Home » Blog » Find the rate of change in volume of a right circular cylinder

Find the rate of change in volume of a right circular cylinder

Given a right circular cylinder whose radius increases at a constant rate and whose altitude is a linear function of the radius. Also, given that the altitude is increasing at a rate three times that of the radius. The volume is increasing at a rate of 1 cubic feet per second when the radius is 6 feet. When the radius is 36 feet the volume is increasing at a rate of n cubic feet per second. Find the value of the integer n.


The following diagram illustrates the setup:

Rendered by QuickLaTeX.com

Since the altitude (which we denote h) is a linear function of the radius and increases three times as quickly, we have

    \[ h = 3r + c \implies \frac{dh}{dr} = 3. \]

When r=1, we are given h=6. Thus, we solve for c and get c = 3.
When r=6, we have

    \[ \frac{dV}{dt} = 1 \text{ cubic foot per second}. \]

So, since r = 6 implies h = 3(6) + 3 = 21 and

    \[  V = \pi r^2 h \implies \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} + 2 \pi r h \frac{dr}{dt} = 1, \]

we have,

    \[ \pi (36) \left( 3 \frac{dr}{dt} \right) + 2 \pi (6) (21) \left( \frac{dr}{dt} \right) = 1. \]

Solving for \frac{dr}{dt} we obtain,

    \[ \frac{dr}{dt} = \frac{1}{360  \pi}. \]

Then, when r = 36 we have

    \begin{align*}  n &= \pi (36)^2 \left( \frac{1}{120 \pi} \right) + 2 \pi (36) (111) \left( \frac{1}{360 \pi} \right) &(r = 36 \implies h = 111) \\  &= \frac{54}{5} + \frac{111}{5} \\  &= 33. \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):