Calculus is essential for Industrial Mechatronics Technology professionals in Kenya to analyze and optimize dynamic systems, control processes, and mechanical components. Differentiation enables precise determination of rates of change, which is crucial in sensor signal processing, actuator control, and system modeling. This chapter develops skills in differentiation up to the third order, equipping learners to solve real-world problems involving velocity, acceleration, jerk, and more complex system behaviors.

6.1 Differentiation up to third order

Differentiation is a fundamental calculus operation that measures how a function changes as its input changes. In mechatronics, it is used to analyze motion, electrical signals, and control systems. This section covers functions relevant to mechatronics, the rules of differentiation, and practical applications involving first, second, and third derivatives.

6.1.1 Functions

Differentiation requires a clear understanding of functions, mathematical expressions that relate input variables to output variables. In Industrial Mechatronics Technology, functions often represent physical quantities such as position, velocity, force, or voltage as functions of time or space.

The derivative of a function gives the instantaneous rate of change of the output relative to the input. The first derivative corresponds to velocity or rate, the second derivative to acceleration or curvature, and the third derivative to jerk or rate of acceleration change.

The general formula for the derivative of a function \( f(x) \) with respect to \( x \) is:

$$ \frac{d}{dx} f(x) = f'(x) $$

Worked Examples

Example 1: A robotic arm position is given by \( s(t) = 5t^2 \) mm, where \( t \) is time in seconds. Find the velocity as a function of time.

Given: \( s(t) = 5t^2 \) mm

$$ v(t) = \frac{d}{dt} s(t) $$

$$ v(t) = \frac{d}{dt} (5t^2) = 5 \times 2t = 10t $$

Answer: \( v(t) = 10t \) mm/s

Example 2: The voltage across a sensor varies as \( V(t) = 3t^3 - 2t + 4 \) volts. Find the rate of change of voltage at any time \( t \).

Given: \( V(t) = 3t^3 - 2t + 4 \) V

$$ \frac{dV}{dt} = \frac{d}{dt} (3t^3 - 2t + 4) $$

$$ = 3 \times 3t^2 - 2 \times 1 + 0 = 9t^2 - 2 $$

Answer: \( \frac{dV}{dt} = 9t^2 - 2 \) volts per second

Example 3: The displacement of a conveyor belt is \( x(t) = 4t^3 - 6t^2 + t - 5 \) meters. Find the velocity and acceleration functions.

Given: \( x(t) = 4t^3 - 6t^2 + t - 5 \) m

Velocity:

$$ v(t) = \frac{dx}{dt} = 12t^2 - 12t + 1 $$

Acceleration:

$$ a(t) = \frac{d^2x}{dt^2} = \frac{d}{dt} v(t) = 24t - 12 $$

Answer: \( v(t) = 12t^2 - 12t + 1 \) m/s, \( a(t) = 24t - 12 \) m/s²

Example 4: The current through a motor coil varies as \( I(t) = 2e^{0.5t} \) amperes. Find the rate of change of current.

Given: \( I(t) = 2e^{0.5t} \) A

$$ \frac{dI}{dt} = 2 \times \frac{d}{dt} e^{0.5t} = 2 \times 0.5 e^{0.5t} = e^{0.5t} $$

Answer: \( \frac{dI}{dt} = e^{0.5t} \) amperes per second

Example 5: The temperature \( T \) in a machine part varies with time as \( T(t) = \ln(t^2 + 1) \) °C. Find the rate of temperature change.

Given: \( T(t) = \ln(t^2 + 1) \) °C

$$ \frac{dT}{dt} = \frac{1}{t^2 + 1} \times 2t = \frac{2t}{t^2 + 1} $$

Answer: \( \frac{dT}{dt} = \frac{2t}{t^2 + 1} \) °C/s

6.1.2 Rules

Differentiation follows specific rules that simplify finding derivatives of complex functions. These rules include the power rule, product rule, quotient rule, and chain rule. Mastery of these rules is essential for analyzing mechatronic systems involving multiple interacting variables.

