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May 21st, 2015, 03:36 PM   #1
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delta r

A truncated cone has smaller radius of 5 feet, larger radius of 10 feet, and a depth of 1 foot. Total volume is approximately 183 cubic feet. If I have water in there at x inches how much does the radius at the surface of the water increase per inch? If I know this then I can just multiply it by the number of inches a particular volume takes up in a truncated cone with a particular smaller radius(in this case 5 feet). I would imagine that this requires me to do this derivative:

$\displaystyle \frac{dr}{dx} \pi r^2 x |0\leq x\leq 12\, and\, 5 \leq r \leq 10$

where x is the height from the bottom of the truncated cone to the surface of the water

Last edited by caters; May 21st, 2015 at 03:38 PM.
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May 21st, 2015, 10:36 PM   #2
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However I think the rate of change is constant since the slope of the metal surface is constant so that would require that I take the derivative twice. That would give me the derivative of 2pirx which since everything but the x is a constant relative to the derivative operator would be 1. But it can't change by 1 foot per inch or it would have to have a larger radius of 17 ft.

In fact I think I know the answer without the derivative but I am not positive on this. I think the change in radius per inch of water is 5 inches.
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