Sand is falling on a pile , having shape pf a cone, at the rate 1.5 m3/min. assime diameter of base is 3 times the altitiude, at what rate is the altitude increasing when the altitude is 2m?
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OpenStudy (anonymous):
@ganeshie8 @ParthKohli @Loser66
OpenStudy (anonymous):
whats is "r" equal too ?
OpenStudy (anonymous):
the question says 3d=h
OpenStudy (anonymous):
radius is half so 2/3r=h?
OpenStudy (anonymous):
other way around. d = 3h
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OpenStudy (anonymous):
so r = 3h/2
OpenStudy (anonymous):
and V = πr²h/3, so make the substitution for r then take the derivative
OpenStudy (anonymous):
okey i did eberyhing i am getting 1/9pi and u ?
OpenStudy (anonymous):
i did somthing wrong
OpenStudy (anonymous):
yeah I got something different? What was your derivative?
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OpenStudy (anonymous):
did u get 1/6pi?
OpenStudy (anonymous):
yes
jimthompson5910 (jim_thompson5910):
If you mean \[\Large \frac{1}{6\pi}\] then you are correct
OpenStudy (anonymous):
how ?
OpenStudy (anonymous):
oh, that's what I thought you meant
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jimthompson5910 (jim_thompson5910):
what did you get for dV/dt ?
OpenStudy (anonymous):
my der was 9/12pih3 dh/dt?
OpenStudy (anonymous):
after putiing in 3/2h
jimthompson5910 (jim_thompson5910):
the h^3 part is incorrect
OpenStudy (anonymous):
how
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jimthompson5910 (jim_thompson5910):
you will have h^3 in the V volume formula
but after you derive, it turns into 3h^2
OpenStudy (anonymous):
the formula for cone is x=1/3r2h
OpenStudy (anonymous):
when 3/2h goes in it become h2 then h3 by timising the outside h