By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems by understanding how things change and accumulate, making you a valuable problem-solver in your trade.
In industrial control and installation, calculus is essential for analyzing and controlling dynamic engineering systems. Professionals use calculus to determine rates of change, optimize system performance, and predict behavior in processes such as temperature regulation, fluid flow, and electrical circuits. Understanding derivatives is especially important for troubleshooting, automation, and maintenance of modern industrial equipment in Kenyan manufacturing, processing, and building services sectors.
Calculus, and specifically the derivative, is fundamental for describing how engineering quantities change with respect to another variable, such as how pressure changes with time or how current varies with resistance. The derivative is a mathematical tool for finding the instantaneous rate of change of one quantity relative to another. In industrial control, this helps in analyzing sensor signals, predicting process trends, and designing control responses.
The derivative of a function \( f(x) \) at a point \( x \) measures how \( f(x) \) changes as \( x \) changes. It is defined as the limit:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$
This formula calculates the instantaneous rate of change of \( f(x) \) with respect to \( x \) and is the foundation for all applications of differential calculus.
Example 1: Find the derivative of \( f(x) = 3x + 2 \).
Given: \( f(x) = 3x + 2 \)
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$
First, compute \( f(x+h) \):
$$ f(x+h) = 3(x+h) + 2 = 3x + 3h + 2 $$
Now, substitute into the formula:
$$ f'(x) = \lim_{h \to 0} \frac{[3x + 3h + 2] - [3x + 2]}{h} $$
$$ = \lim_{h \to 0} \frac{3h}{h} $$
$$ = \lim_{h \to 0} 3 $$
Answer: \( f'(x) = 3 \)
Example 2: Find the derivative of \( f(x) = x^2 \).
Given: \( f(x) = x^2 \)
$$ f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} $$
Expand \( (x+h)^2 \):
$$ (x+h)^2 = x^2 + 2xh + h^2 $$
Substitute:
$$ f'(x) = \lim_{h \to 0} \frac{[x^2 + 2xh + h^2] - x^2}{h} $$
$$ = \lim_{h \to 0} \frac{2xh + h^2}{h} $$
$$ = \lim_{h \to 0} (2x + h) $$
$$ = 2x $$
Answer: \( f'(x) = 2x \)
Example 3: Find the derivative of \( f(x) = 5x^2 + 4x \).
Given: \( f(x) = 5x^2 + 4x \)
$$ f'(x) = \lim_{h \to 0} \frac{[5(x+h)^2 + 4(x+h)] - [5x^2 + 4x]}{h} $$
Expand \( (x+h)^2 \):
$$ (x+h)^2 = x^2 + 2xh + h^2 $$
So,
$$ 5(x+h)^2 = 5x^2 + 10xh + 5h^2 $$
And
$$ 4(x+h) = 4x + 4h $$
Now, sum:
$$ 5x^2 + 10xh + 5h^2 + 4x + 4h - 5x^2 - 4x = 10xh + 5h^2 + 4h $$
So,
$$ f'(x) = \lim_{h \to 0} \frac{10xh + 5h^2 + 4h}{h} $$
$$ = \lim_{h \to 0} (10x + 5h + 4) $$
$$ = 10x + 4 $$
Answer: \( f'(x) = 10x + 4 \)
Example 4: The position \( s \) (in meters) of a conveyor belt is given by \( s(t) = 2t^2 + 3t \), where \( t \) is in seconds. Find the velocity at time \( t \).
Given: \( s(t) = 2t^2 + 3t \)
Velocity is the derivative of position:
$$ v(t) = s'(t) = \frac{d}{dt}(2t^2 + 3t) $$
$$ = 4t + 3 $$
Answer: \( v(t) = 4t + 3 \) m/s
Example 5: In a temperature control system, the temperature \( T \) (in \( ^\circ \)C) at time \( t \) (in minutes) is \( T(t) = 20 + 5t, t^2 \). Find the rate of change of temperature at \( t = 2 \) minutes.
Given: \( T(t) = 20 + 5t, t^2 \)
$$ \frac{dT}{dt} = 5 - 2t $$
At \( t = 2 \):
$$ \frac{dT}{dt} = 5 - 2 \times 2 = 5 - 4 = 1 $$
Answer: The rate of change is \( 1 \ ^\circ \mathrm{C} /\mathrm{min} \) at \( t=2 \) minutes
The derivative represents the instantaneous rate of change of a variable, such as speed, flow rate, or current. In industrial control, derivatives help engineers understand how fast a process variable is changing, which is crucial for tuning controllers, predicting system behavior, and ensuring safety.
Example 1: The current \( I \) in an electrical circuit changes with time according to \( I(t) = 4t^3 \) amperes. Find the rate of change of current at \( t = 2 \) seconds.
