JEE Mathematics/ Lakshmikanta Satapathy/ Differential calculus/ Questions on Application of derivative as rate measurer involving sand cone and slipping ladder problem

1. Physics Helpline
L K Satapathy
Derivative as Rate Measurer 1

2. Physics Helpline
L K Satapathy
Application of Derivative 1
Derivative as Rate Measurer
1 1
( ) / ( ) / ( ) / ( ) /
48 48 24 24
a cm s b cm s c cm s d cm s
 
 
2. A ladder of length 5m is leaning against a vertical wall. The bottom of the ladder is
pulled along the horizontal floor away from the wall at the rate of 2 cm/s. Then find the
rate at which the height of the ladder on the wall is decreasing when the foot of the
ladder is 4m away from the wall.
4 4 8 8
( ) / ( ) / ( ) / ( ) /
3 5 3 5
a cm s b cm s c cm s d cm s
1. Sand is pouring from a pipe at the rate of 12 cm / s. The falling sand forms a cone
on the horizontal ground in such a way that the height of the cone is always one-sixth
of the radius of its base. Find the rate at which the height of the sand cone is
increasing when its height is 4cm.
3

3. Physics Helpline
L K Satapathy
Application of Derivative 1
Derivative as Rate Measurer
CONCEPTS Consider two variables x and y such that y = f (x)
Rate of change of y w. r. to x is ( )
dy
f x
dx

Rate of change of x w. r. to time
dx
dt

Rate of change of y w. r. to time
dy
dt

Chain Rule ( )
dy dy dx dx
f x
dt dx dt dt
    
For a given value of x = c ( )
x c x c
dy dx
f c
dt dt 
   
    
   

4. Physics Helpline
L K Satapathy
Application of Derivative 1
Derivative as Rate Measurer
Answer -1
3
12 /
dV
cm s
dt
 
2 21 1
(6 )
3 3
V r h h h  
A
B CD
h
r
Sand cone
Volume of cone is
Given that sand is pouring at the rate of 12 cm /s3
 Volume of cone increases at the rate of 12 cm /s3
3
12V h 
6
6
r
h r h   
Height of cone = (Radius of base)
We are required to find when 4h cm
[ since r = 6h ]
dh
dt
1
6

5. Physics Helpline
L K Satapathy
Application of Derivative 1
Derivative as Rate Measurer
.
dV dV dh
dt dh dt

2
2
1
12 36 .
3
dh dh
h
dt dt h


   
 The correct option is (b)
3 2
(12 ) 36
dV d
h h
dh dh
   
2
36 .
dV dh
h
dt dt
 
By chain rule we have
Given that
3
12 /
dV
cm s
dt

4
1 1
/
3 16 48h
dh
cm s
dt  
 
   
 
Differentiating w.r.to h

6. Physics Helpline
L K Satapathy
Application of Derivative 1
Derivative as Rate Measurer
Answer -2
2 /
dx
cm s
dt
 
Given :
We are required to find 4
dy
when x m
dt

2 2 2
. . . (1)x y l  
l
x
y
A
O B
The length of the ladder AB = l = 5 m
Bottom B is pulled towards right with speed 2 cm/s
Let at any instant , OB = x and OA = y
As a result , the top of the ladder A moves down with speed
dy
dt
In  OAB , we have
2 2 2
( ) ( ) ( )OA OB AB 

7. Physics Helpline
L K Satapathy
Application of Derivative 1
Derivative as Rate Measurer
2 2 2
( ) ( ) ( )
d d d
x y l
dt dt dt
 
dy x dx
dt y dt
 
    
 
2 2 2 2
4 , 5 4 3 , 2 /
dx
When x m y l x m also cm s
dt
      
4 8
2 / /
3 3
dy m
cm s cm s
dt m
     
 The correct option is (c)
 Height decreases at the rate of (8/3) cm/s
Differentiating eqn. (1) with respect to time
2 2 0
dx dy
x y
dt dt
  
2 2 2
( ). ( ). ( )
d dx d dy d
x y l
dx dt dy dt dt
  
[ Since l is const. ]

8. Physics Helpline
L K Satapathy
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