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A trough is 12 feet long and 3 feet across at the top. It ends are isosceles triangles with a height of 3 feet. If water is being pumped into the trough at 2.5 cubic feet per minute, how fast is the water level rising when the water is 1 foot deep?

  • calculus -

    At a given time of t minutes,
    let the width of the water level be x ft
    let the height of the water level be h ft
    but by ratios, x/h = 3/3
    x = h

    V = area of triangle x 12
    = (1/2)xh(12
    = 6xh
    = 6h^2

    given : dV/dt = 2.5 ft^3/min
    find dh/dt when h = 1

    dV/dt = 12h dh/dt
    2.5 = 12(1) dh/dt
    dh/dt = 2.5/12 = 5/24 ft/min or appr. .208 ft/min

  • calculus -

    Dezz nuts

  • calculus -


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