# Problem 66 the volume v (in cubic meters) of the hot-air ballon described in problem 65 is given by v(r)=4/3 pi r3. If the radius r is the same function of t as in problem 65 find the volume...

**Problem 66 the volume v (in cubic meters) of the hot-air ballon described in problem 65 is given by v(r)=4/3 pi r3. If the radius r is the**

same function of t as in problem 65 find the volume as a function of the time t.

**Problem 65 Surface Area of a Balloon** The surface area *S* (in square meters) of a hot-air balloon is given by

where *r* is the radius of the balloon (in meters). If the radius *r* is increasing with time *t* (in seconds) according to the formula find the surface area *S* of the balloon as a function of the time *t*. Just need answer to problem#66.

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### 1 Answer

Some of the question is missing, so I hope I answer the right question. We are using the chain rule.

`S(r)=4pir^2`

`(dS(r))/(dt)=4pi(2r)(dr)/(dt) = 8pir(dr)/(dt)`

And the answer to your question is

`(dV(r))/(dt)=4/3 pi (3r^2) (dr)/(dt) = 4pir^2(dr)/(dt)`