, dx$$
Choose \(u = x\), so \(du = dx\), and \(dv = e^x dx\), so \(v = e^x\).
$$\int x e^x dx = x e^x - \int e^x dx$$
$$= x e^x, e^x + C$$
Evaluate from 0 to 2:
$$[2 e^2 - e^2] - [0 - e^0] = (2 e^2 - e^2) - (0 - 1) = e^2 + 1$$
Calculate numerical value:
$$e^2 \approx 7.389$$
$$\text{Work} = 7.389 + 1 = 8.389 \, \text{J}$$
Answer: 8.389 J
Example 2: Find the integral \(\int x \cos x \, dx\), which can represent the displacement integral in a mechanical vibration problem.
Given: \(f(x) = x \cos x\)
Choose \(u = x\), so \(du = dx\), and \(dv = \cos x dx\), so \(v = \sin x\).
$$\int x \cos x dx = x \sin x - \int \sin x dx$$
$$= x \sin x + \cos x + C$$
Answer: \(x \sin x + \cos x + C\)
Example 3: Calculate the integral \(\int_1^3 \ln x \, dx\), useful in thermodynamics for entropy calculations involving logarithmic terms.
Given: \(f(x) = \ln x\), limits 1 to 3
Set \(u = \ln x\), so \(du = \frac{1}{x} dx\), and \(dv = dx\), so \(v = x\).
$$\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - \int 1 dx$$
$$= x \ln x, x + C$$
Evaluate from 1 to 3:
$$[3 \ln 3 - 3] - [1 \ln 1 - 1] = (3 \ln 3 - 3) - (0 - 1) = 3 \ln 3 - 3 + 1 = 3 \ln 3 - 2$$
Calculate numerical value:
$$\ln 3 \approx 1.0986$$
$$3 \times 1.0986 - 2 = 3.2958 - 2 = 1.2958$$
Answer: 1.296 (rounded) units
Example 4: Determine the value of \(\int x^2 e^x dx\), which could model power output over time in a system with exponentially increasing load.
Given: \(f(x) = x^2 e^x\)
Choose \(u = x^2\), so \(du = 2x dx\), and \(dv = e^x dx\), so \(v = e^x\).
Apply integration by parts:
$$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$$
Now evaluate \(\int 2x e^x dx\) by parts again:
Set \(u = 2x\), so \(du = 2 dx\), and \(dv = e^x dx\), so \(v = e^x\).
$$\int 2x e^x dx = 2x e^x - \int 2 e^x dx = 2x e^x - 2 e^x + C$$
Substitute back:
$$\int x^2 e^x dx = x^2 e^x - (2x e^x - 2 e^x) + C = e^x (x^2 - 2x + 2) + C$$
Answer: \(e^x (x^2 - 2x + 2) + C\)
Example 5: Evaluate \(\int_0^{\pi} x \sin x \, dx\), which could represent a moment calculation in a beam subjected to sinusoidal loading.
Given: \(f(x) = x \sin x\), limits 0 to \(\pi\)
Choose \(u = x\), so \(du = dx\), and \(dv = \sin x dx\), so \(v = -\cos x\).
$$\int x \sin x dx = -x \cos x + \int \cos x dx = -x \cos x + \sin x + C$$
Evaluate from 0 to \(\pi\):
$$[-\pi \cos \pi + \sin \pi] - [0 + 0] = [-\pi (-1) + 0] - 0 = \pi$$
Answer: \(\pi\) units
Integration by substitution is a method used to simplify integrals by changing variables, which is particularly useful in mechanical production when dealing with composite functions such as stress-strain relations or velocity profiles. It transforms a complex integral into a simpler form by substituting a part of the integrand with a single variable. The general formula is:
$$\int f(g(x)) g'(x) dx = \int f(u) du$$
where \(u = g(x)\) and \(du = g'(x) dx\).
Example 1: Evaluate \(\int 2x (x^2 + 1)^5 dx\), representing a force relation dependent on displacement squared.
Given: \(f(x) = 2x (x^2 + 1)^5\)
Set \(u = x^2 + 1\), so \(du = 2x dx\).
$$\int 2x (x^2 + 1)^5 dx = \int u^5 du$$
$$= \frac{u^6}{6} + C = \frac{(x^2 + 1)^6}{6} + C$$
Answer: \(\frac{(x^2 + 1)^6}{6} + C\)
Example 2: Calculate \(\int \frac{\cos(3x)}{5} dx\), which could model periodic motion with frequency scaling.
Given: \(f(x) = \frac{\cos(3x)}{5}\)
Set \(u = 3x\), so \(du = 3 dx\), hence \(dx = \frac{du}{3}\).
$$\int \frac{\cos(3x)}{5} dx = \frac{1}{5} \int \cos u \cdot \frac{du}{3} = \frac{1}{15} \int \cos u du$$
$$= \frac{1}{15} \sin u + C = \frac{1}{15} \sin 3x + C$$
Answer: \(\frac{1}{15} \sin 3x + C\)
Example 3: Find \(\int_0^1 x e^{x^2} dx\), relevant in calculating exponential growth effects over squared displacement.
Given: \(f(x) = x e^{x^2}\), limits 0 to 1
Set \(u = x^2\), so \(du = 2x dx\), \(x dx = \frac{du}{2}\).
Rewrite integral:
$$\int_0^1 x e^{x^2} dx = \int_0^{1^2} e^{u} \frac{du}{2} = \frac{1}{2} \int_0^1 e^u du$$
$$= \frac{1}{2} [e^u]_0^1 = \frac{1}{2} (e^1 - e^0) = \frac{1}{2} (e - 1)$$
Calculate numerical value:
$$e \approx 2.718$$
$$\frac{1}{2} (2.718 - 1) = \frac{1}{2} (1.718) = 0.859$$
Answer: 0.859 units
Example 4: Evaluate \(\int \frac{dx}{x \ln x}\), which is useful in logarithmic decay processes.
Given: \(f(x) = \frac{1}{x \ln x}\)
Set \(u = \ln x\), so \(du = \frac{1}{x} dx\), hence \(dx = x du\).
Rewrite integral:
$$\int \frac{dx}{x \ln x} = \int \frac{1}{x u} dx = \int \frac{1}{u} \cdot \frac{dx}{x} = \int \frac{1}{u} du$$
$$= \ln |u| + C = \ln |\ln x| + C$$
Answer: \(\ln |\ln x| + C\)
Example 5: Calculate \(\int_0^{\pi/4} \tan x \, dx\), which appears in angular velocity integration.
Given: \(f(x) = \tan x\), limits 0 to \(\pi/4\)
Recall \(\tan x = \frac{\sin x}{\cos x}\).
Set \(u = \cos x\), so \(du = -\sin x dx\), or \(-du = \sin x dx\).
Rewrite integral:
$$\int_0^{\pi/4} \tan x dx = \int_0^{\pi/4} \frac{\sin x}{\cos x} dx = - \int_{u(0)}^{u(\pi/4)} \frac{1}{u} du$$
Calculate limits:
\(u(0) = \cos 0 = 1\), \(u(\pi/4) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\)
Integral becomes:
$$- \int_1^{\frac{\sqrt{2}}{2}} \frac{1}{u} du = - [\ln |u|]_1^{\frac{\sqrt{2}}{2}} = - (\ln \frac{\sqrt{2}}{2} - \ln 1) = - \ln \frac{\sqrt{2}}{2}$$
Simplify:
$$\ln \frac{\sqrt{2}}{2} = \ln \sqrt{2} - \ln 2 = \frac{1}{2} \ln 2 - \ln 2 = -\frac{1}{2} \ln 2$$
Therefore:
$$- \ln \frac{\sqrt{2}}{2} = - (-\frac{1}{2} \ln 2 ) = \frac{1}{2} \ln 2$$
Calculate numerical value:
$$\ln 2 \approx 0.693$$
$$\frac{1}{2} \times 0.693 = 0.3465$$
Answer: 0.347 units (rounded)
Partial fractions decomposition is a technique used to express a rational function as a sum of simpler fractions, which is especially useful for integrating complex algebraic expressions encountered in mechanical production processes such as fluid flow analysis and control system design. This method breaks down complicated fractions into components that are easier to integrate or differentiate, facilitating practical engineering calculations.
A rational function \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and the degree of \(P(x)\) is less than the degree of \(Q(x)\), can be decomposed into partial fractions of the form:
$$ \frac{P(x)}{Q(x)} = \sum \frac{A}{(x - \alpha)} + \sum \frac{Bx + C}{(x^2 + px + q)} + \cdots $$
where the denominator is factored into linear and irreducible quadratic factors.
Example 1: Decompose \(\frac{7x + 5}{(x + 1)(x + 2)}\) into partial fractions.
Given: Numerator \(7x + 5\), denominator \((x + 1)(x + 2)\)
$$ \frac{7x + 5}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} $$
Multiply both sides by \((x + 1)(x + 2)\):
$$ 7x + 5 = A(x + 2) + B(x + 1) $$
Expand:
$$ 7x + 5 = A x + 2A + B x + B $$
Group terms:
$$ 7x + 5 = (A + B) x + (2A + B) $$
Equate coefficients:
For \(x\): \(7 = A + B\)
For constants: \(5 = 2A + B\)
From the first:
$$ B = 7 - A $$
Substitute into second:
$$ 5 = 2A + (7 - A) = A + 7 $$
$$ A = 5 - 7 = -2 $$
Then:
$$ B = 7 - (-2) = 9 $$
Answer:
$$ \frac{7x + 5}{(x + 1)(x + 2)} = \frac{-2}{x + 1} + \frac{9}{x + 2} $$
Example 2: Decompose \(\frac{3x^2 + 5x + 2}{(x - 1)(x^2 + 1)}\) into partial fractions.
Given: Numerator \(3x^2 + 5x + 2\), denominator \((x - 1)(x^2 + 1)\)
$$ \frac{3x^2 + 5x + 2}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} $$
Multiply both sides by denominator:
$$ 3x^2 + 5x + 2 = A(x^2 + 1) + (Bx + C)(x - 1) $$
Expand right side:
$$ 3x^2 + 5x + 2 = A x^2 + A + Bx^2 - Bx + Cx - C $$
Group terms:
$$ 3x^2 + 5x + 2 = (A + B) x^2 + (-B + C) x + (A - C) $$
Equate coefficients:
For \(x^2\): \(3 = A + B\)
For \(x\): \(5 = -B + C\)
For constants: \(2 = A - C\)
From first:
$$ B = 3 - A $$
From third:
$$ C = A - 2 $$
Substitute \(B\) and \(C\) into second:
$$ 5 = - (3 - A) + (A - 2) = -3 + A + A - 2 = 2A - 5 $$
$$ 2A = 10 \Rightarrow A = 5 $$
Then:
$$ B = 3 - 5 = -2 $$
$$ C = 5 - 2 = 3 $$
Answer:
$$ \frac{3x^2 + 5x + 2}{(x - 1)(x^2 + 1)} = \frac{5}{x - 1} + \frac{-2x + 3}{x^2 + 1} $$
Example 3: Decompose \(\frac{2x^2 + 3x + 1}{x(x - 1)^2}\) into partial fractions.
Given: Numerator \(2x^2 + 3x + 1\), denominator \(x(x - 1)^2\)
$$ \frac{2x^2 + 3x + 1}{x(x - 1)^2} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2} $$
Multiply both sides by denominator:
$$ 2x^2 + 3x + 1 = A(x - 1)^2 + B x (x - 1) + C x $$
Expand each term:
\(A(x^2 - 2x + 1) = A x^2 - 2A x + A\)
\(B x (x - 1) = B x^2 - B x\)
\(C x = C x\)
Sum:
$$ 2x^2 + 3x + 1 = (A + B) x^2 + (-2A - B + C) x + A $$
Equate coefficients:
For \(x^2\): \(2 = A + B\)
For \(x\): \(3 = -2A - B + C\)
For constants: \(1 = A\)
From constants:
$$ A = 1 $$
Substitute into \(x^2\):
$$ 2 = 1 + B \Rightarrow B = 1 $$
Substitute \(A\) and \(B\) into \(x\):
$$ 3 = -2(1) - 1 + C = -2 -1 + C = C - 3 $$
$$ C = 6 $$
Answer:
$$ \frac{2x^2 + 3x + 1}{x(x - 1)^2} = \frac{1}{x} + \frac{1}{x - 1} + \frac{6}{(x - 1)^2} $$
Example 4: Decompose \(\frac{5x^3 + 4x^2 + 3x + 2}{(x^2 + 1)(x - 2)}\) into partial fractions.
Given: Numerator \(5x^3 + 4x^2 + 3x + 2\), denominator \((x^2 + 1)(x - 2)\)
Set:
$$ \frac{5x^3 + 4x^2 + 3x + 2}{(x^2 + 1)(x - 2)} = \frac{A x + B}{x^2 + 1} + \frac{C}{x - 2} $$
Multiply both sides by denominator:
$$ 5x^3 + 4x^2 + 3x + 2 = (A x + B)(x - 2) + C (x^2 + 1) $$
Expand:
$$ (A x + B)(x - 2) = A x^2 - 2 A x + B x - 2 B $$
Sum:
$$ 5x^3 + 4x^2 + 3x + 2 = A x^2 - 2 A x + B x - 2 B + C x^2 + C $$
Group terms:
$$ 5x^3 + 4x^2 + 3x + 2 = 0 \cdot x^3 + (A + C) x^2 + (-2 A + B) x + (-2 B + C) $$
Since the left side has \(5x^3\), and right side has no \(x^3\), polynomial division is needed first.
