Application of Derivatives Question Answers: NCERT Class 12 Mathematics

The area of a circle (A) with radius (r) is given by,

A = πr²

Now, the rate of change of the area with respect to its radius is given by,

\begin{align} \frac{dA}{dr} = \frac{d}{dr}(πr^2) = 2πr  \end{align}

1. When r = 3 cm,

\begin{align} \frac{dA}{dr} = 2π (3) = 6π  \end{align}

Hence, the area of the circle is changing at the rate of 6π cm²/s when its radius is 3 cm.

2. When r = 4 cm,

             \begin{align} \frac{dA}{dr} = 2π (4) = 8π \end{align}

Hence, the area of the circle is changing at the rate of 8π cm²/s when its radius is 4 cm.

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x maway from the wall.

Then, by Pythagoras theorem, we have:

x² + y² = 25 [Length of the ladder = 5 m]

\begin{align}\Rightarrow y = \sqrt{25 - x^2}\end{align}

Then, the rate of change of height (y) with respect to time (t) is given by,

\begin{align}\frac{dy}{dx} = \frac{-x}{\sqrt{25 - x^2}}.\frac{dx}{dt}\end{align}

It is given that

\begin{align}\frac{dx}{dt}= 2 \; cm/s.\end{align}

\begin{align}\therefore\frac{dy}{dt} = \frac{-2x}{\sqrt{25 - x^2}}\end{align}

Now, when x = 4 m, we have:

\begin{align}\therefore\frac{dy}{dt} = \frac{-2 \times 4}{\sqrt{25 - 4^2}}=-\frac{8}{3}\end{align}

Hence, the height of the ladder on the wall is decreasing at the rate of

\begin{align}\frac{8}{3}\end{align}

The equation of the curve is given as:

6y = x³ + 2

The rate of change of the position of the particle with respect to time (t) is given by,

\begin{align}6\frac{dy}{dt} = 3x^2\frac{dx}{dt}+0\end{align}

\begin{align}\Rightarrow 2\frac{dy}{dt} = x^2\frac{dx}{dt}\end{align}

When the y-coordinate of the particle changes 8 times as fast as the

\begin{align}x-coordinate\; i.e.,\left(\frac{dy}{dt} = 8\frac{dx}{dt}\right),  we \;have:\end{align}

\begin{align}2\left(8.\frac{dx}{dt}\right) = x^2.\frac{dx}{dt}\end{align}

\begin{align}\Rightarrow 16.\frac{dx}{dt} = x^2.\frac{dx}{dt}\end{align}

\begin{align}\Rightarrow (x^2 - 16).\frac{dx}{dt} =0 \end{align}

\begin{align}\Rightarrow x^2=16 \end{align}

\begin{align}\Rightarrow x=\pm 4 \end{align}

\begin{align}When\; x = 4, y = \frac{4^3 + 2}{6}=\frac{66}{6}=11\end{align}

\begin{align}When\; x = -4, y = \frac{(-4)^3 + 2}{6}=\frac{-62}{6}=\frac{-31}{3}\end{align}

Hence, the points required on the curve are 

\begin{align} (4,11)\; and \;(-4,\frac{-31}{3}).\end{align}

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble (V) with radius (r) is given by,

\begin{align} V = \frac{4}{3}\pi r^3 \end{align}

The rate of change of volume (V) with respect to time (t) is given by,

\begin{align} \frac{dV}{dt} = \frac{4}{3}\pi \frac{d}{dr}(r^3).\frac{dr}{dt} \;\;\;[By\; Chain\; Rule] \end{align}

\begin{align} = \frac{4}{3}\pi (3r^2).\frac{dr}{dt} \end{align}

\begin{align} = \frac{4}{3}\pi r^2.\frac{dr}{dt} \end{align}

It is given that

\begin{align} \frac{dr}{dt}=\frac{1}{2} cm/s .\end{align}

Therefore, when r = 1 cm,

\begin{align} \frac{dV}{dt}=4\pi(1)^2.(\frac{1}{2})=2\pi\; cm^3/s \end{align}

Hence, the rate at which the volume of the bubble increases is 2π cm³/s.

