Derivatives, rates of change (cone)

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1. Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 ft high?

Homework Equations


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

The Attempt at a Solution



Diameter = height, so $$\frac{h}{2}=r$$
$$V=\frac{\pi}{3}\frac{h^2}{4}h = \frac{\pi}{12}h^3$$
$$\frac{dV}{dt}=\frac{\pi}{12}(3×h)\frac{dh}{dt}$$
$$30=\frac{\pi}{12}(3×10)\frac{dh}{dt}$$
$$\frac{dh}{dt}=\frac{12}{\pi}$$

The textbook's answer is $$\frac{6}{5\pi}$$ What did I do wrong?
 
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You incorrectly assumed that the diameter of the pile = half of the radius.
 
Sorry that was just a typo. The work should still follow r=h/2.
 
Check your differentiation.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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