Apply calculus in Automotive Mechatronics involves understanding how changing quantities, such as velocity, acceleration, and sensor signals, vary with respect to time or other variables. Derivatives provide the mathematical tool to describe rates of change essential for diagnosing engine behavior, control system tuning, and dynamic response analysis. This chapter focuses on the fundamental concept of derivatives and how to compute them from first principles, laying the groundwork for more advanced applications in automotive systems.

6.1 Definition of derivatives of a function exercises

Derivatives express the instantaneous rate of change of a function with respect to its independent variable, which in automotive mechatronics often relates to how parameters like speed, torque or voltage change over time or position. The derivative of a function \( f(x) \) at a point \( x \) is defined mathematically as the limit of the average rate of change over an interval shrinking to zero.

6.1.1 Meaning of the derivative and its formula

The derivative of a function \( f \) at a point \( x \), denoted by \( f'(x) \), measures the slope of the tangent line to the curve of \( f \) at that point. It is defined by the limit:

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

This formula calculates how much \( f(x) \) changes per unit change in \( x \), as the increment \( h \) becomes infinitesimally small.

Worked Examples

Example 1:
Given the function \( f(x) = 3x^2 \), find \( f'(2) \).

Given: \( f(x) = 3x^2 \), \( x = 2 \)

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

$$ = \lim_{h \to 0} \frac{3(x+h)^2 - 3x^2}{h} $$

$$ = \lim_{h \to 0} \frac{3(x^2 + 2xh + h^2) - 3x^2}{h} $$

$$ = \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 3x^2}{h} = \lim_{h \to 0} \frac{6xh + 3h^2}{h} $$

$$ = \lim_{h \to 0} (6x + 3h) = 6x $$

At \( x=2 \):

$$ f'(2) = 6 \times 2 = 12 $$

Answer: \( f'(2) = 12 \)

Example 2:
Find the derivative at \( x = 1 \) for \( f(x) = 5x^3 - x \).

Given: \( f(x) = 5x^3 - x \), \( x=1 \)

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

$$ = \lim_{h \to 0} \frac{5(x+h)^3 - (x+h) - [5x^3 - x]}{h} $$

Expand \( (x+h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3 \):

$$ = \lim_{h \to 0} \frac{5(x^3 + 3x^2 h + 3x h^2 + h^3) - x, h - 5x^3 + x}{h} $$

Simplify numerator:

$$ = \lim_{h \to 0} \frac{5x^3 + 15x^2 h + 15x h^2 + 5 h^3 - x, h - 5x^3 + x}{h} $$

$$ = \lim_{h \to 0} \frac{15x^2 h + 15x h^2 + 5 h^3 - h}{h} $$

Divide numerator and denominator by \( h \):

$$ = \lim_{h \to 0} (15x^2 + 15x h + 5 h^2 - 1) $$

At \( h \to 0 \):

$$ = 15x^2 - 1 $$

At \( x=1 \):

$$ f'(1) = 15 \times 1^2 - 1 = 15 - 1 = 14 $$

Answer: \( f'(1) = 14 \)

Example 3:
Calculate the derivative of \( f(x) = \sqrt{x} \) at \( x = 4 \).

Given: \( f(x) = \sqrt{x} = x^{1/2} \), \( x=4 \)

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} $$

Rationalize numerator:

$$ = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} $$

$$ = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} $$

Simplify:

$$ = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} $$

At \( h = 0 \):

$$ = \frac{1}{2 \sqrt{x}} $$

At \( x = 4 \):

$$ f'(4) = \frac{1}{2 \times 2} = \frac{1}{4} $$

Answer: \( f'(4) = \frac{1}{4} \)

6.1.2 Interpretation of derivatives in automotive mechatronics

In automotive mechatronics, the derivative represents physical quantities such as velocity as the rate of change of displacement, or acceleration as the rate of change of velocity. Understanding the derivative helps in sensor signal processing, engine control, and system diagnostics.

Worked Examples

Example 1:
A car's position along a straight road is described by \( s(t) = 4t^2 + 2t \) meters, where \( t \) is time in seconds. Find the velocity at \( t = 3 \) seconds.

