Chapter 6: Applications of Derivatives

This chapter deals with applications of derivatives in various disciplines. We will see how derivatives can be used to determine rate of change of quantities, to find the equations of tangent and normal to a curve at a point, to find turning points on the graph of a function.

Rate of Change of Quantities

If a quantity $y$ varies with another quantity $x$, satisfying some rule $y = f(x)$, then $\frac{dy}{dx}$ (or $f’(x)$) represents the rate of change of $y$ with respect to $x$.

Increasing and Decreasing Functions

Definition: Let $I$ be an open interval contained in the domain of a real valued function $f$. Then $f$ is said to be:

• Strictly increasing on $I$ if $x_1 < x_2$ in $I \implies f(x_1) < f(x_2)$ for all $x_1, x_2 \in I$.
• Strictly decreasing on $I$ if $x_1 < x_2$ in $I \implies f(x_1) > f(x_2)$ for all $x_1, x_2 \in I$.

Theorem: Let $f$ be continuous on $[a, b]$ and differentiable on the open interval $(a, b)$. Then

• $f$ is strictly increasing in $[a, b]$ if $f’(x) > 0$ for each $x \in (a, b)$.
• $f$ is strictly decreasing in $[a, b]$ if $f’(x) < 0$ for each $x \in (a, b)$.
• $f$ is a constant function in $[a, b]$ if $f’(x) = 0$ for each $x \in (a, b)$.

Maxima and Minima

First Derivative Test: Let $f$ be a function defined on an open interval $I$. Let $f$ be continuous at a critical point $c$ in $I$. Then:

• If $f’(x)$ changes sign from positive to negative as $x$ increases through $c$, then $c$ is a point of local maxima.
• If $f’(x)$ changes sign from negative to positive as $x$ increases through $c$, then $c$ is a point of local minima.

Second Derivative Test: Let $f$ be a function defined on an interval $I$ and $c \in I$. Let $f$ be twice differentiable at $c$. Then:

• $x = c$ is a point of local maxima if $f’(c) = 0$ and $f’’(c) < 0$. The value $f(c)$ is local maximum value of $f$.
• $x = c$ is a point of local minima if $f’(c) = 0$ and $f’’(c) > 0$. The value $f(c)$ is local minimum value of $f$.
• The test fails if $f’(c) = 0$ and $f’’(c) = 0$.

Competency-Based Questions

1. [CBSE 2020] A circular disc of radius $3 , \text{cm}$ is being heated. Due to expansion, its radius increases at the rate of $0.05 , \text{cm/s}$. Find the rate at which its area is increasing when radius is $3.2 , \text{cm}$.

Solution:

Let $r$ be the radius of the circular disc and $A$ be its area at any time $t$.

Given: $\frac{dr}{dt} = 0.05 , \text{cm/s}$

We know, $A = \pi r^2$

Differentiating w.r.t $t$:

$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$

When $r = 3.2 , \text{cm}$,

$\frac{dA}{dt} = 2 \pi (3.2)(0.05) = 2 \pi (0.16) = 0.32 \pi , \text{cm}^2/\text{s}$.

2. [Sample Paper 2022] Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing or decreasing.

Solution:

$f(x) = 2x^3 - 3x^2 - 36x + 7$

$f’(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2)$

To find intervals, solve $f’(x) = 0 \implies x = 3, -2$.

These points divide the real line into three disjoint intervals: $(-\infty, -2), (-2, 3), \text{ and } (3, \infty)$.

• In interval $(-\infty, -2)$, say $x = -3$, $f’(-3) = 6(-6)(-1) > 0$. Strictly increasing.

• In interval $(-2, 3)$, say $x = 0$, $f’(0) = 6(-3)(2) < 0$. Strictly decreasing.

• In interval $(3, \infty)$, say $x = 4$, $f’(4) = 6(1)(6) > 0$. Strictly increasing.

Thus, $f$ is strictly increasing in $(-\infty, -2) \cup (3, \infty)$ and strictly decreasing in $(-2, 3)$.

3. [CBSE 2018] Find the maximum and minimum values of the function $f(x) = x + \sin 2x$ on $[0, 2\pi]$.

Solution:

$f(x) = x + \sin 2x$

$f’(x) = 1 + 2\cos 2x$

For extreme values, $f’(x) = 0 \implies \cos 2x = -1/2 = \cos(\pi - \pi/3) = \cos(2\pi/3)$

$2x = 2n\pi \pm 2\pi/3$. In the interval $[0, 2\pi]$, for $x$:

$2x = 2\pi/3, 4\pi/3, 8\pi/3, 10\pi/3$

$x = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$.

Now calculate $f(x)$ at these critical points and also at the endpoints $x=0, 2\pi$:

$f(0) = 0$

$f(\pi/3) = \pi/3 + \sin(2\pi/3) = \pi/3 + \sqrt{3}/2 \approx 1.91$

$f(2\pi/3) = 2\pi/3 + \sin(4\pi/3) = 2\pi/3 - \sqrt{3}/2 \approx 1.22$

$f(4\pi/3) = 4\pi/3 + \sin(8\pi/3) = 4\pi/3 + \sqrt{3}/2 \approx 5.05$

$f(5\pi/3) = 5\pi/3 + \sin(10\pi/3) = 5\pi/3 - \sqrt{3}/2 \approx 4.37$

$f(2\pi) = 2\pi \approx 6.28$

Maximum value = $2\pi$ at $x = 2\pi$.

Minimum value = $0$ at $x = 0$.