JEE Application of Derivatives

This page will teach you how to master JEE Applications of Derivatives . We highlight the main concepts, provide a list of examples with solutions, and include problems for you to try. Once you are confident, you can take the quiz to establish your mastery.

As per JEE syllabus, the mains concepts are Derivative as Rate Measurement, Approximation, Rolle's Theorem and Lagrange's Mean Value Theorem.

Derivative as Rate Measurement

• $y=f(x)$ then $\frac{dy}{dt}=f'(x) \frac{dx}{dt}$
• Applications

Approximation

• Differential of a function : $df=f'(x)dx$
• Linearisation of a function : $f(x)=f(x_0)+f'(x_0)(x-x_0)$
• Geometrical significance of approximation
• Errors

Rolle's Theorem

• Rolle's theorem and its applications
• Geometrical significance

Lagrange's Theorem

• Lagrange's theorem and its applications
• Geometrical significance

A ladder $\SI[per-mode=symbol]{10}{\meter}$ long rests against a vertical wall. If the bottom of the ladder slides away from the wall at the rate of $\SI[per-mode=symbol]{1}{\meter\per\second}$. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is $\SI[per-mode=symbol]{5}{\meter\per\second}$ away from the wall ?

$$\begin{array}{lll} A) \, \SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second} & \quad \quad \quad \quad \quad \quad \quad & B) \, -\SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second} \\ C) \, \SI[per-mode=symbol]{\sqrt{3}}{\meter\per\second} & & D) \, -\SI[per-mode=symbol]{\sqrt{3}}{\meter\per\second} \end{array}$$

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Concepts tested: Rate measurement

Answer: A) $\SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second}$

Solution: We first draw the diagram for the given situation which will look like:

Now, relationship between $x$ and $y$ can be established by Pythagoras Theorem: $x^2+y^2=100$. Differentiating both sides with respect to $t$ (time), we get : $2x \frac{dx}{dt}+2y\frac{dy}{dt}=0 \Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$. When $x=5,$ the Pythagoras Theorem gives $y=5\sqrt{3}$. Now putting the values of $x,y$ and $\frac{dx}{dt}$, we get: $\frac{dy}{dt}=-\frac{5}{5\sqrt{3}}(1)=\SI[per-mode=symbol]{-\frac{1}{\sqrt{3}}}{\meter\per\second}.$

Point to be noted here is that $\frac{dy}{dt}$ is negative means top of the ladder is sliding down the wall at the rate of $\SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second}$. So, the correct answer is $\SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second}$.

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Common mistakes:

• If you considered the negative sign, then it will represent the rate with which top of the ladder will be sliding upwards the wall, while the question asks sliding rate down the wall.

• If you tried to insert the values before differentiating, it will lead to zero rate, which is off course, wrong.

Let $P(x)$ be a polynomial with real coefficients. Let $a$ and $b$ $(a<b)$, be two consecutive zeroes of $P(x)$. If $a<c<b$, then find the value of $P'(c)+10P(c)$.

$$\begin{array}{lll} A) \, 0 & \quad \quad \quad \quad \quad \quad \quad & B) \, -1 \\ C) \, 10 & & D) \, -10 \end{array}$$

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Concepts tested: Applications of Rolle's theorem

Answer: A) $0$

Solution: Let us assume $f(x)=e^{10x}P(x)$. Now $f(a)=f(b)=0$ (as $P(a)=P(b)=0$ ) Also as $P(x)$ is a polynomial, $f(x)$ is continuous in $[a,b]$ and differentiable in $(a,b)$. Hence Rolle's theorem can be applied to $f(x)$. Hence there exists $c \in (a,b)$ such that $f'(c)=0$.
Now, $f'(x)=e^{10x}[P'(x)+10P(x)]$
$\Rightarrow e^{10c}[P'(c)+10P(c)]=0$
$\Rightarrow P'(c)+10P(c)=0 \text{ for some } c \in (a,b)$

The following is a problem from JEE-Mains 2014:

A man runs along a straight path at the speed of $\SI[per-mode=symbol]{4}{\meter\per\second}$. A searchlight is located on the ground $\SI[per-mode=symbol]{20}{\meter}$ from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is $\SI[per-mode=symbol]{15}{\meter}$ from the point on the path closest to the searchlight ?

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Concepts tested: Rate measurement

Answer: $\SI[per-mode=symbol]{0.128}{\radian\per\second}$

Solution:
Step 1: Drawing figure and variable selection
We draw the figure and let $x$ be the distance from the man to the point on the path closest to the searchlight and $\theta$ be the distance from the man to the point on the path closest to the searchlight and perpendicular to the path. The figure will look like: Step 2: Establishing relation between $x$ and $\theta$ and finding the required rate
The equation that relates $x$ and $\theta$ can be written from the figure: $\frac{x}{20}=\tan\theta \Rightarrow x=20 \tan\theta$. Now differentiating both sides with respect to $t$ (time), we get : $\frac{dx}{dt}=\SI[per-mode=symbol]{20}{\second\squared}\theta \frac{d\theta}{dt} \Rightarrow \frac{d\theta}{dt}=\frac{\cos^2\theta}{20} \frac{dx}{dt}$

Step 3: Putting the values
We're given with $\frac{dx}{dt}=\SI[per-mode=symbol]{4}{\meter\per\second}$ and when $x=\SI[per-mode=symbol]{15}{\meter}$, the length of the beam becomes $\SI[per-mode=symbol]{25}{\meter}$ using Pythagoras theorem. So, $\cos\theta=\frac{4}{5}$.

$\Rightarrow \frac{d\theta}{dt}=\frac{1}{20} (\frac{4}{5})^2 (4)=\SI[per-mode=symbol]{0.128}{\radian\per\second}$

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Common mistakes:

• If you didn't the value of $\frac{dx}{dt}$, it will give you linear rate, not the rate at which searchlight is rotating.