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Calculus/Calculus - Related Rates

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Expert: - 10/28/2009


Question
Hi here is the problem:

Let V be the volume and S the total surface area of a solid right circular cylinder that is 5 feet high and has a radius r feet.

Find the rate of change of volume with respect to surface area, dV/dS , when r = 3 ft.

I know that V = (π)(r^2)(h) and that SA = (4)(π)(r^2). When I differentiate I get:

dV/dt = πr^2(1) + (h)2πr(dR/dt) and dS/dt = 8πr(dR/dt)

Want dV/dS when r = 3.

I'm confused about how to proceed from here. Thanks for the help!

Answer
V = volume = πr²h; S = surface area = 2π(r²+h); h = 5.
This means that V = 5πr² and S = 2π(r²+5).  Find dV/dS when r is 3.
It is known that h is 5 and is not changing.

I see dV/dr = 5π•2r, r is 3, so dV/dr = 30π.
dS/dt = 2π•2r = 4πr, r is 3, so 4π.

Want dV/dS when r = 3.

To get dV/dS, the notation is sort of like fractions.  
In other words, in (dV/dr)/(dS/dr), both are over dr, and they can be cancelled.
It can be seen that all that is left is dV/dS.

This said, (dV/dr)/(dS/dr) = 30π / 4π = 7.5.

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