RD Sharma Class 12 Exercise 12 FBQ Derivative as a rate measure Solutions Maths

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

Edited By Lovekush kumar saini | Updated on Jan 20, 2022 07:14 PM IST

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RD Sharma Class 12 Solutions Chapter 12 FBQ Derivative as a Rate Measure - Other Exercise
Derivative as a Rate Measurer Excercise: FBQ
Derivatives as a rate measure exercise fill in the blanks question 5
RD Sharma Chapter-Wise Solutions:

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

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

Derivative as a Rate Measurer Excercise: FBQ

Derivatives as a rate measure exercise fill in the blanks question 1

Answer: $\frac{-12}{5}$
Hint: Here, we use the basic concept of $\frac{dy}{dx}$ and $\frac{dz}{dx}$
Given: $\sqrt{x^{2}+16}$ with respect to $\left [ \frac{x}{x-1} \right ]$ at $x=3$
Solution: Let $y=\sqrt{x^{2}+16}$
$z=\sqrt{\frac{x}{\left ( x-1 \right )}}$
$\frac{dy}{dx}=x\left ( x^{2}+16 \right )^{\frac{-1}{2}}$
$\frac{dz}{dx}=\left [ \frac{\left \{ \left ( x-1 \right ) -x\right \}}{\left \{ \left ( x-1 \right )^{2} \right \}} \right ]$
$\therefore \frac{dy}{dx}=x\left ( x^{2}+16 \right )^{\frac{-1}{2}}$ and $\left ( \frac{dz}{dx} \right )=\frac{-1}{\left ( x-1 \right )^{2}}$
Hence,
$\frac{dy}{dx}=\left [ \frac{\left ( \frac{dy}{dx} \right )}{\left ( \frac{dz}{dx} \right )} \right ]$
$\therefore \frac{dy}{dx}=\frac{\left [ x\left ( x^{2}+16 \right ) \right ]^{\frac{-1}{2}}}{-1/(x-1)^{2}}$
$\frac{dy}{dz}_{x=3}=\left [ 3\times \frac{1}{5} /\frac{-1}{4}\right ]=\left ( \frac{3}{-5} \right )\times 4$
$=\frac{-12}{5}$

Derivatives as a rate measure exercise fill in the blanks question 2

Answer: $r$
Hint: Here, we use the of surface area $S=4\pi r^{2}$
Given: $\frac{dr}{dt}=2\; cm/s$
Solution: $\frac{ds}{dt}=8\pi r\frac{dr}{dt}$
$\frac{ds}{dt}=8\pi r\; \times 2$
$\frac{ds}{dt}=16\pi r\;$
Rate of change of surface area $\infty$ radius of sphere.

Derivatives as a rate measure exercise fill in the blanks question 3

Answer: $10\sqrt{2}\; or\; 14.14cm^{2}/s$
Hint: Here, we use the length of diagonal $\frac{ds}{dt}=0.5\; cm/s$ and $l=\sqrt{2}a$
Given: $\frac{dl}{dt}=0.5\; cm/s$ ,Area = 400
Solution: $\frac{d\left ( \sqrt{2}a \right )}{dt}=0.5$
$\frac{da}{dt}=\frac{0.5}{\sqrt{2}}$
$a^{2}=400$
So, a = 20
$\frac{da^{2}}{dt}=2\times \frac{20\times 0.5}{\sqrt{2}}=> 2\times \frac{10}{\sqrt{2}}$
$=10\sqrt{2}cm^{2}/sec$

Derivatives as a rate measure exercise fill in the blanks question 4

Answer: $1\; cm^{^{3}}/cm^{2}$
Hint: Here, we use the volume of surface area of a cone $V=\frac{4}{3}\pi r^{3}$
Given: Radius is 2 cm
Solution: $V=\frac{4}{3}\pi r^{3}$
$\Rightarrow \frac{dV}{dr}=3\times \frac{4}{3}\pi r^{2}$
$S=4\pi r^{2}$
$\frac{dS}{dr}=2\times 4\pi r$
$=8\pi r$
$\frac{dV}{dS}=\frac{dV}{dr}\times \frac{dr}{dS}$
$=> \frac{r}{2}=1\; at\; r=2$
So, $1\; cm^{^{3}}/cm^{2}$ is the answer.

Derivatives as a rate measure exercise fill in the blanks question 5

Answer: $\theta =\frac{\pi }{3}$
Hint: Here, we use the concept of $\theta$ angle
Given: Let $\theta ,\frac{d\theta }{dt}=\frac{d\left ( \sin \theta \right )}{dt}$
Solution: Let $\theta$ increase twice as fast as it’s sine
$\frac{d\theta }{dt}=\frac{2d\left ( \sin \theta \right )}{dt}$
$\frac{d\theta }{dt}=2\cos \theta \frac{d\theta }{dt}$
$1=2\cos \theta$
$\cos \theta = \frac{1}{2}$
$\theta = \frac{\pi }{3}$

