Exercise-6.1, Class 12th, Maths, Chapter 6, NCERT

Question 1

Find the rate of change of the area of a circle with respect to its radius $r$ when
(a) $r=3$
(b) $r=4$

Answer

Area of a circle:

$$A=\pi r^2$$

Differentiate w.r.t. $r$:

$$\frac{dA}{dr}=2\pi r$$

(a) When $r=3$:

$$\frac{dA}{dr}=2\pi (3)=6\pi {\text{cm}}^2\mathrm{/}\text{cm}$$

(b) When $r=4$:

$$\frac{dA}{dr}=2\pi (4)=8\pi {\text{cm}}^2\mathrm{/}\text{cm}$$

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Question 2

The volume of a cube is increasing at the rate of $8{\text{ cm}}^3\mathrm{/}\text{s}$.
How fast is the surface area increasing when the edge is $12$ cm?

Answer

Let edge = $x$

Volume:

$$V=x^3$$
$$\frac{dV}{dt}=8$$

Differentiate:

$$\frac{dV}{dt}=3x^2\frac{dx}{dt}$$
$$8=3x^2\frac{dx}{dt}$$
$$\frac{dx}{dt}=\frac{8}{3(12{)}^2}=\frac{8}{432}=\frac{1}{54}\text{ cm/s}$$

Surface Area:

$$S=6x^2$$

Differentiate:

$$\frac{dS}{dt}=12x\frac{dx}{dt}$$

Substitute $x=12$ and $\frac{dx}{dt}=\frac{1}{54}$:

$$\frac{dS}{dt}=12(12)\left(\frac{1}{54}\right)=\frac{144}{54}=\frac{8}{3}{\text{ cm}}^2\mathrm{/}\text{s}$$

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Question 3

The radius of a circle is increasing uniformly at the rate of $3$ cm/s.
Find the rate at which the area is increasing when the radius is $10$ cm.

Answer

$$A=\pi r^2$$
$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$

Given:

$$\frac{dr}{dt}=3,r=10$$
$$\frac{dA}{dt}=2\pi (10)(3)=60\pi {\text{ cm}}^2\mathrm{/}\text{s}$$

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Question 4

An edge of a variable cube is increasing at $3$ cm/s.
How fast is the volume increasing when the edge is $10$ cm?

Answer

$$V=x^3$$

Differentiate:

$$\frac{dV}{dt}=3x^2\frac{dx}{dt}$$
$$\frac{dV}{dt}=3(10{)}^2(3)=900{\text{ cm}}^3\mathrm{/}\text{s}$$

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Question 5

A stone is dropped into a lake and waves move in circles at 5 cm/s.
When radius is 8 cm, how fast is the enclosed area increasing?

Answer

$$A=\pi r^2$$
$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$

Given:

$$\frac{dr}{dt}=5,r=8$$
$$\frac{dA}{dt}=2\pi (8)(5)=80\pi {\text{ cm}}^2\mathrm{/}\text{s}$$

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Question 6

The radius of a circle is increasing at the rate of 0.7 cm/s.
What is the rate of increase of its circumference?

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Answer

Let radius be $r$

Circumference of a circle:

$$C=2\pi r$$

Differentiate w.r.t. time $t$:

$$\frac{dC}{dt}=2\pi \frac{dr}{dt}$$

Given:

$$\frac{dr}{dt}=0.7\text{ cm/s}$$

Substitute:

$$\frac{dC}{dt}=2\pi (0.7)$$
$$\frac{dC}{dt}=1.4\pi \text{ cm/s}$$

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Question 7

The length $x$ of a rectangle is decreasing at the rate of 5 cm/min and the width $y$ is increasing at the rate of 4 cm/min.
When $x=8$ and $y=6$, find the rates of change of
(a) the perimeter, and (b) the area of the rectangle.

Given

$$\frac{dx}{dt}=-5\text{ cm/min}(\text{decreasing})$$
$$\frac{dy}{dt}=+4\text{ cm/min}(\text{increasing})$$

(a) Rate of change of Perimeter

Perimeter of rectangle:

$$P=2(x+y)$$

Differentiate w.r.t. time $t$:

$$\frac{dP}{dt}=2(\frac{dx}{dt}+\frac{dy}{dt})$$

Substitute values:

$$\frac{dP}{dt}=2(-5+4)=2(-1)=-2\text{ cm/min}$$

Answer (a)

$$\boxed{{\displaystyle \frac{dP}{dt}=-2\text{ cm/min}}}$$

So, the perimeter is decreasing at 2 cm/min.

(b) Rate of change of Area

Area:$$A=x\cdot y$$

Differentiate:

$$\frac{dA}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$

Substitute values: $x=8,y=6$

$$\frac{dA}{dt}=8(-5)+6(4)$$
$$\frac{dA}{dt}=-40+24=-16{\text{ cm}}^2\mathrm{/}\text{min}$$

Answer (b)

$$\boxed{{\displaystyle \frac{dA}{dt}=-16{\text{ cm}}^2\mathrm{/}\text{min}}}$$

So, the area is decreasing at 16 cm²/min.

