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Rate of change  cones 
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#1
Mar1412, 04:00 PM

P: 614

1. The problem statement, all variables and given/known data
I have attached a link along with my working please can someone help me http://s359.photobucket.com/albums/o...rrent=Math.png 2. Relevant equations 3. The attempt at a solution 


#2
Mar1412, 04:20 PM

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#3
Mar1412, 04:37 PM

P: 614

dv/dt = 4 cm^{3} min^{1} tan(60) = r/h r = √3* h V = (1/3) π r^{2}h dv/dh = (1/3) π r^{2} dh/dt = (dh/dv) * (dv/dt) = ^{1}/_{(1/3) π r2} * 4 = 12/(pi r^{2}) for h = 4 dh/dt = 0.079577 The answer given is 0.0265 cm/min. why? I don't know how do use latex  see post for a clearer solution if you get lost! 


#4
Mar1412, 04:45 PM

P: 614

Rate of change  cones
question was:
A hollow cone with a semivertical angle of 60 degrees is held vertex down with its axis vertical. Water drips into the cone at 4 cm^{3}/min Find the rate at which the depth of water is increasing when the water is 4 cm deep 


#5
Mar1412, 05:40 PM

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P: 21,214

Substitute for r in your volume equation. Then you'll have V purely as a function of h. What you have is not correct, because V is a function of r and h. 


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