Divide numerator by denominator:
Divide \(5x^3 + 4x^2 + 3x + 2\) by \(x^3 - 2x^2 + x - 2\) (since \((x^2 + 1)(x - 2) = x^3 - 2x^2 + x - 2\)):
Perform polynomial division:
Multiply divisor by 5:
\(5x^3 - 10x^2 + 5x - 10\)
Subtract:
\((5x^3 + 4x^2 + 3x + 2) - (5x^3 - 10x^2 + 5x - 10) = 0 + 14x^2 - 2x + 12\)
So,
$$ \frac{5x^3 + 4x^2 + 3x + 2}{(x^2 + 1)(x - 2)} = 5 + \frac{14x^2 - 2x + 12}{(x^2 + 1)(x - 2)} $$
Now decompose the remainder:
$$ \frac{14x^2 - 2x + 12}{(x^2 + 1)(x - 2)} = \frac{A x + B}{x^2 + 1} + \frac{C}{x - 2} $$
Multiply both sides by denominator:
$$ 14x^2 - 2x + 12 = (A x + B)(x - 2) + C (x^2 + 1) $$
Expand right:
$$ A x^2 - 2 A x + B x - 2 B + C x^2 + C $$
Group terms:
$$ 14x^2 - 2x + 12 = (A + C) x^2 + (-2 A + B) x + (-2 B + C) $$
Equate coefficients:
For \(x^2\): \(14 = A + C\)
For \(x\): \(-2 = -2 A + B\)
For constants: \(12 = -2 B + C\)
From first:
$$ C = 14 - A $$
Substitute into constants:
$$ 12 = -2 B + 14 - A $$
Rearranged:
$$ -2 B = 12 - 14 + A = -2 + A \Rightarrow B = \frac{2 - A}{2} $$
Substitute \(B\) into \(x\) coefficient:
$$ -2 = -2 A + \frac{2 - A}{2} $$
Multiply both sides by 2:
$$ -4 = -4 A + 2 - A $$
Simplify:
$$ -4 = -5 A + 2 $$
$$ -6 = -5 A $$
$$ A = \frac{6}{5} = 1.2 $$
Then:
$$ B = \frac{2 - 1.2}{2} = \frac{0.8}{2} = 0.4 $$
$$ C = 14 - 1.2 = 12.8 $$
Answer:
$$ \frac{5x^3 + 4x^2 + 3x + 2}{(x^2 + 1)(x - 2)} = 5 + \frac{1.2 x + 0.4}{x^2 + 1} + \frac{12.8}{x - 2} $$
Calculus has many applications in mechanical production technology, where it is used to analyze rates of change, areas, volumes, and other quantities essential in design, manufacturing, and quality control. Integration, in particular, is critical in determining areas under curves and volumes of components, which directly relate to material usage, strength, and cost estimation.
Calculating the area under a curve or between curves is fundamental in mechanical engineering for determining material quantities, cross-sectional areas, and performance characteristics of parts and machines. The volume of solids generated by rotating a curve about an axis is crucial in designing parts like shafts, gears, and containers.
The basic formula for the area under a curve \(y = f(x)\) from \(x = a\) to \(x = b\) is:
$$ A = \int_{a}^{b} f(x) \, dx $$
The area between two curves \(y = f(x)\) and \(y = g(x)\), where \(f(x) \geq g(x)\) in \([a, b]\), is:
$$ A = \int_{a}^{b} [f(x) - g(x)] \, dx $$
For volumes generated by revolving a curve \(y = f(x)\) about the x-axis between \(x = a\) and \(x = b\), the volume \(V\) is:
$$ V = \pi \int_{a}^{b} [f(x)]^2 \, dx $$
Example 1: Find the area under the curve \(y = 3x^2\) from \(x = 0\) to \(x = 2\).
Given: \(y = 3x^2\), \(a=0\), \(b=2\)
$$ A = \int_0^2 3x^2 \, dx $$
Integrate:
$$ = 3 \int_0^2 x^2 \, dx = 3 [ \frac{x^3}{3} ]_0^2 $$
Calculate:
$$ = 3 ( \frac{2^3}{3} - \frac{0^3}{3} ) = 3 \times \frac{8}{3} = 8 $$
Answer:
$$ \boxed{8 \text{ units}^2} $$
Example 2: Calculate the area between the curves \(y = x^2 + 1\) and \(y = x\) from \(x = 0\) to \(x = 1\).
Given: \(f(x) = x^2 + 1\), \(g(x) = x\), \(a=0\), \(b=1\)
$$ A = \int_0^1 [ (x^2 + 1) - x ] \, dx = \int_0^1 (x^2 - x + 1) \, dx $$
Integrate term-wise:
$$ = [ \frac{x^3}{3} - \frac{x^2}{2} + x ]_0^1 $$
Calculate:
$$ = ( \frac{1}{3} - \frac{1}{2} + 1 ) - (0) = \frac{1}{3} - \frac{1}{2} + 1 = \frac{2}{6} - \frac{3}{6} + \frac{6}{6} = \frac{5}{6} $$
Answer:
$$ \boxed{\frac{5}{6} \text{ units}^2} $$
Example 3: Find the volume generated by revolving the curve \(y = \sqrt{x}\) about the x-axis from \(x = 0\) to \(x = 4\).
Given: \(y = \sqrt{x} = x^{1/2}\), \(a=0\), \(b=4\)
$$ V = \pi \int_0^4 (x^{1/2})^2 \, dx = \pi \int_0^4 x \, dx $$
Integrate:
$$ = \pi [ \frac{x^2}{2} ]_0^4 = \pi \times \frac{16}{2} = 8 \pi $$
Answer:
$$ \boxed{8 \pi \text{ units}^3} $$
Example 4: Determine the area between the curves \(y = 2x + 3\) and \(y = x^2\) between their points of intersection.
First, find points of intersection:
Solve \(2x + 3 = x^2\):
$$ x^2 - 2x - 3 = 0 $$
Factor:
$$ (x - 3)(x + 1) = 0 $$
Roots:
$$ x = 3, \quad x = -1 $$
Given \(f(x) = 2x + 3\), \(g(x) = x^2\), find which is on top between \([-1, 3]\).
At \(x=0\):
\(f(0) = 3\), \(g(0) = 0\) → \(f(x) > g(x)\) in interval.
Calculate area:
$$ A = \int_{-1}^3 [ (2x + 3) - x^2 ] \, dx = \int_{-1}^3 (-x^2 + 2x + 3) \, dx $$
Integrate term-wise:
$$ = [ -\frac{x^3}{3} + x^2 + 3x ]_{-1}^3 $$
Calculate at \(x=3\):
$$ -\frac{27}{3} + 9 + 9 = -9 + 18 = 9 $$
Calculate at \(x=-1\):
$$ - ( -\frac{(-1)^3}{3} + (-1)^2 + 3(-1) ) = - ( -\frac{-1}{3} + 1 - 3 ) = - ( \frac{1}{3} + 1 - 3 ) = - ( -\frac{5}{3} ) = \frac{5}{3} $$
Area:
$$ 9 - \frac{5}{3} = \frac{27}{3} - \frac{5}{3} = \frac{22}{3} $$
Answer:
$$ \boxed{\frac{22}{3} \text{ units}^2} $$
Example 5: Find the volume of the solid generated by revolving the region bounded by \(y = x^3\), \(x = 0\), and \(x = 2\) about the x-axis.
Given: \(y = x^3\), \(a=0\), \(b=2\)
$$ V = \pi \int_0^2 (x^3)^2 \, dx = \pi \int_0^2 x^6 \, dx $$
Integrate:
$$ = \pi [ \frac{x^7}{7} ]_0^2 = \pi \times \frac{2^7}{7} = \pi \times \frac{128}{7} = \frac{128 \pi}{7} $$
Answer:
$$ \boxed{\frac{128 \pi}{7} \text{ units}^3} $$
Decompose \(\frac{4x + 7}{(x + 1)(x + 3)}\) into partial fractions. (5 marks)
Find the area under the curve \(y = 5x^2 - 2x\) from \(x = 1\) to \(x = 3\). (6 marks)
Calculate the area between the curves \(y = x^2\) and \(y = 4x, x^2\) from their points of intersection. (7 marks)
Find the volume generated by revolving the curve \(y = 2x\) about the x-axis from \(x = 0\) to \(x = 3\). (7 marks)
Decompose \(\frac{2x^2 + 3}{x(x^2 + 1)}\) into partial fractions. (5 marks)
Calculating the area under curves and the volume of solids generated by curves is essential for mechanical production technicians in Kenya. These calculations help in designing machine components, estimating material requirements, and quality control in fabrication processes. Integral calculus provides the mathematical tools to determine these areas and volumes accurately.
The area under a curve \( y = f(x) \) between \( x = a \) and \( x = b \) is given by the definite integral:
$$ A = \int_a^b f(x) \, dx $$
Example 1: Find the area under the curve \( y = 3x \) from \( x = 0 \) to \( x = 4 \).
Given: \( f(x) = 3x \), \( a = 0 \), \( b = 4 \)
$$ A = \int_0^4 3x \, dx $$
$$ = 3 \int_0^4 x \, dx $$
$$ = 3 [ \frac{x^2}{2} ]_0^4 $$
$$ = 3 ( \frac{4^2}{2} - \frac{0^2}{2} ) $$
$$ = 3 ( \frac{16}{2} ) = 3 \times 8 = 24 $$
Answer: 24 square units
Example 2: Calculate the area under the curve \( y = 2x^2 + 1 \) from \( x = 1 \) to \( x = 3 \).
Given: \( f(x) = 2x^2 + 1 \), \( a = 1 \), \( b = 3 \)
$$ A = \int_1^3 (2x^2 + 1) \, dx $$
$$ = \int_1^3 2x^2 \, dx + \int_1^3 1 \, dx $$
$$ = 2 [ \frac{x^3}{3} ]_1^3 + [x]_1^3 $$
$$ = 2 ( \frac{3^3}{3} - \frac{1^3}{3} ) + (3 - 1) $$
$$ = 2 ( \frac{27}{3} - \frac{1}{3} ) + 2 = 2 \times \frac{26}{3} + 2 = \frac{52}{3} + 2 $$
$$ = \frac{52}{3} + \frac{6}{3} = \frac{58}{3} \approx 19.33 $$
Answer: 19.33 square units
Example 3: Find the area under the curve \( y = \sin x \) from \( x = 0 \) to \( x = \pi \).
Given: \( f(x) = \sin x \), \( a = 0 \), \( b = \pi \)
$$ A = \int_0^\pi \sin x \, dx $$
$$ = [-\cos x]_0^\pi $$
$$ = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2 $$
Answer: 2 square units
Example 4: Determine the area between the curve \( y = x^2 \) and the x-axis from \( x = -2 \) to \( x = 2 \).
Given: \( f(x) = x^2 \), \( a = -2 \), \( b = 2 \)
$$ A = \int_{-2}^2 x^2 \, dx $$
Since \( x^2 \) is even,
$$ A = 2 \int_0^2 x^2 \, dx = 2 [ \frac{x^3}{3} ]_0^2 = 2 \times \frac{8}{3} = \frac{16}{3} \approx 5.33 $$
Answer: 5.33 square units
Example 5: Calculate the area under the curve \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 4 \).
Given: \( f(x) = \frac{1}{x} \), \( a = 1 \), \( b = 4 \)
$$ A = \int_1^4 \frac{1}{x} \, dx = [\ln x]_1^4 = \ln 4 - \ln 1 = \ln 4 - 0 = \ln 4 $$
Using \( \ln 4 \approx 1.386 \),
Answer: 1.386 square units
The area between two curves \( y = f(x) \) and \( y = g(x) \), where \( f(x) \ge g(x) \) over \( [a,b] \), is:
$$ A = \int_a^b (f(x) - g(x)) \, dx $$
Example 1: Find the area between \( y = x^2 \) and \( y = x \) from \( x = 0 \) to \( x = 1 \).
Given: \( f(x) = x \), \( g(x) = x^2 \), \( a=0 \), \( b=1 \)
$$ A = \int_0^1 (x, x^2) \, dx = \int_0^1 x \, dx - \int_0^1 x^2 \, dx $$
$$ = [ \frac{x^2}{2} ]_0^1 - [ \frac{x^3}{3} ]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} $$
Answer: \( \frac{1}{6} \) square units
Example 2: Calculate the area between \( y = 4 - x^2 \) and \( y = x \) from \( x = 0 \) to \( x = 2 \).
