Math

Water is flowing into a trough at a rate of 100 cubic centimeters per second. The trough has a length of 3 meters and cross sections in the form of a trapezoid, whose height is 50 cm, whose lower base is 25 cm and whose upper base is 1 meter. At what rate is the water rising when the depth of the water is 25 cm?

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asked by Hao
  1. when the water has depth y cm, the width of the surface has width

    25+(75/50)y cm

    So, at depth y, the area of the trapezoidal cross-section is

    1/2 (25 + (75/50)y)(y) = 3/4 y^2 + 25/2 y cm^2

    So, the volume is
    v = 9/4 y^2 + 75/2 y cm^3

    So, we want dy/dt when y=25, and we know that dv/dt=100:

    9/4 (2y dy/dt) + 75/2 dy/dt = 100
    9/4 (50 + 75/2) dy/dt = 100
    dy/dt = 32/63 cm/s

    An unusual answer; better check my math.

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    posted by Steve
  2. I tried a slightly different approach

    I looked at the trapezoid, and drew two verticals upwards from the base,
    creating a rectangle of 25 by 50 cm and two right-angled triangles ,
    each of height 50 and base 75/2

    I drew a water line, of height h and width of r in each of the triangles
    by ratios: r/h = (75/2) / 50 = 3/4
    r = (3/4)h

    V = 300( 25h + 2(1/2)rh)
    = 300(25h + rh)
    = 300(25h + 3/4)h^2)

    dV/dt = 300( 25dh/dt + (3/2) h dh/dt)
    = 300(dh/dt)(25 + (3/2)h)

    when h = 25 , and dV/dt = 100

    100 = 300 (dh/dt) (25 + 37.5)

    dh/dt = 100/(300(62.5)) = 100/18750 = 2/375 cm/s or appr .00533 cm/s

    As usual, my arithmetic should be checked also

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    posted by Reiny
  3. Well, my most obvious mistake was using 3 as the length, not 300 (cm).

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    posted by Steve

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