# Rate of change problem

1. Apr 19, 2008

### Incog

1. The problem statement, all variables and given/known data

A water tank is built in the shape of a circular cone with a height of 6 m and a diameter of 10 m at the top. Water is being pumped into the tank at a rate of 2m$$^{3}$$
per minute. Find the rate at which the water level is rising when the water is 2 m deep.

2. Relevant equations

Volume of a cone - $$\frac{1}{3}$$ $$\Pi$$ r$$^{2}$$ h

Surface area of a cone - $$\Pi$$ r s + $$\Pi$$ r$$^{2}$$

3. The attempt at a solution

$$\frac{dV}{dt}$$ = 2m$$^{3}$$/min

I think I have to find $$\frac{dh}{dt}$$ but other than that I'm completely lost.

2. Apr 19, 2008

### ice109

what is h as a function of v?

3. Apr 20, 2008

### Incog

I'm not sure I understand.

4. Apr 20, 2008

### TMM

Can you relate r and h with the knowledge that the tank is a cone?

5. Apr 20, 2008

### Feldoh

What both ice109 and TMM are trying to say is that you have two variables since

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

Both the radius and height are effecting the volume. So before you can find dh/dt you need to find a way to relate r in terms of h.

6. Apr 21, 2008

### HallsofIvy

Draw a triangle, vertex at the bottom, base horizontal, with height 6 and base length 10, representing the water tank. Draw a horizontal line representing the water line in the tank, with length 2r (since the diameter is twice the radius) and height above the vertex h. Use "similar triangles" to connect h and r.