NCERT Solution Class – 12th Math Chapter – 6 Application of Derivatives Exercise 6.1

June 18, 2022

NCERT Solution Class – 12th Math Chapter – 6 Application of Derivatives Exercise 6.1

Textbook	NCERT
class	Class – 12th
Subject	Mathematics
Chapter	Chapter – 16
Chapter Name	Application of Derivatives
grade	Class 12th Maths solution
Medium	English
Source	last doubt

NCERT Solution Class – 12th Math Chapter – 6 Application of Derivatives Exercise 6.1

?Chapter – 16?

✍ Application of Derivatives✍

?Exercise 16.1?

1. Find the rate of change of the area of a circle with respect to its radius r when r = 3 cm

?‍♂️solution – (i)

The area of a circle

(A)

with radius

(r)

is given by,

A = π r^{2}

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

d A/dr = d/ d r (π r^{2}) = 2 π r

r = 3

cm,

d A/dr = 2 π (3) = 6 π

(ii)

The area of a circle $(A)$ with radius $(r)$ is given by,
$A = π r^{2}$
Now, the rate of change of the area with respect to its radius is given by,
$d A/dr = d / d r (π r^{2}) = 2 π r$
At $r = 4$ cm,
$d A = 2 π (4) = 8 π$

2. The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

?‍♂️solution – Let $x$ be the length of a side, $V$ be the volume, and $S$ be the surface area of the cube.

Then,

V = x^{3}

and

S = 6 x^{2}

It is given that

d V/dt = 8 c m^{3} / s

Then, by using the chain rule, we have –

∴ 8 = d V = d d t (x^{3}) \cdot d/ d x = 3 x^{2} \cdot d x/dt

\Rightarrow d x/dt = 8/ 3 x ^{2} . . . . . . . . . (1)

Now,

d S/dt = d / d t (6 x^{2}) \cdot d / d x = (12 x) \cdot d x

[By chain rule]

= 12 x \cdot d x = 12 x \cdot (8/ 3 x ^{2}) = x 32

Thus, when

x = 12

cm,

d S = 32/ 12 c m^{2} / s = 38 c m^{2} / s .

Hence, if the length of the edge of the cube is

12

cm, then the surface area is increasing at the rate of

38 c m^{2} / s

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

?‍♂️solution – The area of a circle $(A)$ with radius $(r)$ is given by,
$A = π r^{2}$
$d A = d / d t (π r^{2}) \cdot d t d r = 2 π r, d r$ , [By chain rule]
It is given that,
$d r = 3 c m / s$ .
$∴ d A/dt = 2 π r (3) = 6 π r$
Thus, when $r = 10 c m$ ,
$∴ d A/dt = 6 π (10) = 60 π c m^{2} / s$ .

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

?‍♂️solution – Let at any time $^{'} t^{'}$ length of edge of cube is $x$ and its volume is V, then

V = x^{3}

…..(i)

and this edge is increasing at the rate of 3 cm/s.

∴ d x = 3 c m / s

We have to find, rate of change of volume

V

When

x = 10 c m

∴ d V = 3 x^{2} d x/dt

\Rightarrow d V/dt = 3 x^{2} . 3 = 9 x^{2}

When

x = 10 c m, d V/dt = 9 (10)^{2} = 900 c m^{3} / s

Hence, Rate of change of Volume is

900 c m^{3} / s e c

5. A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

?‍♂️solution – Correct option is B)

The area of a circle

A

with radius

r

is given by

A = π r^{2}

Therefore, the rate of change of area

(A)

with respect to time

(t)

is given by,

d A/dt = d/ d t (π r^{2}) = d/ d t (π r^{2}) . d r = 2 π r, dr/ d t [B y C h a i n R u l e]

It is given that:

d r = 5 c m

Thus, when

r = 8 c m

\Rightarrow d A = 2 π (8) (5) = 80 π

Hence, when the radius of the circular wave is

8

cm, the enclosed area is increasing at the rate of

80 π

cm^{$^{2}$}/s.

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?

