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