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Math Help - Related Rates, water entering trough problem

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    Related Rates, water entering trough problem

    A trough is 15ft long and 4ft across the top. Its ends are isosceles triangles with height 3 ft. Water runs into the trough at a rate of 2.5 cubic feet per minute. How fast is the water rising when it is 2 feet deep?

    The answer is .0625 but I'm pretty lost on how to get there.
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    Quote Originally Posted by Porcelain View Post
    A trough is 15ft long and 4ft across the top. Its ends are isosceles triangles with height 3 ft. Water runs into the trough at a rate of 2.5 cubic feet per minute. How fast is the water rising when it is 2 feet deep?

    The answer is .0625 but I'm pretty lost on how to get there.
    \frac{dV}{dt} = 2.5 \, ft^3/min

    volume of water in the tank ...

    V = \frac{1}{2} \cdot b \cdot h \cdot 15

    using similar triangles ...

    \frac{b}{h} = \frac{4}{3}

    solve for b, substitute the result for b in the volume formula to get V as a function of h.

    take the time derivative and determine \frac{dh}{dt} when h = 2 ft
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    Skeeter you always answer my questions. Thanks SOO much
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