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

Derivative as a Rate Measurer Excercise: 12.2

Derivative As a Rate Measure exercise 12.2 question 1

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 $dx/dt=3\mathrm{cm}/\sec$... … (i)
Now, the formula is $V=x^3{\mathrm{cm}}^3$
By applying derivative,
$\frac{dV}{dt}=\frac{d{x}^3}{dt}=3x^2\frac{dx}{dt}=3x^2\times 3=9x^2$ $(\frac{dx}{dt}=3){\text{ from eq}}^n(i)$
Lets put the value of x in above equation
$\begin{array}{rl}& \frac{dV}{dt}=9x^2{\mathrm{cm}}^3/\sec \\ \\ & \frac{dV}{dt}=9\times 10\times 10{\mathrm{cm}}^3/\sec (x=10)\text{ Given }\\ \\ & \frac{dV}{dt}=900{\mathrm{cm}}^3/\sec \end{array}$
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 $\frac{dx}{dt}=0.2cm/sec$ … (i)
By applying derivative in perimeter,
$\frac{dp}{dt}=\frac{d(4x)}{dt}=\frac{dx}{dt}\times 4=4\times 0.2=0.8cm/sec$

$\frac{dx}{dt}=0.2cm/sec$ (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,
$\frac{dC}{dt}=\frac{d(2\pi r)}{dt}=2\pi \frac{dr}{dt}=2\pi \times 0.7=1.4\pi$ cm/sec
$\{\frac{dr}{dt}=0.7(\text{ from }e{q}^n(i))\text{ and }C=2\pi r(\text{ Given })\}$
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
$\frac{ds}{dt}=\frac{d\left(4\pi r^2\right)}{dt}$
$\begin{array}{rl}& \frac{ds}{dt}=\frac{4\pi \left(d{r}^2\right)}{dt}\\ \\ & \frac{ds}{dt}=4\pi \times 2r\times \frac{dr}{dt}\end{array}$
$\frac{ds}{dt}=4\pi \times 2r\times 0.2$ $(\frac{dr}{dt}=0.2)(\text{ from }e{q}^n(i))$
$\frac{dS}{dt}=1.6\pi r$
lets put value of r in above formula $r=7$(given)
$\frac{dS}{dt}=1.6\pi \times 7=11.2\pi$ 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: $\frac{dr}{dt}=\frac{1}{\pi }\mathrm{cm}/\sec$
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,
$\frac{\mathrm{d}\mathrm{V}}{\mathrm{dt}}=900{\mathrm{cm}}^3/\sec$
$\frac{dV}{dt}=\frac{d\left(\frac{4}{3}\pi r^3\right)}{dt}$ $\therefore (V=\frac{4}{3}\pi r^3)$
lets put value of $\frac{dV}{dt}=900$(given)
Let’s differentiate $\left(\frac{4}{3}\pi r^3\right)$ w.r.t time
$\begin{array}{c}900=\frac{4\times 3}{3}\times \pi r^2\frac{dr}{dt}\\ \\ 900=4\pi r^2\frac{dr}{dt}\end{array}$
So, lets put value of $r=15$ (given)
$\begin{array}{rl}& \Rightarrow 900=4\pi \times 15\times 15\times \frac{dr}{dt}\\ \\ & \Rightarrow 900=900\pi \times \frac{dr}{dt}\\ \\ & \Rightarrow \frac{900}{900\pi }=\frac{dr}{dt}\end{array}$
$\begin{array}{rl}& \Rightarrow \frac{1}{\pi }=\frac{dr}{dt}\\ \\ & \text{ So, }\frac{dr}{dt}=\frac{1}{\pi }\end{array}$

Derivative As a Rate Measure exercise 12.2 question 7

Answer: $\frac{dV}{dt}=2\pi {\mathrm{cm}}^3/\sec$
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 $\frac{\mathrm{dr}}{\mathrm{dt}}=0.5\mathrm{cm}/\sec$

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

Let’s differentiate $d\left(\frac{4}{3}\pi r^3\right)$
$\begin{array}{rl}& \frac{dV}{dt}=\frac{4\times 3}{3}\times \pi r^2\frac{dr}{dt}\\ \\ & \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\end{array}$

Let’s put value of $\frac{dr}{dt}=0.5$ (given)
$\frac{dV}{dt}=4\pi r^2\times 0.5$

Let’s put value of $r=1$
$\begin{array}{rl}& \frac{dV}{dt}=4\pi \times 1^2\times 0.5\\ \\ & \frac{dV}{dt}=2\pi {\mathrm{cm}}^3/\sec \end{array}$

