Asked by Vince

Air is being pumped into a spherical balloon so that its volume increases at a rate of 90{cm}^3/s. How fast is the surface area of the balloon increasing when its radius is 7{cm}? Recall that a ball of radius r has volume \displaystyle V={4}/{3}pie r^3 and surface area S=4\pi r^2

Answers

Answered by bobpursley
dV/dt= 4/3 PI 3 r^2 dr/dt
solve for dr/dt given dv/dt

Then dS/dt=4PI 2r dr/dt, knowing dr/dt solve for dS/dt

and it is done.
Answered by Vince
ok thanks alot for clearing it up
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