Rate of Change of Bodies

The rate of change of one variable with respect to another is a fundamental concept in calculus. If two variables \(x\) and \(y\) depend on a third variable \(t\), such that \(x = f(t)\) and \(y = g(t)\), then by the Chain Rule, the rate of change of \(y\) with respect to \(x\) is given by:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \quad \text{provided } \frac{dx}{dt} \neq 0 \]

This means the rate of change of \(y\) with respect to \(x\) can be computed using their rates of change with respect to \(t\).

When \(y\) is a function of \(x\), say \(y = f(x)\), then \(\frac{dy}{dx}\) represents the instantaneous rate of change of \(y\) with respect to \(x\) at a particular value \(x = \alpha\).

Worked Illustration

Example 1: Find the rate of change of the area of a circle with respect to its radius \(r\) when \(r = 5\) cm.

Solution:

The area of a circle is \( A = \pi r^2 \).

Differentiate \(A\) with respect to \(r\):

\[ \frac{dA}{dr} = \frac{d}{dr} (\pi r^2) = 2 \pi r \]

At \(r = 5\) cm,

\[ \frac{dA}{dr} = 2 \pi \times 5 = 10 \pi \]

Thus, the area is increasing at a rate of \(10 \pi\) cm² per unit increase in radius.

Practice Set

• Level 1 – Easy: Find the rate of change of the circumference of a circle with respect to its radius.
• Level 2 – Moderate: If the radius of a sphere increases at a rate of 3 cm/s, find the rate of change of its volume when the radius is 4 cm.
• Level 3 – Challenging: A ladder leaning against a wall slides down at a rate of 2 m/s. Find the rate of change of the distance of the foot of the ladder from the wall when the top of the ladder is 5 m above the ground.

Answer Key

• Level 1: \( \frac{dC}{dr} = 2 \pi \)
• Level 2: Volume \( V = \frac{4}{3} \pi r^3 \), \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} = 4 \pi \times 16 \times 3 = 192 \pi \) cm³/s
• Level 3: Using Pythagoras and related rates, \( \frac{dx}{dt} = \frac{5}{x} \times 2 \) m/s (detailed steps required)

Quick Reference

Concept	Formula
Rate of change of \(y\) w.r.t \(x\)	\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)
Area of circle	\( A = \pi r^2 \)
Rate of change of area	\( \frac{dA}{dr} = 2 \pi r \)

Glossary

• Rate of Change: How one quantity changes in relation to another.
• Chain Rule: A rule to differentiate composite functions.
• Instantaneous Rate: The rate of change at a specific instant.

Increasing and Decreasing Functions

A function \(f(x)\) is said to be increasing on an interval \([a,b]\) if for any \(\alpha, \beta \in [a,b]\) with \(\alpha > \beta\), we have \(f(\alpha) > f(\beta)\). If \(f'(x) \geq 0\) for all \(x \in (a,b)\) and \(f\) is continuous at \(a\) and \(b\), then \(f\) is increasing on \([a,b]\).

Similarly, \(f(x)\) is decreasing on \([a,b]\) if for any \(\alpha, \beta \in [a,b]\) with \(\alpha > \beta\), \(f(\alpha) < f(\beta)\). If \(f'(x) \leq 0\) for all \(x \in (a,b)\) and \(f\) is continuous at \(a\) and \(b\), then \(f\) is decreasing on \([a,b]\).

If \(f'(x) = 0\) for all \(x \in (a,b)\), then \(f\) is constant on \([a,b]\).

A function is called monotonic on an interval if it is either entirely increasing or entirely decreasing on that interval.

Algorithm to find intervals of increase/decrease

1. Find \(f'(x)\).
2. Solve \(f'(x) = 0\) to find critical points.
3. Determine the sign of \(f'(x)\) in intervals defined by critical points.
4. Where \(f'(x) > 0\), \(f\) is increasing; where \(f'(x) < 0\), \(f\) is decreasing.

Worked Example

Example 2: Show that \(f(x) = x^3 - 3x^2 + 4x\) is increasing on \(\mathbb{R}\).

Solution:

Compute the derivative:

\[ f'(x) = 3x^2 - 6x + 4 = 3(x^2 - 2x + \frac{4}{3}) \]

Complete the square:

\[ x^2 - 2x + 1 + \frac{1}{3} = (x-1)^2 + \frac{1}{3} > 0 \]

Therefore, \(f'(x) = 3\left((x-1)^2 + \frac{1}{3}\right) > 0\) for all \(x\).

Hence, \(f\) is strictly increasing on \(\mathbb{R}\).

Practice Set

• Level 1 – Easy: Determine intervals where \(f(x) = 2x + 3\) is increasing or decreasing.
• Level 2 – Moderate: Find intervals of increase and decrease for \(f(x) = x^3 - 6x^2 + 9x + 1\).
• Level 3 – Challenging: For \(f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1\), find intervals where \(f\) is increasing and decreasing.

Answer Key

• Level 1: \(f'(x) = 2 > 0\), so \(f\) is increasing everywhere.
• Level 2: \(f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)\). Increasing on \((-,1) \cup (3, )\), decreasing on \((1,3)\).
• Level 3: \(f'(x) = 4x^3 - 12x^2 + 12x - 4 = 4(x-1)^3\). Increasing on \((1, )\), decreasing on \((-,1)\).

