NCERT 12th maths

NCERT Solutions for Class 12

Mathematics

Chapter 6 - Application of Derivatives


EXERCISE- 6.1

1. Find the rate of change of the area of a circle with respect to its radius r when

(a) r  3 cm

Ans: We know that A   r 2

Differentiate w.r.t r


dA d

dr dr
 
 r 2  2 r


When r  3 cm ,

dA
 2 (3)  6
dr

The area is changing at 6 cm 2 / s when radius is 3 cm .

(b) r  4 cm

Ans:

We know that A   r 2

Differentiate w.r.t r


dA d

dr dr
 
 r 2  2 r


When r  4 cm ,



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,dA
 2 (4)  8 x
dr

The area is changing at 8 cm 2 / s when radius is 4 cm .



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 its edge is 12 cm ?

Ans: Let the side length, volume and surface area respectively be equal to \[x,v\]
and S . V  x 3

S  6x2

dV
 8 cm 3 / s
dt


8 
dV d 3

dt dt
x  


dx
 
d 3 dx
x
dt
 3x 2
dx
dt


8 
dV d 3

dt dt
x   
d 3 dx
dx
x
dt
 
 3x 2
dx
dt

dx 8
 2
dt 3x

dS d

dt dt
6x2  


d
dx
 
6x2
dx
dt

 8  32
  6x2    6x2 
dS d d dx dx
 12 x  12 x  2  
dt dt dx dt dt  3x  x

So, when x  12 cm,

dS 32
 cm 2 / s
dt 12



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, dS 32 8
x  12 cm,  cm 2 / s  cm 2 / s.
dt 12 3




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 / s

Ans: We know that A   r 2 
dA d

dt dr

 r2
dr
dt

 2 r
dr
dt

dr
 3 cm / s
dt

dA
  2 r (3)  6 r
dt

So, when r  10 cm ,

dA
 6 (10)  60 cm 2 / s
dt




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?

Ans: Let the length and the volume of the cube respectively be x and V .

V  x3



dV d 3

dt dt
x   
d 3 dx
dx
x
dt
 3x 2 
dx
dt
dx
dt
 3 cm / s


dV
  3x 2 (3)  9 x 2
dt

So, when x  10 cm ,

dV
 9(10) 2  900 cm3 / s
dt


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, 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?
Ans: We know that A   r 2
dA d

dt dt

 r2 
d
dr

 r2 
dr
dt

 2 r
dr
dt

dr
 5 cm / s
dt

So, when r  8 cm ,

dA
 2 (8)(5)
dt

 80 cm 2 / s.



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?
Ans: We know that C  2 r .
dC dC dr d dr dr
   (2 r )  2
dt dr dt dr dt dt

dr
 0.7 cm / s
dt

dC
  2 (0.7)  1.4 cm / s
dt



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



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