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2.10.5 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 + 7 h , 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 + 7 h d V d h = 3 5 h 2 + 7 = 3 h 2 + 35 5 Given  d V d t = 15 ,  h = 3 Rate of change of the height of water = d h d t d h d t = d h d V × d V d t ← Chain rule d h d t = 5 3 h 2 + 35 × 15 d h d t = 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 = 2 π r d L d r = 2 π

(a)
Given  d L d t = 0.3 Rate of change in the radius of the circle = d r d t d r d t = d r d L × d L d t d r d t = 1 2 π × 0.3 d r d t = 0.0477  cms − 1

(b)
2 π r = 88 r = 88 2 π = 44 π Hence, the radius of the circle after  5 s = 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 d V d h = ( 3 ) 4 27 π h 2 d V d h = 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 = d h d t . d h d t = d h d V × d V d t ← Chain rule d h d t = 9 4 π h 2 × 0.02 ← Given  d V d t = 0.02 d h d t = 9 4 π ( 0.5 ) 2 × 0.02 d h d t = 0.0572   m s − 1

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