1. ## Related Rate problem

Gravel is being dumped from a conveyor belt at a rate of 11 m^3/min The coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are in the ratio of 5 to 2 . How fast is the height of the pile increasing when the pile is 6 metres high?

$\displaystyle \displaystyle v=\frac{1}{3}\pi*r^2h$

$\displaystyle \displaystyle v=\frac{1}{3}\pi(\frac{5}{4}h)^2h$

$\displaystyle \displaystyle \frac{dv}{dt}=\frac{\pi}{3}(\frac{5}{4}*h)^2+h(\fr ac{\pi}{3}*2(\frac{5}{2}*h*\frac{5}{4}*\frac{dh}{d t}))$

$\displaystyle \displaystyle \frac{dv}{dt}=11$

$\displaystyle \displaystyle \frac{dh}{dt}=\frac{11-\frac{\pi}{3}(\frac{5}{4}*h)^2}{\frac{11}{3}*h(\fr ac{10}{4}*h)*\frac{5}{4}}$

When I finally plug in 6 for h, I get −.0587184655608
which appears to be wrong!

What am I doing wrong here?

2. $\displaystyle \displaystyle \frac{dv}{dt}=\frac{\pi}{3}(\frac{5}{4}*h)^2+h(\fr ac{\pi}{3}*2(\frac{5}{2}*h*\frac{5}{4}*\frac{dh}{d t}))$

This is not dv/dt, this is dv/dh as you differentiated V with respect to h.

And you needn't apply the product rule here...

$\displaystyle V=\dfrac{1}{3}\pi(\frac{5}{4}h)^2h = \dfrac{25}{48} \pi h^3$

$\displaystyle \dfrac{dV}{dh} = \dfrac{25}{16} \pi h^2$

Then use the chain rule:

$\displaystyle \dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

And solve for dh/dt at h = 6