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Rates of change

  1. Mar 26, 2006 #1
    A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3ft across at the top and have a height of 1ft. if the trough is filled with water at a rate of 12ft^3/min, how fast does teh water level rise when the water is 6inches deep?

    so far i have: dv/dt= 12ft^3/min and need to find dh/dt.
    V= base* height*length ? where length=10 ft.
    what i don't understand is how come in my notes, the next step is that b becomes 0.5*b*h??
     
    Last edited: Mar 26, 2006
  2. jcsd
  3. Mar 26, 2006 #2

    Integral

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    I have no way to help you with what is written in your notes. But...

    The expression:

    V= b*h*L

    Gives the volume of a solid with a rectangular cross section. You have a solid with a triangular cross section so the the expression

    V = .5*b*h*L

    Would be correct.

    can you complete the problem from there?
     
  4. Mar 27, 2006 #3

    HallsofIvy

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    V= base*height*length for a RECTANGULAR trough. This is TRIANGULAR. The area of a triangle is 0.5*b*h.
     
  5. Apr 2, 2006 #4
    I think I get it now. since it's a triangular prism, you take the area of the triangle and multiply it by the base of the rectangle. I think that's correct now.
     
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