Water Tank Rate of Change Problem: Find the Rate at Which Water Level is Rising

[Incog]

Homework Statement

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.

Homework Equations

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

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

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.

 

[ice109]

what is h as a function of v?

 

[Incog]

I'm not sure I understand.

 

[TMM]

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

 

[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.

 

[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.