RD Sharma Class 12 Exercise 12.2 Derivative as a rate measure Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 12.2 Derivative as a rate measure Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 12:40 PM IST

The RD Sharma Class 12 Solution of Derivatives As A Rate Measure Exercise 12.2 is the most recommended book by the CBSE board institutions to their students. Not only do the teachers suggest this book, but many previous batch students also recommend it to their juniors.

## Derivative as a Rate Measurer Excercise: 12.2

Derivative As a Rate Measure exercise 12.2 question 1

Answer: The area is increasing at the rate of 64 cm2/min
Hint: Here, we use the formula of the square sheet in (cm) is, $A=x^{2} \mathrm{~cm}^{2}$
Given: The side of square sheet is increasing at the rate of 4 cm per minute, rate of area increasing when the side is 8 cm long. So Here, our x =8.
Solution:
Suppose the side of given square sheet be x cm at any instant time.
Now according to the question,
The rate of side of sheet increasing is $\frac{d x}{d t}=4\: cm/min$ cm/min
Now the area of square sheet at any time t will be $A=x^{2} \mathrm{~cm}^{2}$... (i)
On differentiating (i) with respect to time both side,
$\frac{d A}{d t}=\frac{d x^{2}}{d t}=2 x \frac{d x}{d t}=2 x \times 4=8 x$ [ $\frac{d x}{d t}=4$ From equation number (i)]
Therefore, $\frac{d A}{d t}=8 x$
Let’s Put value of x = 8
$\Rightarrow \frac{d A}{d t}=8 \times 8=64 \quad(x=8\{\text { given }\})$
$d A / d t=64 \mathrm{~cm}^{2} / \mathrm{min}$
The area is increasing at the rate of $64\; cm^{2}/min$ when the side is 8 cm long.

Derivative As a Rate Measure exercise 12.2 question 2

Answer: The volume of the cube increasing at the rate of 900cm3/sec
Hint: Here, we use the formula of cube’s volume $V=x^{3} \mathrm{~cm}^{3}$
Given: An edge of a variable cube is increasing at the rate of 3 cm per second, the volume of the cube increasing when the edge is 10 cm long, So Here x =10
Solution: Suppose the edge of the given cube be x cm at any instant time.
Now according to the question,
The rate of edge of the cube increasing is $d x / d t=3 \mathrm{~cm} / \mathrm{sec}$... … (i)
Now, the formula is $V=x^{3} \mathrm{~cm}^{3}$
By applying derivative,
$\frac{d V}{d t}=\frac{d x^{3}}{d t}=3 x^{2} \frac{d x}{d t}=3 x^{2} \times 3=9 x^{2}$ $\left(\frac{d x}{d t}=3\right) \text { from eq}^{n}(i)$
Lets put the value of x in above equation
\begin{aligned} &\frac{d V}{d t}=9 x^{2} \mathrm{~cm}^{3} / \mathrm{sec} \\\\ &\frac{d V}{d t}=9 \times 10 \times 10 \mathrm{~cm}^{3} / \mathrm{sec} \quad(x=10) \text { Given } \\\\ &\frac{d V}{d t}=900 \mathrm{~cm}^{3} / \mathrm{sec} \end{aligned}
Thus the volume of cube increasing at the rate of 900 cm3/sec

Derivative As a Rate Measure exercise 12.2 question 3

Answer: The rate of increasing of the perimeter of the square will be 0.8 cm/sec
Hint: Here, the perimeter of the square at any time t will be p = 4x cm
Given: The side of square is increasing at the rate of 0.2 cm/sec.
Solution: Suppose the edge of the given cube be x cm at any instant time.
Now according to the question,
The rate of side of the square is increasing is $\frac{d x}{d t}=0.2\: cm/sec$ … (i)
By applying derivative in perimeter,
$\frac{d p}{d t}=\frac{d(4 x)}{d t}=\frac{d x}{d t} \times 4=4 \times 0.2=0.8\: cm/sec$

$\frac{d x}{d t}=0.2\: cm/sec$ (from eqn 1) and p = 4x (given)
So, Thus the rate of increasing of the perimeter of the square will be 0.8 cm/sec.

Derivative As a Rate Measure exercise 12.2 question 4

Answer: The rate of increasing of the circle’s circumference will be $1.4\pi$ cm/sec
Hint: The circumference of circle at any time t will be $C=2\pi r$ cm
Given: The radius of a circle is increasing at the rate of 0.7 cm/sec.
Solution: Suppose the radius of the given circle be r cm at any instant time.
Now according to the question,
The rate of radius of a circle is increasing is $\frac{dr}{dt}=0.7$ cm/sec ...… (i)
Now the circumference of the circle at any time t will be $C=2\pi r$ cm
By applying derivative on circumference,
$\frac{d C}{d t}=\frac{d(2 \pi r)}{d t}=2 \pi \frac{d r}{d t}=2 \pi \times 0.7=1.4 \pi$ cm/sec
$\left\{\frac{d r}{d t}=0.7\left(\text { from } e q^{n}(i)\right) \text { and } C=2 \pi r(\text { Given })\right\}$
Thus the rate of increasing of circle circumference will be $1.4\pi$ cm/sec.

Derivative As a Rate Measure exercise 12.2 question 5

Answer: The rate of increase of its surface area, when the radius is 7 cm is $11.2\pi$ cm2/sec
Hint: The surface area of spherical soap bubble at any time t will be $S=4\pi r^{2}$ cm2
Given: The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec.
Solution: Suppose the radius of the given spherical soap bubble be $r$ cm at any instant time.
Now according to the question,
$\frac{dr}{dt}=0.2$ cm/sec ….... (i)
By applying derivative on surface area
$\frac{d s}{d t}=\frac{d\left(4 \pi r^{2}\right)}{d t}$
\begin{aligned} &\frac{d s}{d t}=\frac{4 \pi\left(d r^{2}\right)}{d t} \\\\ &\frac{d s}{d t}=4 \pi \times 2 r \times \frac{d r}{d t} \end{aligned}
$\frac{d s}{d t}=4 \pi \times 2 r \times 0.2$ $\left(\frac{d r}{d t}=0.2\right)\left(\text { from } e q^{n}(i)\right)$
$\frac{d S}{d t}=1.6 \pi r$
lets put value of r in above formula $r=7$(given)
$\frac{d S}{d t}=1.6 \pi \times 7=11.2 \pi$ cm/sec
Thus the rate of increase of its surface area, when the radius is 7 cm is $11.2\pi$ cm2/sec