The volume of a sphere (V) with radius (r) is given by,

\begin{align} V=\frac{4}{3}\pi r^3 \end{align}

It is given that:

\begin{align} Diameter =\frac{3}{2}(2x+1) \end{align}

\begin{align} \Rightarrow r =\frac{3}{4}(2x+1) \end{align}

\begin{align} \therefore V =\frac{4}{3}\pi(\frac{3}{4})^3(2x+1)^3=\frac{9}{16}\pi\times(2x+1)^3 \end{align}

Hence, the rate of change of volume with respect to x is as

\begin{align} \frac{dV}{dx}=\frac{9}{16}\pi\frac{d}{dx}(2x+1)^3=\frac{9}{16}\pi\times3(2x+1)^2 \times2=\frac{27}{8}\pi(2x+1)^2\end{align}

The volume of a cone (V) with radius (r) and height (h) is given by,

\begin{align}V=\frac{1}{3}\pi r^2h\end{align}

It is given that,

\begin{align}h=\frac{1}{6} r\Rightarrow r =6h\end{align}

\begin{align}\therefore V=\frac{1}{3}\pi (6h)^2.h = 12\pi h^3\end{align}

The rate of change of volume with respect to time (t) is given by,

\begin{align} \frac{dV}{dt}=12 \pi \frac{d}{dh}(h^3).\frac{dh}{dt}[By\; Chain\; Rule]\end{align}

\begin{align}=12 \pi (3h^2).\frac{dh}{dt}\end{align}

\begin{align}=36 \pi h^2.\frac{dh}{dt}\end{align}

It is also given that

\begin{align}\frac{dV}{dt}=12\;cm^3/s \end{align}

Therefore, when h = 4 cm, we have:

\begin{align}12=36\pi (4)^2.\frac{dh}{dt}\end{align}

\begin{align}\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi (16)}=\frac{1}{48\pi}\end{align}

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of

\begin{align}\frac{1}{48\pi}.\end{align}

Marginal cost is the rate of change of total cost with respect to output.

∴Marginal cost 

\begin{align}MC=\frac{dC}{dx}=0.007 (3x^2) - 0.003(2x) + 15\end{align}

MC = 0.021 x² - 0.006x + 15

When x = 17, MC = 0.021 (17²) − 0.006 (17) + 15

= 0.021(289) − 0.006(17) + 15

= 6.069 − 0.102 + 15

= 20.967

Hence, when 17 units are produced, the marginal cost is Rs. 20.967.

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

\begin{align}MR=\frac{dR}{dx}\end{align}

∴Marginal Revenue

MR = 13(2x) + 26 = 26x + 26

When x = 7,

MR = 26(7) + 26 = 182 + 26 = 208

Hence, the required marginal revenue is Rs 208.

The area of a circle (A) with radius (r) is given by,

A = πr²

Therefore, the rate of change of the area with respect to its radius r is

\begin{align}\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r\end{align}

∴When r = 6 cm,

\begin{align}\frac{dA}{dr} = 2\pi \times 6 =12 \pi\; cm^2/s\end{align}

Hence, the required rate of change of the area of a circle is 12π cm²/s.

The correct answer is B.

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

\begin{align}MR=\frac{dR}{dx} \end{align}

∴Marginal Revenue (MR)= 3(2x) + 36 = 6x + 36

∴When x = 15,

MR = 6(15) + 36 = 90 + 36 = 126

Hence, the required marginal revenue is Rs 126.

The correct answer is D.

Let x be the length of a side, V be the volume, and s be the surface area of the cube.

Then, V = x³ and S = 6x² where x is a function of time t.

It is given that

\begin{align} \frac{dV}{dt} = 8 cm^3 / s \end{align}

Then, by using the chain rule, we have:

∴ \begin{align} \frac{dV}{dt} = \frac{d}{dt} (x^3) . \frac{dx}{dt} = 3x^2 . \frac{dx}{dt} =8 \end{align}

⇒ \begin{align} \frac{dx}{dt} = \frac{8}{3 x^2} ……… (1) \end{align}

Now \begin{align} \frac{dS}{dt} = \frac{d}{dx}(6x^2).\frac{dx}{dt} [By Chain Rule] \end{align}

                    \begin{align} =12x .\frac{dx}{dt} =12x.(\frac{8}{3x^2}) = \frac{32}{x}  \end{align}

Thus, when x = 12 cm, \begin{align} \frac{dS}{dt} = \frac{32}{12} cm^2 / s = 8 cm^2 / s \end{align}

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \begin{align} \frac{8}{3} cm^2 / s \end{align}.

The area of a circle (A) with radius (r) is given by,

A = πr²

Now, the rate of change of area (A) with respect to time (t) is given by,

\begin{align} \frac{dA}{dt}=\frac{d}{dt}(\pi r^2).\frac{dr}{dt}=2\pi r\frac{dr}{dt}\;\;\;[By\; Chain \;Rule]\end{align}

It is given that,

\begin{align} \frac{dr}{dt}= 3\; cm/s\end{align}

\begin{align} \therefore \frac{dA}{dt}= 2\pi r(3)=6 \pi r \end{align}

Thus, when r = 10 cm,

\begin{align} \frac{dA}{dt}= 6\pi(10)=60 \pi\; cm^2/s \end{align}

Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm²/s.

