# related rates help

• Feb 22nd 2007, 03:49 AM
related rates help
WAAAAAAHHHHHH!!!! please move this to calculus section... am very sorry for this....
A water tank in the form of an inverted cone is being emptied at the rate of 6m^3 / min. The altitude of the cone is 24m, and the radius is 12m. Find how fast the water level is lowering when the water is 10m deep.
• Feb 22nd 2007, 05:06 AM
Jhevon
We first want to draw a diagram to see what's going on. And fill in everything we know, and want to know. Notice the rate dV/dt is negative, since the volume is decreasing.

We realize from the second diagram (the cross section of the tank) that we have a similar triangle relationship between the radius and the height. We will rewrite the height in terms of the radius so we can use it, since we are not given the rate the radius is changing. The formula for the volume of the tank is V=(1/3)pi*r^2*h, so here goes.

From similar triangles we realize that r/h = 12/24 => r = h/2

Now V=(1/3)pi*r^2*h
=> V=(1/3)pi*(h/2)^2*h ..............plugged in the formula for r from above
V=(1/3)pi*[(h^2)/4]*h
V = (1/12)pi*h^3
=> dV/dt = (pi/12)*3h^2 dh/dt
=> dV/dt = (pi*h^2/4)dh/dt
=> dh/dt = (4/pi*h^2) dV/dt

plugging in the values we want: h=10, dV/dt= -6

so dh/dt = - 6/25pi m/min................which is the rate of change of the height of the water when h=10 m, note it is negative since the height is decreasing

So that's the way to do it, at least the only way I know. Will one of you fine mathematicians check my calculations, I'm late for school, so i was in a rush typing this.