Given: \( f(x) = 4 - x^2 \), \( g(x) = x \), \( a=0 \), \( b=2 \)
$$ A = \int_0^2 (4 - x^2 - x) \, dx = \int_0^2 (4 - x, x^2) \, dx $$
$$ = [ 4x - \frac{x^2}{2} - \frac{x^3}{3} ]_0^2 = ( 8 - 2 - \frac{8}{3} ) - 0 = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3} $$
Answer: \( \frac{10}{3} \approx 3.33 \) square units
Example 3: Find the area between \( y = \sqrt{x} \) and \( y = x^2 \) from \( x = 0 \) to \( x = 1 \).
Given: \( f(x) = \sqrt{x} = x^{1/2} \), \( g(x) = x^2 \), \( a=0 \), \( b=1 \)
$$ A = \int_0^1 (x^{1/2} - x^2) \, dx = \int_0^1 x^{1/2} \, dx - \int_0^1 x^2 \, dx $$
$$ = [ \frac{2}{3} x^{3/2} ]_0^1 - [ \frac{x^3}{3} ]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} $$
Answer: \( \frac{1}{3} \) square units
Example 4: Determine the area between \( y = \sin x \) and \( y = \cos x \) from \( x = 0 \) to \( x = \frac{\pi}{4} \).
Given: \( f(x) = \cos x \), \( g(x) = \sin x \), \( a=0 \), \( b=\frac{\pi}{4} \)
$$ A = \int_0^{\pi/4} (\cos x - \sin x) \, dx = [\sin x + \cos x]_0^{\pi/4} $$
$$ = ( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} ) - (\sin 0 + \cos 0) = ( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} ) - (0 + 1) = \sqrt{2} - 1 $$
Using \( \sqrt{2} \approx 1.414 \),
Answer: 0.414 square units
Example 5: Calculate the area between \( y = e^x \) and \( y = x + 1 \) from \( x = 0 \) to \( x = 1 \).
Given: \( f(x) = e^x \), \( g(x) = x + 1 \), \( a=0 \), \( b=1 \)
$$ A = \int_0^1 (e^x, x - 1) \, dx = \int_0^1 e^x \, dx - \int_0^1 x \, dx - \int_0^1 1 \, dx $$
$$ = [e^x]_0^1 - [ \frac{x^2}{2} ]_0^1 ☑_0^1 = (e - 1) - \frac{1}{2} - 1 = e - 2.5 $$
Using \( e \approx 2.718 \),
$$ A = 2.718 - 2.5 = 0.218 $$
Answer: 0.218 square units
The volume of a solid formed by revolving the curve \( y = f(x) \) about the x-axis from \( x = a \) to \( x = b \) is:
$$ V = \pi \int_a^b [f(x)]^2 \, dx $$
Example 1: Find the volume generated by revolving \( y = x \) about the x-axis from \( x = 0 \) to \( x = 2 \).
Given: \( f(x) = x \), \( a=0 \), \( b=2 \)
$$ V = \pi \int_0^2 x^2 \, dx = \pi [ \frac{x^3}{3} ]_0^2 = \pi \times \frac{8}{3} = \frac{8\pi}{3} $$
Using \( \pi \approx 3.142 \),
$$ V = \frac{8 \times 3.142}{3} = 8.38 $$
Answer: 8.38 cubic units
Example 2: Calculate the volume generated by revolving \( y = \sqrt{x} \) about the x-axis from \( x = 0 \) to \( x = 4 \).
Given: \( f(x) = \sqrt{x} = x^{1/2} \), \( a=0 \), \( b=4 \)
$$ V = \pi \int_0^4 (x^{1/2})^2 \, dx = \pi \int_0^4 x \, dx = \pi [ \frac{x^2}{2} ]_0^4 = \pi \times \frac{16}{2} = 8\pi $$
Using \( \pi \approx 3.142 \),
$$ V = 8 \times 3.142 = 25.14 $$
Answer: 25.14 cubic units
Example 3: Find the volume generated by revolving \( y = 3 - x \) about the x-axis from \( x = 0 \) to \( x = 3 \).
Given: \( f(x) = 3 - x \), \( a=0 \), \( b=3 \)
$$ V = \pi \int_0^3 (3 - x)^2 \, dx $$
Expanding the square:
$$ (3 - x)^2 = 9 - 6x + x^2 $$
Then,
$$ V = \pi \int_0^3 (9 - 6x + x^2) \, dx = \pi [ 9x - 3x^2 + \frac{x^3}{3} ]_0^3 $$
$$ = \pi ( 27 - 27 + 9 ) = \pi \times 9 = 9\pi $$
Using \( \pi \approx 3.142 \),
$$ V = 9 \times 3.142 = 28.28 $$
Answer: 28.28 cubic units
Example 4: Calculate the volume of the solid formed by revolving \( y = \sin x \) about the x-axis from \( x = 0 \) to \( x = \pi \).
Given: \( f(x) = \sin x \), \( a=0 \), \( b=\pi \)
$$ V = \pi \int_0^\pi \sin^2 x \, dx $$
Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
$$ V = \pi \int_0^\pi \frac{1 - \cos 2x}{2} \, dx = \frac{\pi}{2} \int_0^\pi (1 - \cos 2x) \, dx $$
$$ = \frac{\pi}{2} [ x - \frac{\sin 2x}{2} ]_0^\pi = \frac{\pi}{2} ( \pi - 0 ) = \frac{\pi^2}{2} $$
Using \( \pi \approx 3.142 \),
$$ V = \frac{(3.142)^2}{2} = \frac{9.869}{2} = 4.934 $$
Answer: 4.934 cubic units
Example 5: Find the volume generated by revolving \( y = e^x \) about the x-axis from \( x = 0 \) to \( x = 1 \).
Given: \( f(x) = e^x \), \( a=0 \), \( b=1 \)
$$ V = \pi \int_0^1 (e^x)^2 \, dx = \pi \int_0^1 e^{2x} \, dx = \pi [ \frac{e^{2x}}{2} ]_0^1 = \pi \times \frac{e^{2} - 1}{2} $$
Using \( e^2 \approx 7.389 \), \( \pi \approx 3.142 \),
$$ V = 3.142 \times \frac{7.389 - 1}{2} = 3.142 \times \frac{6.389}{2} = 3.142 \times 3.194 = 10.04 $$
Answer: 10.04 cubic units
Differential equations describe relationships involving rates of change and are fundamental for modeling mechanical systems such as vibrations, heat transfer, and fluid flow. Mechanical production technicians use differential equations to analyze and predict system behavior in design and troubleshooting.
A first-order differential equation with separable variables can be written as:
$$ \frac{dy}{dx} = g(x) h(y) $$
Separating variables gives:
$$ \frac{1}{h(y)} dy = g(x) dx $$
Integrating both sides yields the solution.
Example 1: Solve \( \frac{dy}{dx} = xy \).
Given: \( \frac{dy}{dx} = xy \)
Separate variables:
$$ \frac{1}{y} dy = x dx $$
Integrate both sides:
$$ \int \frac{1}{y} dy = \int x dx $$
$$ \ln |y| = \frac{x^2}{2} + C $$
Solve for \( y \):
$$ y = Ce^{\frac{x^2}{2}} $$
Answer: \( y = Ce^{x^2/2} \), where \( C \) is constant
Example 2: Solve \( \frac{dy}{dx} = \frac{2x}{y} \) with initial condition \( y=1 \) when \( x=0 \).
Given: \( \frac{dy}{dx} = \frac{2x}{y} \), \( y(0) = 1 \)
Separate variables:
$$ y \, dy = 2x \, dx $$
Integrate both sides:
$$ \int y \, dy = \int 2x \, dx $$
$$ \frac{y^2}{2} = x^2 + C $$
Apply initial condition \( x=0, y=1 \):
$$ \frac{1^2}{2} = 0 + C \Rightarrow C = \frac{1}{2} $$
$$ \frac{y^2}{2} = x^2 + \frac{1}{2} $$
Multiply both sides by 2:
$$ y^2 = 2x^2 + 1 $$
Solve for \( y \):
$$ y = \pm \sqrt{2x^2 + 1} $$
Since \( y=1 \) when \( x=0 \), take positive root.
Answer: \( y = \sqrt{2x^2 + 1} \)
Example 3: Solve \( \frac{dy}{dx} = \frac{x + y}{x} \).
Given: \( \frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x} \)
Rewrite as:
$$ \frac{dy}{dx} - \frac{y}{x} = 1 $$
This is a linear first-order ODE.
Integrating factor:
$$ \mu(x) = e^{-\int \frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} $$
Multiply both sides by \( \frac{1}{x} \):
$$ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{1}{x} $$
Rewrite left side as derivative:
$$ \frac{d}{dx} ( \frac{y}{x} ) = \frac{1}{x} $$
Integrate both sides:
$$ \int \frac{d}{dx} ( \frac{y}{x} ) dx = \int \frac{1}{x} dx $$
$$ \frac{y}{x} = \ln |x| + C $$
Multiply both sides by \( x \):
$$ y = x \ln |x| + Cx $$
Answer: \( y = x \ln |x| + Cx \)
Example 4: Solve \( \frac{dy}{dx} = y \sin x \) with initial condition \( y(0) = 2 \).
Given: \( \frac{dy}{dx} = y \sin x \), \( y(0) = 2 \)
Separate variables:
$$ \frac{1}{y} dy = \sin x \, dx $$
Integrate both sides:
$$ \int \frac{1}{y} dy = \int \sin x \, dx $$
$$ \ln |y| = -\cos x + C $$
Solve for \( y \):
$$ y = Ce^{-\cos x} $$
Apply initial condition \( x=0, y=2 \):
$$ 2 = C e^{-\cos 0} = C e^{-1} $$
$$ C = 2 e $$
Final solution:
$$ y = 2 e \cdot e^{-\cos x} = 2 e^{1 - \cos x} $$
Answer: \( y = 2 e^{1 - \cos x} \)
Example 5: Solve \( \frac{dy}{dx} = \frac{y}{x} + x^2 \) with \( y(1) = 2 \).
Given: \( \frac{dy}{dx} - \frac{y}{x} = x^2 \), \( y(1) = 2 \)
Integrating factor:
$$ \mu(x) = e^{-\int \frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} $$
Multiply both sides by \( \frac{1}{x} \):
$$ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = x $$
Rewrite left side as derivative:
$$ \frac{d}{dx} ( \frac{y}{x} ) = x $$
Integrate both sides:
$$ \int \frac{d}{dx} ( \frac{y}{x} ) dx = \int x \, dx $$
$$ \frac{y}{x} = \frac{x^2}{2} + C $$
Multiply both sides by \( x \):
$$ y = \frac{x^3}{2} + Cx $$
Apply initial condition \( x=1, y=2 \):
$$ 2 = \frac{1^3}{2} + C \times 1 = \frac{1}{2} + C $$
$$ C = 2 - \frac{1}{2} = \frac{3}{2} $$
Final solution:
$$ y = \frac{x^3}{2} + \frac{3}{2} x $$
Answer: \( y = \frac{x^3}{2} + \frac{3}{2} x \)
A first-order differential equation is homogeneous if it can be written as:
$$ \frac{dy}{dx} = F(\frac{y}{x}) $$
Use substitution \( v = \frac{y}{x} \), so \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Rewrite and solve for \( v \).
Example 1: Solve \( \frac{dy}{dx} = \frac{y + x}{x} \).
Given:
$$ \frac{dy}{dx} = \frac{y + x}{x} = \frac{y}{x} + 1 $$
Substitute \( v = \frac{y}{x} \), so \( y = vx \), \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Rewrite:
$$ v + x \frac{dv}{dx} = v + 1 $$
Subtract \( v \) from both sides:
$$ x \frac{dv}{dx} = 1 $$
Separate variables:
$$ dv = \frac{1}{x} dx $$
Integrate:
$$ \int dv = \int \frac{1}{x} dx $$
$$ v = \ln |x| + C $$
Recall \( v = \frac{y}{x} \), so:
$$ \frac{y}{x} = \ln |x| + C $$
Multiply both sides by \( x \):
$$ y = x \ln |x| + Cx $$
Answer: \( y = x \ln |x| + Cx \)
Example 2: Solve \( \frac{dy}{dx} = \frac{x + y}{x, y} \).
Given:
$$ \frac{dy}{dx} = \frac{x + y}{x, y} $$
Substitute \( v = \frac{y}{x} \), so \( y = vx \), \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Express numerator and denominator:
$$ x + y = x + vx = x(1 + v) $$
$$ x, y = x, vx = x(1 - v) $$
So,
$$ v + x \frac{dv}{dx} = \frac{1 + v}{1 - v} $$
Subtract \( v \):
$$ x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v, v(1 - v)}{1 - v} = \frac{1 + v, v + v^2}{1 - v} = \frac{1 + v^2}{1 - v} $$
$$ \frac{dv}{dx} = \frac{1 + v^2}{x (1 - v)} $$
Separate variables:
$$ \frac{1 - v}{1 + v^2} dv = \frac{1}{x} dx $$
Rewrite left side:
$$ \frac{1 - v}{1 + v^2} = \frac{1}{1 + v^2} - \frac{v}{1 + v^2} $$
Integrate both sides:
$$ \int ( \frac{1}{1 + v^2} - \frac{v}{1 + v^2} ) dv = \int \frac{1}{x} dx $$
Calculate integrals:
$$ \int \frac{1}{1 + v^2} dv = \tan^{-1} v $$
$$ \int \frac{v}{1 + v^2} dv = \frac{1}{2} \ln (1 + v^2) $$
$$ \tan^{-1} v - \frac{1}{2} \ln (1 + v^2) = \ln |x| + C $$
Recall \( v = \frac{y}{x} \):
$$ \tan^{-1} ( \frac{y}{x} ) - \frac{1}{2} \ln ( 1 + ( \frac{y}{x} )^2 ) = \ln |x| + C $$
Answer: Implicit solution above.
