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The surface area of a sphere is increasing at a rate of 14\pi14π14, pi square meters per hour. At a certain instant, the surface area is 36\pi36π36, pi square meters. What is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?

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6 votes

Answer:


21\pi cubic meters per hour

Explanation:

We know that the surface area of a sphere is,


A=4\pi r^2 ...........(1),

Where,

r = radius of the sphere,

We have, A =
36\pi square meters,


36\pi = 4\pi r^2


9 = r^2


\implies r = 3\text{ meters}

Now, differentiating equation (1) with respect to t ( time ),


(dA)/(dt) = 8\pi r(dr)/(dt)

We have,


(dA)/(dt)=14\pi\text{ square meters per hour}, r = 3\text{ meters}


14\pi = 8\pi (3) (dr)/(dt)


(14)/(24)=(dr)/(dt)


(7)/(12) = (dr)/(dt)

Also, the volume of a sphere is,


V = (4)/(3)\pi r^3

Differentiating with respect to t(time)


(dV)/(dt)=4\pi r^2(dr)/(dt)

By substituting the values,


(dV)/(dt)=4\pi (3)^2 ((7)/(12))=36\pi (7)/(12)=21\pi\text{ cube meters per hour}

Hence, the rate of change of the volume of the sphere at that instant would be
21\pi cubic meters per hour

User Jayaram
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