The main differentiation rules are:

• Power rule: For \( f(x) = x^n \), \( \frac{d}{dx} x^n = nx^{n-1} \)
• Constant multiple rule: \( \frac{d}{dx} [cf(x)] = c \frac{d}{dx} f(x) \)
• Sum/difference rule: \( \frac{d}{dx} [f(x) \pm g(x)] = \frac{d}{dx} f(x) \pm \frac{d}{dx} g(x) \)
• Product rule: \( \frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \)
• Quotient rule: \( \frac{d}{dx} [\frac{f(x)}{g(x)}] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \)
• Chain rule: \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

Worked Examples

Example 1: Differentiate \( f(x) = 7x^4 \)

Given: \( f(x) = 7x^4 \)

Using power and constant multiple rule:

$$ \frac{d}{dx} 7x^4 = 7 \times \frac{d}{dx} x^4 = 7 \times 4x^3 = 28x^3 $$

Answer: \( f'(x) = 28x^3 \)

Example 2: Differentiate \( y = (3x^2)(2x + 1) \)

Given: \( y = (3x^2)(2x + 1) \)

Using product rule:

$$ \frac{dy}{dx} = \frac{d}{dx} (3x^2) \times (2x + 1) + 3x^2 \times \frac{d}{dx} (2x + 1) $$

$$ = 6x (2x + 1) + 3x^2 \times 2 = 6x (2x + 1) + 6x^2 $$

$$ = 12x^2 + 6x + 6x^2 = 18x^2 + 6x $$

Answer: \( \frac{dy}{dx} = 18x^2 + 6x \)

Example 3: Differentiate \( z = \frac{5x^3 + 1}{x^2} \)

Given: \( z = \frac{5x^3 + 1}{x^2} \)

Using quotient rule:

$$ \frac{dz}{dx} = \frac{(15x^2)(x^2) - (5x^3 + 1)(2x)}{(x^2)^2} $$

$$ = \frac{15x^4 - 2x(5x^3 + 1)}{x^4} = \frac{15x^4 - (10x^4 + 2x)}{x^4} $$

$$ = \frac{15x^4 - 10x^4 - 2x}{x^4} = \frac{5x^4 - 2x}{x^4} = 5 - \frac{2}{x^3} $$

Answer: \( \frac{dz}{dx} = 5 - \frac{2}{x^3} \)

Example 4: Differentiate \( y = \sin(3x^2) \)

Given: \( y = \sin(3x^2) \)

Using chain rule:

$$ \frac{dy}{dx} = \cos(3x^2) \times \frac{d}{dx} (3x^2) = \cos(3x^2) \times 6x = 6x \cos(3x^2) $$

Answer: \( \frac{dy}{dx} = 6x \cos(3x^2) \)

Example 5: Differentiate \( f(t) = e^{4t^3 + 2t} \)

Given: \( f(t) = e^{4t^3 + 2t} \)

Using chain rule:

$$ \frac{df}{dt} = e^{4t^3 + 2t} \times \frac{d}{dt} (4t^3 + 2t) = e^{4t^3 + 2t} \times (12t^2 + 2) $$

Answer: \( \frac{df}{dt} = e^{4t^3 + 2t} (12t^2 + 2) \)

6.1.3 Applications

Higher order derivatives describe physical phenomena in mechatronics such as velocity, acceleration, and jerk in motion control, and rates of change in electrical signals. The first derivative can represent speed, the second derivative acceleration, and the third derivative jerk, which affects mechanical stress and system smoothness.

The general forms of derivatives are:

• First derivative: \( \frac{dy}{dx} \)
• Second derivative: \( \frac{d^2y}{dx^2} \)
• Third derivative: \( \frac{d^3y}{dx^3} \)

Worked Examples

Example 1: The position of a piston in a hydraulic actuator is \( s(t) = 2t^3 - 9t^2 + 12t \) mm. Find velocity, acceleration, and jerk.

Given: \( s(t) = 2t^3 - 9t^2 + 12t \) mm

Velocity:

$$ v(t) = \frac{ds}{dt} = 6t^2 - 18t + 12 $$

Acceleration:

$$ a(t) = \frac{d^2s}{dt^2} = \frac{d}{dt} v(t) = 12t - 18 $$

Jerk:

$$ j(t) = \frac{d^3s}{dt^3} = \frac{d}{dt} a(t) = 12 $$

Answer:

\( v(t) = 6t^2 - 18t + 12 \) mm/s

\( a(t) = 12t - 18 \) mm/s²

\( j(t) = 12 \) mm/s³

Example 2: The voltage across a sensor varies as \( V(t) = 4 \sin(2t) \) volts. Find the rate of change of voltage, acceleration of voltage change, and jerk.