Example 3: Solve \( \frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + ( \frac{y}{x} )^2} \).
Given:
$$ \frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + ( \frac{y}{x} )^2} $$
Substitute \( v = \frac{y}{x} \), so \( y = vx \), \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Rewrite:
$$ v + x \frac{dv}{dx} = v + \sqrt{1 + v^2} $$
Subtract \( v \):
$$ x \frac{dv}{dx} = \sqrt{1 + v^2} $$
Separate variables:
$$ \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} $$
Integrate both sides:
$$ \int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x} $$
Left integral:
$$ \sinh^{-1} v $$
Right integral:
$$ \ln |x| + C $$
$$ \sinh^{-1} v = \ln |x| + C $$
Recall \( v = \frac{y}{x} \):
$$ \sinh^{-1} ( \frac{y}{x} ) = \ln |x| + C $$
Answer: Implicit solution above.
Example 4: Solve \( \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \).
Given:
$$ \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} $$
Rewrite numerator and denominator dividing numerator and denominator by \( x^2 \):
$$ = \frac{1 + ( \frac{y}{x} )^2}{2 \times \frac{y}{x}} = \frac{1 + v^2}{2v} $$
Substitute \( v = \frac{y}{x} \), \( y = vx \), \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Rewrite:
$$ v + x \frac{dv}{dx} = \frac{1 + v^2}{2v} $$
Subtract \( v \):
$$ x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2 v^2}{2 v} = \frac{1 - v^2}{2 v} $$
Separate variables:
$$ \frac{2 v}{1 - v^2} dv = \frac{dx}{x} $$
Rewrite left side:
$$ \frac{2 v}{1 - v^2} = \frac{-2 v}{v^2 - 1} $$
Integrate both sides:
$$ \int \frac{-2 v}{v^2 - 1} dv = \int \frac{dx}{x} $$
Let \( u = v^2 - 1 \), so \( du = 2 v dv \).
$$ \int \frac{-2 v}{v^2 - 1} dv = - \int \frac{du}{u} = - \ln |u| + C = - \ln |v^2 - 1| + C $$
Right side integral:
$$ \ln |x| + C $$
Equate:
$$ - \ln |v^2 - 1| = \ln |x| + C $$
Rearranged:
$$ \ln |v^2 - 1| = - \ln |x| + C_1 = \ln ( \frac{C_2}{x} ) $$
$$ v^2 - 1 = \frac{C}{x} $$
Recall \( v = \frac{y}{x} \):
$$ ( \frac{y}{x} )^2 - 1 = \frac{C}{x} $$
Multiply both sides by \( x^2 \):
$$ y^2 - x^2 = C x $$
Answer: \( y^2 - x^2 = C x \)
Example 5: Solve \( \frac{dy}{dx} = \frac{x + 2y}{x} \).
Given:
$$ \frac{dy}{dx} = \frac{x + 2y}{x} = 1 + 2 \frac{y}{x} $$
Substitute \( v = \frac{y}{x} \), so \( y = vx \), \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Rewrite:
$$ v + x \frac{dv}{dx} = 1 + 2v $$
Subtract \( v \):
$$ x \frac{dv}{dx} = 1 + 2v, v = 1 + v $$
Separate variables:
$$ \frac{dv}{1 + v} = \frac{dx}{x} $$
Integrate both sides:
$$ \int \frac{1}{1 + v} dv = \int \frac{1}{x} dx $$
$$ \ln |1 + v| = \ln |x| + C $$
Exponentiate:
$$ 1 + v = C x $$
Recall \( v = \frac{y}{x} \):
$$ 1 + \frac{y}{x} = C x $$
Multiply both sides by \( x \):
$$ x + y = C x^2 $$
Solve for \( y \):
$$ y = C x^2 - x $$
Answer: \( y = C x^2 - x \)
A linear first-order differential equation has the standard form:
$$ \frac{dy}{dx} + P(x) y = Q(x) $$
The integrating factor is:
$$ \mu(x) = e^{\int P(x) dx} $$
The solution is:
$$ y \cdot \mu = \int Q(x) \mu \, dx + C $$
Example 1: Solve \( \frac{dy}{dx} + 2y = 4 \).
Given: \( P(x) = 2 \), \( Q(x) = 4 \)
Calculate integrating factor:
$$ \mu = e^{\int 2 dx} = e^{2x} $$
Multiply entire equation by \( \mu \):
$$ e^{2x} \frac{dy}{dx} + 2 e^{2x} y = 4 e^{2x} $$
Left side is derivative:
$$ \frac{d}{dx} (y e^{2x}) = 4 e^{2x} $$
Integrate both sides:
$$ y e^{2x} = \int 4 e^{2x} dx + C = 4 \int e^{2x} dx + C $$
$$ = 4 \times \frac{e^{2x}}{2} + C = 2 e^{2x} + C $$
Solve for \( y \):
$$ y = 2 + C e^{-2x} $$
Answer: \( y = 2 + C e^{-2x} \)
Example 2: Solve \( \frac{dy}{dx} + y \tan x = \sin x \), \( 0 < x < \frac{\pi}{2} \).
Given: \( P(x) = \tan x \), \( Q(x) = \sin x \)
Calculate integrating factor:
$$ \mu = e^{\int \tan x dx} = e^{-\ln |\cos x|} = \frac{1}{\cos x} $$
Multiply entire equation by \( \frac{1}{\cos x} \):
$$ \frac{1}{\cos x} \frac{dy}{dx} + y \frac{\tan x}{\cos x} = \frac{\sin x}{\cos x} $$
Rewrite left side as derivative:
$$ \frac{d}{dx} ( \frac{y}{\cos x} ) = \tan x $$
Integrate both sides:
$$ \int \frac{d}{dx} ( \frac{y}{\cos x} ) dx = \int \tan x \, dx $$
$$ \frac{y}{\cos x} = - \ln |\cos x| + C $$
Multiply both sides by \( \cos x \):
$$ y = \cos x ( C - \ln |\cos x| ) $$
Answer: \( y = \cos x ( C - \ln |\cos x| ) \)
Example 3: Solve \( \frac{dy}{dx} - \frac{2}{x} y = x^2 \), \( x > 0 \).
Given: \( P(x) = -\frac{2}{x} \), \( Q(x) = x^2 \)
Calculate integrating factor:
$$ \mu = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = x^{-2} $$
Multiply entire equation by \( x^{-2} \):
$$ x^{-2} \frac{dy}{dx} - \frac{2}{x} x^{-2} y = x^{-2} x^2 = 1 $$
Rewrite left side as derivative:
$$ \frac{d}{dx} (y x^{-2}) = 1 $$
Integrate both sides:
$$ y x^{-2} = \int 1 \, dx + C = x + C $$
Multiply both sides by \( x^{2} \):
$$ y = x^{3} + C x^{2} $$
Answer: \( y = x^{3} + C x^{2} \)
Example 4: Solve \( \frac{dy}{dx} + \frac{1}{x} y = x^2 \), \( x > 0 \).
Given: \( P(x) = \frac{1}{x} \), \( Q(x) = x^2 \)
Calculate integrating factor:
$$ \mu = e^{\int \frac{1}{x} dx} = e^{\ln x} = x $$
Multiply entire equation by \( x \):
$$ x \frac{dy}{dx} + y = x^{3} $$
Rewrite left side as derivative:
$$ \frac{d}{dx} (y x) = x^{3} $$
Integrate both sides:
$$ y x = \int x^{3} dx + C = \frac{x^{4}}{4} + C $$
Solve for \( y \):
$$ y = \frac{x^{3}}{4} + \frac{C}{x} $$
Answer: \( y = \frac{x^{3}}{4} + \frac{C}{x} \)
Example 5: Solve \( \frac{dy}{dx} + 3 y = 6 e^{-3x} \).
Given: \( P(x) = 3 \), \( Q(x) = 6 e^{-3x} \)
Calculate integrating factor:
$$ \mu = e^{\int 3 dx} = e^{3x} $$
Multiply entire equation by \( e^{3x} \):
$$ e^{3x} \frac{dy}{dx} + 3 e^{3x} y = 6 e^{0} = 6 $$
Left side is derivative:
$$ \frac{d}{dx} (y e^{3x}) = 6 $$
Integrate both sides:
$$ y e^{3x} = 6x + C $$
Solve for \( y \):
$$ y = 6 x e^{-3x} + C e^{-3x} $$
Answer: \( y = 6 x e^{-3x} + C e^{-3x} \)
Find the area under the curve \( y = 5x^2 \) from \( x = 1 \) to \( x = 3 \). (5 marks)
Calculate the area between the curves \( y = x^2 \) and \( y = 2x \) from \( x = 0 \) to \( x = 2 \). (6 marks)
Determine the volume of the solid formed by revolving the curve \( y = \sqrt{x} \) about the x-axis from \( x = 1 \) to \( x = 4 \). (7 marks)
Solve the differential equation \( \frac{dy}{dx} = 3 x y \) given \( y = 1 \) when \( x = 0 \). (6 marks)
Solve the linear differential equation \( \frac{dy}{dx} + \frac{2}{x} y = x^2 \), \( x > 0 \). (6 marks)
Find the solution to the homogeneous differential equation \( \frac{dy}{dx} = \frac{y}{x} + \frac{x}{y} \). (7 marks)
Double and triple integrals are essential tools in mechanical production for calculating volumes, masses, and other quantities distributed over areas and volumes. For a mechanical production technician in Kenya, these integrals help determine material quantities, heat distribution in casting molds, and the center of mass of complex components. The double integral calculates over a two-dimensional region, while the triple integral extends this to three dimensions.
Double integrals over rectangular regions calculate the volume under a surface \(z = f(x,y)\) above a rectangular domain \(a \le x \le b\), \(c \le y \le d\). The double integral is expressed as
$$ \iint_R f(x,y) \, dA = \int_c^d \int_a^b f(x,y) \, dx \, dy $$
where \(dA = dx \, dy\).
Example 1: Find the volume under the surface \(z = 4 - x, y\) over the rectangular region \(0 \le x \le 1\), \(0 \le y \le 2\).
Given: \(f(x,y) = 4 - x, y\), \(a=0\), \(b=1\), \(c=0\), \(d=2\)
$$ V = \int_0^2 \int_0^1 (4 - x, y) \, dx \, dy $$
Integrate with respect to \(x\):
$$ \int_0^1 (4 - x, y) \, dx = [4x - \frac{x^2}{2} - yx ]_0^1 = 4(1) - \frac{1^2}{2} - y(1) = 4 - 0.5 - y = 3.5 - y $$
Integrate with respect to \(y\):
$$ \int_0^2 (3.5 - y) \, dy = [3.5y - \frac{y^2}{2}]_0^2 = 3.5(2) - \frac{2^2}{2} = 7 - 2 = 5 $$
Answer: Volume = 5 cubic units
Example 2: Calculate the volume under \(z = x^2 + y^2\) over the square region \(0 \le x \le 1\), \(0 \le y \le 1\).
Given: \(f(x,y) = x^2 + y^2\), \(a=0\), \(b=1\), \(c=0\), \(d=1\)
$$ V = \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy $$
Integrate with respect to \(x\):
$$ \int_0^1 (x^2 + y^2) \, dx = [\frac{x^3}{3} + y^2 x]_0^1 = \frac{1}{3} + y^2 $$
Integrate with respect to \(y\):
$$ \int_0^1 (\frac{1}{3} + y^2) dy = [\frac{y}{3} + \frac{y^3}{3}]_0^1 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} $$
Answer: Volume = \(\frac{2}{3}\) cubic units
Example 3: Find the volume under \(z = e^{-(x+y)}\) over the rectangle \(0 \le x \le 1\), \(0 \le y \le 1\).
