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 May 21st, 2015, 03:36 PM #1 Senior Member   Joined: Nov 2013 Posts: 247 Thanks: 2 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. May 21st, 2015, 10:36 PM #2 Senior Member   Joined: Nov 2013 Posts: 247 Thanks: 2 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. Tags delta, derivative, height, radius, volume Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Pumaftw Elementary Math 2 October 13th, 2014 10:07 AM daigo Calculus 2 June 23rd, 2013 02:52 PM vypyr Calculus 2 February 9th, 2013 02:30 PM mymathgeo Real Analysis 17 January 26th, 2012 10:00 PM sikwn32 Calculus 1 June 24th, 2010 09:51 AM

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