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Question 8

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer

Let the radius of the balloon = $r$

Volume of a sphere:

$$V=\frac{4}{3}\pi r^3$$

Differentiate w.r.t. time $t$:

$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$

Given:

$$\frac{dV}{dt}=900{\text{ cm}}^3\mathrm{/}\text{s},r=15\text{ cm}$$

Substitute in the formula:

$$900=4\pi (15{)}^2\frac{dr}{dt}$$
$$900=4\pi \cdot 225\frac{dr}{dt}$$
$$900=900\pi \frac{dr}{dt}$$
$$\frac{dr}{dt}=\frac{900}{900\pi }=\frac{1}{\pi }\text{ cm/s}$$

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Question 9

A balloon, which always remains spherical, has a variable radius.
Find the rate at which its volume is increasing with respect to the radius when the radius is 10 cm.

Answer

Let the radius of the balloon = $r$

Volume of a sphere:

$$V=\frac{4}{3}\pi r^3$$

We need:

$$\frac{dV}{dr}$$Differentiate with respect to $r$:

$$\frac{dV}{dr}=4\pi r^2$$

Now substitute $r=10\text{ cm}$:

$$\frac{dV}{dr}=4\pi (10{)}^2=4\pi \cdot 100=400\pi {\text{ cm}}^3\mathrm{/}\text{cm}$$

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Question 10

A ladder 5 m long is leaning against a wall.
The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s.
How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer

Let:

• $x$ = distance of the bottom of the ladder from the wall (ground level)

• $y$ = height of the ladder on the wall

• Length of ladder = $5$ m (constant)

From the geometry of the right triangle:

$$x^2+y^2=5^2$$

$$x^2+y^2=25\text{(1)}$$

Differentiate w.r.t. time $t$:

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$

Divide by 2:

$$x\frac{dx}{dt}+y\frac{dy}{dt}=0$$

Given:

$$\frac{dx}{dt}=2\text{ cm/s}=0.02\text{ m/s}$$

When $x=4$, from (1):

$$4^2+y^2=25$$

$$16+y^2=25$$
$$y^2=9\Rightarrow y=3\text{ m}$$

Now substitute into differentiation equation:

$$4(0.02)+3\frac{dy}{dt}=0$$

$$0.08+3\frac{dy}{dt}=0$$

$$3\frac{dy}{dt}=-0.08$$
$$\frac{dy}{dt}=-\frac{0.08}{3}=-0.0267\text{ m/s}$$

Convert to cm/s:

$$-0.0267\times 100=-2.67\text{ cm/s}$$

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Question 11

A particle moves along the curve

$$6y=x^3+2$$

Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer

Given:

$$6y=x^3+2$$

Differentiate both sides with respect to $t$ (time):

$$6\frac{dy}{dt}=3x^2\frac{dx}{dt}$$

Divide both sides by 3:

$$2\frac{dy}{dt}=x^2\frac{dx}{dt}$$

Now divide both sides by $\frac{dx}{dt}$ (assuming $\frac{dx}{dt}\mathrm{\ne }0$):

$$2\frac{dy}{dt}/\frac{dx}{dt}=x^2$$

$$2\frac{dy}{dx}=x^2$$

Given condition:

$$\frac{dy}{dx}=8$$

Substitute:

$$2(8)=x^2$$
$$16=x^2$$
$$x=\pm 4$$

Find corresponding $y$-values

Use original equation:

$$6y=x^3+2$$

For $x=4$:

$$6y=(4{)}^3+2=64+2=66$$

$$y=\frac{66}{6}=11$$

For $x=-4$:

$$6y=(-4{)}^3+2=-64+2=-62$$

$$y=\frac{-62}{6}=-\frac{31}{3}$$

Final Answer

$$\boxed{{\displaystyle \text{The required points on the curve are }(4,11)\text{ and }(-4,-\frac{31}{3})\mathrm{.}}}$$

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Question 12

The radius of an air bubble is increasing at the rate of

$$\frac{dr}{dt}=\frac{1}{2}\text{ cm/s}$$

At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer

Let $r$ = radius of the spherical bubble.

Volume of a sphere:

$$V=\frac{4}{3}\pi r^3$$

Differentiate with respect to time $t$:

$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$

Given:

$$\frac{dr}{dt}=\frac{1}{2},r=1\text{ cm}$$

Substitute values:

$$\frac{dV}{dt}=4\pi (1{)}^2\left(\frac{1}{2}\right)$$

$$\frac{dV}{dt}=4\pi \cdot \frac{1}{2}=2\pi {\text{ cm}}^3\mathrm{/}\text{s}$$

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Question 13

A balloon, which always remains spherical, has a variable diameter

$$D=\frac{3}{2}(2x+1)$$

Find the rate of change of its volume with respect to $x$.