Given: \( s(t) = 4t^2 + 2t \), \( t=3 \, s \)

Velocity \( v(t) \) is the derivative of position:

$$ v(t) = s'(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h} $$

Calculate derivative directly:

$$ s'(t) = \lim_{h \to 0} \frac{4(t+h)^2 + 2(t+h) - [4t^2 + 2t]}{h} $$

Expand numerator:

$$ = \lim_{h \to 0} \frac{4(t^2 + 2t h + h^2) + 2t + 2h - 4t^2 - 2t}{h} $$

Simplify numerator:

$$ = \lim_{h \to 0} \frac{4t^2 + 8t h + 4 h^2 + 2t + 2h - 4t^2 - 2t}{h} $$

$$ = \lim_{h \to 0} \frac{8t h + 4 h^2 + 2h}{h} = \lim_{h \to 0} (8t + 4 h + 2) $$

At \( h=0 \):

$$ v(t) = 8t + 2 $$

At \( t=3 \):

$$ v(3) = 8 \times 3 + 2 = 24 + 2 = 26 \text{ m/s} $$

Answer: Velocity at 3 s is 26 m/s

Example 2:
The voltage output \( V(t) \) from a sensor is \( V(t) = 12t, t^3 \) volts. Find the rate of change of voltage at \( t=2 \) seconds.

Given: \( V(t) = 12t, t^3 \), \( t=2 \, s \)

Calculate derivative:

$$ V'(t) = \lim_{h \to 0} \frac{12(t+h) - (t+h)^3 - [12t, t^3]}{h} $$

Expand \( (t+h)^3 = t^3 + 3 t^2 h + 3 t h^2 + h^3 \):

$$ = \lim_{h \to 0} \frac{12t + 12h, t^3 - 3 t^2 h - 3 t h^2 - h^3 - 12t + t^3}{h} $$

Simplify numerator:

$$ = \lim_{h \to 0} \frac{12 h - 3 t^2 h - 3 t h^2 - h^3}{h} $$

Divide numerator and denominator by \( h \):

$$ = \lim_{h \to 0} (12 - 3 t^2 - 3 t h, h^2) $$

At \( h=0 \):

$$ V'(t) = 12 - 3 t^2 $$

At \( t=2 \):

$$ V'(2) = 12 - 3 \times 2^2 = 12 - 3 \times 4 = 12 - 12 = 0 \text{ V/s} $$

Answer: Rate of change of voltage at 2 s is 0 V/s

Example 3:
A vehicle's acceleration \( a(t) \) is the derivative of velocity \( v(t) = 5t^2 + 3t \) m/s. Find the acceleration at \( t=4 \) seconds.

Given: \( v(t) = 5t^2 + 3t \), \( t=4 \, s \)

Calculate acceleration \( a(t) = v'(t) \):

$$ a(t) = \lim_{h \to 0} \frac{v(t+h) - v(t)}{h} $$

Expand numerator:

$$ = \lim_{h \to 0} \frac{5(t+h)^2 + 3(t+h) - [5t^2 + 3t]}{h} $$

Expand \( (t+h)^2 = t^2 + 2 t h + h^2 \):

$$ = \lim_{h \to 0} \frac{5(t^2 + 2 t h + h^2) + 3 t + 3 h - 5 t^2 - 3 t}{h} $$

Simplify numerator:

$$ = \lim_{h \to 0} \frac{5 t^2 + 10 t h + 5 h^2 + 3 t + 3 h - 5 t^2 - 3 t}{h} $$

$$ = \lim_{h \to 0} \frac{10 t h + 5 h^2 + 3 h}{h} = \lim_{h \to 0} (10 t + 5 h + 3) $$

At \( h=0 \):

$$ a(t) = 10 t + 3 $$

At \( t=4 \):

$$ a(4) = 10 \times 4 + 3 = 40 + 3 = 43 \text{ m/s}^2 $$

Answer: Acceleration at 4 s is 43 m/s²

6.1.3 Derivatives of polynomial functions

Polynomial functions are common in automotive system modelling, such as engine torque curves or sensor calibration curves. Their derivatives follow power rules simplifying the process of differentiation.

The derivative of \( f(x) = a x^n \) is:

$$ f'(x) = a n x^{n-1} $$

Worked Examples

Example 1:
Find the derivative of \( f(x) = 7x^4 \) at \( x = 1 \).