Given: \(f(x,y) = e^{-(x+y)}\), \(a=0\), \(b=1\), \(c=0\), \(d=1\)
$$ V = \int_0^1 \int_0^1 e^{-(x+y)} \, dx \, dy $$
Integrate with respect to \(x\):
$$ \int_0^1 e^{-(x+y)} \, dx = e^{-y} \int_0^1 e^{-x} \, dx = e^{-y} [-e^{-x}]_0^1 = e^{-y} (1 - e^{-1}) $$
Integrate with respect to \(y\):
$$ \int_0^1 e^{-y} (1 - e^{-1}) \, dy = (1 - e^{-1}) \int_0^1 e^{-y} \, dy = (1 - e^{-1}) [-e^{-y}]_0^1 = (1 - e^{-1})(1 - e^{-1}) = (1 - e^{-1})^2 $$
Calculate numerical value:
\(e^{-1} \approx 0.3679\)
\(1 - e^{-1} = 0.6321\)
\((1 - e^{-1})^2 = 0.6321^2 = 0.399\)
Answer: Volume ≈ 0.399 cubic units
When the region is circular or better described in terms of radius and angle, double integrals are evaluated using polar coordinates \((r, \theta)\). The double integral becomes
$$ \iint_R f(r,\theta) \, r \, dr \, d\theta $$
where \(r\) is the radius and \(\theta\) the angle.
Example 1: Find the area of a quarter circle of radius 2 in the first quadrant using a double integral.
Given: \(f(r,\theta) = 1\), \(0 \le r \le 2\), \(0 \le \theta \le \frac{\pi}{2}\)
$$ A = \int_0^{\pi/2} \int_0^2 r \, dr \, d\theta $$
Integrate with respect to \(r\):
$$ \int_0^2 r \, dr = [\frac{r^2}{2}]_0^2 = \frac{4}{2} = 2 $$
Integrate with respect to \(\theta\):
$$ \int_0^{\pi/2} 2 \, d\theta = 2 \theta \bigg|_0^{\pi/2} = 2 \times \frac{\pi}{2} = \pi $$
Answer: Area = \(\pi\) square units
Example 2: Calculate the volume under \(z = r\) over the quarter circle \(r \le 3\), \(0 \le \theta \le \frac{\pi}{2}\).
Given: \(f(r,\theta) = r\), \(0 \le r \le 3\), \(0 \le \theta \le \frac{\pi}{2}\)
$$ V = \int_0^{\pi/2} \int_0^3 r \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_0^3 r^2 \, dr \, d\theta $$
Integrate with respect to \(r\):
$$ \int_0^3 r^2 \, dr = [\frac{r^3}{3}]_0^3 = \frac{27}{3} = 9 $$
Integrate with respect to \(\theta\):
$$ \int_0^{\pi/2} 9 \, d\theta = 9 \theta \bigg|_0^{\pi/2} = 9 \times \frac{\pi}{2} = \frac{9\pi}{2} $$
Answer: Volume = \(\frac{9\pi}{2}\) cubic units
Example 3: Find the volume under \(z = e^{-r^2}\) over the full circle of radius 1.
Given: \(f(r,\theta) = e^{-r^2}\), \(0 \le r \le 1\), \(0 \le \theta \le 2\pi\)
$$ V = \int_0^{2\pi} \int_0^1 e^{-r^2} r \, dr \, d\theta $$
Integrate with respect to \(r\) using substitution \(u = r^2\), \(du = 2r dr\):
$$ \int_0^1 e^{-r^2} r \, dr = \frac{1}{2} \int_0^{1^2} e^{-u} du = \frac{1}{2} [-e^{-u}]_0^1 = \frac{1}{2} (1 - e^{-1}) $$
Integrate with respect to \(\theta\):
$$ \int_0^{2\pi} \frac{1}{2} (1 - e^{-1}) \, d\theta = \frac{1}{2} (1 - e^{-1}) \times 2\pi = \pi (1 - e^{-1}) $$
Calculate numerical value:
\(1 - e^{-1} = 0.6321\)
\(\pi \times 0.6321 = 1.985\)
Answer: Volume ≈ 1.985 cubic units
Triple integrals extend integration to three dimensions, useful for finding volumes and masses of solids with density functions. For a region \(V\) and density \(\rho(x,y,z)\), the mass is
$$ M = \iiint_V \rho(x,y,z) \, dV $$
where \(dV = dx \, dy \, dz\).
Example 1: Find the volume of a cube with edges 2 units, bounded by \(0 \le x,y,z \le 2\).
Given: \(V\) is cube with edges 2 units, \(\rho = 1\) (uniform density)
$$ V = \int_0^2 \int_0^2 \int_0^2 1 \, dx \, dy \, dz $$
Integrate with respect to \(x\):
$$ \int_0^2 1 \, dx = [x]_0^2 = 2 $$
Integrate with respect to \(y\):
$$ \int_0^2 2 \, dy = 2y \bigg|_0^2 = 4 $$
Integrate with respect to \(z\):
$$ \int_0^2 4 \, dz = 4z \bigg|_0^2 = 8 $$
Answer: Volume = 8 cubic units
Example 2: Calculate the mass of a solid cube \(0 \le x,y,z \le 1\) with density \(\rho = x + y + z\).
Given: \(\rho = x + y + z\), cube edges 1 unit
$$ M = \int_0^1 \int_0^1 \int_0^1 (x + y + z) \, dx \, dy \, dz $$
Integrate with respect to \(x\):
$$ \int_0^1 (x + y + z) \, dx = [\frac{x^2}{2} + yx + zx]_0^1 = \frac{1}{2} + y + z $$
Integrate with respect to \(y\):
$$ \int_0^1 (\frac{1}{2} + y + z) dy = [\frac{y}{2} + \frac{y^2}{2} + zy]_0^1 = \frac{1}{2} + \frac{1}{2} + z = 1 + z $$
Integrate with respect to \(z\):
$$ \int_0^1 (1 + z) dz = [z + \frac{z^2}{2}]_0^1 = 1 + \frac{1}{2} = \frac{3}{2} $$
Answer: Mass = \(\frac{3}{2}\) units
Example 3: Find the volume of the solid bounded by \(x^2 + y^2 + z^2 \le 1\) (unit sphere) using spherical coordinates.
Given: Sphere radius \(r \le 1\), volume element in spherical coordinates \(dV = r^2 \sin\phi \, dr \, d\phi \, d\theta\), where \(0 \le r \le 1\), \(0 \le \phi \le \pi\), \(0 \le \theta \le 2\pi\).
$$ V = \int_0^{2\pi} \int_0^\pi \int_0^1 r^2 \sin\phi \, dr \, d\phi \, d\theta $$
Integrate with respect to \(r\):
$$ \int_0^1 r^2 \, dr = [\frac{r^3}{3}]_0^1 = \frac{1}{3} $$
Integrate with respect to \(\phi\):
$$ \int_0^\pi \sin\phi \, d\phi = [-\cos\phi]_0^\pi = -( \cos\pi - \cos0) = -(-1 - 1) = 2 $$
Integrate with respect to \(\theta\):
$$ \int_0^{2\pi} d\theta = 2\pi $$
Multiply results:
$$ V = \frac{1}{3} \times 2 \times 2\pi = \frac{4\pi}{3} $$
Answer: Volume = \(\frac{4\pi}{3}\) cubic units
Triple and double integrals are used to determine the mass, center of gravity, and moments of inertia of irregular mechanical parts. For example, calculating the mass of a casting with variable density or the heat distribution in a furnace chamber. These integrals allow precise engineering design and quality control.
Laplace transforms convert differential equations common in mechanical systems into algebraic equations, simplifying analysis and solution of dynamic systems such as vibration, heat transfer, and control systems. For a mechanical production technician, Laplace transforms help model machine tool dynamics and system responses.
The Laplace transform \(F(s)\) of a function \(f(t)\) defined for \(t \ge 0\) is
$$ F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt $$
where \(s\) is a complex number. This integral transform changes functions of time \(t\) into functions of complex frequency \(s\).
Example 1: Find the Laplace transform of \(f(t) = 1\).
Given: \(f(t) = 1\)
$$ F(s) = \int_0^\infty e^{-st} \cdot 1 \, dt $$
Integrate:
$$ F(s) = [-\frac{1}{s} e^{-st}]_0^\infty = 0 + \frac{1}{s} = \frac{1}{s}, \quad \text{for } \Re(s) > 0 $$
Answer: \(\mathcal{L}\{1\} = \frac{1}{s}\)
Example 2: Find the Laplace transform of \(f(t) = t\).
Given: \(f(t) = t\)
$$ F(s) = \int_0^\infty e^{-st} t \, dt $$
Use integration by parts:
Let \(u = t\), \(dv = e^{-st} dt\), then \(du = dt\), \(v = -\frac{1}{s} e^{-st}\).
$$ F(s) = uv \bigg|_0^\infty - \int_0^\infty v \, du = [-\frac{t}{s} e^{-st}]_0^\infty + \frac{1}{s} \int_0^\infty e^{-st} dt $$
The first term evaluates to 0, the second integral is
$$ \frac{1}{s} \times \frac{1}{s} = \frac{1}{s^2} $$
Answer: \(\mathcal{L}\{t\} = \frac{1}{s^2}\)
Example 3: Find the Laplace transform of \(f(t) = e^{at}\), where \(a\) is a constant.
Given: \(f(t) = e^{at}\)
$$ F(s) = \int_0^\infty e^{-st} e^{at} dt = \int_0^\infty e^{-(s, a)t} dt $$
Integrate:
$$ F(s) = [-\frac{1}{s, a} e^{-(s, a)t}]_0^\infty = 0 + \frac{1}{s, a}, \quad \text{for } \Re(s) > \Re(a) $$
Answer: \(\mathcal{L}\{e^{at}\} = \frac{1}{s, a}\)
The Laplace transform converts derivatives into algebraic terms: for \(f(t)\) with Laplace transform \(F(s)\),
$$ \mathcal{L}\{f'(t)\} = sF(s) - f(0) $$
$$ \mathcal{L}\{f''(t)\} = s^2 F(s) - s f(0) - f'(0) $$
This property is used to solve ordinary differential equations (ODEs).
Example 1: Solve the ODE \(f'(t) + 3f(t) = 0\) with initial condition \(f(0) = 4\).
Take Laplace transform on both sides:
$$ \mathcal{L}\{f'(t)\} + 3 \mathcal{L}\{f(t)\} = 0 $$
Using the derivative property:
$$ sF(s) - f(0) + 3F(s) = 0 $$
Substitute \(f(0) = 4\):
$$ s F(s) - 4 + 3 F(s) = 0 $$
Group terms:
$$ (s + 3) F(s) = 4 $$
Solve for \(F(s)\):
$$ F(s) = \frac{4}{s + 3} $$
Find inverse Laplace transform:
$$ f(t) = 4 e^{-3t} $$
Answer: \(f(t) = 4 e^{-3t}\)
Example 2: Solve \(f''(t) - 2 f'(t) + f(t) = 0\), with \(f(0) = 1\), \(f'(0) = 0\).
Take Laplace transforms:
$$ \mathcal{L}\{f''(t)\} - 2 \mathcal{L}\{f'(t)\} + \mathcal{L}\{f(t)\} = 0 $$
Using derivative properties:
$$ (s^2 F(s) - s f(0) - f'(0)) - 2 (s F(s) - f(0)) + F(s) = 0 $$
Substitute initial conditions:
$$ s^2 F(s) - s(1) - 0 - 2 (s F(s) - 1) + F(s) = 0 $$
Simplify:
$$ s^2 F(s) - s - 2 s F(s) + 2 + F(s) = 0 $$
Group \(F(s)\) terms:
$$ F(s) (s^2 - 2s + 1) = s - 2 $$
Note \(s^2 - 2s + 1 = (s - 1)^2\):
$$ F(s) = \frac{s - 2}{(s - 1)^2} $$
Use partial fraction or known transforms to find inverse Laplace transform:
Rewrite numerator:
$$ s - 2 = (s - 1) - 1 $$
Hence,
$$ F(s) = \frac{s - 1}{(s - 1)^2} - \frac{1}{(s - 1)^2} = \frac{1}{s - 1} - \frac{1}{(s - 1)^2} $$
Inverse transforms:
$$ \mathcal{L}^{-1} \{\frac{1}{s - 1}\} = e^{t} $$
$$ \mathcal{L}^{-1} \{\frac{1}{(s - 1)^2}\} = t e^{t} $$
$$ f(t) = e^{t} - t e^{t} = e^{t} (1 - t) $$
Answer: \(f(t) = e^{t} (1 - t)\)
Knowledge of Laplace transforms of standard functions enables quick solution of engineering problems. Common transforms include unit step, sine, cosine, and exponential functions.
Example 1: Find the Laplace transform of \(f(t) = \sin \omega t\).
Given: \(f(t) = \sin \omega t\)
$$ F(s) = \int_0^\infty e^{-st} \sin \omega t \, dt = \frac{\omega}{s^2 + \omega^2} $$
Answer: \(\mathcal{L}\{\sin \omega t\} = \frac{\omega}{s^2 + \omega^2}\)
Example 2: Find the Laplace transform of \(f(t) = \cos \omega t\).
Given: \(f(t) = \cos \omega t\)
$$ F(s) = \int_0^\infty e^{-st} \cos \omega t \, dt = \frac{s}{s^2 + \omega^2} $$
Answer: \(\mathcal{L}\{\cos \omega t\} = \frac{s}{s^2 + \omega^2}\)
Example 3: Find the inverse Laplace transform of
$$ F(s) = \frac{5}{s^2 + 9} $$
Recognize denominator corresponds to \(\omega = 3\), numerator constant 5:
$$ f(t) = \frac{5}{3} \sin 3t $$
Answer: \(f(t) = \frac{5}{3} \sin 3t\)
Mechanical vibration in production machinery can be modeled by second-order ODEs. Laplace transforms simplify solving these equations with given initial conditions.
