geerky42
  • geerky42
A trough is 12 feet long and 3 feet across the top. Its end are isolates triangles with altitudes of 3 feet. If water is being pumped into the trough at 2 cubic feet per min, how fast is the water level rising when the water is 1 foot deep?
Calculus1
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SOLVED
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chestercat
  • chestercat
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geerky42
  • geerky42
@hartnn @Hero @helder_edwin @jiteshmeghwal9
jiteshmeghwal9
  • jiteshmeghwal9
sorry @geerky42 no idea :( i'm not good in geometry
geerky42
  • geerky42
Any idea? @rajathsbhat

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More answers

anonymous
  • anonymous
Yes, actually. But tell me, is the trough like this:|dw:1352120844035:dw|?
anonymous
  • anonymous
i don't get the "12 feet long" part.
phi
  • phi
Is isolates triangles supposed to be isosceles triangles?
geerky42
  • geerky42
|dw:1352121448125:dw|
geerky42
  • geerky42
I think this is what question means, but i honestly have no idea...
anonymous
  • anonymous
where are the triangles? :\
geerky42
  • geerky42
|dw:1352121575320:dw|
geerky42
  • geerky42
triangle prism, idk.
anonymous
  • anonymous
yes! phi's got it right!
geerky42
  • geerky42
So it's like trapezoid prism or what?
geerky42
  • geerky42
@mahmit2012 @mayankdevnani
anonymous
  • anonymous
I think it's a triangular prism.
geerky42
  • geerky42
An isosceles triangular prism?
geerky42
  • geerky42
where it has base of 3ft and heigh of 3 ft?
phi
  • phi
the only thing that makes sense is |dw:1352123776310:dw|
phi
  • phi
The ratio of base to altitude of the triangular base is 3/3 = 1 when the water is at height h, the base (width of the water) is also h |dw:1352123882327:dw| the area as a function of h is 1/2 h^2 volume is 12 * 1/2 *h^2
phi
  • phi
\[ v= 6 h^2\] \[\frac{dv}{dt}= 12 h \frac{dh}{dt} \] plug in for h and dh/dt to find dv/dt
geerky42
  • geerky42
I see, thanks!
phi
  • phi
I still don't know what Its end are isolates triangles means....

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