Quick Reference

Condition	Function Behavior
\(f'(x) > 0\)	Increasing
\(f'(x) < 0\)	Decreasing
\(f'(x) = 0\)	Constant or critical point

Glossary

• Increasing Function: Function values rise as \(x\) increases.
• Decreasing Function: Function values fall as \(x\) increases.
• Critical Point: Point where \(f'(x) = 0\) or derivative does not exist.
• Monotonic Function: Function that is either entirely increasing or decreasing.

Maxima and Minima

Maxima and minima are points where a function attains its highest or lowest values locally or globally.

Let \(f(x)\) be defined on an interval \(I\).

• Maximum value: \(f(c)\) is a maximum if \(f(c) \geq f(x)\) for all \(x \in I\).
• Minimum value: \(f(c)\) is a minimum if \(f(c) \leq f(x)\) for all \(x \in I\).
• Extreme value: Either a maximum or minimum value.

Local maxima/minima: Points where \(f(c)\) is maximum/minimum in a neighborhood around \(c\).

Critical points: Points where \(f'(c) = 0\) or \(f'\) does not exist.

Worked Example

Example 3: Find the minimum and maximum values of \(f(x) = x^2\) on \(\mathbb{R}\).

Solution:

\(f'(x) = 2x\). Setting \(f'(x) = 0\) gives \(x = 0\) as critical point.

Since \(f(x) = x^2 \geq 0\) for all \(x\), minimum value is 0 at \(x=0\).

There is no maximum value as \(f(x) \to \infty\) as \(x \to \pm \infty\).

First Derivative Test

1. Find \(f'(x)\).
2. Find critical points by solving \(f'(x) = 0\).
3. Check sign changes of \(f'(x)\) around critical points:
• If \(f'\) changes from positive to negative, local maximum.
• If \(f'\) changes from negative to positive, local minimum.
• If no sign change, point of inflection.

Second Derivative Test

For twice differentiable \(f\), at critical point \(c\):

• If \(f'(c) = 0\) and \(f''(c) < 0\), local maximum at \(c\).
• If \(f'(c) = 0\) and \(f''(c) > 0\), local minimum at \(c\).
• If \(f''(c) = 0\), test is inconclusive; use first derivative test.

Absolute Maxima and Minima

If \(f\) is continuous on a closed interval \([a,b]\), then \(f\) attains absolute maximum and minimum values at critical points or endpoints.

Algorithm to find absolute extrema

1. Find critical points in \([a,b]\).
2. Evaluate \(f\) at critical points and endpoints.
3. Compare values to identify absolute maximum and minimum.

Worked Example

Example 4: Find local maxima and minima of \(f(x) = 3x^4 + 4x^3 - 12x^2 + 12\).

Solution:

Compute first derivative:

\[ f'(x) = 12x^3 + 12x^2 - 24x = 12x(x-1)(x+2) \]

Critical points at \(x = 0, 1, -2\).

Compute second derivative:

\[ f''(x) = 36x^2 + 24x - 24 = 12(3x^2 + 2x - 2) \]

Evaluate at critical points:

• \(f''(0) = -24 < 0\) → local maximum at \(x=0\), \(f(0) = 12\).
• \(f''(1) = 36 > 0\) → local minimum at \(x=1\), \(f(1) = 7\).
• \(f''(-2) = 72 > 0\) → local minimum at \(x=-2\), \(f(-2) = -20\).

Practice Set

• Level 1 – Easy: Find local maxima and minima of \(f(x) = x^2 - 4x + 3\).
• Level 2 – Moderate: Find local extrema of \(f(x) = x^3 - 3x + 1\).
• Level 3 – Challenging: Find absolute maxima and minima of \(f(x) = x^3 - 6x^2 + 9x + 1\) on \([0,4]\).

Answer Key

• Level 1: \(f'(x) = 2x - 4 = 0 \Rightarrow x=2\). \(f''(2) = 2 > 0\) local minimum at \(x=2\).
• Level 2: \(f'(x) = 3x^2 - 3 = 0 \Rightarrow x= \pm 1\). \(f''(1) = 6 > 0\) local minimum, \(f''(-1) = -6 < 0\) local maximum.
• Level 3: Critical points at \(x=1,3\). Evaluate \(f\) at 0,1,3,4. Absolute max at \(x=4\), absolute min at \(x=1\).

Quick Reference

Test	Condition	Result
First Derivative	\(f'\) changes + to -	Local maximum
First Derivative	\(f'\) changes - to +	Local minimum
Second Derivative	\(f'(c)=0, f''(c) < 0\)	Local maximum
Second Derivative	\(f'(c)=0, f''(c) > 0\)	Local minimum

Glossary

• Local Maximum: Highest value in a neighborhood.
• Local Minimum: Lowest value in a neighborhood.
• Absolute Maximum: Highest value on entire domain.
• Absolute Minimum: Lowest value on entire domain.
• Critical Point: Point where derivative is zero or undefined.
• Point of Inflexion: Point where concavity changes and derivative does not change sign.