Hello, Tascja!

Sand is being poured onto a conical pile at the rate of 9 m³/hr.

Friction forces in the sand are such that the slope of the sides of the conical pile is always 2/3. Code:

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The volume of a cone is: .$\displaystyle V \:=\:\frac{\pi}{3}r^2 h$

We are told that the ratio: .$\displaystyle \frac{h}{r} \,=\,\frac{2}{3}\quad\Rightarrow\quad\begin{Bmatri x}h = \frac{2}{3}r \\ r = \frac{3}{2}h\end{Bmatrix}$

a. How fast is the altitude increasing when the radius of the base of the pile is 6 m?

We have: .$\displaystyle V \:=\:\frac{\pi}{3}r^2 h$

Since $\displaystyle r = \frac{3}{2}h\!:\;\;V \:=\:\frac{\pi}{3}\left(\frac{3}{2}h\right)\!^2h \:=\:\frac{3\pi}{4}h^3$

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt}\:=\:\frac{9\pi}{4}h^2\left(\frac{dh} {dt}\right)$ **[1]**

When $\displaystyle r = 6\!:\;h = \frac{2}{3}(6) = 4$ . . . and we are given: .$\displaystyle \frac{dV}{dt} = 9$

Substitute into **[1]**: .$\displaystyle 9 \:=\:\frac{9\pi}{4}(4^2)\left(\frac{dh}{dt}\right) $

Therefore: .$\displaystyle \frac{dh}{dt} \:=\:\frac{1}{4\pi}$ m/hr.

b. How fast is the radius of the base increasing when the height of the pile is 10 m?

We have: .$\displaystyle V \:=\:\frac{\pi}{3}r^2 h$

Since $\displaystyle h = \frac{2}{3}r\!:\;\;V \:=\:\frac{\pi}{3}r^2\left(\frac{2}{3}r\right)\:=\ :\frac{2\pi}{9}r^3$

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt}\:=\:\frac{2\pi}{3}r^2\left(\frac{dr} {dt}\right)$ **[2]**

When $\displaystyle h = 10\!:\;r = \frac{3}{2}(10) = 15$ . . . and we are given: .$\displaystyle \frac{dV}{dt} = 9$

Substitute into **[2]**: .$\displaystyle 9 \:=\:\frac{2\pi}{3}(15^2)\left(\frac{dr}{dt}\right ) $

Therefore: .$\displaystyle \frac{dr}{dt} \:=\:\frac{3}{50\pi}$ m/hr.