Example 1: Solve \(m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + k x = 0\) with \(m=2\) kg, \(c=4\) Ns/m, \(k=10\) N/m, initial conditions \(x(0)=0.1\) m, \(\frac{dx}{dt}(0)=0\).
Take Laplace transform:
$$ 2 (s^2 X(s) - s x(0) - x'(0)) + 4 (s X(s) - x(0)) + 10 X(s) = 0 $$
Substitute initial values:
$$ 2 (s^2 X(s) - 0.1 s - 0) + 4 (s X(s) - 0.1) + 10 X(s) = 0 $$
Expand:
$$ 2 s^2 X(s) - 0.2 s + 4 s X(s) - 0.4 + 10 X(s) = 0 $$
Group \(X(s)\) terms:
$$ X(s)(2 s^2 + 4 s + 10) = 0.2 s + 0.4 $$
Solve for \(X(s)\):
$$ X(s) = \frac{0.2 s + 0.4}{2 s^2 + 4 s + 10} $$
Simplify denominator by dividing numerator and denominator by 2:
$$ X(s) = \frac{0.1 s + 0.2}{s^2 + 2 s + 5} $$
Complete the square in denominator:
$$ s^2 + 2 s + 5 = (s + 1)^2 + 4 $$
Rewrite numerator to match form for inverse transform:
$$ 0.1 s + 0.2 = 0.1 (s + 1) + 0.1 $$
So,
$$ X(s) = \frac{0.1 (s + 1) + 0.1}{(s + 1)^2 + 4} = 0.1 \cdot \frac{s + 1}{(s + 1)^2 + 4} + 0.1 \cdot \frac{1}{(s + 1)^2 + 4} $$
Inverse Laplace transforms:
$$ \mathcal{L}^{-1} \{ \frac{s + a}{(s + a)^2 + b^2} \} = e^{-a t} \cos b t $$
$$ \mathcal{L}^{-1} \{ \frac{b}{(s + a)^2 + b^2} \} = e^{-a t} \sin b t $$
Adjust second term to match numerator \(b=2\):
Multiply numerator and denominator by 2:
$$ 0.1 \cdot \frac{1}{(s + 1)^2 + 4} = 0.05 \cdot \frac{2}{(s + 1)^2 + 4} $$
$$ x(t) = 0.1 e^{-t} \cos 2 t + 0.05 e^{-t} \sin 2 t $$
Answer: \(x(t) = e^{-t} (0.1 \cos 2 t + 0.05 \sin 2 t)\) meters
Example 2: Solve \( \frac{d^2 y}{dt^2} + 9 y = \sin 3 t \) with \(y(0)=0\), \(y'(0)=0\).
Take Laplace transforms:
$$ s^2 Y(s) - s y(0) - y'(0) + 9 Y(s) = \frac{3}{s^2 + 9} $$
Simplify:
$$ s^2 Y(s) + 9 Y(s) = \frac{3}{s^2 + 9} $$
Factor \(Y(s)\):
$$ Y(s) (s^2 + 9) = \frac{3}{s^2 + 9} $$
Solve for \(Y(s)\):
$$ Y(s) = \frac{3}{(s^2 + 9)^2} $$
Inverse Laplace transform corresponds to
$$ \mathcal{L}^{-1} \{ \frac{3}{(s^2 + 3^2)^2} \} = \frac{t}{6} \sin 3 t $$
Answer: \(y(t) = \frac{t}{6} \sin 3 t\)
Calculate the volume under the surface \(z = 5 - 2x, y\) over the rectangle \(0 \le x \le 2\), \(0 \le y \le 1\) using double integrals. (5 marks)
Find the volume of the solid bounded by \(z = x^2 + y^2\) and the square \(0 \le x,y \le 1\). (6 marks)
Compute the volume under \(z = e^{-x^2 - y^2}\) over the circular region \(x^2 + y^2 \le 1\) using polar coordinates. (8 marks)
Determine the mass of a solid cube \(1 \times 1 \times 1\) with density \(\rho = 2x + y + 3z\). (7 marks)
Find the Laplace transform of \(f(t) = t^2\). (5 marks)
Solve the ODE \(f''(t) + 4 f(t) = 0\) with \(f(0) = 1\), \(f'(0) = 0\) using Laplace transforms. (8 marks)
Find the inverse Laplace transform of \(F(s) = \frac{2s}{s^2 + 16}\). (4 marks)
Solve the vibration equation \(m \frac{d^2x}{dt^2} + k x = 0\) for \(m=1\) kg, \(k=9\) N/m, with \(x(0)=0.05\) m, \(x'(0)=0\). (8 marks)
Fourier series are essential in mechanical production engineering for analyzing periodic signals such as vibrations in machinery, heat distribution in furnaces, and stress variations in rotating shafts. In Kenya, mechanical technicians use Fourier series to decompose complex waveforms into simpler sine and cosine components, facilitating diagnosis and design improvements. This topic covers the mathematical foundation of Fourier series and its application to periodic functions relevant in mechanical production.
A periodic function \( f(x) \) with period \( 2L \) can be expressed as an infinite sum of sine and cosine terms. This decomposition enables analysis of mechanical vibrations and cyclic loading in components.
The Fourier series for \( f(x) \) is given by:
$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} ( a_n \cos \frac{n \pi x}{L} + b_n \sin \frac{n \pi x}{L} ) $$
where coefficients are
$$ a_0 = \frac{1}{L} \int_{-L}^{L} f(x) \, dx $$
$$ a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos \frac{n \pi x}{L} \, dx $$
$$ b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{n \pi x}{L} \, dx $$
Example 1: A mechanical technician measures a periodic vibration displacement \( f(x) = x \) over the interval \( -\pi \le x \le \pi \). Find the Fourier series coefficients \( a_0 \), \( a_n \), and \( b_n \).
Given: \( f(x) = x \), \( L = \pi \)
Calculate:
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \, dx $$
$$ a_0 = \frac{1}{\pi} [ \frac{x^2}{2} ]_{-\pi}^{\pi} = \frac{1}{\pi} ( \frac{\pi^2}{2} - \frac{\pi^2}{2} ) = 0 $$
$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos(nx) \, dx $$
Since \( x \cos(nx) \) is an odd function (product of odd and even), integral over symmetric limits is zero:
$$ a_n = 0 $$
$$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx $$
Since \( x \sin(nx) \) is even (product of odd and odd), integrate:
Use integration by parts:
Let \( u = x \), \( dv = \sin(nx) dx \)
Then \( du = dx \), \( v = -\frac{\cos(nx)}{n} \)
$$ b_n = \frac{1}{\pi} [ -\frac{x \cos(nx)}{n} \Big|_{-\pi}^{\pi} + \frac{1}{n} \int_{-\pi}^{\pi} \cos(nx) dx ] $$
Calculate boundary term:
$$ -\frac{\pi \cos(n\pi)}{n} + \frac{\pi \cos(-n\pi)}{n} = -\frac{\pi (-1)^n}{n} + \frac{\pi (-1)^n}{n} = 0 $$
Integral of cosine over full period:
$$ \int_{-\pi}^{\pi} \cos(nx) dx = 0 $$
$$ b_n = \frac{1}{\pi} \times \frac{1}{n} \times 0 = 0 $$
But this contradicts the known result for \( b_n \) of \( f(x) = x \). Re-examining:
Actually, the boundary terms cancel, but the integral of cosine is zero, so the integral reduces to:
$$ b_n = \frac{2}{\pi} \int_0^\pi x \sin(nx) dx $$
Calculate:
Integration by parts:
\( u = x \), \( dv = \sin(nx) dx \)
\( du = dx \), \( v = -\frac{\cos(nx)}{n} \)
So,
$$ \int_0^\pi x \sin(nx) dx = -\frac{x \cos(nx)}{n} \Big|_0^\pi + \frac{1}{n} \int_0^\pi \cos(nx) dx $$
Calculate:
$$ -\frac{\pi \cos(n\pi)}{n} + 0 + \frac{1}{n} [ \frac{\sin(nx)}{n} ]_0^\pi = -\frac{\pi (-1)^n}{n} + \frac{1}{n^2} [\sin(n\pi) - 0] = -\frac{\pi (-1)^n}{n} $$
Since \( \sin(n\pi) = 0 \),
$$ b_n = \frac{2}{\pi} \times ( -\frac{\pi (-1)^n}{n} ) = \frac{2 (-1)^{n+1}}{n} $$
Answer:
$$ a_0 = 0, \quad a_n = 0, \quad b_n = \frac{2 (-1)^{n+1}}{n} $$
Example 2: A rotating shaft has a stress variation modeled by \( f(x) = x^2 \) over \( -\pi \le x \le \pi \). Find the first two Fourier cosine coefficients \( a_0 \) and \( a_1 \).
Given: \( f(x) = x^2 \), \( L = \pi \)
Calculate:
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 dx = \frac{1}{\pi} [ \frac{x^3}{3} ]_{-\pi}^{\pi} = \frac{1}{\pi} ( \frac{\pi^3}{3} - (-\frac{\pi^3}{3}) ) = \frac{2 \pi^2}{3} $$
$$ a_1 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos x dx $$
Since \( x^2 \cos x \) is even, integrate from 0 to \( \pi \) and double:
$$ a_1 = \frac{2}{\pi} \int_0^\pi x^2 \cos x dx $$
Use integration by parts twice:
Let \( u = x^2 \), \( dv = \cos x dx \)
Then \( du = 2x dx \), \( v = \sin x \)
So,
$$ \int x^2 \cos x dx = x^2 \sin x - \int 2x \sin x dx $$
Now integrate \( \int 2x \sin x dx \) by parts:
Let \( u = 2x \), \( dv = \sin x dx \)
Then \( du = 2 dx \), \( v = -\cos x \)
So,
$$ \int 2x \sin x dx = -2x \cos x + \int 2 \cos x dx = -2x \cos x + 2 \sin x + C $$
$$ \int x^2 \cos x dx = x^2 \sin x + 2x \cos x - 2 \sin x + C $$
Evaluate definite integral:
$$ \int_0^\pi x^2 \cos x dx = [ x^2 \sin x + 2x \cos x - 2 \sin x ]_0^\pi $$
Calculate terms at \( x = \pi \):
\( \pi^2 \sin \pi = 0 \), \( 2\pi \cos \pi = -2\pi \), \( -2 \sin \pi = 0 \)
Sum: \( -2\pi \)
Calculate terms at \( x=0 \):
\( 0 \), \( 0 \), \( -2 \sin 0 = 0 \)
Sum: \( 0 \)
$$ \int_0^\pi x^2 \cos x dx = -2\pi - 0 = -2\pi $$
Calculate \( a_1 \):
$$ a_1 = \frac{2}{\pi} (-2\pi) = -4 $$
Answer:
$$ a_0 = \frac{2 \pi^2}{3}, \quad a_1 = -4 $$
Example 3: A heat distribution in a rod is described by \( f(x) = x \) for \( 0 \le x \le 2\pi \) with period \( 2\pi \). Find the Fourier sine coefficients \( b_n \).
Given: \( f(x) = x \), \( L = \pi \), interval \( 0 \le x \le 2\pi \) (period \( 2\pi \))
The Fourier sine coefficients are:
$$ b_n = \frac{1}{\pi} \int_0^{2\pi} x \sin \frac{n \pi x}{\pi} dx = \frac{1}{\pi} \int_0^{2\pi} x \sin n x dx $$
Use integration by parts:
Let \( u = x \), \( dv = \sin n x dx \)
Then \( du = dx \), \( v = -\frac{\cos n x}{n} \)
Calculate:
$$ b_n = \frac{1}{\pi} [ -\frac{x \cos n x}{n} \Big|_0^{2\pi} + \frac{1}{n} \int_0^{2\pi} \cos n x dx ] $$
Evaluate boundary term:
$$ -\frac{2\pi \cos 2\pi n}{n} + 0 = -\frac{2\pi \cos 2\pi n}{n} = -\frac{2\pi}{n} $$
since \( \cos 2\pi n = 1 \)
Integral of cosine over full period:
$$ \int_0^{2\pi} \cos n x dx = 0 $$
So,
$$ b_n = \frac{1}{\pi} \times ( -\frac{2\pi}{n} + 0 ) = -\frac{2}{n} $$
Answer:
$$ b_n = -\frac{2}{n} $$
Mechanical signals often exhibit symmetry simplifying Fourier series calculations. Even functions use cosine terms only, odd functions use sine terms only.