Derivative As a Rate Measure exercise 12.2 question 6

Answer: $\frac{d r}{d t}=\frac{1}{\pi} \mathrm{cm} / \mathrm{sec}$
Hint: We know that the volume of the spherical balloon is $V=\frac{4}{3}\pi r^{3}$ .
Given: The spherical balloon inflated by pumping in 900 cubic cm of gas per second.
Solution: Suppose the radius of the given spherical balloon be r cm and let V be the volume of the spherical balloon at any instant time.
The balloon is inflated by pumping 900 cubic cm of gas per second hence the rate of volume of spherical balloon is increased by,
$\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=900 \mathrm{~cm}^{3} / \mathrm{sec}$
$\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{d t}$ $\therefore\left(V=\frac{4}{3} \pi r^{3}\right)$
lets put value of $\frac{d V}{d t}=900$(given)
Let’s differentiate $\left(\frac{4}{3} \pi r^{3}\right)$ w.r.t time
$\begin{gathered} 900=\frac{4 \times 3}{3} \times \pi r^{2} \frac{d r}{d t} \\\\ 900=4 \pi r^{2} \frac{d r}{d t} \end{gathered}$
So, lets put value of $r= 15$ (given)
\begin{aligned} &\Rightarrow 900=4 \pi \times 15 \times 15 \times \frac{d r}{d t} \\\\ &\Rightarrow 900=900 \pi \times \frac{d r}{d t} \\\\ &\Rightarrow \frac{900}{900 \pi}=\frac{d r}{d t} \end{aligned}
\begin{aligned} &\Rightarrow \frac{1}{\pi}=\frac{d r}{d t} \\\\ &\text { So, } \frac{d r}{d t}=\frac{1}{\pi} \end{aligned}

Derivative As a Rate Measure exercise 12.2 question 7

Answer: $\frac{d V}{d t}=2 \pi \mathrm{cm}^{3} / \mathrm{sec}$
Hint: As we know that the volume of bubble is $V=\frac{4}{3} \pi r^{3}$ .
Given: The radius of air bubble is increasing at the rate of 0.5 cm/sec
Solution: Suppose the radius of the given air bubble be $r$ cm and let $V$ be the volume of the air bubble at any instant time.
The rate of increasing in the radius of the air bubble is $\frac{\mathrm{dr}}{\mathrm{dt}}=0.5 \mathrm{~cm} / \mathrm{sec}$

By applying derivative on volume
$\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{d t}$ $\therefore\left(v=\frac{4}{3} \pi r^{3}\right)$

Let’s differentiate $d\left(\frac{4}{3} \pi r^{3}\right)$
\begin{aligned} &\frac{d V}{d t}=\frac{4 \times 3}{3} \times \pi r^{2} \frac{d r}{d t} \\\\ &\frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t} \end{aligned}

Let’s put value of $\frac{d r}{d t}=0.5$ (given)
$\frac{d V}{d t}=4 \pi r^{2} \times 0.5$

Let’s put value of $r = 1$
\begin{aligned} &\frac{d V}{d t}=4 \pi \times 1^{2} \times 0.5 \\\\ &\frac{d V}{d t}=2 \pi \mathrm{cm}^{3} / \mathrm{sec} \end{aligned}

The rate is increasing is $2 \pi \mathrm{cm}^{3} / \mathrm{sec}$ when the radius is 1 cm.

Derivative As a Rate Measure exercise 12.2 question 8

Answer: $\frac{d s}{d t}=\frac{5}{2}=2.5\; km/hr$
Hint: The rate at which the length of the man’s shadow increase will be $\frac{ds}{dt}$ .
Given: A man 2 meters high walk at a uniform speed of 5 km/hr away from lamp post 6 meters high.
Solution:

Suppose AB the lamp post and let MN be of the man of height 2 m.
Suppose AM = l meter and MS be the shadow of the man
Suppose length of the shadow MS=s
Given as man walk at the speed of 5 km/hr
$\frac{d I}{d t}=5\; km/hr$
So considering $\Delta ASB,$
$\tan \theta=\frac{A B}{A S}=\frac{6}{I+s}$................(i)
Then considering $\Delta MNS,$
$\tan \theta=\frac{M N}{M S}=\frac{2}{s}$...................(ii)
From equation (i) and (ii)
\begin{aligned} &\frac{6}{l+s}=\frac{2}{s} \\\\ &6 s=2(I+s) \\\\ &6 s=(2 I+2 s) \end{aligned}
\begin{aligned} &6 s-2 s=2 l \\\\ &4 s=2 l \\\\ &2 s=1 \end{aligned}..............(iii)
By applying derivative, With respect to time on both side
$\frac{d l}{d t}=\frac{d(2 s)}{d t}$ $I=2 s\left(\text { from } e q^{n} \text { iii }\right)$
\begin{aligned} &\frac{d l}{d t}=2 \frac{d s}{d t} \\\\ &5=2 \frac{d s}{d t} \end{aligned} $\left(\frac{d l}{d t}=5\right)(\text { given })$
\begin{aligned} &\frac{5}{2}=\frac{d s}{d t} \\\\ &2.5=\frac{d s}{d t} \end{aligned}
Thus, the rate at which the length of his shadow increases by 2.5 km/hr

Derivative As a Rate Measure exercise 12.2 question 9

Answer:$\frac{d A}{d t}=80 \: \pi \: \mathrm{cm}^{2} / \mathrm{sec}$
Hint: As we know that the area of circle is $\pi r^{2}$ .
Given: As a stone is dropped into a quiet lake and waves move in circle at a speed of 4 cm/sec
Solution: Suppose $r$ be the radius of the circle and $A$ be the area of the circle.
Whenever stone is dropped into the lake waves in circle at the speed of 4 cm/sec.
That is the radius of the circle increases at the rate of 4 cm/sec.
$\frac{dr}{dt}=4$ cm/sec ……………. (i)
Therefore,
$\frac{d A}{d t}=\frac{d\left(\pi r^{2}\right)}{d t}=\frac{\pi d\left(r^{2}\right)}{d t}=\pi \times 2 r \frac{d r}{d t}=2 \pi r \times 4$
So, when the radius of circle is 10 cm,
$\frac{d A}{d t}=2 \pi \times 10 \times 4=80 \pi\; cm^{2}/sec$
Thus the enclosed area is increasing at the rate of $80\pi$ cm2/sec.

