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

Question

Yamilet

Physics

27 May, 18:09

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

+5

Answers (1)

Know the Answer?

Lainey Murillo · Answer 1

Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.

We have been given a value of dr/dt and are required to find dh/dt

Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:

128 = πr²h

h = 128/πr²

Now we differentiate both sides:

dh/dr = - 256/πr³

Applying the chain rule:

dh/dt = dh/dr x dr/dt

dh/dt = (-256/πr³) x - 0.05

dh/dt = 64/5πr³; substituting the value of r

dh/dt = 64/5π (1.5) ³

dh/dt = 1.21 in/sec