A machine lengthens a metal cylinder by rolling it under pressure. The radius of the cylinder decreases at a constant rate of 0.05 inches per second while the volume stays constant at 128π cubic inches. At what rate is the length L changing when the radius r is 1.8 inches? [Note: V = πr 2L.]

Answer 1

V = pi r^2 L

ln V = ln pi + 2 lnr + lnL

d/dt (lnV) = 0 + (2/r) dr/dt + (1/L) dL/dt

dL/dt = - (2L/r) dr/dt

When r = 1.8,
128 pi = pi*(1.8)^2 L
L = 39.51 inch

You know dr/dt. Use the dL/dt formula

Answer 2

thanks that helps alot

Answer 3

I forgot to add this step:
0 = (2/r) dr/dt + (1/L) dL/dt

which is true because V does not change.