find rate of change of depth of water in conical tank

[Math wiz ya rite 09]

The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute.

Find the rate of change of the depth of the water in the tank.

 

[galactus]

We want dh/dt when h = 7 and dV/dt = -3

This is a case for similar triangles.

Let r = radius of liquid at height h.

Then we have \(\displaystyle \L\\\frac{r}{h}=\frac{5}{12}\)

\(\displaystyle \L\\r=\frac{5h}{12}\)

Sub into the equation for volume of cone.

\(\displaystyle \L\\V=\frac{1}{3}{\pi}\left(\frac{5h}{12}\right)^{2}h=\frac{\pi25h^{3}}{432}\)

Now, differentiate and solve for dh/dt.

 

[Math wiz ya rite 09]

do i take the derivative of this?

\(\displaystyle \L\\V=\frac{\pi25h^{3}}{432}\)

 

[Math wiz ya rite 09]

any other support is appreciated & thank you glatacus

 

[Math wiz ya rite 09]

is the answer -0.11225 ft/sec

 

[galactus]

Yes, it is. Good.

\(\displaystyle \L\\\frac{dV}{dt}=\frac{25{\pi}h^{2}}{144}\frac{dh}{dt}\)

Now, enter in the knowns. dV/dt=-3 and h=7. Solve for dh/dt

\(\displaystyle \L\\-3=\frac{1225{\pi}}{144}\frac{dh}{dt}\)

\(\displaystyle \L\\\frac{dh}{dt}=\frac{-432}{1225{\pi}}\approx{-0.112253...}\)

Perhaps leave your answer in the fractional format instead of the decimal. Looks better and most instructors prefer that.