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kekepania

New member
Joined
Dec 13, 2005
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7
the problem I am trying to figure out is this:

A conical water tower has water entering at the rate of 5 cubic feet per minute. how fast is the radius changing at the instant there is 100 cuic feet of water in the tank?

The diameter is 10 ft and the concial water tower if 10ft in height

I got that the equations for a cone is v=(1/3)(pie)r^2h

I know that DV/DT= 5

I also know that
dv/dt= 1/3(pie) (2r (dr/dt)h + r^2 (dh/dt)

from there i was lost. I know that i want to get rid of one of the variables, and I think that it is h that i want to get rid of so that i can find (dr/dt) but I ma not sure. If someone could help that wold be great... thanks
 
What is the orientation of the cone? Is the "peak" (the pointy bit) at the top, where the water is entering, or at the bottom?

Thank you.

Eliz.
 
Using similar triangles is a good way to eliminate a variable:

\(\displaystyle \L\\\frac{h}{r}=\frac{10}{5}\)

\(\displaystyle \L\\h=2r\)

Sub into your volume formula, differentiate, and solve for dr/dt.

Now, they want change of radius when volume is 100. So, using this info

and your newly acquired volume formula, solve for r and sub into your

derivative.
 
I finally got it... thank you so much for your help... i was just getting a little lost in there..... thanks :D
 
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