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.

Re: Finding change of volume in cylinder - related rate

Try this:

$\displaystyle \frac {dV}{dt} = (\frac {dV}{dr})(\frac {dr}{dt}) $ - assuming that h is a constant (is it?)

$\displaystyle V = \pi r^2h$, so $\displaystyle \frac {dV}{dr} = 2 \pi r h$

$\displaystyle \frac {dr}{dt} = 0.1$ mm/min = 0.01 cm/min.

$\displaystyle \frac {dV}{dt} = 2 \pi (\frac {9.5 \ cm } 2 )(15 \ cm ) \times 0.01\ cm/min = 4.47 \ cm^3/min$.

Re: Finding change of volume in cylinder - related rate

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

Re: Finding change of volume in cylinder - related rate

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

Re: Finding change of volume in cylinder - related rate

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

Re: Finding change of volume in cylinder - related rate

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

$\displaystyle \frac {dV}{dt} = \frac {\pi r^2 h} {dt} = \frac {d \pi} {dt} r^2 h + \pi \frac {dr^2} {dt} h + \pi r^2 \frac {dH}{dt} $

Since $\displaystyle \pi$ and h are constants their derivatives are both zero, giving you:

$\displaystyle \frac {dV}{dt} = \pi h \frac {dr^2}{dt} = \pi h 2 r \frac {dr}{dt} $

Re: Finding change of volume in cylinder - related rate

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

Re: Finding change of volume in cylinder - related rate

actually you got it there, I see now. Thanks