Thread: Finding change of volume in cylinder - related rate

Engine cylinder 15 cm deep, being bored out
radius increases by .10 mm/min
How fast does volume change when dia is 9.5 cm?

pi(r^2)h = V
pi(4.25)^2(15) = 851.12 cm^3

dr/dt = .01 cm
dv/dt = ?

f'gh + fg'h + fgh'
f' = pi
g' = 2r(dr/dt)
h' = dh/dt

pi(4.25^2)(15) + (pi)((4.25)(0.01) + (pi)(r^2)(1) = 911.91 cm^3

which is very wrong. Where did I go wrong? thanks.

Try this:

- assuming that h is a constant (is it?)

, so

mm/min = 0.01 cm/min.

.

dv/dt =dr/dt * dv/dr

dv/dr = 2pi(h)(r) = 447.67

dv/dt = .01 * 447.67

ans = 4.4767 cm^3 per minute

but I thought you were supposed to use a chain rule when 3 things multiply together like that

no, look over the chain rule, here we did not multiply three things together, we just used the information they gave us to set up a simple equation

Well, you could use the chain rule, but it's a bit of overkill to apply to constants:



Since and h are constants their derivatives are both zero, giving you:

in a recent thread I was told that pi remains even though the derivative is a constant.

actually you got it there, I see now. Thanks