Derivative As a Rate Measure exercise 12.2 question 10

Answer: $\frac{d s}{d t}=0.4 m / \sec$
Hint: The rate at which the length of the man’s shadow increase will be $\frac{ds}{dt}$ .
Given: A man 160 cm tall walks away from a source of light situated at the top of a pole 6m high, at the rate of 1.1 m/sec.
Solution:

Suppose $AB$ the lamp post and let $MN$ be the man of height 160 cm or 1.6 m.
Suppose AM = l meter and $MS$ be the shadow of the man.
Suppose length of the shadow $MS=s$
So, $\frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}=1.1 \mathrm{~m} / \mathrm{sec}$
Considering $\Delta MNS$
$\tan \theta=\frac{M N}{M S}=\frac{1.6}{s}$.....…(i)
So considering $\Delta ASB,$
$\tan \theta=\frac{A B}{A S}=\frac{6}{I+s}$ ....… (ii)
Therefore, from equation (i) and (ii)
\begin{aligned} &\frac{6}{I+s}=\frac{1.6}{s} \\\\ &6 s=1.6(I+s) \\\\ &6 s=1.6 I+1.6 s \end{aligned}
\begin{aligned} &6 s-1.6 s=1.6 \\\\ &4.4 s=1.6I \\\\ &I=\frac{4.4}{1.6} s \\\\ &I=2.75 s \end{aligned}.................(iii)

By applying derivative with respect to time on both side
$\frac{d I}{d t}=\frac{d(2.75 s)}{d t}$ $I=2.75 s\left(\text { from } e q^{n} \text { iii }\right)$
\begin{aligned} &\frac{d I}{d t}=2.75 \frac{d s}{d t} \\\\ &1.1=2.75 \frac{d s}{d t} \end{aligned} $\ldots\left(\frac{d I}{d t}=1.1\right)(\text { given })$
$\frac{1.1}{2.75}=\frac{d s}{d t}$
$\frac{d s}{d t}=0.4 \mathrm{~m} / \mathrm{hr}$
Thus, the rate at which the length of his shadow increases by 0.4 m/hr.

Derivative As a Rate Measure exercise 12.2 question 11

Answer: $\frac{d s}{d t}=0.5 \mathrm{~m} / \mathrm{sec}$
Hint: The rate at which the length of the man’s shadow increase will be $\frac{ds}{dt}$ .
Given: A man 180 cm tall walks at a rate of 2 m/sec away from a source of light that is 9m above the ground.
Solution: Suppose $AB$ the lamp post and let $MN$ be of height of man.
Suppose AM =I meter and $MS$ be the shadow of the man.
Given as man walk at the speed of 2 m/sec

So, $\frac{d I}{d t}=2$ m/sec …(i)

So considering $\Delta ASB,$\begin{aligned} &\tan \theta=\frac{A B}{A S} \\\\ &\tan \theta=\frac{9}{I+s} \ldots(\mathrm{ii}) \end{aligned}

Then$\Delta MNS,$,
$\tan \theta=\frac{M N}{M S}=\frac{1.8}{s}$

So by (i) and (ii)
$\frac{9}{I+s}=\frac{1.8}{s}$
$\begin{gathered} \Rightarrow 9 s=1.8 I+1.8 s \\\\ \Rightarrow 9 s-1.8 s=1.8 I \\\\ \Rightarrow 7.2 s=1.8 I \end{gathered}$
$\Rightarrow I=4 s$ .......(iii)

By applying derivative with respect to time on both side
\begin{aligned} &\frac{d I}{d t}=4 \frac{d s}{d t} \\\\ &2=4 \frac{d s}{d t} \end{aligned} $\ldots \ldots\left(\frac{d I}{d t}=2\right)(\text { given })$
\begin{aligned} &\frac{2}{4}=\frac{d s}{d t} \\\\ &\frac{d s}{d t}=0.5 \mathrm{~m} / \mathrm{sec} \end{aligned}
Thus, the rate at which the length of his shadow increases by 0.57 m/sec.

Derivative As a Rate Measure exercise 12.2 question 12

Hint: Here we use the Pythagoras theorem,
$A B^{2}+B C^{2}=A C^{2}$
Given: A ladder 13 m leans against at a wall and the foot of the ladder is pulled along the ground away from the wall,
at the rate of 1.5 m/sec.
Solution: Suppose AC be the position of the ladder initially, then h = AC=13 m, DE be the position of the ladder after being pulled at the rate of 1.5 m/sec.

Let the bottom of the ladder be at distance of x m from the wall and it’s top be at a height y m from the ground
Therefore, $\frac{d x}{d t}=1.5$ m/sec .......(i)

By Pythagoras theorem,
Considering, $\Delta A B C, A B^{2}+B C^{2}=A C^{2}$
$y^{2}+x^{2}=h^{2}$ .......(ii)
$\tan \theta=\frac{A B}{B C}=\frac{y}{x}$
$\Rightarrow x \tan \theta=y$ ......(iii)

Put the value of y in equation (ii) , we get
\begin{aligned} &x^{2}+(x \tan \theta)^{2}=(13)^{2} \\\\ &x^{2}+x^{2} \tan ^{2} \theta=169 \\\\ &x^{2}\left(1+\tan ^{2} \theta\right)=169 \\\\ &\sec ^{2} \theta=\frac{169}{x^{2}} \end{aligned}......(iv)

