Question

A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume, V, is 128pi cubic inches. At what rate is the length, h, changing when the radius, r, is 1.5 inches? Note: V=pi(r^2)h

Hi, as above this is the problem I am doing. There is no answer given to check so I was wondering about a couple of things:

Is volume assumed to be a constant in this problem?

Is the height change decreasing (negative rate) or increasing (positive rate) in the final answer?

I worked out dh/dt to be 512/135 in./s

Answer 1

Again, any help is appreciated! Oh and my workings are as follows:

Given: dr/dt = -0.05, V=128pi
Known: V=pi(r^2)h
Find: dh/dt, when r=1.5 inches

\[V= \pi r^2h\]
\[128\pi = 2.25\pi h \]
\[h = 512/9\]

\[V=\pi r^2h \]
\[0=(2\pi rh \times dr/dt) + (\pi r^2 \times dh/dt)\]
\[dh/dt = (-2\pi rh \times dr/dt)/ (\pi r^2h)\]
\[dh/dt = [(-2\pi (1.5)(512/9)) \times (-0.05)]/(2.25\pi)\]
\[dh/dt = 512/135\]

Answer 2

From what i can tell, Volume is indeed constant, Height should increase as radius decreases

Answer 3

you may have taken the radius out of the equation too early, since it's changning with time it's a variable.
\[128\pi = \pi r^2 h\]
\[128=r^2h\]
\[h=\frac{128}{r^2}\]
taking the derivative with respect to time you get:
\[\huge{\frac{dh}{dt}=-2(128)r^{-3}\frac{dr}{dt}}\]

Answer 4

\[\frac{dh}{dt}=-2(128)(1.5)^{-3}(-0.05)\]
\[\frac{dh}{dt}=3.7926\]

Answer 5

Never mind...your implicit product rule does the job fine.

Answer 6

Ok so whenever the radius decreases, the height increases... then that would make sense for the final answer to be positive in this case. Well that clears it up, thanks so much for the help!!!