Question

The volume of spherical balloon is increasing at the rate of 25cm^3 /sec. Find the rate of change of its surface area at the instant when radius is 5 cm.

Answer 1

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

Then, 

V=4/3πr³

⇒dV/dt=4πr²(dr/dt )

⇒dr/dt=1/4πr²×(dV/dt)

⇒dr/dt=25/4π(5)²

 [∵ r=5 cm and dV/dt=25 cm³/sec]

⇒dr/dt=1/4π cm/sec

Now, 

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

Then,

 S= 4πr²

⇒dS/dt=8πr×dr/dt

⇒dS/dt=8π(5)×1/4π 

 [∵ r=5 cm and dr/dt=14π cm/sec]

⇒dS/dt=10 cm²/sec ____answer

I think my answer is capable to clear your confusion

Devil_king ▄︻̷̿┻̿═━一

Answer 2

Answer:

above answer is the right answer