These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.1

Question 1.

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

Solution:

Let A be the area of the circle with radius r.

A = πr²

Differentiating w.r.t. r, \(\frac { dA }{ dr }\) = 2πr

i. r = 3cm, \(\frac { dA }{ dr }\) = 2π(3) = 6πcm²/cm

ii. r = 4cm, \(\frac { dA }{ dr }\) = 2π(4) = 8πcm² /cm

Question 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 the cube, V the volume and S the surface area at time l

∴ V = x³ and S = 6x².

\(\frac { dV }{ dt }\) = 8cm³/s

V = x³

Differentiating both sides, w.r.t. t

When the edge of the cube is 12 cm, area increases at the rate of \(\frac { 8 }{ 3 }\) cm²/s.

Question 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:

Let r be the radius of the circle.

Area of circle = πr² = A

also \(\frac { dr }{ dt }\) = 3 cm/ sec.

r = 10 m

Differentiating A w.r.t. t, we get

\(\frac { dA }{ dt }\) = 2πr.\(\frac { dr }{ dt }\)

= 2π(10)(3) = 60π cm²/sec

Question 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 V be the volume and x be the edge of the cube at time ‘t’.

V = x³

Given \(\frac { dx }{ dt }\) = 3 cm/s

Differentiating both sides, w.r.t. t

\(\frac { dV }{ dt }\) = 3x².\(\frac { dx }{ dt }\) = 3x²(3) = 9x²

When x = 10 cm, \(\frac { dV }{ dt }\) = 9 x (10)² = 900 cm³/s

Question 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:

Let r be the radius of a wave circle:

\(\frac { dx }{ dt }\) = 5cm/sec.

A = πr²

Differentiating w.r.t. t

\(\frac { dA }{ dt }\) = 2πr\(\frac { dr }{ dt }\) = 2πr(5) = 10πr

When r = 8, \(\frac { dA }{ dt }\) = 10π(8) = 80π cm²/s

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?

Solution:

The rate of change of circle w.r.t time t is given to be 0.7 cm/sec. i.e. \(\frac { dr }{ dt }\) = 0.7 cm/sec.

Now, circumference of the circle is c = 2πr

\(\frac{d \mathrm{C}}{d t}=2 \pi \cdot \frac{d r}{d t}\) = 2π(0.7) = 1.4πcm/s

The rate of increase of circumference is 1.4 π cm/s.

Question 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:

The length x of a rectangle is decreasing at dx the rate of 5cm/min. ⇒ \(\frac { dx }{ dt }\) = – 5cm min … (i)

The width y is increasing at the rate of 4cm/min.

\(\frac { dy }{ dt }\) = 4 cm/minute

(a) p = 2x + 2y

Differentiating both sides w.r.t., t,

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

= 2(-5) + 2(4) = – 2cm/minute

Perimeter is decreasing at the rate of 2cm/ minute

(b) A = xy

Differentiating both sides w.r.t., t,

\(\frac { dA }{ dt }\) = x.\(\frac { dy }{ dt }\) + y.\(\frac { dx }{ dt }\)

When x = 8 cm and y = 6 cm

= 32 – 30 = 2 cm²/minute

Area is increasing at the rate of 2 cm²/minute

Question 8.

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

Solution:

Let V be the volume and r be the radius of the balloon at time t.

Given \(\frac { dV }{ dt }\) = 900

V = \(\frac { 4 }{ 3 }\) πr³

Differentiating both sides w.r.t. t

\(\frac { dV }{ dt }\) = \(\frac { d }{ dt }\)(\(\frac { 4 }{ 3 }\) πr³) = \(\frac { d }{ dr }\)(\(\frac { 4 }{ 3 }\) πr³) \(\frac { d }{ dr }\)

900 = 4πr².\(\frac { dr }{ dt }\)

\(\frac { dr }{ dt }\) = \(\frac{900}{4 \pi r^{2}}\) =\(\frac{225}{\pi r^{2}}\)

when r = 15 cm

\(\frac { dr }{ dt }\) = \(\frac{225}{\pi(15)^{2}}\) = \(\frac{1}{\pi}\) cm/sec.

When radius is 15 cm, the rate at which the radius increases is \(\frac { 1 }{ π }\) cm/sec.