Differentiate the above equation w.r.t ‘t’
\begin{aligned} &\frac{d}{d t}\left(\sec ^{2} \theta\right)=\frac{d}{d t}\left(\frac{169}{x^{2}}\right) \\\\ &\Rightarrow 2 \sec \theta \sec \theta \tan \theta \frac{d \theta}{d t}=169 \frac{d}{d t}\left(x^{-2}\right) \end{aligned}
$\left[\because \frac{d}{d x}\left\{f(x)^{n}\right\}=n f(x)^{n-1} \frac{d}{d x} f(x)\right]$
$\Rightarrow 2 \sec ^{2} \theta \tan \theta \frac{d \theta}{d t}=169(-2) x^{-2-1} \frac{d x}{d t} \quad\left[\frac{d}{d x} x^{n}=n x^{n-1}\right]$
\begin{aligned} &\Rightarrow 2 \sec ^{2} \theta \tan \theta \frac{d \theta}{d t}=-338 x^{-3} \frac{d x}{d t} \\\\ &\Rightarrow \frac{d \theta}{d t}=\frac{-338}{2 x^{3} \sec ^{2} \theta \tan \theta} \frac{d x}{d t} \end{aligned}
\begin{aligned} &\Rightarrow \frac{d \theta}{d t}=\frac{-338}{2 x^{3} \sec ^{2} \theta \tan \theta}(1.5) \\\\ &\Rightarrow \frac{d \theta}{d t}=\frac{-169 \times 1.5}{x^{3} \sec ^{2} \theta \tan \theta} \end{aligned}........(v)

Now the foot of the ladder is pulled along the ground away from the wall,
So, x=12m
$\Rightarrow y=\sqrt{h^{2}-x^{2}}$ .......from equation (ii)
$\Rightarrow y=\sqrt{169-144} \quad\left[\because h=13 \Rightarrow h^{2}=169 \& x=12 \Rightarrow x^{2}=144\right]$
$\Rightarrow y=\sqrt{25}=5 m$ .........(iv)

Thus equation (v)
$\Rightarrow \frac{d \theta}{d t}=\frac{-169 \times 1.5}{(12)^{3}\left(\frac{169}{144}\right)\left(\frac{5}{12}\right)}\left[\begin{array}{l} \because \sec ^{2} \theta=\frac{169}{x^{2}}=\frac{169}{144} \\ \tan \theta=\frac{y}{x}=\frac{5}{12} \end{array}\right]$
\begin{aligned} &\Rightarrow \frac{d \theta}{d t}=\frac{-169 \times 1.5}{(12)^{3} \times \frac{169}{\left(12^{2}\right)} \times 5}=\frac{-1.5}{5} \\\\ &=-0.3 \mathrm{rad} / \mathrm{sec} \end{aligned}

Derivative As a Rate Measure exercise 12.2 question 13

Answer:$x=-\frac{1}{2}$ and $y=-\frac{3}{4}$
Hint: Here we use the equation of curve are $y=x^{2}+2x$ .
Given: Given as particle moves along the curve $y=x^{2}+2x$ .
Solution: As to find the points at which the curve are the x and y coordinates of the particle changing at the same rate $y=x^{2}+2x$
\begin{aligned}\\ &\frac{d y}{d x}=\frac{d\left(x^{2}+2 x\right)}{d x} \\\\ &\frac{d y}{d x}=\frac{d\left(x^{2}\right)}{d x}+\frac{d(2 x)}{d x}=2 x+2 \end{aligned} ........(i)
When x and y co-ordinate of the particle are changing at the same rate
\begin{aligned} &\frac{d y}{d t}=\frac{d x}{d t} \\\\ &\frac{d y}{d x}=\frac{d t}{d t} \\\\ &\frac{d y}{d x}=1 \end{aligned}

Substituting value from eq (i)

So, $2x+2=1$

\begin{aligned} &2 x=-1 \\\\ &x=-\frac{1}{2} \end{aligned}
and $y=x^{2}+2x$
\begin{aligned} &y=\frac{1}{4}-1 \\\\ &y=-\frac{3}{4} \end{aligned}
$x=-\frac{1}{2}$ and $y=-\frac{3}{4}$ .

## Derivative As a Rate Measure exercise 12.2 question 14

Answer:$\frac{dM}{dt}=(48)$
Hint: Here we use the equation of given curve.
Given: Equation of the curve $y=7x-x^{3}$ and $x$ increases at the rate of 4 units per second
Solution: As the differentiate, the above equation with respect to $x$ , we get the slope of the curve
$\frac{d y}{d x}=\frac{d\left(7 x-x^{3}\right)}{d x}$
$\frac{d y}{d x}=\frac{d(7 x)}{d x}-\frac{d\left(x^{3}\right)}{d x}=7-3 x^{2}$ …….(i)

Suppose $M$ be the slope of the given curve then the above equation $M=7-3x^{2}$

Given $x$ increases at the rate of 4 units per second therefore,
$\frac{d x}{d t}=4 \text { units } / \mathrm{sec}$ ……… eq (ii)

Differentiate the equation of the slope that is equation (ii)
\begin{aligned} &\frac{d M}{d t}=\frac{d\left(7-3 x^{2}\right)}{d t} \\\\ &\frac{d M}{d t}=\frac{d(7)}{d t}-\frac{d\left(3 x^{2}\right)}{d t} \end{aligned}
\begin{aligned} &\frac{d M}{d t}=0-(3 \times 2 x) \frac{d x}{d t} \\\\ &\frac{d M}{d t}=-6 x \times \frac{d x}{d t} \end{aligned}

Let’s put value of $\frac{d x}{d t}=4 \text { units } / \sec$
$\frac{d M}{d t}=-6 x \times 4$

Let’s put value of $x = 2$(given)
$\frac{d M}{d t}=-6 x \times 4=-6 \times 2 \times 4=-48$

The slope cannot be negative,
Thus, the slope of the curve is changing at the rate of 48 units/sec when $x = 2$

Derivative As a Rate Measure exercise 12.2 question 15

Answer:$\left ( 1,1 \right )$ and $\left ( -1,-1 \right )$
Hint: Here we use the equation of curve $y=x^{3}$
Given: As particle moves along the curve $y=x^{3}$
Solution: Differentiate the above equation with respect to $t$ ,
$\frac{d y}{d t}=\frac{d\left(x^{3}\right)}{d t}=3 x^{2} \frac{d x}{d t}$ …(i)

When y-coordinate change three times more rapidly than x-coordinate that is
$\frac{d y}{d t}=3 \frac{d x}{d t}$ …(ii)

Then equating (i) and (ii)
\begin{aligned} &3 x^{2} \frac{d x}{d t}=3 \frac{d x}{d t} \\\\ &x^{2}=1 \Rightarrow x=\pm 1 \end{aligned}

When $x=1$
$y=x^{3}=(1)^{3}=y=1$

When $x=-1$
$y=x^{3}=(-1)^{3}=y=-1$

$\left ( 1,1 \right )$ and $\left ( -1,-1 \right )$ .