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

Derivative As a Rate Measure exercise 12.2 question 8

Answer: $\frac{ds}{dt}=\frac{5}{2}=2.5km/hr$
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
$\frac{dI}{dt}=5km/hr$
So considering $\mathrm{\Delta }ASB,$
$\tan \theta =\frac{AB}{AS}=\frac{6}{I+s}$................(i)
Then considering $\mathrm{\Delta }MNS,$
$\tan \theta =\frac{MN}{MS}=\frac{2}{s}$...................(ii)
From equation (i) and (ii)
$\begin{array}{rl}& \frac{6}{l+s}=\frac{2}{s}\\ \\ & 6s=2(I+s)\\ \\ & 6s=(2I+2s)\end{array}$
$\begin{array}{rl}& 6s-2s=2l\\ \\ & 4s=2l\\ \\ & 2s=1\end{array}$..............(iii)
By applying derivative, With respect to time on both side
$\frac{dl}{dt}=\frac{d(2s)}{dt}$ $I=2s\left(\text{ from }e{q}^n\text{ iii }\right)$
$\begin{array}{rl}& \frac{dl}{dt}=2\frac{ds}{dt}\\ \\ & 5=2\frac{ds}{dt}\end{array}$ $(\frac{dl}{dt}=5)(\text{ given })$
$\begin{array}{rl}& \frac{5}{2}=\frac{ds}{dt}\\ \\ & 2.5=\frac{ds}{dt}\end{array}$
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:$\frac{dA}{dt}=80\pi {\mathrm{cm}}^2/\sec$
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,
$\frac{dA}{dt}=\frac{d\left(\pi r^2\right)}{dt}=\frac{\pi d\left(r^2\right)}{dt}=\pi \times 2r\frac{dr}{dt}=2\pi r\times 4$
So, when the radius of circle is 10 cm,
$\frac{dA}{dt}=2\pi \times 10\times 4=80\pi c{m}^2/sec$
Thus the enclosed area is increasing at the rate of $80\pi$ cm²/sec.

Derivative As a Rate Measure exercise 12.2 question 10

Derivative As a Rate Measure exercise 12.2 question 11

Then_{$\mathrm{\Delta }MNS,$},
$\tan \theta =\frac{MN}{MS}=\frac{1.8}{s}$

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

By applying derivative with respect to time on both side
$\begin{array}{rl}& \frac{dI}{dt}=4\frac{ds}{dt}\\ \\ & 2=4\frac{ds}{dt}\end{array}$ $\dots \dots (\frac{dI}{dt}=2)(\text{ given })$
$\begin{array}{rl}& \frac{2}{4}=\frac{ds}{dt}\\ \\ & \frac{ds}{dt}=0.5\mathrm{m}/\sec \end{array}$
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

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 $y=x^2+2x$
$\begin{array}{r}\\ & \frac{dy}{dx}=\frac{d(x^2+2x)}{dx}\\ \\ & \frac{dy}{dx}=\frac{d\left(x^2\right)}{dx}+\frac{d(2x)}{dx}=2x+2\end{array}$ ........(i)
When x and y co-ordinate of the particle are changing at the same rate
$\begin{array}{rl}& \frac{dy}{dt}=\frac{dx}{dt}\\ \\ & \frac{dy}{dx}=\frac{dt}{dt}\\ \\ & \frac{dy}{dx}=1\end{array}$

Substituting value from eq (i)

So, $2x+2=1$

$\begin{array}{rl}& 2x=-1\\ \\ & x=-\frac{1}{2}\end{array}$
and $y=x^2+2x$
$\begin{array}{rl}& y=\frac{1}{4}-1\\ \\ & y=-\frac{3}{4}\end{array}$
$x=-\frac{1}{2}$ 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
$\frac{dy}{dx}=\frac{d(7x-x^3)}{dx}$
$\frac{dy}{dx}=\frac{d(7x)}{dx}-\frac{d\left(x^3\right)}{dx}=7-3x^2$ …….(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,
$\frac{dx}{dt}=4\text{ units }/\sec$ ……… eq (ii)

Differentiate the equation of the slope that is equation (ii)
$\begin{array}{rl}& \frac{dM}{dt}=\frac{d(7-3x^2)}{dt}\\ \\ & \frac{dM}{dt}=\frac{d(7)}{dt}-\frac{d\left(3x^2\right)}{dt}\end{array}$
$\begin{array}{rl}& \frac{dM}{dt}=0-(3\times 2x)\frac{dx}{dt}\\ \\ & \frac{dM}{dt}=-6x\times \frac{dx}{dt}\end{array}$

Let’s put value of $\frac{dx}{dt}=4\text{ units }/\sec$
$\frac{dM}{dt}=-6x\times 4$

Let’s put value of $x=2$(given)
$\frac{dM}{dt}=-6x\times 4=-6\times 2\times 4=-48$

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:$(1,1)$ and $(-1,-1)$
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 $t$ ,
$\frac{dy}{dt}=\frac{d\left(x^3\right)}{dt}=3x^2\frac{dx}{dt}$ …(i)

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

Then equating (i) and (ii)
$\begin{array}{rl}& 3x^2\frac{dx}{dt}=3\frac{dx}{dt}\\ \\ & x^2=1\Rightarrow x=\pm 1\end{array}$

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

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

$(1,1)$ and $(-1,-1)$ .