Let x be the length of a side and V be the volume of the cube. Then,

V = x^3.

\begin{align}\therefore \frac{dV}{dt}=3x^2.\frac{dx}{dt}\;\;\;[By\; Chain \;Rule]\end{align}

It is given that,

\begin{align} \frac{dx}{dt}=3 \;cm^2/s\end{align}

\begin{align}\therefore \frac{dV}{dt}=3x^2.(3) = 9x^2\end{align}

Thus, when x = 10 cm,

\begin{align} \frac{dV}{dt}=9 (10)^2=900 \;cm^3/s\end{align}

Hence, the volume of the cube is increasing at the rate of 900 cm³/s when the edge is 10 cm long. 

The area of a circle (A) with radius (r) is given by A = πr².

Therefore, the rate of change of area (A) with respect to time (t) is given by,

\begin{align} \frac{dA}{dt}=\frac{d}{dt}(\pi r^2)=\frac{d}{dt}(\pi r^2).\frac{dr}{dt}=2\pi r\frac{dr}{dt}\;\;\;[By\; Chain \;Rule]\end{align}

It is given that

\begin{align} \frac{dr}{dt}=5\; cm/s\end{align}

Thus, when r = 8 cm,

\begin{align} \frac{dA}{dt}=2\pi(8)(5)=80\pi\end{align}

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm²/s.

The circumference of a circle (C) with radius (r) is given by

C = 2πr.

Therefore, the rate of change of circumference (C) with respect to time (t) is given by,

\begin{align} \frac{dC}{dt}=\frac{dC}{dr}.\frac{dr}{dt}\;\;\; [By\; Chain\; Rule]\end{align}

\begin{align} =\frac{d}{dr}(2\pi r).\frac{dr}{dt}\end{align}
\begin{align} =2\pi.\frac{dr}{dt}\end{align}

It is given that

\begin{align} \frac{dr}{dt}=0.7\; cm/s\end{align}

Hence, the rate of increase of the circumference 2π(0.7)=1.4π cm/s

Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have:

\begin{align} \frac{dx}{dt} = -5 \;cm/min\; and \; \frac{dy}{dt}= 4 \;cm/min\end{align}

(a) The perimeter (P) of a rectangle is given by,

P = 2(x + y)

\begin{align} \therefore\frac{dp}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right)= 2(-5+4)=-2\;cm/min\end{align}

Hence, the perimeter is decreasing at the rate of 2 cm/min.

(b) The area (A) of a rectangle is given by,

A = x⋅ y

\begin{align} \therefore\frac{dA}{dt} = \frac{dx}{dt}.y + x.\frac{dy}{dt}=-5y + 4x \end{align}

When x = 8 cm and y = 6 cm, 

\begin{align} \frac{dA}{dt} = (-5 \times 6 + 4 \times 8)\; cm^2/min = 2\; cm^2/min\end{align}

Hence, the area of the rectangle is increasing at the rate of 2 cm²/min. 

The volume of a sphere (V) with radius (r) is given by,

\begin{align} V=\frac{4}{3}\pi r^2\end{align}

∴Rate of change of volume (V) with respect to time (t) is given by,

\begin{align} \frac{dV}{dt} =\frac{dV}{dr}.\frac{dr}{dt}\;\;\;[By\; Chain\; Rule]\end{align}

\begin{align} =\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right).\frac{dr}{dt}\end{align}

\begin{align} =4\pi r^2.\frac{dr}{dt}\end{align}

It is given that

\begin{align} \frac{dV}{dt}=900\; cm^3/s\end{align}

\begin{align} \therefore 900=4\pi r^2.\frac{dr}{dt}\end{align}

\begin{align} \Rightarrow \frac{dr}{dt}=\frac{900}{4\pi r^2}=\frac{225}{\pi r^2}\end{align}

Therefore, when radius = 15 cm,

\begin{align} \frac{dr}{dt}=\frac{225}{\pi (15)^2}=\frac{1}{\pi }\end{align}

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is 

\begin{align} \frac{1}{\pi }\end{align}

The volume of a sphere (V) with radius (r) is given by

\begin{align} V=\frac{4}{3 }\pi r^3\end{align}

Rate of change of volume (V) with respect to its radius (r) is given by,

\begin{align} \frac{dV}{dr }=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi \left(3r^2\right)=4\pi r^2\end{align}

Therefore, when radius = 10 cm,

\begin{align} \frac{dV}{dr }=4\pi(10)^2=400\pi\end{align}

Hence, the volume of the balloon is increasing at the rate of 400π cm³/s.