Derivative As a Rate Measure exercise 12.2 question 16(i)

Answer: $\frac{7\pi }{6}$

Hint: Here we use the formula of angles

Given: $\frac{d \theta}{d t}=2 \times \frac{d(\cos \theta)}{d t}$

Solution:

\begin{aligned} &\frac{d \theta}{d t}=2 \times \frac{d(\cos \theta)}{d t} \\\\ &\frac{d \theta}{d t}=2 \times \frac{d(\cos \theta)}{d \theta} \times \frac{d \theta}{d t} \end{aligned}

\begin{aligned} &1=2(-\sin \theta) \\\\ &\sin \theta=-\frac{1}{2} \end{aligned}

Hence, $\theta=\frac{7 \pi}{6}$

So the value of angle $\theta$ which increases twice as fast as its cosine is $\frac{7\pi }{6}$

Derivative As a Rate Measure exercise 12.2 question 16(ii)

Answer: $\theta=\frac{\pi}{6}$
Hint: Here we use the given condition, $\frac{d \theta}{d t}=-2 \frac{d(\cos \theta)}{d t}$
Given: Whose rate of increase twice as twice the rate of decrease of its cosine.
Solution: As per the condition
$\frac{d \theta}{d t}=-2 \frac{d(\cos \theta)}{d t}$
As the rate of increase twice as twice the rate of decrease hence the minus sign,
\begin{aligned} &\frac{d \theta}{d t}=-2 \frac{d(\cos \theta)}{d t} \times \frac{d \theta}{d t} \\\\ &1=-2(-\sin \theta) \end{aligned}
\begin{aligned} &\sin \theta=\frac{1}{2} \\\\ &\theta=\frac{\pi}{6} \end{aligned}

Derivative As a Rate Measure exercise 12.2 question 17

Answer:$\frac{d y}{d t}=-\frac{1}{\sqrt{5}} m / s e c$ and $x=3 \sqrt{2} m$
Hint: Here we use the theorem of Pythagoras
Given: The top of ladder is 6 meter long is resting against a vertical wall on a level pavement, when the ladder begins to side outward.
Solution: Let $AC$ be the position of the ladder initially then $AC=6$ m
$DE$ be the position of the ladder after being pulled at the rate of 0.5 m/sec.

So $\frac{d x}{d t}=0.5 m/sec$ …….. (i)
$\Delta ABC$ , let’s apply Pythagoras theorem
$A B^{2}+B C^{2}=A C^{2}$
$y^{2}+x^{2}=h^{2}$ ……… (ii)
Here h = 6 and x=4
\begin{aligned} &y^{2}=6^{2}-4^{2} \\\\ &y^{2}=36-16 \\\\ &y^{2}=20=4 \times 5=2 \sqrt{5} \end{aligned}
Differentiate equation (ii)
\begin{aligned} &y^{2}+x^{2}=h^{2} \\\\ &2 x \frac{d x}{d t}=-2 y \frac{d y}{d t} \end{aligned}
Let’s put value of $x = 4$ and $y = 2\sqrt{5}$ and $\frac{d x}{d t}=0.5$ ........equation (iii)
\begin{aligned} &2 \times 4 \times 0.5=-2 \times 2 \sqrt{5} \times \frac{d y}{d t} \\\\ &\Rightarrow \frac{d y}{d t}=-\frac{1}{\sqrt{5}} \mathrm{~m} / \mathrm{sec} \end{aligned}
From equation (iii) we get,
$2 x \frac{d x}{d t}=-2 y \frac{d y}{d t}$ as we know from question $\left[\frac{d x}{d t}=\frac{d y}{d t}\right]$
$x=-y$
Substituting $x=-y$ in $x^{2}+y^{2}=36$
We get
\begin{aligned} &x^{2}+x^{2}=36 \\\\ &2 x^{2}=36 \\\\ &x^{2}=18 \\\\ &x=3 \sqrt{2} \end{aligned}

Derivative As a Rate Measure exercise 12.2 question 18

Answer: $12 \pi \mathrm{cm}^{3} / \mathrm{sec}$
Hint: Use $V=\frac{4}{3} \pi r^{3}$
Given: A balloon in the form of a right circular cone surrounded by hemisphere having a diameter equal to the height of the cone is being inflated.
Solution: Let, the total height of the balloon, then

then
\begin{aligned} &H=h+r \\\\ &H=2 r+r \quad(h=2 r) \\\\ &\frac{d H}{d t}=\frac{3 d r}{d t} \end{aligned}...........(i)

So, volume of the cone volume of the hemisphere
$V=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}$

Substituting $h =2r$
So, we get
\begin{aligned} &V=\frac{2}{3} \pi r^{3}+\frac{2}{3} \pi r^{3} \\\\ &V=\frac{4}{3} \pi r^{3} \end{aligned}

Let’s take derivative w.r.t to time both side
\begin{aligned} &\frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t} \\\\ &\frac{d V}{d t}=4 \pi r^{2} \times \frac{d H}{3 \times d t} \end{aligned}

Let’s put value of $r = 3$
\begin{aligned} &\frac{d V}{d t}=4 \pi \times 3 \times 3 \times \frac{d H}{3 \times d t} \\\\ &\frac{d V}{d H}=12 \pi \mathrm{cm}^{3} / \mathrm{sec} \end{aligned}

Derivative As a Rate Measure exercise 12.2 question 19

Hint:$\frac{dh}{dt}=\theta$ Here, Volume of the cone is $V=\frac{1}{3} \pi r^{2} h$
Given: The height of the cone is 10 meters and the radius of its base is 5 m and water is running into an inverted cone at a rate of $\pi$ cubic meter per minute
Solution:

Let r be the radius, h be height of the cone and v be the volume of cone at any time of $t$
Let O be the semi vertical angle of the cone CAB whose height CO is 10m and radius is OB = 5m
\begin{aligned} &\therefore \tan \theta=\frac{O B}{O C}=\frac{5}{10} \\\\ &\tan \theta=\frac{1}{2} \end{aligned}........(i)