Given: \( V(t) = 4 \sin(2t) \) V

First derivative:

$$ \frac{dV}{dt} = 4 \times 2 \cos(2t) = 8 \cos(2t) $$

Second derivative:

$$ \frac{d^2V}{dt^2} = -8 \times 2 \sin(2t) = -16 \sin(2t) $$

Third derivative:

$$ \frac{d^3V}{dt^3} = -16 \times 2 \cos(2t) = -32 \cos(2t) $$

Answer:

\( \frac{dV}{dt} = 8 \cos(2t) \) V/s

\( \frac{d^2V}{dt^2} = -16 \sin(2t) \) V/s²

\( \frac{d^3V}{dt^3} = -32 \cos(2t) \) V/s³

Example 3: A motor shaft angle \( \theta(t) = 5t^4 - 3t^3 + 2t \) radians. Find angular velocity, angular acceleration, and angular jerk.

Given: \( \theta(t) = 5t^4 - 3t^3 + 2t \) rad

Angular velocity:

$$ \omega(t) = \frac{d\theta}{dt} = 20t^3 - 9t^2 + 2 $$

Angular acceleration:

$$ \alpha(t) = \frac{d^2\theta}{dt^2} = 60t^2 - 18t $$

Angular jerk:

$$ j(t) = \frac{d^3\theta}{dt^3} = 120t - 18 $$

Answer:

\( \omega(t) = 20t^3 - 9t^2 + 2 \) rad/s

\( \alpha(t) = 60t^2 - 18t \) rad/s²

\( j(t) = 120t - 18 \) rad/s³

Example 4: The displacement of a conveyor belt is given by \( x(t) = 10e^{-0.5t} \) cm. Find velocity, acceleration, and jerk.

Given: \( x(t) = 10e^{-0.5t} \) cm

Velocity:

$$ v(t) = \frac{dx}{dt} = 10 \times (-0.5) e^{-0.5t} = -5 e^{-0.5t} $$

Acceleration:

$$ a(t) = \frac{d^2x}{dt^2} = \frac{d}{dt} (-5 e^{-0.5t}) = -5 \times (-0.5) e^{-0.5t} = 2.5 e^{-0.5t} $$

Jerk:

$$ j(t) = \frac{d^3x}{dt^3} = \frac{d}{dt} (2.5 e^{-0.5t}) = 2.5 \times (-0.5) e^{-0.5t} = -1.25 e^{-0.5t} $$

Answer:

\( v(t) = -5 e^{-0.5t} \) cm/s

\( a(t) = 2.5 e^{-0.5t} \) cm/s²

\( j(t) = -1.25 e^{-0.5t} \) cm/s³

Example 5: The pressure in a pneumatic system is \( P(t) = 3t^5 - 2t^3 + t \) kPa. Find the first, second and third derivatives.

Given: \( P(t) = 3t^5 - 2t^3 + t \) kPa

First derivative:

$$ \frac{dP}{dt} = 15t^4 - 6t^2 + 1 $$

Second derivative:

$$ \frac{d^2P}{dt^2} = 60t^3 - 12t $$

Third derivative:

$$ \frac{d^3P}{dt^3} = 180t^2 - 12 $$

Answer:

\( \frac{dP}{dt} = 15t^4 - 6t^2 + 1 \) kPa/s

\( \frac{d^2P}{dt^2} = 60t^3 - 12t \) kPa/s²

\( \frac{d^3P}{dt^3} = 180t^2 - 12 \) kPa/s³

Chapter Summary

This chapter covered the application of calculus focusing on differentiation and integration techniques relevant to engineering contexts. Differentiation was explored up to the third order, starting with various types of functions including linear, trigonometric, logarithmic, and exponential functions. The chapter detailed key differentiation rules such as the power, product, chain, and quotient rules, enabling the calculation of derivatives for complex expressions. Practical applications of differentiation included finding stationary points of functions and determining rates of change in engineering problems. Integration was introduced through standard and definite integrals, with techniques such as integration by parts, substitution, and partial fractions to solve more challenging integrals. The chapter concluded with applications of integration, specifically calculating areas between and under curves as well as volumes of solids, essential for many engineering design and analysis tasks. Overall, the chapter provided a comprehensive foundation in calculus methods necessary for solving real-world engineering problems.