Given: \( I(t) = 4t^3 \)
$$ \frac{dI}{dt} = 12t^2 $$
At \( t = 2 \):
$$ \frac{dI}{dt} = 12 \times (2)^2 = 12 \times 4 = 48 $$
Answer: The rate of change is \( 48 \) A/s at \( t = 2 \) s
Example 2: The pressure \( P \) in a hydraulic system is given by \( P(x) = 50x^2 + 10x \) kPa, where \( x \) is the valve opening in mm. Find the rate at which pressure changes when \( x = 3 \) mm.
Given: \( P(x) = 50x^2 + 10x \)
$$ \frac{dP}{dx} = 100x + 10 $$
At \( x = 3 \):
$$ \frac{dP}{dx} = 100 \times 3 + 10 = 300 + 10 = 310 $$
Answer: The rate of change is \( 310 \) kPa/mm at \( x = 3 \) mm
Example 3: In a cooling system, water flows at a rate \( Q(t) = 12 - 2t \) litres per minute, where \( t \) is time in minutes. Find the rate at which the flow rate is changing at \( t = 4 \) minutes.
Given: \( Q(t) = 12 - 2t \)
$$ \frac{dQ}{dt} = -2 $$
Answer: The flow rate is decreasing by 2 L/min each minute
Example 4: The voltage across a sensor is \( V(t) = 10e^{-0.5t} \) volts. Find the rate of change of voltage at \( t = 2 \) seconds.
Given: \( V(t) = 10e^{-0.5t} \)
$$ \frac{dV}{dt} = 10 \times (-0.5) e^{-0.5t} = -5e^{-0.5t} $$
At \( t = 2 \):
$$ \frac{dV}{dt} = -5e^{-1} \approx -5 \times 0.3679 \approx -1.8395 $$
Answer: The rate is approximately \( -1.84 \) V/s at \( t = 2 \) s
Example 5: The speed \( v \) of a conveyor is given by \( v(t) = 3t^2 + 2t \) m/s. Find the acceleration at \( t = 1 \) s.
Given: \( v(t) = 3t^2 + 2t \)
$$ a(t) = \frac{dv}{dt} = 6t + 2 $$
At \( t = 1 \):
$$ a(1) = 6 \times 1 + 2 = 6 + 2 = 8 $$
Answer: The acceleration is \( 8 \) m/s\(^2\) at \( t = 1 \) s
Common rules make finding derivatives of most functions straightforward. The most used are:
Example 1: Differentiate \( f(x) = 7 \).
Given: \( f(x) = 7 \)
$$ f'(x) = 0 $$
Answer: \( f'(x) = 0 \)
Example 2: Differentiate \( f(x) = 8x^5 \).
Given: \( f(x) = 8x^5 \)
$$ f'(x) = 8 \times 5 x^{5-1} = 40x^4 $$
Answer: \( f'(x) = 40x^4 \)
Example 3: Differentiate \( f(x) = 6x^2 - 4x \).
Given: \( f(x) = 6x^2 - 4x \)
$$ f'(x) = 12x - 4 $$
Answer: \( f'(x) = 12x - 4 \)
Example 4: Differentiate \( f(x) = 2x^3 + 5x^2 - 7 \).
Given: \( f(x) = 2x^3 + 5x^2 - 7 \)
$$ f'(x) = 6x^2 + 10x $$
Answer: \( f'(x) = 6x^2 + 10x \)
Example 5: Differentiate \( f(x) = 9x - 2x^4 \).
Given: \( f(x) = 9x - 2x^4 \)
$$ f'(x) = 9 - 8x^3 $$
Answer: \( f'(x) = 9 - 8x^3 \)
In industrial control, derivatives are used to set controller parameters, monitor process changes, and analyze dynamic responses. For example, the derivative action in a PID controller predicts future errors, improving system stability.
Example 1: A temperature controller uses the error signal \( E(t) = 4t^2 - 6t \) (degrees Celsius). Find the rate at which the error is changing at \( t = 3 \) seconds.
Given: \( E(t) = 4t^2 - 6t \)
$$ \frac{dE}{dt} = 8t - 6 $$
At \( t = 3 \):
$$ \frac{dE}{dt} = 8 \times 3 - 6 = 24 - 6 = 18 $$
Answer: The error is increasing at \( 18 \ ^\circ \mathrm{C}/\mathrm{s} \) at \( t = 3 \) s
Example 2: The water level in a storage tank is \( h(t) = 15 - 0.3t \) meters. Find how fast the level is dropping.