Let v be the volume of water in the cone then,
\begin{aligned} &v=\frac{1}{3} \pi\left(O^{\prime} B^{\prime}\right)^{2}\left(C O^{\prime}\right) \\\\ &=\frac{1}{3} \pi(h \tan \theta)^{2}(h) \\\\ &=\frac{1}{3} \pi h^{3} \tan ^{2} \theta \end{aligned}
\begin{aligned} &\Rightarrow \frac{\pi h^{3}}{3}(\tan \theta)^{2} \\\\ &\Rightarrow \frac{\pi h^{3}}{3}\left(\frac{1}{2}\right)^{2}=\frac{\pi h^{3}}{3}\left(\frac{1}{4}\right) \quad\left[\tan \theta=\frac{1}{2}\right] \\\\ &\Rightarrow \frac{\pi h^{3}}{12} \end{aligned}

Differentiate w.r.t 't' then
$\frac{d v}{d t}=\frac{\pi}{12} \frac{d}{d t}\left(h^{3}\right)=\frac{\pi}{12} 3 h^{2} \frac{d h}{d t}$
$\Rightarrow \pi=\frac{\pi}{12} 3 h^{2} \frac{d h}{d t}$ $\left[\frac{d v}{d t}=\pi \frac{m^{3}}{\min }\right]$
\begin{aligned} &\Rightarrow 1=\frac{h^{2}}{4} \frac{d h}{d t} \\\\ &\Rightarrow \frac{4}{h^{2}}=\frac{d h}{d t} \Rightarrow \frac{d h}{d t}=\frac{4}{h^{2}} \end{aligned}

Now, when the water stands 7.5 m below the base
\begin{aligned} &\text { i.e } 10-7.5=2.5 \mathrm{~m} \\\\ &\frac{d h}{d t}=\frac{4}{h^{2}}=\frac{4}{(2.5)^{2}}=0.64 m/min\end{aligned}

Derivative As a Rate Measure exercise 12.2 question 20

Answer: $\frac{d y}{d t}=\frac{3 k m}{h r}$
Hint: The rate at which the length of the man’s shadow increases will be $\frac{d y}{d t}$
Given: A man 2 m tall walks at a speed of 6 km/hr away from a source of light that is 6 m above the ground.
Solution: Let $AB$ the lamp post and let at any time t .The man CD be at a distance of x km from the lamp post and y be the length of his shadow

Since triangle $\Delta ABE$ and $\Delta CDE$ are similar

\begin{aligned} &\frac{A B}{C D}=\frac{B E}{D E} \\\\ &\frac{6}{2}=\frac{x+y}{y} \\\\ &\frac{x}{y}=\frac{6}{2}-1=2 \end{aligned}
Let’s take derivative with respect to time both side
$\frac{d y}{d t}=\frac{1}{2} \times \frac{d x}{d t}$

Let’s put value of $\frac{d x}{d t}=6 \mathrm{~km}(\text { given })$

$\frac{d y}{d t}=\frac{1}{2} \times 6=3 \mathrm{~km} / \mathrm{hr}$

Derivative As a Rate Measure exercise 12.2 question 21

Answer:$6 \mathrm{~cm}^{3} / \mathrm{sec}$
Hint: The surface area of the bubble will be, $S A=4 \pi r^{2}$ .
Given: The surface area of a spherical bubble is increasing at the rate of 2 cm2/sec
Solution: To find rate of increase of its volume, when the radius is 6 cm.
Let the radius of the given spherical bubble be $r$ cm at any instant time.
It is given that the surface area of a spherical bubble is increasing at the rate of $2 \mathrm{~cm}^{2} / \sec \frac{d s}{d t}=2$
\begin{aligned} &S=4 \pi r^{2} \\\\ &\frac{d S}{d t}=8 \pi r \times \frac{d r}{d t} \end{aligned}
\begin{aligned} &\frac{d r}{d t}=\frac{1}{8 \pi r} \times \frac{d S}{d t} \\\\ &\frac{d r}{d t}=\frac{1}{8 \pi r} \times 2 \quad\left(\frac{d S}{d t}=2(\text { given })\right) \end{aligned}
\begin{aligned} &\frac{d r}{d t}=\frac{1}{8 \pi \times 6} \times 2 \quad(r=6(\text { given })) \\\\ &\frac{d r}{d t}=\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec} \end{aligned}
Now, Volume of sphere $=V=\frac{4}{3} \pi r^{3}$

$\frac{d V}{d t}=4 \pi r^{2} \times \frac{d r}{d t}$
Let’s put value of $r =6$ and $\frac{d r}{d t}=\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec}$

$\frac{d V}{d t}=4 \pi \times 6^{2} \times \frac{1}{24 \pi}=6 \mathrm{~cm}^{3} / \mathrm{sec}$

## Derivative As a Rate Measure exercise 12.2 question 22

Answer:$33 \pi \mathrm{cm}^{3} / \mathrm{sec}$
Hint: we know the volume of the cylinder is $V=\pi r^{2} h$
Given: The radius of a cylinder is increasing at the rate 2cm/sec and its altitude is increasing at rate of 3cm
Solution: The radius of a cylinder is increasing at the rate 2 cm/sec and its altitude is decreasing at the rate of 3 cm/sec
To find the rate of change of volume when radius is 3 cm and altitude 5 cm
Let V be the volume of the cylinder, r be its radius and h be its altitude at any instant of time ‘t’.
We know volume of the cylinder is $V=\pi r^{2} h$
Differentiating this with respect to time we get
\begin{aligned} &\frac{d V}{d t}=\frac{d\left(\pi r^{2} h\right)}{d t} \\\\ &\frac{d v}{d t}=\pi\left[\frac{d\left(r^{2} h\right)}{d t}\right] \end{aligned}
Now will apply the product rule of differentiation
$\frac{d(u v)}{d x}=v \frac{d(u)}{d x}+u \frac{d(v)}{d x}$
So above equation becomes
\begin{aligned} &\frac{d V}{d t}=\pi\left[h \frac{d\left(r^{2}\right)}{d t}+r^{2} \frac{d(h)}{d t}\right] \\\\ &\frac{d V}{d t}=\pi\left[h \times 2 r \times \frac{d r}{d t}+r^{2} \frac{d h}{d t}\right] \end{aligned}
But given of a cylinder is increasing at the rate 2 cm/sec, i.e., $\frac{d r}{d t}=2\; cm/sec$and its altitude is decreasing at the rate of 3 cm/sec, i.e., $\frac{d h}{d t}=-3\: cm/sec$
by substituting the above values in equation we get
\begin{aligned} &\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\pi\left[\mathrm{h} \times 2 \mathrm{r}(2)+\mathrm{r}^{2}(-3)\right] \\\\ &\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\pi\left[4 \mathrm{hr}-3 \mathrm{r}^{2}\right] \end{aligned}
When radius of the cylinder, $r = 3$ cm and its altitude,$h = 5$ cm, the above equation becomes
\begin{aligned} &\mathrm{d} \mathrm{V} / \mathrm{dt}=\pi\left[4 \times 5 \times 3-3 \times 3^{2}\right] \\\\ &\frac{d V}{d t}=\pi[60-27] \\\\ &\frac{d V}{d t}=33 \pi \mathrm{cm}^{3} / \mathrm{sec} \end{aligned}
Hence the rate of change of volume when radius is 3 cm and altitude 5cm is 33$\pi$ cm3/sec