Derivatives as a rate measure exercise fill in the blanks question 6

Answer: $10\sqrt{3}cm^{2}/s$
Hint: Let the side of $cm$ equilateral triangle be $x\; cm$
Given: $\frac{dx}{dt}=2cm^{2}/s$
Area of equilateral triangle $A=\frac{\sqrt{3}}{4}x^{2}$ ....... $(i)$
Solution: Differentiating equation $\left ( i \right )$ with respect to $x$ we get
$\frac{dA}{dt}=\frac{\sqrt{3}}{4}\times 2x\times \frac{dx}{dt}$
$= \frac{\sqrt{3}x}{2}\frac{dx}{dt}$
When $x= 10$
$\frac{dA}{dt}=\frac{10\sqrt{3}}{2}\times 2=10\sqrt{3}\; cm^{2}/s$

Derivatives as a rate measure exercise fill in the blanks question 7

Answer: $\frac{1}{30\pi }ft/min$
Hint: Here we use the formula of volume, $V=\frac{4}{3}\pi r^{3}$
Given: $\frac{dV}{dt}=30\; cm^{3}/sec$
Solution: $V=\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt}=4\pi r^{2}\left ( \frac{dr}{dt} \right )$
$\frac{dr}{dt}= \left [ \frac{dV}{dt} \right ]/4\pi r^{2}$
$\frac{dr}{dt}= \left [ \frac{30}{\left [ 4\times \pi \times 15\times 15 \right ]} \right ]$
$= \frac{1}{30\pi }ft/min$

Derivatives as a rate measure exercise fill in the blanks question 8

Answer: $s$
Hint: Here we use the concept and formula of acceleration and velocity
Given: $S= ae^{t}+\left ( \frac{b}{e^{t}} \right )$
Solution: Acceleration $= d^{2}s/dt^{2}$
$S= ae^{t}+\left ( \frac{b}{e^{t}} \right )$
$\frac{dS}{dt}=ae^{t}+be^{-t}\left ( -1 \right )$
$=ae^{t}-be^{-t}$
$\frac{\mathrm{d} ^{2}S}{\mathrm{d} t^{2}}=ae^{t}+be^{-t}$
$a=ae^{t}+be^{-t}=s$
So, $a=s$

Derivatives as a rate measure exercise fill in the blanks question 9

Answer: $\frac{15}{13}cm/s$
Hint: Here we use the concept of acceleration and velocity
Given: $V= 5x-\left ( x^{2}/6 \right )$
Solution: So,
$\frac{dV}{dt}=5\times \frac{dx}{dt}-\frac{x}{3}\times \frac{dx}{dt}$
$\frac{dx}{dt}=\frac{\frac{dV}{dt}}{5-\left ( \frac{x}{3} \right )}$
When $x=2$ and $\frac{dV}{dt}=5cm^{3}/sec$
$\frac{dx}{dt}=\frac{5}{5-\frac{2}{3}}=\frac{15}{13}cm/s$

Derivatives as a rate measure exercise fill in the blanks question 10

Answer: $\frac{2}{\pi }ft/minute$
Hint: Here we use the formula of volume of vertical cylindrical tank
Given: Radius=2 ft as the r of 8 cubic ft/min
Solution: $V=\pi r^{2}h$
$\frac{dV}{dt}=4\pi \left ( \frac{dh}{dt} \right )$
$\left ( \frac{dh}{dt} \right )= \left ( \frac{1}{4\pi } \right )\left ( \frac{dV}{dt} \right )$
$\frac{dV}{dt}=\left ( \frac{1}{4\pi } \right )\times 8$
$=\frac{2}{\pi }ft/minute$

Derivatives as a rate measure exercise fill in the blanks question 11

Answer: $\frac{dC}{dt}\pi \; cm/sec$
Hint: Here we use the circumference of circle, $C=2\pi r$
Given: The radius of a circle is increasing at the rate of 0.5 cm/sec
Solution: $C=2\pi r$
$\frac{dC}{dt}=2\pi \frac{dr}{dt}$
$= 2\pi \times 0.5$ $\left [ \because \frac{dr}{dt}=0.5 \right ]$
$\frac{dC}{dt}=\pi \; cm/sec$

Derivatives as a rate measure exercise fill in the blanks question 12

Answer: $\frac{dA}{dt}=12\pi \; cm^{2}/sec$
Hint: Here we use the formula of area of circle (A) with radius r is given by
Given: $A=\pi r^{2}$
Solution: $\frac{dA}{dt}=\frac{d\left ( \pi r^{2} \right )}{dt}\times \frac{dr}{dt}=2\pi r\frac{dr}{dt}$ .......[by chain rule]
It is given that
$\frac{dr}{dt}=3\; cm/s$
$\therefore \frac{dA}{dt}=2\pi r(3)=6\pi r$
Thus, when r=2 cm
$\therefore \frac{dA}{dt}=6\pi(2)=12\pi \; cm^{2}/s$

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RD Sharma Class 12th exercise FBQ has 12 questions and It includes the following concepts

Definition of derivative as a rate measurer.
Rate of increase of the perimeter, circumference of any figure
Rate of Change in Marginal cost and Marginal Revenue.
Application based question on derivative as a rate measurer

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RD Sharma Chapter-Wise Solutions:

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