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Math Help - Related Rates Problems (two)

  1. #1
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    Related Rates Problems (two)

    Alright, I will need some assistance for two problems that I face. Thanks in advance.

    A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet.

    (a) If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.9 foot deep?

    (b) If the water is rising at a rate of inch per minute when h = 2.5, determine the rate at which water is being pumped into the trough.

    Honestly I'm not sure how to do this one at all, I need an explanation.

    AND

    All edges of a cube are expanding at a rate of 6 centimeters per second. (a) How fast is the volume changing when each edge is 3 centimeter(s)?

    -I set dV/dt as 6 cm/s
    -We must find dx/dt when x=3cm

    The eq. is V=x^3

    So dV/dt = 3x^2 (dr/dt)

    so I solved for dr/dt, dr/dt = dV/dt*1/3x^2. Then I plugged in 3 for x, but this wasn't correct, where did I go wrong? Thanks a lot guys, much appreciated.
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  2. #2
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    look at the small triangle of water at the end of the trough ... it is similar to the triangle that forms the end base of the entire trough.

    \frac{b}{h} = \frac{3}{3} = \frac{1}{1}

    base equals height.

    the volume of water in the trough is ...

    V = \frac{1}{2}bh \cdot 12 = 6h^2

    now you have a water volume formula strictly in terms of h ... take the time derivative, substitute your given info, then solve for what you need.


    for the second problem, the edge is increasing at 6 cm/sec, not the volume.

    \frac{dx}{dt} = 6

    V = x^3

    take the time derivative, sub in your given values, calculate \frac{dV}{dt}

    btw, where did "r" come from?
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