RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives Ex 8.1 are part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives Ex 8.1.

## Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.1

Question 1.

Find the rate of change of the area of a circle with respect to its radius r, when r = 3 cm and r = 4 cm.

Solution:

Let, area of circle = A

Question 2.

A particle moves along y = \(\frac { 2 }{ 3 } \) x^{3} + 1. Find the points on the curve at which y – coordinate is changing twice as fast as the x – coordinate.

Solution:

Given equation

y = \(\frac { 2 }{ 3 } \) x^{3} + 1 …..(i)

Let position of particle at time ’t’ is p(x, y) and hence p(x, y) is situated on curve (i).

y = \(\frac { 2 }{ 3 } \) x^{3} + 1

Question 3.

A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec. How fast is the angle θ between the ladder and the ground changing when the foot of the ladder is 12 m away from the wall.

Solution:

Let AB is a ladder, its end ‘A’ is x distance far from wall and upper end is at y from the ground. Let slope between ladder and ground is θ.

Hence, angle between ground and ladder is decreasing at the rate of \(\frac { 3 }{ 10 } \) rad/s

Question 4.

An edge of a variable cube is increasing at the rate of 3 cm/sec. How 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.

∴ \(\frac { dx }{ dt } \) = 3 cm/s.

We have to find, rate of change of volume V

When x = 10cm

Question 5.

A Balloon which is always spherical, is being inflated by pumping in 900 cm^{3} of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm.

Solution:

Let at any time t, radius of balloon is r and its volume is V.

Question 6.

A balloon which is always spherical, has diameter \(\frac { 3 }{ 2 } \) (2x +1) Find rate of change of its volume w.r.t. x.

Solution:

Let volume of balloon is V.

According to question,

Question 7.

The total cost C(x) associated with the production of x units of a product is given by

C(x) = 0.005x^{3} – 0.02x^{2} + 30x + 5000

Find the marginal cost when 3 units are produced.

Solution:

Let cost of production of x units is C(x), where

C(x) = 0.005x^{3} – 0.02x^{2} + 30x + 5000

Marginal cost = MC

= 0.005 × 3x^{2} – 0.02 × 2x + 30 × 1

= 0.005 × 3x^{2} – 0.02 × 2x + 30

For x = 3,

MC = 0.005 × 3 × (3)^{2} – o.02 × 2 × (3) + 30

= 0.005 × 27 – 0.02 × 6 + 30

= 0.135 – 0.12 + 30 = 30.015 or 30.02 (Approx.)

Hence, marginal cost is ₹ 30.02 (Approx).

Question 8.

The radius of a soap bubble is increasing at the rate of 0.2 cm/sec. At what rate is the surface area of a bubble increasing when the radius is 7 cm and at what rate is the volume of that bubble increasing when the radius is 5 cm.

Solution:

Let radius of soap bubble is r and surface area is S.

According to question,

Hence, rate of increasing surface area of soap bubble is 11.2π cm^{2}/s.

Again, let at any time t, radius of soap bubble is r and volume is V.

According to question,

Question 9.

Sand is pouring from a pipe at the rate of 12 cm^{3}/sec. The falling sand forms a cone of 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 height of the sand cone increasing when the height is 4 cm ?

Solution:

Let at any time “t” volume of cone of sand is V, height h and radius r.

According to question,

Question 10.

The total revenue received from the sale of x units of a product is given by

R(x) = 13x^{2} + 26x + 15

Find marginal cost when x = 15

Solution:

Given,

R(x) = 13x^{2} + 26x + 15

Marginal cost MR(x) = \(\frac { d }{ dx } \) {R(x)}

= \(\frac { d }{ dx } \) (13x^{2} + 26X + 15)

= 26x + 26

Putting x = 15

Then MR(x) = 26 × 15 + 26

= 390 + 26

MR (7) =416

Hence,

Marginal cost = ₹ 416

## Leave a Reply