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Question
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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