?‍♂️solution – Radius of circle is increasing at the rate of $0.7 c m / s$

i.e.

d R/dt = 0.7 c m / s

Circumference of circle

= 2 π R

t h e r e f o r e

Rate of change of Circumference,

C

\Rightarrow d c/dt = 2 π, dR/ d t

\Rightarrow d c/dt = 2 π (0.7) = 1.4 π

\Rightarrow d c/dt = 14 10 \times 7/22 = 4.4 c m / s

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

?‍♂️solution – (i)

It is given that length $(x)$ is decreasing at the rate of $5$ cm/min
and the width $(y)$ is increasing at the rate of $4$ cm/min,
$\Rightarrow d x = - 5$ cm/min and $d y/dt = 4$ cm/min
Thus the perimeter $(P)$ of a rectangle is, $P = 2 (x + y)$
$d P/dt = 2 (d x + d y/dt) = 2 (- 5 + 4) = - 2$ cm/min
Hence, the perimeter is decreasing at the rate of $2$ cm/min.

(ii)

It is given that length

(x)

is decreasing at the rate of

5

cm/min
and the width

(y)

is increasing at the rate of

4

cm/min,

\Rightarrow d x = - 5

cm/min and

d y/dt = 4

cm/min
Thus the area

(A)

of a rectangle is,

A = x y

∴ (d A/dt)_{x = 8, y = 6} = (y, d x/dt + x, d y)_{x = 8, y = 6} = 6 (- 5) + 8 (4) = 2

cm^{$^{2}$}/min
Hence, the area is increasing at the rate of

2

cm^{$^{2}$}/min.

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.

?‍♂️solution – The volume of a sphere $(V)$ with radius $(r)$ is given by,
$V = 4/ 3 π r^{3}$ $∴$ Rate of change of volume $(V)$ with respect to time $(t)$ is given by,
$d V/dt = d V/dr \cdot d r/dt = 4 π r^{2} \cdot d r/dt$
It is given that $d V = 900 c m^{3} / s$ .
$∴ 900 = 4 π r^{2} \cdot d t d r$
$\Rightarrow d r =^{2} 900/ 4 π r = 225/ π r ^{2}$
Therefore, when radius $= 15$ cm,
$d r = 225/ π ( 15 ) ^{2} = 1/ π$ cm/s

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

?‍♂️solution – The volume of a sphere $(V)$ with radius $(r)$ is given by $V = 4/ 3 π r^{3}$ .
Rate of change of volume $(V)$ with respect to its radius $(r)$ is given by,
$d V/dr = d/dr (4/ 3 π r^{3}) = 4/ 3 π (3 π r^{2}) = 4 π r^{2}$
Therefore, when radius $= 10 c m$ ,
$d V/ = 4 π (10)^{2} = 400 π c m^{2}$

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?

?‍♂️solution – Let $y m$ be the height of the wall at which the ladder touches. Also, let the foot of the ladder be $x m$ away from the wall.
Then, by Pythagoras theorem, we have –
$x^{2} + y^{2} = 25$ [Length of the ladder $= 5 m$ ]
$\Rightarrow y = \sqrt 25 - x^{2}$
Then, the rate of change of height $(y)$ with respect to time $(t)$ is given by,
$d y = - x,\sqrt 25 - x ^{2} \cdot d x/dt$
It is given that $d t d x = 2 c m / s$
$∴ d y/dt = -2 x, \sqrt 25 - x ^{2}$
Now, when $x = 4 m$ , we have –
$d y/dt = -2 \times 4 25 - 4 ^{2} = - 38$
Hence, the height of the ladder on the wall is decreasing at the rate of $38$ cm/s.

11. A particle moves along the curve 6y = x³ +2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

?‍♂️solution - Step 1 - Differentiating given curve to find rate of change of y w.r.t t .

$The equation of the curve is given as,$ $6 y = x^{3} + 2 . . . . . . . . . . . (1)$
$Differentiating (1) both sides w.r.t. t$
$\Rightarrow 6 d y/dt = 3 x^{2} d x/dt + 0$
$\Rightarrow 2 d y = x^{2} d x/dt . . . . . (2)$

Step 2 - Solve to find the required point on the given curve.

∵ It is given that, y coordinate is changing 8 times as fast as the x-coordiante

\Rightarrow d y/ = 8, d x . . . . . . . . . . . . . . (3)

Thus we have from (2) and (3),

2 (8, d x/dt) = x^{2} d x/dt

\Rightarrow 16 d t d x = x^{2} d t d x

\Rightarrow (x^{2} - 16) d x = 0

\Rightarrow x^{2} - 16 = 0 \Rightarrow x = \pm 4

[∵ d x/ d t \neq = 0]

When, x = 4 \Rightarrow y = 4 ^{3} + 2/6 = 62/6 = 11

[from(1)]

when, x = (- 4) \Rightarrow y = 6 ( -4 ) ^{3} + 2 = - 662 = - 31/3

[from(1)]

Hence, the points required on the curve are (4, 11)and (-4, -31 3) .

