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Calculus/Related Rates

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Glenninator wrote at 2008-12-07 19:58:23
He's kind of right, however the answer s flawed because a cone's volume formula is (((pi)r^2)h))/3. He solved the problem for a cylinder. Do the work again put with the afforementioned formula from the top. I'm not gonna give you the answer put you should be able to follow this guy's work with the new formula.


Alexis wrote at 2011-10-11 05:41:53
Relations:



From similar triangles, h=4r [Since a full pool has h=4r, so does any partially full pool.]



V= (pi/3) r^2h



Diff:



dv/dt= (pi/3)*(h/4)^2 h



dv/dt= pi/16h^2 dh/dt



Answer:



dh/dt= 12/pi


Jothi wrote at 2014-10-15 01:20:49
Part 2 How fast the radius changing.



We have 4r=h



v=1/3pir^2 h



v=(1/3)pi r^2 (4r)



v=(1/3) pi4 =r^3



dv/dt =(4/3) pi 3r^2 dr/dt     h=4r so at h=2  r=1/2





so dv/dt =(4/3) pi 3/4   dr /dt

3=pi dr /dt



dr/dt =3/pi  meter/minute









Or take h=4r

differentiate dh/dt=4 dr/dt  we know that dh/dt =12/pi

         

         12/pi=4dr/dt



         dr/dt=3/pi meter/minute  


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Paul Klarreich

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All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

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