Thread: Cones

I need help solving this question... i dont even know where to start... so confused:

Sand is being poured onto a conical pile at the rate of 9 m^3/h. Friction forces in the sand are such that the slope of the sides of the conical pile is always 2/3.

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

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

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.

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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.