Differentiation Short Questions (Question 19 - 21)


Question 19:
The volume of water V cm3, in a container is given by   V= 1 5 h 3 +7h , where h cm is the height of the water in the container. Water is poured into the container at the rate of 15cm3s-1. Find the rate of change of the height of water in cms-1, at the instant when its height is 3cm. 

Solution:

V= 1 5 h 3 +7h dV dh = 3 5 h 2 +7= 3 h 2 +35 5 Given  dV dt =15h=3 Rate of change of the height of water= dh dt dh dt = dh dV × dV dt Chain rule dh dt = 5 3 h 2 +35 ×15 dh dt = 75 62  cms 1



Question 20:
A wire of length 88 cm is bent to form a circle. When the wire is heated, the length increases at the rate of 0.3 cms-1.
(a) Calculate the rate of change in the radius of the circle.
(b) Hence, calculate the radius of the circle after 5s.

Solution:
Length of circumference of a circle,  L=2πr dL dr =2π
(a)
Given  dL dt =0.3 Rate of change in the radius of the circle= dr dt dr dt = dr dL × dL dt dr dt = 1 2π ×0.3 dr dt =0.0477  cms 1
(b)
2πr=88 r= 88 2π = 44 π Hence, the radius of the circle after 5s = 44 π +5( 0.0477 ) =14.24 cm



Question 21:

The diagram shows a conical container with diameter 0.8m and height 0.6m. Water is poured into the container at a constant rate of 0.02m3s-1. Calculate the rate of change of the height of the water level when the height of water level is 0.5m.

Solution:

Let,  h = height of the water level r = radius of the water surface V = volume of the water r h = 0.4 0.6 Concept of similar triangles r h = 2 3 r= 2 3 h Volume of water, V= 1 3 π r 2 h V= 1 3 π ( 2 3 h ) 2 h V= 4 27 π h 3 dV dh =( 3 ) 4 27 π h 2 dV dh = 4 9 π h 2 The rate of change of the height of the water level when the height of water level is 0.5 m= dh dt . dh dt = dh dV × dV dt Chain rule dh dt = 9 4π h 2 ×0.02 Given  dV dt =0.02 dh dt = 9 4π ( 0.5 ) 2 ×0.02 dh dt =0.0572 m s 1