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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.

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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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