# The Volume of a Sphere is Increasing at the Rate of 8 Cm3/S. Find the Rate at Which Its Surface Area is Increasing When the Radius of the Sphere is 12 Cm. - Mathematics

The volume of a sphere is increasing at the rate of 8 cm3/s. Find the rate at which its surface area is increasing when the radius of the sphere is 12 cm.

#### Solution

Let r be the radius and V be the volume of the sphere at any time t. Then,

V = 4/3pir^3

=> (dV)/(dt) = 4pir^2 (dr)/(dt)

=> (dr)/(dt) = 1/(4pir^2) (dV)/(dT)

=> (dr)/(dt) = 8/(4pi(12)^2)       [∵ r = 12 cm and (dV)/(dt) = 8 cm^3"/sec"]

=> (dr)/(dt) =  1/(72pi) "cm/sec"

Now, let S be the surface area of the sphere at any time t. Then,

S = 4πr2

=> (dS)/(dt) = 8pir (dr)/(dt)

=> (dS)/(dt) = 8pi(12)xx 1/(72pi)   [∵  r = 12 cm and (dr)/(dt) = 1/(72pi) "cm/sec"]

=> (dS)/(dt) = 4/3 cm^2"/sec"

Hence, the surface area of the sphere is increasing at the rate of 4/3 cm^2"/sec"

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