Derivative As a Rate Measure exercise 12.2 question 23

Answer: $\frac{d R}{d t}=0.25\; cm/sec$
Hint: The volume of the hallow sphere is $V=\frac{4}{3} \pi\left(R^{3}-r^{3}\right)$
Given: radius $=1$ cm/sec and outer radius are 4 cm and 8 cm
Solution: Differentiating the volume with respect to time,
$\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi\left(R^{3}-r^{3}\right)\right)}{d t}$
This is the rate of the volume of the hollow sphere
$0=\frac{4}{3} \pi \frac{d\left(R^{3}-r^{3}\right)}{d t}=\frac{d\left(R^{3}\right)}{d t}-\frac{d\left(r^{3}\right)}{d t}=0$
$\frac{d r}{d t}=1\: cm/sec$
\begin{aligned} &3 R^{2} \frac{d R}{d t}-3 r^{2}(1)=0 \\\\ &3 R^{2} \frac{d R}{d t}=3 r^{2} \end{aligned}
Here radius are 4 cm and 8 cm
\begin{aligned} &3(8)^{2} \frac{d R}{d t}=3(4)^{2} \\\\ &\frac{d R}{d t}=\frac{3(8)^{2}}{3(4)^{2}} \\\\ &=0.25 \: \mathrm{cm} / \mathrm{sec} \end{aligned}

Derivative As a Rate Measure exercise 12.2 question 24

Answer: $\frac{d h}{d t}=\frac{1}{2 \pi}cm/min$
Hint: We know, volume of the cone is $V=\frac{1}{3} \pi r^{2} h$
Given: Sand is being poured into a conical pile at the constant rate of 50 cm3/min
Solution: Let the volume be V , height be $h$ and radius be $r$ of the cone at any instant of time.
We know volume of the cone is $V=\frac{1}{3} \pi r^{2} h$ .
$r=2 h$
So, the new volume becomes,
$V=\frac{1}{3} \pi(2 h)^{2} h=\frac{4}{3} \pi h^{3}$
Differentiate the above equation with respect to time,
$\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi h^{3}\right)}{d t}=\frac{4}{3} \pi \frac{d\left(h^{3}\right)}{d t}=\frac{4}{3} \pi \times 3 h^{2} \frac{d h}{d t}$
Here, $\frac{d V}{d t}=50\: cm^{3}/min$
$50=\frac{4}{3} \pi \times 3 h^{2} \frac{d h}{d t}$
$\frac{d h}{d t}=\frac{50 \times 3}{4 \times 3 \pi h^{2}}=\frac{50 \times 3}{12 \pi(5)^{2}}=\frac{1}{2 \pi}$ (height =5 ).

Derivative As a Rate Measure exercise 12.2 question 25

Answer:$\frac{d y}{d t}=20m/sec$
Hint: Here, we use the Pythagoras theorem $P R^{2}=P Q^{2}+Q R^{2}$
Given: A kite is 120 m high and 130 m string is out.
Solution:

\begin{aligned} &Q R=x \\ &P R=y \end{aligned}
Then from figure by applying Pythagoras theorem,
\begin{aligned} &P R^{2}=P Q^{2}+Q R^{2} \\\\ &y^{2}=(120)^{2}+x^{2} \end{aligned}.......(i)
Differentiating the above equation with respect to time,
\begin{aligned} &\frac{d y^{2}}{d t}=\frac{d\left((120)^{2}+x^{2}\right)}{d t} \\\\ &2 y \frac{d y}{d t}=0+2 x \frac{d x}{d t} \end{aligned}
So , $\frac{d x}{d t}=52m/sec$
$2 y \frac{d y}{d t}=2 x \times 52$ ......(ii)\begin{aligned} &y^{2}=x^{2}+(120)^{2} \\\\ &130^{2}=x^{2}+14400 \end{aligned}

Therefore, $x=50$m

So, let use equation (i) and (ii)

\begin{aligned} &2 y \frac{d y}{d t}=2 x \times 52 \\\\ &2(130) \frac{d y}{d t}=2(50) \times 52 \\\\ &\frac{d y}{d t}=\frac{2(50) \times 52}{2 \times 130}=20m/sec \end{aligned}

Derivative As a Rate Measure exercise 12.2 question 26

Answer:$\left ( 1,\frac{5}{3} \right )$ and $\left ( -1,\frac{1}{3} \right )$
Hint: Here we use the equation of curve.
Given: As particle moves along the curve $y=\frac{2}{3} x^{3}+1$
Solution: Equation of curve is $y=\frac{2}{3} x^{3}+1$
Differentiating the above equation,
$\frac{d y}{d t}=\frac{d\left(\frac{2}{3} x^{3}+1\right)}{d t}$
$=\frac{d\left(\frac{2}{3} x^{3}+1\right)}{d t}+\frac{d(1)}{d t}=\frac{2}{3} \times 3 x^{2} \frac{d x}{d t}$ .......…(i)
When y-coordinate is changing twice as fast as x coordinate
$\frac{d y}{d t}=2 \frac{d x}{d t}$ …......(ii)
Equating equation (i) and (ii)
$2 x^{2} \frac{d x}{d t}=2 \frac{d x}{d t}$
$x=\pm 1$
When $x=1; y=\frac{2}{3}(1)^{3}+1$
\begin{aligned} &y=\frac{2+3}{3} \\\\ &y=\frac{5}{3} \end{aligned}
When $x=-1; y=\frac{2}{3}(-1)^{3}+1$

\begin{aligned} &y=\frac{-2+3}{3} \\\\ &y=\frac{1}{3} \end{aligned}
$\left ( 1,\frac{5}{3} \right )$ and $\left ( -1,\frac{1}{3} \right )$