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

Solution:

$\begin{array}{rl}& \frac{d\theta }{dt}=2\times \frac{d(\cos \theta )}{dt}\\ \\ & \frac{d\theta }{dt}=2\times \frac{d(\cos \theta )}{d\theta }\times \frac{d\theta }{dt}\end{array}$

$\begin{array}{rl}& 1=2(-\sin \theta )\\ \\ & \sin \theta =-\frac{1}{2}\end{array}$

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, $\frac{d\theta }{dt}=-2\frac{d(\cos \theta )}{dt}$
Given: Whose rate of increase twice as twice the rate of decrease of its cosine.
Solution: As per the condition
$\frac{d\theta }{dt}=-2\frac{d(\cos \theta )}{dt}$
As the rate of increase twice as twice the rate of decrease hence the minus sign,
$\begin{array}{rl}& \frac{d\theta }{dt}=-2\frac{d(\cos \theta )}{dt}\times \frac{d\theta }{dt}\\ \\ & 1=-2(-\sin \theta )\end{array}$
$\begin{array}{rl}& \sin \theta =\frac{1}{2}\\ \\ & \theta =\frac{\pi }{6}\end{array}$

Derivative As a Rate Measure exercise 12.2 question 17

Derivative As a Rate Measure exercise 12.2 question 18

Answer: $12\pi {\mathrm{cm}}^3/\sec$
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
$\begin{array}{rl}& H=h+r\\ \\ & H=2r+r(h=2r)\\ \\ & \frac{dH}{dt}=\frac{3dr}{dt}\end{array}$...........(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
$\begin{array}{rl}& V=\frac{2}{3}\pi r^3+\frac{2}{3}\pi r^3\\ \\ & V=\frac{4}{3}\pi r^3\end{array}$

Let’s take derivative w.r.t to time both side
$\begin{array}{rl}& \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\ \\ & \frac{dV}{dt}=4\pi r^2\times \frac{dH}{3\times dt}\end{array}$

Let’s put value of $r=3$
$\begin{array}{rl}& \frac{dV}{dt}=4\pi \times 3\times 3\times \frac{dH}{3\times dt}\\ \\ & \frac{dV}{dH}=12\pi {\mathrm{cm}}^3/\sec \end{array}$

Derivative As a Rate Measure exercise 12.2 question 19

Derivative As a Rate Measure exercise 12.2 question 20

Since triangle $\mathrm{\Delta }ABE$ and $\mathrm{\Delta }CDE$ are similar

$\begin{array}{rl}& \frac{AB}{CD}=\frac{BE}{DE}\\ \\ & \frac{6}{2}=\frac{x+y}{y}\\ \\ & \frac{x}{y}=\frac{6}{2}-1=2\end{array}$
Let’s take derivative with respect to time both side
$\frac{dy}{dt}=\frac{1}{2}\times \frac{dx}{dt}$

Let’s put value of $\frac{dx}{dt}=6\mathrm{km}(\text{ given })$

$\frac{dy}{dt}=\frac{1}{2}\times 6=3\mathrm{km}/\mathrm{hr}$

Derivative As a Rate Measure exercise 12.2 question 21

Derivative As a Rate Measure exercise 12.2 question 22

Answer:$33\pi {\mathrm{cm}}^3/\sec$
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
$\begin{array}{rl}& \frac{dV}{dt}=\frac{d\left(\pi r^{2}h\right)}{dt}\\ \\ & \frac{dv}{dt}=\pi \left[\frac{d\left(r^{2}h\right)}{dt}\right]\end{array}$
Now will apply the product rule of differentiation
$\frac{d(uv)}{dx}=v\frac{d(u)}{dx}+u\frac{d(v)}{dx}$
So above equation becomes
$\begin{array}{rl}& \frac{dV}{dt}=\pi [h\frac{d\left(r^2\right)}{dt}+r^2\frac{d(h)}{dt}]\\ \\ & \frac{dV}{dt}=\pi [h\times 2r\times \frac{dr}{dt}+r^2\frac{dh}{dt}]\end{array}$
But given of a cylinder is increasing at the rate 2 cm/sec, i.e., $\frac{dr}{dt}=2cm/sec$and its altitude is decreasing at the rate of 3 cm/sec, i.e., $\frac{dh}{dt}=-3cm/sec$
by substituting the above values in equation we get
$\begin{array}{rl}& \frac{\mathrm{d}\mathrm{V}}{\mathrm{dt}}=\pi [\mathrm{h}\times 2\mathrm{r}(2)+{\mathrm{r}}^2(-3)]\\ \\ & \frac{\mathrm{d}\mathrm{V}}{\mathrm{dt}}=\pi [4\mathrm{hr}-3{\mathrm{r}}^2]\end{array}$
When radius of the cylinder, $r=3$ cm and its altitude,$h=5$ cm, the above equation becomes
$\begin{array}{rl}& \mathrm{d}\mathrm{V}/\mathrm{dt}=\pi [4\times 5\times 3-3\times 3^2]\\ \\ & \frac{dV}{dt}=\pi [60-27]\\ \\ & \frac{dV}{dt}=33\pi {\mathrm{cm}}^3/\sec \end{array}$
Hence the rate of change of volume when radius is 3 cm and altitude 5cm is 33$\pi$ cm³/sec