Question

Help with related rates problems?1). At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at the rate of 13 cubic feet/min. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when it is 15 ft high? Answer in units of ft/min
- I got 52/2025*pi or 0.08067299653

This one has two parts to it.

Assume that x and y are both differentiable functions of t and xy=4
a). Find dy/dt when x=7.8 and dx/dt=10.5
- -.690
b). Find dx/dt when x=1.3 and dy/dt=-6.4
- 2.703

Answer 1

If you guys could also post your work, that would be great :)

Answer 2

Let the radius of your cone be r. Let its altitude be h. It's volume V is given by the formula
\[V=\frac{ 1 }{ 3 } \pi r^2h\]
We know that V is changing at a constant rate of 13 cubic feet per minute.
Hence,
\[\frac{ dV }{ dt }= 13 \frac{ ft }{ \min }\]
We also know that base diameter = 2r = 3h, or r = (3/2) h. If we replace this into the expression for the volume, we get: \[V=\frac{ 3}{ 4 } \pi h^3\]

Note that we have length cubed, so the units are consistent for volume. We would like to compute the rate of change of the altitude h of the cone, or dh/dt. Use the "chain rule," keeping in mind that \[\frac{ dy }{ dx }= \frac{ 1 }{ \frac{ dx }{ dy } }\]
so: \[\frac{ dV }{ dt} = (\frac{ dV }{ dh })^{-1} \frac{ dV }{ dt }\]

You should be able to figure it out from here, but remember to plug in 15 for h into this formula!

Answer 3

A solution using Mathematica 8 Home Edition is attached.