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Triangular Prism

  1. Nov 26, 2006 #1
    A trinagular prism is 10 m long and has an equilaterlal trinagle for its base. WAter is added at a rate of 2m^3/min. Determine the rate of change of the water level when the water is [itex] \sqrt{3} [/itex] m deep.

    ok so the volume of a prism is
    [tex] V = \frac{1}{2} lwh [/tex] .... (1)
    l is the length
    w is the width
    h is the height

    now dl/dt = 0 because the length of the column of water is constant

    to find a relation between h and w i got this because the triangle is an equilaterla triangle

    [tex] h = \frac{\sqrt{3}}{2} w[/tex] ... (2)

    and it follos that
    [tex] \frac{dh}{dt} = \frac{\sqrt{3}}{2}\frac{dx}{dt} [/tex] ... (3)

    now subbing 1 into 2

    [tex] V = \frac{1}{\sqrt{3}} lh^2 [/tex]

    [tex] \frac{dV}{dt} = \frac{l}{\sqrt{3}} 2h \frac{dh}{dt} [/tex]

    now here's the problem ... what is h??
    h does not represent the depth of the water, does it??
    it reprsnts the height of the remainder of the prism that has not been filled iwht water. so really

    [tex] \frac{dh_{water}}{dt} = \frac{dh_{empty part}}{dt} [/tex]

    is it reasonable to say that??

    thank you for all your input!!
     
  2. jcsd
  3. Nov 27, 2006 #2

    HallsofIvy

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    h is the height of the triangle so it is the depth of the water- you triangle has its vertex downward, remember?
     
  4. Nov 27, 2006 #3
    ahhh true

    the book was rather deceptive in taht it drew the triangle right side up

    but that wouldnt make a difference??
     
  5. Jul 23, 2011 #4
    Yes, the depth of water is the height of the prism and which is root 3 as given.
     
  6. Jul 23, 2011 #5
    I doubt stunner5000pt is still working on this problem after 5 years.
     
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