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Depth of a cone rate problem (question about the equation I'm using)

  • Thread starter KatieLynn
  • Start date
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1. Homework Statement

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

2. Homework Equations

Cone: V= (Ah/3)

Right Circular Cone: V=[(pi)(r^2)(h)]/(3)

3. The Attempt at a Solution

So on this on I'm confused about the equation I need to derive to begin with..

I looked up the volume of a cone and it had two choices (stated above)

So I think I need to use just the cone one because the problem doesn't say its a right circular cone. Is that correct?

Assuming I do in fact need to use V= (Ah/3)

Do I then plug in A=(pi)(r^2) which is the area of a circle? the book says A equalls the area of the base, which appears to be a circle to me...
 
Last edited:

Answers and Replies

1,631
4
ok the volume of the cone is

[tex]V=\frac{1}{3}r^{2}\pi H[/tex]

The info that you know are:

[tex]\frac{dV}{dt}=10[/tex], [tex] r=5[/tex]

We need to find [tex]\frac{dH}{dt}=?[/tex] when H=8(be carefule here, you are not to plug this value for H in, immediately on the equation for the volume, since H (depth) is changing). Now try to find a relation between the radius of the circle on the cone when H=8, using these information.
Draw a picture it usually helps.
 

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