RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives Ex 8.1

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

RBSE Solutions For Class 12 Maths Chapter 8 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

RBSE Class 12 Maths Chapter 8 Question 2.
A particle moves along y = 2/3 x³ + 1. Find the points on the curve at which y – coordinate is changing twice as fast as the x – coordinate.
Solution:
Given equation
y = 2/3 x³ + 1 …..(i)
Let position of particle at time ’t’ is p(x, y) and hence p(x, y) is situated on curve (i).
y = 2/3 x³ + 1

Class 12 Maths RBSE Solutions 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 3/10 rad/s

Class 12 Math RBSE Solutions 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 = x³ …..(i)
and this edge is increasing at the rate of 3 cm/s.
∴ dx/dt = 3 cm/s.
We have to find, rate of change of volume V
When x = 10cm

Class 12 RBSE Solutions Maths Question 5.
A Balloon which is always spherical, is being inflated by pumping in 900 cm³ 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.

RBSE Maths Solution Class 12 Question 6.
A balloon which is always spherical, has diameter 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,

RBSE Solution Class 12 Maths Question 7.
The total cost C(x) associated with the production of x units of a product is given by
C(x) = 0.005x³ – 0.02x² + 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) = 0.005x³ – 0.02x² + 30x + 5000
Marginal cost = MC

= 0.005 × 3x² – 0.02 × 2x + 30 × 1
= 0.005 × 3x² – 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).

RBSE Class 12 Maths Solutions 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 cm²/s.
Again, let at any time t, radius of soap bubble is r and volume is V.
According to question,

RBSE Solutions Class 12 Maths Question 9.
Sand is pouring from a pipe at the rate of 12 cm³/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,

Class 12th RBSE Maths Solution Question 10.
The total revenue received from the sale of x units of a product is given by
R(x) = 13x² + 26x + 15
Find marginal cost when x = 15
Solution:
Given,
R(x) = 13x² + 26x + 15
Marginal cost MR(x) = d/dx {R(x)}
= d/dx (13x² + 26X + 15)
= 26x + 26
Putting x = 15
Then MR(x) = 26 × 15 + 26
= 390 + 26
MR (7) =416
Hence,
Marginal cost = ₹ 416