# Composite function problem

• Sep 6th 2010, 05:44 PM
Tubaman
Composite function problem
Here is the problem I'm trying to solve:

The volume of a conical pile of sand is increasing at a rate of 243(pi)ft^3/min, and the height of the pile always equals the radius r of the base. Express r as a function of time t (in minutes), assuming that r = 0 when t = 0.

I'm not even sure how to set this one up! Please help!
• Sep 6th 2010, 06:18 PM
Soroban
Hello, Tubaman!

Quote:

$\displaystyle \text{The volume of a conical pile of sand is increasing at }243\pi\text{ft}^3\!/\text{min}$
$\displaystyle \text{The height }h\text{ of the pile always equals the radius }r\text{ of the base.}$
$\displaystyle \text}Express }r\text{ as a function of time }t\text{, assuming that }r\,\!=\,\!0 \text{ when }t\,\!=\,\!0.$

The volume of a cone is: .$\displaystyle V \;=\;\frac{\pi}{3}r^2h$

Since $\displaystyle r = h$, we have: .$\displaystyle V \;=\;\frac{\pi}{3}r^3$

Differentiate with respect to time: .$\displaystyle \dfrac{dV}{dt} \;=\;\pi r^2\dfrac{dr}{dt}$

We are told that: $\displaystyle \dfrac{dV}{dt} \,=\,243\pi\text{ ft}^3\!/\text{min}$

Hence: .$\displaystyle \pi r^2\dfrac{dr}{dt} \;=\;243\pi \quad\Rightarrow\quad r^2\,dr \;=\;243\,dt$

Integrate: .$\displaystyle \frac{1}{3}r^3 \;=\;243t + c \quad\Rightarrow\quad r^3 \;=\;729t + C$

When $\displaystyle r=0,\:t=0\!:\;\;0^3 \:=\:729(0) + C \quad\Rightarrow\quad C \,=\,0$

So we have: .$\displaystyle r^3 \:=\:729t$

Therefore: .$\displaystyle r \;=\;9\sqrt[3]{t}$
• Sep 6th 2010, 06:31 PM
Tubaman