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?

- anonymous

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- anonymous

@ganeshie8 @ParthKohli @Loser66

- anonymous

whats is "r" equal too ?

- anonymous

the question says 3d=h

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## More answers

- anonymous

radius is half so 2/3r=h?

- anonymous

other way around. d = 3h

- anonymous

so r = 3h/2

- anonymous

and V = πr²h/3, so make the substitution for r then take the derivative

- anonymous

okey i did eberyhing i am getting 1/9pi and u ?

- anonymous

i did somthing wrong

- anonymous

yeah I got something different? What was your derivative?

- anonymous

did u get 1/6pi?

- anonymous

yes

- jim_thompson5910

If you mean \[\Large \frac{1}{6\pi}\] then you are correct

- anonymous

how ?

- anonymous

oh, that's what I thought you meant

- jim_thompson5910

what did you get for dV/dt ?

- anonymous

my der was 9/12pih3 dh/dt?

- anonymous

after putiing in 3/2h

- jim_thompson5910

the h^3 part is incorrect

- anonymous

how

- jim_thompson5910

you will have h^3 in the V volume formula
but after you derive, it turns into 3h^2

- anonymous

the formula for cone is x=1/3r2h

- anonymous

when 3/2h goes in it become h2 then h3 by timising the outside h

- anonymous

|dw:1433903048615:dw|

- anonymous

|dw:1433903137752:dw|

- anonymous

ok what are the steps take the derv fist then sub?

- anonymous

Sub for r first because we don't have a value for dr/dt, then take the derivative

- anonymous

wait so (3h/2) doesnt equal 9h2/4? it eiqls 9h/4?

- anonymous

oh wait i think i get it now

- anonymous

you good?

- anonymous

h is 2 right

- anonymous

yes

- anonymous

then 9(2)^2 = 36

- anonymous

hold on the 1/3 is gone beacuse of derv right

- anonymous

well the 1/3 was gone before the derivative. It was 3/4 after the substitution, and 9/4 after the derivative

- anonymous

damn i am going to take 5 min brake

- dumbcow

\[V = \frac{1}{3} \pi r^2 h\]
\[r = \frac{3}{2} h\]
\[\rightarrow V = \frac{1}{3} \pi (\frac{3}{2}h)^2 h = \frac{3}{4} \pi h^3\]
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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