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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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?

Calculus1
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whats is "r" equal too ?
the question says 3d=h

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Other answers:

radius is half so 2/3r=h?
other way around. d = 3h
so r = 3h/2
and V = πr²h/3, so make the substitution for r then take the derivative
okey i did eberyhing i am getting 1/9pi and u ?
i did somthing wrong
yeah I got something different? What was your derivative?
did u get 1/6pi?
yes
If you mean \[\Large \frac{1}{6\pi}\] then you are correct
how ?
oh, that's what I thought you meant
what did you get for dV/dt ?
my der was 9/12pih3 dh/dt?
after putiing in 3/2h
the h^3 part is incorrect
how
you will have h^3 in the V volume formula but after you derive, it turns into 3h^2
the formula for cone is x=1/3r2h
when 3/2h goes in it become h2 then h3 by timising the outside h
|dw:1433903048615:dw|
|dw:1433903137752:dw|
ok what are the steps take the derv fist then sub?
Sub for r first because we don't have a value for dr/dt, then take the derivative
wait so (3h/2) doesnt equal 9h2/4? it eiqls 9h/4?
oh wait i think i get it now
you good?
h is 2 right
yes
then 9(2)^2 = 36
hold on the 1/3 is gone beacuse of derv right
well the 1/3 was gone before the derivative. It was 3/4 after the substitution, and 9/4 after the derivative
damn i am going to take 5 min brake
\[V = \frac{1}{3} \pi r^2 h\] \[r = \frac{3}{2} h\] \[\rightarrow V = \frac{1}{3} \pi (\frac{3}{2}h)^2 h = \frac{3}{4} \pi h^3\] Take derivative with respect to time \[\frac{dV}{dt}= \frac{9}{4} \pi h^2 \frac{dh}{dt}\] plug in dV = 1.5, h =2 , then solve for dh/dt \[\frac{dh}{dt} = \frac{1.5}{9 \pi}\]

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