12. The radius of an air bubble is increasing at the rate 12 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

?‍♂️solution – The air bubble is in the shape of a sphere.
Now, the volume of an air bubble $(V)$ with radius $(r)$ is given by,
$V = 4/ 3 π r^{3}$
The rate of change of volume $(V)$ with respect to time $(t)$ is given by,
$d V/dt = 4/ 3 π d/dr (r^{3}) \cdot d r/dt 4/ 3 π (3 r^{2}) \cdot d r/dt = 4 π r^{2} d r/dt$
It is given that $d r/dt = 1/ 2$ cm/s
Therefore, when $r = 1$ cm,
$d V/dt = 4 π (1)^{2} (1/ 2) = 2 π c m^{3} / s$
Hence, the rate at which the volume of the bubble increases is $2 π c m^{3} / s$ .

13. The radius of an air bubble is increasing at the rate 12 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

?‍♂️solution – The air bubble is in the shape of a sphere.
Now, the volume of an air bubble $(V)$ with radius $(r)$ is given by,
$V = 34 π r^{3}$
The rate of change of volume $(V)$ with respect to time $(t)$ is given by,
$d V = 4/ 3 π, d/ d r (r^{3}) \cdot d r 3 π (3 r^{2}) \cdot d r/ = 4 π r^{2} d r$
It is given that $d r/dt =1/ 2$ cm/s
Therefore, when $r = 1$ cm,
$d V/dt = 4 π (1)^{2} (1/ 2) = 2 π c m^{3} / s$
Hence, the rate at which the volume of the bubble increases is $2 π c m^{3} / s$ .

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

?‍♂️solution – Let $r = r a d i u s; h = h e i g h t; V =$ Volume of sand cone, $t = t i m e$

Given

h = 4 c m; d V = 12 c m^{3} / s

and

h = 1/ 6 r

r = 6 h

∴ V = 1/ 3 π r^{2} h

= 1/ 3 π (6 h)^{2} h = 3/36 h ^{3} π = 12 π h^{3}

(diff w.r.t.

t

)

∴ d V/dt = 12 π 3. h^{2} d h/dt

12 = 12 π . 3 (4)^{2} d h/dt

∴ d h = 1/ 48 π c m / s

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

C(x) = 0.007x³ – 0.003x² + 15x + 4000.

Find the marginal cost when 17 units are produced

?‍♂️solution – Marginal cost is the rate of change of total cost with respect to output.

∴

Marginal cost (MC)

= d C = 0.007 (3 x^{2}) - 0.003 (2 x) + 15

= 0.021 x^{2} - 0.006 x + 15

Now When

x = 17

,
MC

= 0.021 (1 7^{2}) - 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 .

16. The total revenue in Rupees received from the sale of x units of a product is given by

R(x) = 13x² + 26x + 15.

Find the marginal revenue when x = 7.

?‍♂️solution – Marginal revenue is the rate of change of total revenue with respect to the number of units sold.
$\Rightarrow$ Marginal Revenue (MR) $= d R = 13 (2 x) + 26 = 26 x + 26$
Thus when $x = 7$ , MR $= 26 (7) + 26 = 182 + 26 = 208$
Hence, the required marginal revenue is Rs $208$ .

17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is

(A) 10π

(B) 12π

(C) 8π

(D) 11π

?‍♂️solution – Correct option is B)

The area

(A)

of circle with radius

r

is,

A = π r^{2}

Thus rate of change of area of circle with radius is,
$\Rightarrow d A/dr = 2 π r$ .
Hence at $r = 6$ cm $, d A = 12 π$ cm

18. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is

(A) 116

(B) 96

(C) 90

(D) 126

?‍♂️solution – Correct option is D)

We have, $R (x) = 3 x^{2} + 36 x + 5$
$\Rightarrow Marginal revenue = d R ( x )/ d x = 6 x + 36$
Thus at $x = 15, Marginal revenue = 6 (15) + 36 = 126$