For an even function \( f(x) = f(-x) \), the Fourier series reduces to:
$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos \frac{n \pi x}{L} $$
with
$$ a_n = \frac{2}{L} \int_0^L f(x) \cos \frac{n \pi x}{L} dx $$
For an odd function \( f(-x) = -f(x) \), the series is:
$$ f(x) = \sum_{n=1}^\infty b_n \sin \frac{n \pi x}{L} $$
with
$$ b_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n \pi x}{L} dx $$
Example 1: A mechanical vibration displacement is an odd function \( f(x) = x \) defined on \( -\pi \le x \le \pi \). Find the first three non-zero sine coefficients \( b_1, b_2, b_3 \).
Given: \( f(x) = x \), \( L = \pi \)
Calculate:
$$ b_n = \frac{2}{\pi} \int_0^\pi x \sin n x dx $$
Use integration by parts as before:
$$ b_n = \frac{2}{\pi} ( -\frac{\pi \cos n \pi}{n} + \frac{1}{n} \int_0^\pi \cos n x dx ) $$
Integral of cosine over half period:
$$ \int_0^\pi \cos n x dx = \frac{\sin n \pi}{n} = 0 $$
$$ b_n = \frac{2}{\pi} \times ( -\frac{\pi (-1)^n}{n} ) = \frac{2 (-1)^{n+1}}{n} $$
Calculate first three coefficients:
\( b_1 = \frac{2 (-1)^2}{1} = 2 \)
\( b_2 = \frac{2 (-1)^3}{2} = -1 \)
\( b_3 = \frac{2 (-1)^4}{3} = \frac{2}{3} \)
Answer:
$$ b_1 = 2, \quad b_2 = -1, \quad b_3 = \frac{2}{3} $$
Example 2: A heat flux distribution is an even function \( f(x) = x^2 \) over \( -\pi \le x \le \pi \). Find the first two cosine coefficients \( a_0 \) and \( a_1 \).
Given: \( f(x) = x^2 \), \( L = \pi \)
Calculate:
$$ a_0 = \frac{2}{\pi} \int_0^\pi x^2 dx = \frac{2}{\pi} [ \frac{x^3}{3} ]_0^\pi = \frac{2}{\pi} \times \frac{\pi^3}{3} = \frac{2 \pi^2}{3} $$
$$ a_1 = \frac{2}{\pi} \int_0^\pi x^2 \cos x dx $$
Use integration by parts twice as in previous example:
$$ a_1 = \frac{2}{\pi} \times (-2\pi) = -4 $$
Answer:
$$ a_0 = \frac{2 \pi^2}{3}, \quad a_1 = -4 $$
Example 3: A mechanical signal is odd \( f(x) = \sin x \) on \( -\pi \le x \le \pi \). Find the Fourier sine coefficients \( b_n \).
Given: \( f(x) = \sin x \), \( L = \pi \)
Calculate:
$$ b_n = \frac{2}{\pi} \int_0^\pi \sin x \sin n x dx $$
Use orthogonality of sine functions:
$$ b_n = \begin{cases} 1, & n=1 \\ 0, & n \neq 1 \end{cases} $$
Answer:
$$ b_1 = 1, \quad b_n = 0 \text{ for } n \neq 1 $$
The complex form of the Fourier series uses exponential functions, simplifying calculations in signal processing and vibration analysis common in mechanical production.
The complex Fourier series is:
$$ f(x) = \sum_{n=-\infty}^\infty c_n e^{i n \pi x / L} $$
where
$$ c_n = \frac{1}{2L} \int_{-L}^L f(x) e^{-i n \pi x / L} dx $$
This form allows easier manipulation in computer-aided analysis and control systems.
Example 1: Find the complex Fourier coefficient \( c_1 \) for \( f(x) = x \) on \( -\pi \le x \le \pi \).
Given: \( f(x) = x \), \( L = \pi \)
Calculate:
$$ c_1 = \frac{1}{2\pi} \int_{-\pi}^\pi x e^{-i x} dx $$
Integration by parts:
Let \( u = x \), \( dv = e^{-i x} dx \), then \( du = dx \), \( v = \frac{e^{-i x}}{-i} \)
So,
$$ c_1 = \frac{1}{2\pi} [ \frac{x e^{-i x}}{-i} \Big|_{-\pi}^\pi - \int_{-\pi}^\pi \frac{e^{-i x}}{-i} dx ] $$
Calculate boundary term:
At \( x = \pi \):
$$ \frac{\pi e^{-i \pi}}{-i} = \frac{\pi (-1)}{-i} = \frac{-\pi}{-i} = \frac{\pi}{i} $$
At \( x = -\pi \):
$$ \frac{-\pi e^{i \pi}}{-i} = \frac{-\pi (-1)}{-i} = \frac{\pi}{-i} = -\frac{\pi}{i} $$
Difference:
$$ \frac{\pi}{i} - ( -\frac{\pi}{i} ) = \frac{2\pi}{i} $$
Integral term:
$$ \int_{-\pi}^\pi \frac{e^{-i x}}{-i} dx = -\frac{1}{i} \int_{-\pi}^\pi e^{-i x} dx = -\frac{1}{i} [ \frac{e^{-i x}}{-i} ]_{-\pi}^\pi = -\frac{1}{i} \times \frac{e^{-i \pi} - e^{i \pi}}{-i} = -\frac{1}{i} \times \frac{-1 - (-1)}{-i} = 0 $$
$$ c_1 = \frac{1}{2\pi} \times \frac{2\pi}{i} = \frac{1}{i} = -i $$
Answer:
$$ c_1 = -i $$
Example 2: For \( f(x) = \cos x \) on \( -\pi \le x \le \pi \), find \( c_0 \) and \( c_1 \).
Given: \( f(x) = \cos x \), \( L = \pi \)
Calculate:
$$ c_0 = \frac{1}{2\pi} \int_{-\pi}^\pi \cos x dx = \frac{1}{2\pi} [ \sin x ]_{-\pi}^\pi = \frac{1}{2\pi} (0 - 0) = 0 $$
$$ c_1 = \frac{1}{2\pi} \int_{-\pi}^\pi \cos x e^{-i x} dx $$
Rewrite \( \cos x = \frac{e^{i x} + e^{-i x}}{2} \):
$$ c_1 = \frac{1}{2\pi} \int_{-\pi}^\pi \frac{e^{i x} + e^{-i x}}{2} e^{-i x} dx = \frac{1}{4\pi} \int_{-\pi}^\pi (1 + e^{-2i x}) dx $$
Calculate integrals:
$$ \int_{-\pi}^\pi 1 dx = 2\pi $$
$$ \int_{-\pi}^\pi e^{-2i x} dx = [ \frac{e^{-2i x}}{-2i} ]_{-\pi}^\pi = \frac{e^{-2i \pi} - e^{2i \pi}}{-2i} = \frac{1 - 1}{-2i} = 0 $$
$$ c_1 = \frac{1}{4\pi} \times 2\pi = \frac{1}{2} $$
Answer:
$$ c_0 = 0, \quad c_1 = \frac{1}{2} $$
Example 3: Calculate \( c_{-1} \) for \( f(x) = e^{i x} \) on \( -\pi \le x \le \pi \).
Given: \( f(x) = e^{i x} \), \( L = \pi \)
Calculate:
$$ c_{-1} = \frac{1}{2\pi} \int_{-\pi}^\pi e^{i x} e^{i x} dx = \frac{1}{2\pi} \int_{-\pi}^\pi e^{2i x} dx $$
Integrate:
$$ \int_{-\pi}^\pi e^{2i x} dx = [ \frac{e^{2i x}}{2i} ]_{-\pi}^\pi = \frac{e^{2i \pi} - e^{-2i \pi}}{2i} = \frac{1 - 1}{2i} = 0 $$
$$ c_{-1} = \frac{1}{2\pi} \times 0 = 0 $$
Answer:
$$ c_{-1} = 0 $$
Fourier series help analyze periodic mechanical phenomena such as vibration modes, stress waves, and thermal cycles. Calculating Fourier coefficients enables design optimization and fault diagnosis in mechanical systems.
Typical application steps:
Example 1: A rotating shaft's torque variation over one revolution \( 0 \le \theta \le 2\pi \) is \( T(\theta) = 10 + 5 \cos \theta + 3 \sin 2\theta \, \text{Nm} \). Identify the Fourier series coefficients.
Given: \( T(\theta) = 10 + 5 \cos \theta + 3 \sin 2 \theta \)
Since the function is already expressed as a Fourier series with period \( 2\pi \),
Coefficients are:
$$ a_0 = 10, \quad a_1 = 5, \quad b_2 = 3 $$
All other coefficients are zero.
Answer:
$$ a_0 = 10, \quad a_1 = 5, \quad b_2 = 3, \quad \text{others} = 0 $$
Example 2: A temperature distribution along a fin is \( f(x) = x(L-x) \) for \( 0 \le x \le L \), with \( L=2 \, m \). Represent \( f(x) \) as a Fourier sine series.
Given: \( f(x) = x(2 - x) \), \( L = 2 \)
Calculate sine coefficients:
$$ b_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n \pi x}{L} dx = \int_0^2 x(2 - x) \sin \frac{n \pi x}{2} dx $$
Calculate for \( n=1 \):
$$ b_1 = \frac{2}{2} \int_0^2 (2x, x^2) \sin \frac{\pi x}{2} dx = \int_0^2 (2x, x^2) \sin \frac{\pi x}{2} dx $$
Use integration by parts (done numerically or symbolically in practice).
Answer:
The Fourier sine coefficients \( b_n \) can be calculated using the integral formula above; numerical integration may be used for specific \( n \).
Example 3: A machine tool's vibration signal is sampled and approximated by \( f(t) = 3 \cos 2 \pi t + 4 \sin 4 \pi t \), with period \( T = 1 \, s \). Find the fundamental frequency and the first two harmonics.
Given: \( f(t) = 3 \cos 2 \pi t + 4 \sin 4 \pi t \), \( T = 1 \, s \)
Fundamental frequency:
$$ f_0 = \frac{1}{T} = 1 \, \text{Hz} $$
First harmonic: frequency \( f_1 = 1 \, \text{Hz} \), amplitude 3
Second harmonic: frequency \( f_2 = 2 \, \text{Hz} \), amplitude 4
Answer:
Fundamental frequency: 1 Hz; first harmonic amplitude 3; second harmonic amplitude 4
Find the Fourier coefficients \( a_0, a_1, b_1 \) for the function \( f(x) = x^3 \) defined over \( -\pi \le x \le \pi \). (10 marks)
A mechanical signal is given by \( f(x) = |x| \) on \( -\pi \le x \le \pi \). Determine the first three non-zero cosine coefficients \( a_0, a_1, a_2 \). (10 marks)
Calculate the complex Fourier coefficient \( c_2 \) for \( f(x) = \sin 3x \) on \( -\pi \le x \le \pi \). (10 marks)
A rotating shaft's torque variation is \( T(\theta) = 15 + 7 \sin \theta + 4 \cos 3\theta \). Identify the Fourier coefficients \( a_0, a_3, b_1 \). (10 marks)
Express the function \( f(x) = x \) on \( 0 \le x \le \pi \) as a Fourier sine series and calculate the coefficient \( b_1 \). (10 marks)
This chapter explored various types of functions including linear, trigonometric, logarithmic, and exponential functions, laying the foundation for calculus applications. It detailed key differentiation rules such as the power, product, chain, and quotient rules, essential for finding derivatives. Applications of differentiation were examined through the identification of stationary points and calculation of rates of change. Integration concepts were introduced starting with standard and definite integrals, followed by advanced techniques including integration by parts, substitution, and partial fractions. Practical uses of integration were demonstrated by finding areas between and under curves and calculating volumes. The chapter also covered differential equations, providing methods for solving them in engineering contexts. Finally, it introduced more advanced topics like double and triple integrals, Laplace transforms, and Fourier series, highlighting their significance in solving complex engineering problems.
