Question
Question: The volume of a sphere increases at the rate of \[20\,{\text{c}}{{\text{m}}^{\text{3}}}{{\text{s}}^{...
The volume of a sphere increases at the rate of 20cm3s - 1 . Find the rate of change of its surface area when its radius is 5cm .
Solution
Hint : We are asked to find the rate of change of surface area of the sphere. First, recall the formula for sphere area of a sphere. Recall how rate of change of a quantity is found with respect to time, use this to find the rate of surface area of the sphere using the value of radius and rate of change of volume
** Complete step-by-step answer** :
Given, rate of change volume is dtdV=20cm3s - 1
Radius of the sphere is r=5cm
We are asked to find the rate of change of surface area of the sphere.
Surface area of a sphere is written as,
S=4πr2
To find the rate of change of surface area, differentiate S with respect to time t .
dtdS=dtd(4πr2)
⇒dtdS=4π×2r×dtdr
⇒dtdS=8πrdtdr (i)
Volume of a sphere is written as,
V=34πr3
For rate of change of volume, we differentiate V with respect to time t .