Question 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:

Let r be the variable radius of the balloon which is in the form of sphere Vol. of the sphere

Differentiating w.r.t. r, \(\frac { dV }{ dr }\) = 4π²

When r = 10, \(\frac { dV }{ dr }\) = 4π(10)² = 400πcm³/cm

i.e., The rate at which volume of the bal¬loon is increasing with respect to the radius is 400πcm³/cm

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 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from die wall ?

Solution:

Let AB be the ladder and OB be the wall. At an instant,

let OA = x, OB = y,

x² + y² = 25 …(i)

Differentiating both sides w.r.t., t,

2x.\(\frac { dx }{ dt }\) + 2y.\(\frac { dy }{ dt }\) = 0 … (1)

x² + y² = 25 ⇒ y = \(\sqrt{25-x^{2}}\)

When x = 4m, y = \(\sqrt{25-16}\) = 3m

When x = 400 cm, y = 300 cm

(1) ⇒ (2 x 400 x 2) + (2 x 300 x \(\frac { dy }{ dt }\)) = 0

1600 + 600\(\frac { dy }{ dt }\) = 0

∴\(\frac { dy }{ dt }\) = \(\frac { -16 }{ 6 }\) = \(\frac { -8 }{ 3 }\) cm/s

∴ Height of the ladder on the wall is de-creasing at the rate of \(\frac { 8 }{ 3 }\) cm/s.

Question 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:

We have

6y = x³ + 2 … (i)

Differentiating both sides w.r.t., t,

Hence the required points on the curve are (4, 11) and (- 4, \(\frac { -31 }{ 3 }\))

Question 12.

The radius of an air bubble is increasing at the rate of \(\frac { 1 }{ 2 }\) cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Solution:

Let r be the radius then V = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

\(\frac { dr }{ dt } =\frac { 1 }{ 2 } \)cm/sec

\(\frac { dv }{ dt } =\frac { d }{ dt } \left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) =\frac { 4 }{ 3 } { \pi .3r }^{ 2 }.\frac { dr }{ dt } \) = 2πr²

Hence, the rate of increase of volume when radius is 1 cm = 2π x 1² = 2π cm^{3}/sec.

Question 13.

A balloon, which always remains spherical, has a variable diameter \(\frac { 3 }{ 2 } \)(2x+1). Find the rate of change of its volume with respect to x.

Solution:

Let r be the radius and V be the volume at time t.

Question 14.

Sand is pouring from a pipe at the rate of 12 cm^{3}/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 V be the volume, r be the base radius and h be the height of the sand cone at time t.

When h = 4 cm, \(\frac { dh }{ dt }\) = \(\frac { 1 }{ 48π }\) cm/sec

When the height is 4 cm, the rate at which the height of the sand cone increases is \(\frac { 1 }{ 48π }\) cm/sec.

Question 15.

The total cost C (x) in Rupees associated with the production of x units of an item is given by C (x) = 0.007 x^{3} – 0.003 x^{2} + 15x + 4000. Find the marginal cost when 17 units are produced.

Solution:

C(x) = 0.007x³ – 0.003x² + 15x + 4000

Marginal cost, MC = \(\frac { d }{ dx }\) C(x)

= 3(0.007)x² – 2(0.003)x + 15

= 0.021 x² – 0.006x + 15

When x = 17

MC = 0.021(17²) – (0.006)17 + 15

= 6.069 – 0.102 + 15 = 20.967

When 17 units are produced,

Marginal cost = ₹ 20.967

∴ The required marginal cost ₹ 30.02 (approx)

Question 16.

The total revenue in rupees received from the sale of JC units of a product is given by

R (x) = 13x² + 26x + 15.

Find the marginal revenue when x = 7.

Solution:

Marginal Revenue (MR)

= Rate of change of total revenue w.r.t. the

number of items sold at an instant = \(\frac { dR }{ dx } \)

We know R(x) = 13x² + 26x + 15,

MR = \(\frac { dR }{ dx } \) = 26x + 26 = 26(x+1)

Now x = 7, MR = 26 (x + 1) = 26 (7 + 1) = 208

Hence, Marginal Revenue = Rs 208.

Question 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:

(b) ∵ A = πr² ⇒ \(\frac { dA }{ dr } \) = 2π x 6 = 12π cm²/radius

Question 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:

(d) R(x) = 3x² + 36x + 5 ,

MR = \(\frac { dR }{ dx } \) = 6x + 36 ,

At x = 15; \(\frac { dR }{ dx } \) = 6 x 15 + 36 = 90 + 36 = 126