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apbaldasare

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Jan 3, 2012
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Hey guys! So, I'm rather new to this page. Anyway, I have a take home quiz that I was absent from school on the day it was handed out. I'm in Ap Calculus. Anyway, it's one of the "filling a conical tank" exercises.

A container has the shape of an open right circular cone. The height of the container is 10 cm. The diameter of the opening is also 10 cm. The water in the container is evaporating so that it's depth(h) is changing at the constant rate of -3/10 cm/hr.

Basically, I need to find the rate of change of the volume of water in the container, with respect to time, when the depth= 5cm. I also need to indicate any units of measure.

I really appreciate anyone who can try to help me out on this! Thanks guys!
 
You need the Volume of the cone in terms of the height of the cone - "height" meaning apex (the bottom) to the top of the water.

You will need the constant ratio of the height to the radius at any depth.

Let's see what you get.
 
Volume

Okay, I was absent when we went over this. But I know the formula for volume is V= 1/3 pi r2 h. So, considering the fact that both sides are 10, I'm assuming it's the same for the water. (I'm probably wrong though) So, 1/3 pi times 2.52 times 5 = 32.7. Is this correct, and where would I go from there? Thanks for your help!
 
Okay, that helps me find the volume. How does it help me find the rate of change of the volume, when h=5cm?
 
What did you get for V(h) - the Volume in terms of the Height only?
 
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