At what rate is the length changing when the radius is 1.8 inches? List the given,unknown and constants
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 126π cubic inches. At what rate is the length (h) changing when the radius (r) is 1.8 inches? (hint: V=πr^2h)
1 Answer | Add Yours
We are given a cylinder of volume `126pi`cubic inches . We want to find the rate of change of the height at the moment when the radius is 1.8in.
(1) `V=pir^2h` . If we differentiate with respect to `t` we get:
`2pir(dr)/(dt)h+pir^2(dh)/(dt)=0` using the product rule and that the derivative of a constant (the volume) is 0.
(2) When the radius is 1.8 we can find the height:
(3) Then we have `r=1.8,h~~38.89,(dr)/(dt)=-.05` and we are looking for `(dh)/(dt)` ; substituting we get:
(4) Thus when the radius is 1.8in, the height is changing at a rate of 2.16 in/sec
We’ve answered 318,944 questions. We can answer yours, too.Ask a question