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

Answer:

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

Step-by-step explanation:

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

$h (t)=\int {\frac{dh}{dt} } \, dt = 2t + k$. 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,

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