## Derivative As a Rate Measure exercise 12.2 question 27

Answer: $\left ( 2,4 \right )$
Hint: Here we use the concept of equation curve $y^{2^{}}=8x$
Given: A point on the curve $y^{2^{}}=8x$ .
Solution: Differentiating the above equation,
\begin{aligned} &\frac{d\left(y^{2}\right)}{d t}=\frac{d(8 x)}{d t} \\\\ &2 y \frac{d y}{d t}=8 \frac{d x}{d t} \\\\ &\frac{y}{4} \frac{d y}{d t}=\frac{d y}{d t} \end{aligned}..........(i)
$\frac{d y}{d t}=\frac{d x}{d t}$ ..........(ii)
Let use equation (i) and (ii)
\begin{aligned} &\frac{y}{4} \frac{d y}{d t}=\frac{d y}{d t} \\\\ &y=4 \end{aligned}
When $y^{2}=8 x \Rightarrow(4)^{2}=8 x$
$x=2$
Hence the point on the curve at which the abscissa and coordinate change at the same rate is $\left ( 2,4 \right )$ .

Derivative As a Rate Measure exercise 12.2 question 28

Answer: $\frac{d s}{d t}=3.6\: cm^{2}/sec$
Hint: We know that $V=x^{3}$
Given: The volume of a cube is increasing at the rate of 9 cm3/sec
Solution: Differentiating the equation,
$\frac{d V}{d t}=\frac{d x^{3}}{d t}=3 x^{2} \frac{d x}{d t}$........ …(i)
The cube is increasing at the rate of 9 cm3/sec
So the above equation becomes,
$9=3 x^{2} \frac{d x}{d t}$
$\frac{d x}{d t}=\frac{9}{3 x^{2}}$........(ii)
So, $s=6 x^{2}$
Again differentiating the above equation with respect to time is,
$\frac{d s}{d t}=\frac{d\left(6 x^{2}\right)}{d t}=6 \times 2 x \frac{d x}{d t}$
Substitute equation (i) in the above equation,
$\frac{d s}{d t}=6 \times 2 x \times \frac{9}{3 x^{2}}=\frac{36}{x}$
So, length 10 cm,
$\frac{d s}{d t}=\frac{36}{x}=\frac{36}{10}=3.6\; cm^{2}/sec$

Derivative As a Rate Measure exercise 12.2 question 29

Answer:$\frac{d s}{d t}=10 \mathrm{~cm}^{2} / \mathrm{sec}$
Hint: $V=\frac{4}{3} \pi r^{3}$
Given: The volume of spherical balloon is increasing at the rate of 25 cm3/sec
Solution: Let the radius of the given spherical balloon be $r$ cm and $V$ be its volume at any instant time.
Then according to the given criteria,
The rate of volume of the spherical balloon is increasing is $\frac{d V}{d t}=25 \: cm^{3}/sec$ ....…(i)
But volume of balloon is $\frac{4}{3} \pi r^{3}$
Applying derivative
\begin{aligned} &\frac{d V}{d t}=\frac{d\left[\frac{4}{3} \pi r^{3}\right]}{d t} \\\\ &\frac{d V}{d t}=\frac{4}{3} \pi \times 3 r^{2} \frac{d r}{d t} \end{aligned}
Substituting the value from equation (i)
\begin{aligned} &25=\frac{4}{3} \pi \times 3 r^{2} \frac{d r}{d t}\\\\ &\frac{d r}{d t}=\frac{25}{4 \pi r^{2}} \end{aligned}.......(ii)
Now the surface area of spherical balloon at any time $t$ will be $S=4 \pi r^{2} \mathrm{~cm}^{2}$
Applying derivative,
$\frac{d S}{d t}=\frac{d\left(4 \pi r^{2}\right)}{d t}=4 \pi \times 2 r \frac{d r}{d t}$
Substituting the value from equation (ii) we get,
$\frac{d S}{d t}=4 \pi \times 2 r \times \frac{25}{4 \pi r^{2}}$
So, when the radius is 5 cm,
$\frac{d S}{d t}=\frac{25 \times 2}{5}=10 \mathrm{~cm}^{2} / \mathrm{sec}$

Derivative As a Rate Measure exercise 12.2 question 30

Answer: (i) -2 cm/min (ii) 2 cm2/min
Hint: We know that, $p=2(x+y)$
Given: The length of the rectangle is $x$ cm and width is $y$ cm.
Solution: As per the given criteria, the length is decreasing at the rate of 5 cm/min
(i) $\frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}$
And width is increasing at the rate of 4 cm/min
$\frac{d x}{d t}=4 \mathrm{cm} / \mathrm{min}$
Differentiating both sides,
$\frac{d p}{d x}=\frac{d(2(x+y))}{d t}=2\left[\frac{d x}{d t}+\frac{d y}{d t}\right]$
So, $\frac{d p}{d x}=2[-5+4]=-2\; cm/min$

(ii) Let A be the area of the rectangle, $A=xy$
\begin{aligned} &\frac{d A}{d t}=\frac{d(x y)}{d t} \\\\ &\frac{d(u v)}{d x}=v \frac{d u}{d x}+u \frac{d v}{d x} \end{aligned}
So the above equation becomes,
$\frac{d A}{d t}=y \frac{d x}{d t}+x \frac{d y}{d t}$
Substituting the values from equation (i) and (ii)
$\frac{d A}{d t}=y(-5)+x(4)$
When $x=8$ cm and $y=6$ cm the above equation becomes,
$\frac{d A}{d t}=6(-5)+8(4)=-30+32=2\: cm^{2}/min$

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## RD Sharma Class 12 Solutions Chapter 12 Derivative as a Rate Measure - Other Exercise

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There are 18 questions present in this exercise. The RD Sharma Class 12th exercise 12.2 comprises of the solved sums for those questions. Along with the solved sums, there are numerous practice questions asked in the RD Sharma Class 12th Exercise 12.2 reference material. Some of the important concepts that are present in this exercise are concepts of derivate. Rate of increase of perimeter and circumference of the given figure, and other application-based questions on the derivatives topic.

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