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Math Help - [SOLVED] Related rates: trapezoidal water trough

  1. #1
    Member sinewave85's Avatar
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    [SOLVED] Related rates: trapezoidal water trough

    A water trough is 2 ft deep and 10 ft long. It has a trapezoidal cross section with base lengths 2 and 5 feet.
    • Find a relationship between the volume of water in the tank at any given time and the depth of the water at that time.

    Where V= volume, L = length, D = depth of water
    V = L [(b1 + b2)/2] D
    V = 35D

    • If water enters the trough at a rate of 10 ft^3/min, how fast is the water level rising (to the nearest 0.5 in./min) when the water is 1 ft deep.

    V' = 35

    Obviously, that doesn't work. Any advice?
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  2. #2
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    We want \frac{dh}{dt} when h=1, given \frac{dV}{dt}=10.

    Let b=length of top of cross-section of the water.

    h=depth of water.

    By the area of a trapezoid, we have the volume of the trough:

    V=h((b+2)/2)(10)

    By similar triangles, think of the two triangles on the ends as one isosceles triangle.

    (b-2)/h=3/1

    b=3h+2

    V=5(3h^2+4h)

    Now, can you finish up by differentiating impicitly wrt time?.

    Enter in your givens, h, dV/dt, and solve for dh/dt.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sinewave85 View Post
    A water trough is 2 ft deep and 10 ft long. It has a trapezoidal cross section with base lengths 2 and 5 feet.
    • Find a relationship between the volume of water in the tank at any given time and the depth of the water at that time.

    Where V= volume, L = length, D = depth of water
    V = L [(b1 + b2)/2] D
    V = 35D
    • If water enters the trough at a rate of 10 ft^3/min, how fast is the water level rising (to the nearest 0.5 in./min) when the water is 1 ft deep.

    V' = 35
    Obviously, that doesn't work. Any advice?
    Similar triangles


    EDIT: oops, missed Galactus's post

    Sorry
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  4. #4
    Member sinewave85's Avatar
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    Quote Originally Posted by galactus View Post
    We want \frac{dh}{dt} when h=1, given \frac{dV}{dt}=10.

    Let b=length of top of cross-section of the water.

    h=depth of water.

    By the area of a trapezoid, we have the volume of the trough:

    V=h((b+2)/2)(10)

    By similar triangles, think of the two triangles on the ends as one isosceles triangle.

    (b-2)/h=3/1

    b=3h+2

    V=5(3h^2+4h)

    Now, can you finish up by differentiating impicitly wrt time?.

    Enter in your givens, h, dV/dt, and solve for dh/dt.

    Do'h -- I was using top of the trough instead of the top of the water for the second b! No wonder I was a variable short of an equation, so to speak. Thanks so much for the help!
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  5. #5
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    i have a question in galactus post, in similar triangles, is it really really (b-2)/h=3/1 or should it be (b-2)/h=3/2 ?? please rep.. im really confused... tnx
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