Question

A cylinder is being flattened so that its volume does not change. Find the rate of change of radius when r = 2 inches and h = 5 inches, if the height is decreasing at 0.7 in/sec. Hint: what is the rate of change of volume?

Answer 1

`(dr)/(dt) = 0.14 in/s`

Step-by-step explanation:

As the volume of the cylinder is constant here so we can say that its rate of change in volume must be zero

so here we can say

`(dV)/(dt) = 0`

now we have

`V = \pi r^2 h`

now find its rate of change in volume with respect to time

`(dV)/(dt) = 2\pi rh(dr)/(dt) + \pi r^2(dh)/(dt)`

now we know that

`(dV)/(dt) = 0 = \pi r(2h (dr)/(dt) + r(dh)/(dt))`

given that

h = 5 inch

r = 2 inch

`(dh)/(dt) = - 0.7 in/s`

now we have

`0 = 2(5) (dr)/(dt) + 2(-0.7)`

`(dr)/(dt) = 0.14 in/s`