Given: \( h(t) = 15 - 0.3t \)
$$ \frac{dh}{dt} = -0.3 $$
Answer: The level drops at \( 0.3 \) m/min
Example 3: The output voltage of a signal transducer is \( V(t) = 5t^2 + 2 \). Find the rate at which voltage increases when \( t = 4 \) seconds.
Given: \( V(t) = 5t^2 + 2 \)
$$ \frac{dV}{dt} = 10t $$
At \( t = 4 \):
$$ \frac{dV}{dt} = 10 \times 4 = 40 $$
Answer: The voltage increases at \( 40 \) V/s at \( t = 4 \) s
Example 4: The displacement of a robotic arm is \( s(t) = 3t^3 - 2t^2 + t \). Find the velocity when \( t = 2 \) seconds.
Given: \( s(t) = 3t^3 - 2t^2 + t \)
$$ v(t) = \frac{ds}{dt} = 9t^2 - 4t + 1 $$
At \( t = 2 \):
$$ v(2) = 9 \times 4 - 4 \times 2 + 1 = 36 - 8 + 1 = 29 $$
Answer: The velocity is \( 29 \) m/s at \( t = 2 \) s
Example 5: The current through a heating element is \( I(t) = 2t^2 + 3t + 1 \). Find the rate of change of current at \( t = 5 \) seconds.
Given: \( I(t) = 2t^2 + 3t + 1 \)
$$ \frac{dI}{dt} = 4t + 3 $$
At \( t = 5 \):
$$ \frac{dI}{dt} = 4 \times 5 + 3 = 20 + 3 = 23 $$
Answer: The current increases at \( 23 \) A/s at \( t = 5 \) s
The displacement \( s \) (in meters) of a conveyor belt is given by \( s(t) = 3t^2 + 2t \), where \( t \) is in seconds. Find the velocity at \( t = 3 \) seconds. (4 marks)
For the function \( f(x) = 4x^3 - 7x^2 + 5x \), find \( f'(x) \). (4 marks)
The temperature in a heating process is \( T(t) = 25 + 2t - 0.5t^2 \). Find the rate of change of temperature at \( t = 4 \) minutes. (4 marks)
The voltage across a resistor is given by \( V(t) = 12 - 0.8t \). What is the rate of change of voltage? (4 marks)
The water level in a tank is \( h(t) = 10 - t^2 \) meters. Find the rate at which the level is changing at \( t = 2 \) minutes. (4 marks)
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Create a free accountThis chapter introduced the concept of derivatives, explaining how they represent the rate at which a function changes. Differentiation from first principles was demonstrated with functions such as \(\sin x\), \(\cos x\), \(x^n\), and \(\ln x\), followed by a review of common derivatives using tables. Key rules of differentiation, including the product, chain, quotient, sum, and implicit differentiation, were applied to various mathematical problems. The relationship between derivatives and rates of change, as well as small changes in engineering contexts, was explored. Techniques for finding derivatives of inverse functions and identifying stationary points in functions of two variables were also covered. The chapter then shifted to integration, discussing its meaning and distinguishing between indefinite and definite integrals. Methods of integration, including integration by parts, substitution, polynomials, and inverse functions, were explained with practical applications. Finally, integrals involving hyperbolic and inverse functions were addressed, consolidating the calculus tools necessary for advanced engineering analysis.
Step 1: Formula
The derivative is$$f'(x) = \frac{d}{dx}(4x^2 + 7x)$$
Step 2: Differentiate$$f'(x) = 8x + 7$$
Step 3: Substitute \(x = 3\)$$f'(3) = 8 \times 3 + 7$$$$f'(3) = 24 + 7$$$$f'(3) = 31$$
Final Answer:
31
Step 1: Formula$$f'(x) = \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h}$$
Step 2: Use sine addition formula$$\sin(x + h) = \sin x \cos h + \cos x \sin h$$ So$$f'(x) = \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$$$$= \lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}$$
Step 3: Separate terms$$= \lim_{h \to 0} [ \sin x \frac{\cos h - 1}{h} + \cos x \frac{\sin h}{h} ]$$
As \(h \to 0\), \(\frac{\cos h - 1}{h} \to 0\), \(\frac{\sin h}{h} \to 1\).
Step 4: Substitute limits$$= \sin x \times 0 + \cos x \times 1$$$$= \cos x$$
Final Answer:
\(\cos x\)
The control panel at a Nairobi manufacturing plant is monitored by a sensor whose position changes with time according to the function \( s(t) = 4t^3 - 2t^2 + 5t \), where \( s \) is in centimetres and \( t \) in seconds. Find the instantaneous velocity of the sensor at \( t = 2 \) seconds. (4 marks)
Using first principles, differentiate \( f(x) = \sin x \) with respect to \( x \). (4 marks)
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