Given: \( f(x) = 7x^4 \), \( x=1 \)

$$ f'(x) = 7 \times 4 x^{3} = 28 x^{3} $$

At \( x=1 \):

$$ f'(1) = 28 \times 1^3 = 28 $$

Answer: \( f'(1) = 28 \)

Example 2:
Calculate the derivative of \( f(x) = 2x^5 - 4x^3 + x \) at \( x=2 \).

Given: \( f(x) = 2x^5 - 4x^3 + x \), \( x=2 \)

$$ f'(x) = 2 \times 5 x^{4} - 4 \times 3 x^{2} + 1 = 10 x^{4} - 12 x^{2} + 1 $$

At \( x=2 \):

$$ f'(2) = 10 \times 2^{4} - 12 \times 2^{2} + 1 = 10 \times 16 - 12 \times 4 + 1 = 160 - 48 + 1 = 113 $$

Answer: \( f'(2) = 113 \)

Example 3:
Differentiate \( f(x) = 3x^3 + 5x^2 - 2x + 7 \) and find \( f'(3) \).

Given: \( f(x) = 3x^3 + 5x^2 - 2x + 7 \), \( x=3 \)

$$ f'(x) = 3 \times 3 x^{2} + 5 \times 2 x - 2 = 9 x^{2} + 10 x - 2 $$

At \( x=3 \):

$$ f'(3) = 9 \times 3^{2} + 10 \times 3 - 2 = 9 \times 9 + 30 - 2 = 81 + 30 - 2 = 109 $$

Answer: \( f'(3) = 109 \)

6.1.4 Derivatives of trigonometric functions relevant to automotive systems

Trigonometric functions such as sine and cosine model oscillations in sensors and rotating systems. Their derivatives help analyze signal changes and mechanical vibrations.

The derivatives are:

• \( \frac{d}{dx} \sin x = \cos x \)
• \( \frac{d}{dx} \cos x = -\sin x \)

Worked Examples

Example 1:
Find the derivative of \( f(x) = \sin x \) at \( x = \frac{\pi}{2} \).

Given: \( f(x) = \sin x \), \( x = \frac{\pi}{2} \)

$$ f'(x) = \cos x $$

At \( x = \frac{\pi}{2} \):

$$ f'(\frac{\pi}{2}) = \cos \frac{\pi}{2} = 0 $$

Answer: \( f'(\frac{\pi}{2}) = 0 \)

Example 2:
Calculate the derivative of \( f(x) = 4 \cos x \) at \( x = 0 \).

Given: \( f(x) = 4 \cos x \), \( x=0 \)

$$ f'(x) = 4 \times (-\sin x) = -4 \sin x $$

At \( x=0 \):

$$ f'(0) = -4 \times \sin 0 = 0 $$

Answer: \( f'(0) = 0 \)

Example 3:
If \( f(x) = 3 \sin x + 2 \cos x \), find \( f'(\frac{\pi}{3}) \).

Given: \( f(x) = 3 \sin x + 2 \cos x \), \( x = \frac{\pi}{3} \)

$$ f'(x) = 3 \cos x - 2 \sin x $$

At \( x = \frac{\pi}{3} \), \( \cos \frac{\pi}{3} = \frac{1}{2} \), \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \):

$$ f'(\frac{\pi}{3}) = 3 \times \frac{1}{2} - 2 \times \frac{\sqrt{3}}{2} = \frac{3}{2} - \sqrt{3} $$

Numerical approximation:

$$ \frac{3}{2} = 1.5, \quad \sqrt{3} \approx 1.732 $$

$$ f'(\frac{\pi}{3}) \approx 1.5 - 1.732 = -0.232 $$

Answer: \( f'(\frac{\pi}{3}) \approx -0.232 \)

Chapter Summary

This chapter introduced the concept of derivatives by defining them and practicing their calculation through exercises. It then explained differentiation from first principles to establish a fundamental understanding. Common derivatives were presented in tabular form to aid quick reference. Various rules of differentiation were covered, including sum and difference, product, quotient, chain, parametric, and implicit differentiation, enabling the handling of diverse functions. The chapter explored the concepts of rate of change and small change, crucial for practical applications. It also addressed stationary points of functions of two variables to analyze maxima, minima, and saddle points. Integration was defined and differentiated between indefinite and definite integrals. Several methods of integration were taught, including substitution, reduction formula, integration by parts, and partial fractions, along with integrals involving hyperbolic and inverse functions. Finally, the chapter demonstrated the application of integration in solving engineering problems.