1. ## Related Rates

A growing sand pile Sand falls from a conveyor belt at the rate of 10 m^3 / min onto the top of a conical pile. The height of the pile is always three-eighths of the base of the diameter. How fast are the (a) height and (b) radius changing when the pile is 4m high? Answer in cm/min

I've set up the picture as best as I can. I'm trying to understand the next step.

I first need to find the the rate of change of the height, or dh/dt... so would I differentiate both sides, giving me dV/dt = 1/3π[ 2r(dr/dt) + r^2(dh/dt)] ? I am not sure if this is right or how to proceed.

2. ## Re: Related Rates

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

$h = \dfrac{3}{8}d = \dfrac{3}{8}(2r) = \dfrac{3}{4}r$

$\dfrac{d}{dt}(h) = \dfrac{d}{dt}\left(\dfrac{3}{4} r \right)$

$\dfrac{dh}{dt} = \dfrac{3}{4} \dfrac{dr}{dt}$

Now, substitute in.

$V = \dfrac{1}{3} \pi r^2 \left(\dfrac{3}{4} r \right) = \dfrac{1}{4} \pi r^3$

$\dfrac{dV}{dt} = \dfrac{3}{4} \pi r^2 \dfrac{dr}{dt}$

Substitute in for $r = \dfrac{4}{3} h$ and $\dfrac{dV}{dt} = 10$:

$10 = \dfrac{3}{4} \pi \left( \dfrac{4}{3} h \right)^2 \dfrac{dr}{dt}$

$10 = \dfrac{4}{3} \pi h^2 \dfrac{dr}{dt} = \dfrac{4^3}{3} \pi \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \dfrac{15\pi}{32}$ m/min. Since the answer should be in cm/min, multiply by $\dfrac{100 \text{ cm} }{1 \text{ m}}$:

$\dfrac{dr}{dt} = \dfrac{375 \pi}{8}$ cm/min

Now, above, we found $\dfrac{dh}{dt} = \dfrac{3}{4} \dfrac{dr}{dt} = \dfrac{1125\pi}{32}$ cm/min.

3. ## Re: Related Rates

Unless I'm calculating this wrong, these aren't the right answer.. book shows a) dh/dt = 11.19 cm/min and b) 14.92 cm/min

4. ## Re: Related Rates

Sorry, pi should be in the denominator. My mistake.

$\dfrac{dr}{dt}=\dfrac{375}{8\pi}, \dfrac{dh}{dt}=\dfrac{1125}{32\pi}$