Answer

Diameter:

$$D=\frac{3}{2}(2x+1)$$

Radius is:

$$r=\frac{D}{2}=\frac{1}{2}\cdot \frac{3}{2}(2x+1)=\frac{3}{4}(2x+1)$$

Volume of a sphere:

$$V=\frac{4}{3}\pi r^3$$

Substitute $r=\frac{3}{4}(2x+1)$:

$$V=\frac{4}{3}\pi {[\frac{3}{4}(2x+1)]}^3$$

$$V=\frac{4}{3}\pi \cdot \frac{27}{64}(2x+1{)}^3$$

$$V=\frac{108\pi }{192}(2x+1{)}^3$$

$$V=\frac{9\pi }{16}(2x+1{)}^3$$

Differentiate with respect to $x$

$$\frac{dV}{dx}=\frac{9\pi }{16}\cdot 3(2x+1{)}^2\cdot \frac{d}{dx}(2x+1)$$

$$\frac{d}{dx}(2x+1)=2$$

So,

$$\frac{dV}{dx}=\frac{9\pi }{16}\cdot 3\cdot 2(2x+1{)}^2$$

$$\frac{dV}{dx}=\frac{54\pi }{16}(2x+1{)}^2$$

$$\frac{dV}{dx}=\frac{27\pi }{8}(2x+1{)}^2$$

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Question 14

Sand is pouring from a pipe at the rate of 12 cm³/s.
The sand forms a cone such that the height is always one-sixth of the radius.
How fast is the height of the cone increasing when the height is 4 cm?

Answer

Let:

• $V$ = volume of the cone

• $r$ = radius of base

• $h$ = height of the cone

Given:

$$\frac{dV}{dt}=12{\text{cm}}^3\mathrm{/}\text{s}$$

Also,

$$h=\frac{1}{6}r\Rightarrow r=6h$$

Volume of cone

$$V=\frac{1}{3}\pi r^{2}h$$

Substitute $r=6h$:

$$V=\frac{1}{3}\pi (6h{)}^{2}h$$

$$V=\frac{1}{3}\pi (36h^2)h$$

$$V=12\pi h^3$$

Differentiate w.r.t. time $t$

$$\frac{dV}{dt}=36\pi h^2\frac{dh}{dt}$$

Given $\frac{dV}{dt}=12$ and $h=4$:

$$12=36\pi (4{)}^2\frac{dh}{dt}$$

$$12=36\pi \cdot 16\cdot \frac{dh}{dt}$$

$$12=576\pi \frac{dh}{dt}$$

$$\frac{dh}{dt}=\frac{12}{576\pi }$$

$$\frac{dh}{dt}=\frac{1}{48\pi }\text{ cm/s}$$

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Question 15

The total cost $C(x)$ in Rupees associated with the production of $x$ units of an item is given by:

$$C(x)=0.007x^3-0.003x^2+15x+4000$$

Find the marginal cost when 17 units are produced.
(Marginal cost means the instantaneous rate of change of total cost, i.e., $\frac{dC}{dx}$)

Answer

Given:

$$C(x)=0.007x^3-0.003x^2+15x+4000$$

Differentiate with respect to $x$:

$$\frac{dC}{dx}=0.021x^2-0.006x+15$$

We need the marginal cost at $x=17$:

$$MC=0.021(17{)}^2-0.006(17)+15$$

Calculate step-by-step:

$${17}^2=289$$
$$0.021\times 289=6.069$$
$$0.006\times 17=0.102$$

Now substitute:

$$MC=6.069-0.102+15$$

$$MC=20.967$$

Final Answer

$$\boxed{{\displaystyle \text{The marginal cost when 17 units are produced is }\approx \text{ ₹ }20.97}}$$

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Question 16

The total revenue in Rupees received from the sale of $x$ units of a product is:

$$R(x)=13x^2+26x+15$$

We need to find the marginal revenue when $x=7$.
(Marginal revenue = derivative of revenue w.r.t. $x$)

Solution

Differentiate:

$$\frac{dR}{dx}=26x+26$$

Substitute $x=7$:

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

Final Answer

So, the marginal revenue when 7 units are sold is ₹ 208.

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Choose the correct answer for questions 17 and 18.

Question 17

The rate of change of the area of a circle with respect to its radius $r$ at $r=6$ is:

Differentiate w.r.t. $r$:

Substitute $r=6$:

Correct Option

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Question 18

The total revenue received from the sale of $x$ units of a product is:

$$R(x)=3x^2+36x+5$$

We must find the marginal revenue when $x=15$.
(Marginal revenue = $\frac{dR}{dx}$)

Solution

Differentiate:

$$\frac{dR}{dx}=6x+36$$

Now substitute $x=15$:

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

Final Answer

$$\boxed{{\displaystyle 126}}$$

Correct Option

$$\boxed{{\displaystyle (D)126}}$$