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At a sand and gravel plant, sand is falling off a conveyor, and onto a conical pile at a rate of 10 cubic feet per minute. 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 the pile is 15 feet high?

My Attempt

Given:

$\frac{dV}{dt}=10 \frac{\mathbb{ft}^3}{\mathbb{min}}$
$d=3h \therefore r=\frac{3}{2}h$
$h=15$

Unknown:

$\frac{dh}{dt}=?$

Equation:

$V=\frac{h}{3}\pi (\frac{3}{2}h)^2$
$V=\frac{3}{4}h^3\pi$

Then from here I would take the derivative, and solve for $\frac{dh}{dt}$ right? For an answer I got .0068, which I think is wrong.

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1 Answer

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1. $V = \pi r^2 \frac{h}{3}$

2. diameter $= 3h \Rightarrow r = \frac{3}{2}h$

3. $V(h) = \pi \cdot \frac{9}{4}h^2 \cdot \frac{h}{3}= \frac{3}{4}\pi h^3$

4. Height is a function of time since the height increases as sand piles on.

5. I.e $V(h) = (V \circ h)(t)$

6. Let $t_0$ be such that $h(t_0) = 15$. You just need to solve:

$$10=\frac{d}{dt}\Bigr|_{t = t_0} ( V \circ h)(t) = V'(h(t_0)) \cdot h'(t_0) $$

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