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I have a practice test for my calculus final and I need to figure out how to do this stuff.... if anyone can do this problem and show the work for me to understand it, I would appreciate it.... it is:

A load of grain is being dumped so that the grain forms a pile in the shape of a cone where the volume of the grain is given by V= (2Pie)/(9) x r^3 where r is the radius of the base of the pile. The radius of the base is changing at the rate of 4.8 ft/min at the moment the radius is 21.2 ft. How fast is the volume of the pile changing at that moment.... to the nearest tenth. Again, the equation is: 2 Pie divided by 9 times r^3 If you are smart, please help me

2007-04-23 18:06:51 · 2 answers · asked by nig_ga_4_life 1 in Science & Mathematics Mathematics

2 answers

First, please use "pi" for 3.1416. That's what they call the small Greek letter. If the volume is the relation given, V=V(r), so to get dV/dt, we use the "chain rule" dV/dr x dr/dt = dV/dt. Then dV/dt= (2 pi/9) 3r^2 dr/dt
Now all you do is substitute 4.8 ft/min for dr/dt, and 21.2 ft for r in the dV/dt equation and solve.

2007-04-23 18:20:23 · answer #1 · answered by cattbarf 7 · 0 0

V = 2pir^3/9
dV/dr = 6/9 pir^2 = 2pir^2/3
So dV = 2pir^2/3 dr
So dV = 2*3.14*21.2^2*4.8 = 13,547.9 ft^3/min

2007-04-23 18:23:08 · answer #2 · answered by ironduke8159 7 · 0 0

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