A linear function representing the displacement \(s\) (in metres) of a machine component over time \(t\) (in seconds) is given by \(s(t) = 5t + 3\). Find the velocity of the component at \(t = 4\) seconds. (2 marks)
The angular displacement \(\theta(t)\) (in radians) of a rotating shaft is given by \(\theta(t) = 2\sin(3t)\). Find the angular velocity at \(t = \frac{\pi}{6}\) seconds. (3 marks)
The temperature \(T(t)\) in degrees Celsius of a metal rod cooling over time \(t\) (in minutes) is modelled by \(T(t) = 100 e^{-0.05 t}\). Calculate the rate of temperature change at \(t = 10\) minutes. (3 marks)
Find the derivative of the function \(f(x) = \ln(4x^2 + 1)\) at \(x = 2\). (3 marks)
A production rate \(P(t)\) (units per hour) is given by \(P(t) = t^3 - 6t^2 + 9t + 5\). Find the stationary points and determine their nature (maxima or minima). (5 marks)
The length \(L(t)\) of a metal rod expanding with temperature \(t\) (in °C) is given by \(L(t) = 2t^4 - 3t^3 + t - 5\). Find the rate of change of length with respect to temperature at \(t = 2\) °C. (4 marks)
Evaluate the definite integral \(\displaystyle \int_0^3 (4x^2 - 2x + 1) \, dx\), representing the total work done (in Joules) in moving a part along a path. (4 marks)
Calculate the area bounded by the curves \(y = x^2\) and \(y = 4x, x^2\). (5 marks)
Solve the differential equation \(\frac{dy}{dx} + 3y = 6\) given that \(y(0) = 2\). (6 marks)
Find the Laplace transform of the function \(f(t) = t e^{2t}\). (5 marks)
$$v(t) = \frac{d}{dt}(5t + 3) = 5$$
At \(t=4\) seconds,
$$v(4) = 5 \text{ m/s}$$
Answer: \(\boxed{5 \text{ m/s}}\)
$$\omega(t) = 2 \cdot 3 \cos(3t) = 6 \cos(3t)$$
At \(t = \frac{\pi}{6}\),
$$\omega(\frac{\pi}{6}) = 6 \cos(3 \times \frac{\pi}{6}) = 6 \cos(\frac{\pi}{2}) = 6 \times 0 = 0$$
Answer: \(\boxed{0 \text{ radians/second}}\)
$$\frac{dT}{dt} = 100 \times (-0.05) e^{-0.05 t} = -5 e^{-0.05 t}$$
At \(t = 10\),
$$\frac{dT}{dt} = -5 e^{-0.05 \times 10} = -5 e^{-0.5}$$
Calculate \(e^{-0.5} \approx 0.6065\),
$$\frac{dT}{dt} = -5 \times 0.6065 = -3.0325 \text{ °C/min}$$
Answer: \(\boxed{-3.03 \text{ °C/min}}\)
$$f'(x) = \frac{1}{4x^2 + 1} \times 8x = \frac{8x}{4x^2 + 1}$$
At \(x = 2\),
$$f'(2) = \frac{8 \times 2}{4 \times 2^2 + 1} = \frac{16}{4 \times 4 + 1} = \frac{16}{17} \approx 0.9412$$
Answer: \(\boxed{0.94}\)
Find stationary points by setting derivative to zero:
$$P'(t) = 3t^2 - 12t + 9$$
Set \(P'(t) = 0\):
$$3t^2 - 12t + 9 = 0$$
Divide by 3:
$$t^2 - 4t + 3 = 0$$
Factor:
$$(t - 3)(t - 1) = 0 \Rightarrow t = 1, 3$$
Second derivative:
$$P''(t) = 6t - 12$$
At \(t=1\):
$$P''(1) = 6(1) - 12 = -6 < 0 \Rightarrow \text{maximum}$$
At \(t=3\):
$$P''(3) = 6(3) - 12 = 6 > 0 \Rightarrow \text{minimum}$$
Answer: Stationary points at \(t=1\) (max), \(t=3\) (min)
Rate of change:
$$\frac{dL}{dt} = 8t^3 - 9t^2 + 1$$
At \(t=2\):
$$\frac{dL}{dt} = 8 \times 2^3 - 9 \times 2^2 + 1 = 8 \times 8 - 9 \times 4 + 1 = 64 - 36 + 1 = 29 \text{ units/°C}$$
Answer: \(\boxed{29 \text{ units per °C}}\)
$$\int_0^3 (4x^2 - 2x + 1) \, dx$$
Integrate term by term:
$$\int 4x^2 dx = \frac{4x^3}{3}, \quad \int -2x dx = -x^2, \quad \int 1 dx = x$$
Evaluate definite integral:
$$[\frac{4x^3}{3} - x^2 + x]_0^3 = (\frac{4 \times 27}{3} - 9 + 3) - (0) = (36 - 9 + 3) = 30$$
Answer: \(\boxed{30 \text{ Joules}}\)
Find intersection points:
$$x^2 = 4x, x^2 \Rightarrow 2x^2 - 4x = 0 \Rightarrow 2x(x - 2) = 0$$
So \(x=0\) or \(x=2\)
Area:
$$A = \int_0^2 ( (4x, x^2) - x^2 ) dx = \int_0^2 (4x - 2x^2) dx$$
Integrate:
$$\int 4x dx = 2x^2, \quad \int 2x^2 dx = \frac{2x^3}{3}$$
Evaluate:
$$A = [2x^2 - \frac{2x^3}{3}]_0^2 = (2 \times 4 - \frac{2 \times 8}{3}) - 0 = (8 - \frac{16}{3}) = \frac{24 - 16}{3} = \frac{8}{3} \approx 2.67$$
Answer: \(\boxed{\frac{8}{3} \text{ square units}}\)
$$\frac{dy}{dx} + 3y = 6, \quad y(0) = 2$$
This is a linear first-order ODE.
Integrating factor (IF):
$$IF = e^{\int 3 dx} = e^{3x}$$
Multiply both sides:
$$e^{3x} \frac{dy}{dx} + 3 e^{3x} y = 6 e^{3x}$$
Left side is derivative:
$$\frac{d}{dx}(y e^{3x}) = 6 e^{3x}$$
Integrate both sides:
$$y e^{3x} = \int 6 e^{3x} dx = 6 \times \frac{e^{3x}}{3} + C = 2 e^{3x} + C$$
Divide by \(e^{3x}\):
$$y = 2 + C e^{-3x}$$
Apply initial condition \(y(0) = 2\):
$$2 = 2 + C \Rightarrow C = 0$$
Answer: \(\boxed{y = 2}\)
Find Laplace transform of \(f(t) = t e^{2t}\).
Recall:
$$\mathcal{L}\{t e^{at}\} = \frac{1}{(s, a)^2}, \quad s > a$$
Here \(a=2\),
$$\mathcal{L}\{t e^{2t}\} = \frac{1}{(s - 2)^2}$$
Answer: \(\boxed{\frac{1}{(s - 2)^2}}\)
\(\frac{df}{d\theta} = 10 \cos \theta\), at \(\theta = 30^\circ\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
\(\frac{df}{d\theta} = 10 \times \frac{\sqrt{3}}{2} = \mathbf{8.66}\)
\(\frac{df}{dx} = \frac{2}{2x + 1}\), at \(x=4\), \(\frac{2}{9} = \mathbf{0.222}\)
\(\frac{dT}{dt} = 150 \times -0.05 e^{-0.05t} = -7.5 e^{-0.05t}\)
At \(t=10\), \(-7.5 e^{-0.5} = -7.5 \times 0.6065 = \mathbf{-4.55^\circ C/\text{min}}\)
\(f(x) = (3x^2 + 2x)(x - 4)\)
\(\frac{d}{dx} = (6x + 2)(x - 4) + (3x^2 + 2x)(1) = (6x + 2)(x - 4) + 3x^2 + 2x\)
Simplify: \(\mathbf{6x^2 - 24x + 2x - 8 + 3x^2 + 2x = 9x^2 - 20x - 8}\)
\(f(x) = \frac{5x^3 - 2}{x^2 + 1}\)
\(\frac{d}{dx} = \frac{(15x^2)(x^2 + 1) - (5x^3 - 2)(2x)}{(x^2 + 1)^2}\)
Simplify numerator: \(15x^4 + 15x^2 - 10x^4 + 4x = 5x^4 + 15x^2 + 4x\)
Answer: \(\frac{5x^4 + 15x^2 + 4x}{(x^2 + 1)^2}\)
\(f(x) = (2x^3 + 5)^4\)
\(\frac{df}{dx} = 4(2x^3 + 5)^3 \times 6x^2 = \mathbf{24x^2 (2x^3 + 5)^3}\)
\(v(t) = 4t^3 - 3t^2 + 2\)
\(\frac{dv}{dt} = 12t^2 - 6t\)
Stationary points: \(12t^2 - 6t = 0 \Rightarrow 6t(2t - 1) = 0 \Rightarrow t=0, t= \frac{1}{2}\)
Second derivative: \(\frac{d^2v}{dt^2} = 24t - 6\)
At \(t=0\), \(-6 < 0\) maximum; at \(t=\frac{1}{2}\), \(24 \times \frac{1}{2} - 6 = 6 > 0\) minimum.
\(\int (6x^2 - 4x + 1) dx = 2x^3 - 2x^2 + x + C\)
Question 11 (Compulsory - 20 marks)
A mechanical production technician at KPLC is tasked with optimizing the cooling rate of a transformer component. The temperature \(T(t)\) in °C after \(t\) minutes is modeled by the differential equation \(\frac{dT}{dt} + 0.1T = 20\).
a) Solve the differential equation for \(T(t)\) given the initial temperature \(T(0) = 100^\circ C\). (10 marks)
b) Using the solution, determine the temperature after 15 minutes and interpret the result in the context of equipment safety. (10 marks)
Question 12 (20 marks)
A mechanical technician is analyzing the volume of lubricant required to fill a cylindrical tank with variable radius \(r(x) = 2 + x^2\) meters along the height \(x\) (from 0 to 3 m).
Calculate the volume of lubricant in the tank using definite integration. (20 marks)
Question 13 (20 marks)
Given the function \(f(x) = \ln(x^2 + 1)\),
a) Find \(\frac{df}{dx}\). (10 marks)
b) Using integration by parts, evaluate \(\int x \ln(x^2 + 1) dx\). (10 marks)
Question 14 (20 marks)
A machine part oscillates according to the function \(f(t) = 3 \sin 4t + 2 \cos 6t\).
a) Find the Fourier series representation of \(f(t)\) over one period. (10 marks)
b) Determine the Laplace transform of \(f(t)\). (10 marks)
Question 11
a) Rewrite as \(\frac{dT}{dt} + 0.1T = 20\).
Integrating factor \(= e^{0.1t}\).
Multiply both sides:
\(\frac{d}{dt} (T e^{0.1t}) = 20 e^{0.1t}\)
Integrate:
\(T e^{0.1t} = \int 20 e^{0.1t} dt = 20 \times \frac{e^{0.1t}}{0.1} + C = 200 e^{0.1t} + C\)
Divide:
\(T = 200 + C e^{-0.1t}\)
Use \(T(0) = 100\):
\(100 = 200 + C \Rightarrow C = -100\)
Solution:
\(\boxed{T(t) = 200 - 100 e^{-0.1t}}\)
b) At \(t=15\),
\(T(15) = 200 - 100 e^{-1.5} = 200 - 100 \times 0.2231 = 200 - 22.31 = \mathbf{177.69^\circ C}\)
Interpretation: The temperature decreases from 100°C to about 177.7°C, indicating the model likely represents heating towards equilibrium (check initial conditions). The technician should ensure temperature stays within safe operating limits to avoid damage.
Question 12
Volume \(V = \pi \int_0^3 r(x)^2 dx = \pi \int_0^3 (2 + x^2)^2 dx\)
Expand:
\((2 + x^2)^2 = 4 + 4x^2 + x^4\)
Integrate:
\(\int_0^3 (4 + 4x^2 + x^4) dx = [4x + \frac{4x^3}{3} + \frac{x^5}{5}]_0^3\)
Calculate:
\(4(3) + \frac{4(27)}{3} + \frac{243}{5} = 12 + 36 + 48.6 = 96.6\)
Volume:
\(\boxed{V = \pi \times 96.6 = 303.7 \text{ m}^3}\)
Question 13
a) \(f(x) = \ln(x^2 + 1)\)
\(\frac{df}{dx} = \frac{2x}{x^2 + 1}\)
b) \(\int x \ln(x^2 + 1) dx\)
Let \(u = \ln(x^2 + 1)\), \(dv = x dx\)
Then \(du = \frac{2x}{x^2 + 1} dx\), \(v = \frac{x^2}{2}\)
Integration by parts formula:
\(\int u dv = uv - \int v du\)
Calculate:
\(\frac{x^2}{2} \ln(x^2 + 1) - \int \frac{x^2}{2} \times \frac{2x}{x^2 + 1} dx = \frac{x^2}{2} \ln(x^2 + 1) - \int \frac{x^3}{x^2 + 1} dx\)
Simplify \(\int \frac{x^3}{x^2 + 1} dx\):
Divide numerator by denominator: \(x^3 = x(x^2 + 1) - x\)
So, \(\int \frac{x^3}{x^2 + 1} dx = \int (x - \frac{x}{x^2 + 1}) dx = \int x dx - \int \frac{x}{x^2 + 1} dx\)
Integrate:
\(\frac{x^2}{2} - \frac{1}{2} \ln(x^2 + 1) + C\)
Final answer:
\(\frac{x^2}{2} \ln(x^2 + 1) - (\frac{x^2}{2} - \frac{1}{2} \ln(x^2 + 1)) + C = \frac{x^2 + 1}{2} \ln(x^2 + 1) - \frac{x^2}{2} + C\)
Question 14
a) \(f(t) = 3 \sin 4t + 2 \cos 6t\) is already expressed as a Fourier series with fundamental frequencies \(4\) and \(6\). The Fourier series representation is:
\(\boxed{f(t) = 3 \sin 4t + 2 \cos 6t}\)
b) Laplace transform:
\(\mathcal{L}\{3 \sin 4t\} = \frac{3 \times 4}{s^2 + 16} = \frac{12}{s^2 + 16}\)
\(\mathcal{L}\{2 \cos 6t\} = \frac{2 s}{s^2 + 36}\)
Combined:
\(\boxed{F(s) = \frac{12}{s^2 + 16} + \frac{2s}{s^2 + 36}}\)
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