Chapter 6 Application of Derivatives Ex 6.1 Solutions
Question - 1 : - Find the rate of change of the area of acircle with respect to its radius r when
(a) r =3 cm (b) r =4 cm
Answer - 1 : -
The area of a circle (A)with radius (r) isgiven by,
Now, the rate of change ofthe area with respect to its radius is given by,
When r = 3 cm,
Hence, the area of the circle is changing at the rate of6Ï€ cm when its radius is 3 cm.
When r = 4 cm,
Hence, the area of the circle is changing at the rate of8Ï€ cm when its radius is 4 cm.
Question - 2 : - The volume of a cube is increasing at therate of 8 cm3/s. How fast is the surface area increasing when thelength of an edge is 12 cm?
Answer - 2 : -
Let x be the length of a side, V bethe volume, and s bethe surface area of the cube.
Then, V = x3 and S = 6x2 where x isa function of time t.
It is given thatThen, by using the chain rule, we have:
∴
Thus, when x =12 cm, 
Hence, if the length of theedge of the cube is 12 cm, then the surface area is increasing at the rateof 
cm2/s.
The radius of a circle is increasing uniformly at the rateof 3 cm/s. Find the rate at which the area of the circle is increasing when theradius is 10 cm.
Answer - 3 : -
The area of a circle (A) with radius (r)is given by,
Now, the rate of change of area (A)with respect to time (t) isgiven by,
It is given that,
Thus, when r = 10 cm,
Hence, the rate at which the area of thecircle is increasing when the radius is 10 cm is 60π cm2/s.
Question - 4 : - An edge of a variable cube is increasing at the rate of 3cm/s. How fast is the volume of the cube increasing when the edge is 10 cmlong?
Answer - 4 : -
Let x be the length of a sideand V bethe volume of the cube. Then,
V = x3.
∴
(By chain rule)
It is given that,∴
Thus, when x = 10 cm,
Hence, the volume of the cube is increasingat the rate of 900 cm3/s when the edge is 10 cm long.
A stone is dropped into a quiet lake and waves move incircles at the speed of 5 cm/s. At the instant when the radius of the circularwave is 8 cm, how fast is the enclosed area increasing?
Answer - 5 : - The area of a circle (A)with radius (r)is given by
Therefore, the rate of change of area (A) withrespect to time (t) isgiven by,
[By chain rule]
It is given that
Thus, when r = 8 cm,
Hence, when the radius of the circular waveis 8 cm, the enclosed area is increasing at the rate of 80π cm2/s.
The radius of a circle is increasing at the rate of 0.7cm/s. What is the rate of increase of its circumference?
Answer - 6 : -
The circumference of a circle (C)with radius (r) isgiven by
C = 2πr.
Therefore, the rate of change of circumference (C) withrespect to time (t) isgiven by,
(By chain rule)
It is given that
Hence, the rate of increaseof the circumference
Question - 7 : - The length x of a rectangle isdecreasing at the rate of 5 cm/minute and the width y is increasing at the rateof 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 therectangle.
Answer - 7 : -
Since the length (x) is decreasing at the rateof 5 cm/minute and the width (y) isincreasing at the rate of 4 cm/minute, we have:
and 
(a) The perimeter (P) of a rectangle is givenby,
P = 2(x + y)
Hence, the perimeter is decreasing at the rate of 2cm/min.
(b) The area (A) ofa rectangle is given by,
A = x⋅ y
When x =8 cm and y =6 cm,
Hence, the area of the rectangle isincreasing at the rate of 2 cm2/min.
Question - 8 : - A balloon, which always remains spherical on inflation, isbeing inflated by pumping in 900 cubic centimetres of gas per second. Find therate at which the radius of the balloon increases when the radius is 15 cm.
Answer - 8 : -
The volume of a sphere (V) with radius (r)is given by,
∴Rate of change of volume (V) with respect to time (t) isgiven by,
[By chain rule]
It is given that
Therefore, when radius = 15 cm,
Hence, the rate at whichthe radius of the balloon increases when the radius is 15 cm is
Question - 9 : - A balloon, which always remains spherical has a variableradius. Find the rate at which its volume is increasing with the radius whenthe later is 10 cm.
Answer - 9 : - The volume of a sphere (V)with radius (r)is given by
Rate of change of volume (V) withrespect to its radius (r) isgiven by,
Therefore, when radius = 10 cm,
Hence, the volume of the balloon isincreasing at the rate of 400π cm2.
A ladder 5 m long is leaning against a wall. The bottom ofthe 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 ladderis 4 m away from the wall?
Answer - 10 : -
Let y m be the height of thewall at which the ladder touches. Also, let the foot of the ladder be x mawayfrom the wall.
Then, byPythagoras theorem, we have:
x2 + y2 = 25 [Length of the ladder = 5 m]
Then, the rate of change of height (y) withrespect to time (t) isgiven by,
It is given that
Now, when x = 4 m, we have:
Hence, the height of theladder on the wall is decreasing at the rate of