Question

Consider a cylindrical object such that its diameter is always half its height. If the cylinder is shrinking in a such a way that its height is reducing at a rate of 2 cm/minute and its initial height (at time t=0) is 40 cm, find a formula for the volume of the cylinder in terms of time t measured in minutes. You may use the volume formula for a cylinder V=πr2h.

Answer 1

`V (t) = \pi (2t + 40) ^ 3/16.`

Explanation:

Assuming that height
`h` is a function of time, we have to
`(dh)/(dt) = 2cm/s`. Integrating
`(dh)/(dt)` and applying the first fundamental theorem of the calculation you get:

`h (t)=\int {(dh)/(dt) } \, dt = 2t + k`. Since
`h (0) = 40`, you have to
`40 = 2 (0) + k` and therefore,
`k = 40cm` and
`h (t) = 2t + 40.`

Now, if diameter is always half its height,
`D = h / 2` or what is equal
`r = h / 4`. With all this,

`V (t) = \pi r ^ 2 h = \pi (h / 4) ^ 2 h = \pi h ^ 3/16 = \pi (2t + 40) ^ 3/16.`