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A trough is triangular in cross section, an isosceles triangle with sides of 12 inches, and a top of 10 inches. The trough is 40 inches long. How fast is the depth changing if you are pumping one cubic foot per minute into the trough?

I am having a hard time getting it in terms of h.

  • Calculus -

    Make a sketch of the cross-section of the isosceles triangle. Draw it some water level, letting the height be h and letting the width of the water level be x
    by similar triangles
    x/h = 10/12
    12x = 10h
    x = 5h/6

    volume of water = 40(area of triangle)
    = 40(1/2)(x)h
    = 20(5h/6)(h) = 50/3 h^3

    V = (50/3) h^3
    dV/dt = 50 h^2 dh/dt
    have to watch our units, 1 cubic foot = 12^3 or 1728 cubic inches

    1728 = 50 h^2 dh/dt
    dh/dt = 1728/(50h^2) = 864/25 inches/minute

  • Calculus -

    wow I was headed that way didn't watch units Thank You very much

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