1. Rates of change

Sand being poured from a conveyor belt forms a cone with height h and a semi-vertical angle of 60 degrees. Let the volume of the sand is $\displaystyle \pi h^3$.

Suppose that the sand is being poured at a constant rate of 0.3m^3/min. Find the rate of increase of the base area.

2. I can suggest

$\displaystyle \frac{dV}{dh} = 3\pi h^2$

$\displaystyle \frac{dV}{dt} = 0.3$

$\displaystyle \frac{dA}{dr} = 2\pi r$

3. Was $\displaystyle V= \pi h^3$ given in the problem?
The volume of a cone, of height h and base radius r, is given by $\displaystyle V= \frac{1}{3}\pi r^2 h$. Here, you are told that the "semi-vertical angle" is 60 degrees so using $\displaystyle \frac{r}{h}= tan(60)= \sqrt{3}$. $\displaystyle r= \sqrt{3}h$, $\displaystyle r^2= 3h^2$ and $\displaystyle V= \pi h^3$. That is where that formula came from.

To change to a "rate of change" formula, differentiate both sides with respect to t, using the chain rule.

$\displaystyle \frac{dV}{dt}= \frac{d\left(\pi h^3\right)}{dt}= 3\pi h^2 \frac{dh}{dt}$
Using the fact that $\displaystyle \frac{dV}{dt}= .3$, you can solve that for $\displaystyle \frac{dh}{dt}$, as a function of h.

The area of the base is, of course, $\displaystyle A= \pi r^2= 3\pi h^2$ so $\displaystyle \frac{dA}{dt}= 6\pi h\frac{dh}{dt}$.

4. Originally Posted by HallsofIvy
Was $\displaystyle V= \pi h^3$ given in the problem?
The volume of a cone, of height h and base radius r, is given by $\displaystyle V= \frac{1}{3}\pi r^2 h$. Here, you are told that the "semi-vertical angle" is 60 degrees so using $\displaystyle \frac{r}{h}= tan(60)= \sqrt{3}$. $\displaystyle r= \sqrt{3}h$, $\displaystyle r^2= 3h^2 and$V= \pi h^3[/tex]. That is where that formula came from.

To change to a "rate of change" formula, differentiate both sides with respect to t, using the chain rule.

$\displaystyle \frac{dV}{dt}= \frac{d\left(\pi h^3\right)}{dt}= 3\pi h^2 \frac{dh}{dt}$
Using the fact that $\displaystyle \frac{dV}{dt}= .3$, you can solve that for $\displaystyle \frac{dh}{dt}$, as a function of h.

The area of the base is, of course, $\displaystyle A= \pi r^2= 3\pi h^2$ so $\displaystyle \frac{dA}{dt}= 6\pi h\frac{dh}{dt}$.
Yes but then, what- I get a figure which still has the pronumeral "h"

5. Hello requal
Originally Posted by requal
Yes but then, what- I get a figure which still has the pronumeral "h"
Unless you are given a specific value of $\displaystyle h$, that's all you will be able to do: express the rate of change of $\displaystyle A$ as a function of $\displaystyle h$.

You could do, by eliminating $\displaystyle h$ between the equations $\displaystyle V= \pi h^3$ and $\displaystyle A = 3\pi h^2$, but you'd then end up with an expression for $\displaystyle \frac{dA}{dt}$ in terms of $\displaystyle V$ (or $\displaystyle A$, or both).