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 = \(\frac { 2 }{ 3 } \) 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 = \(\frac { 2 }{ 3 } \) 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 = \(\frac { 2 }{ 3 } \) 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 \(\frac { 3 }{ 10 } \) 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.
∴ \(\frac { dx }{ dt } \) = 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 \(\frac { 3 }{ 2 } \) (2x +1) Find rate of change of its volume w.r.t. x.
Solution:
Let volume of balloon is V.
According to question,
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) = \(\frac { d }{ dx } \) {R(x)}
= \(\frac { d }{ dx } \) (13x2 + 26X + 15)
= 26x + 26
Putting x = 15
Then MR(x) = 26 × 15 + 26
= 390 + 26
MR (7) =416
Hence,
Marginal cost = ₹ 416