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Math Help - Rate of change

  1. #1
    Junior Member ihmth's Avatar
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    Rate of change

    1.) A swimming pool os 40ft long 20ft wide 8ft deep at the deep end and 3ft deep at a shallow depth. The bottom being rectangular. If the pool is filled with by pumping water into it at the rate of 100 cubic ft per min. How fast is the water level rising, when it is 3ft deep at the deep end?

    2.) At what rate is the shadow of a 6ft tall man shortening as he walks at 8ft per sec. on a level path towards the street light that is 20ft above the ground.

    Here are two problems about related rates that I am having a problem with solving, so can anyone give me the right way to solve them?


    tnx
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  2. #2
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    Hello, ihmth!

    1) A swimming pool is 40 ft long, 20 ft wide, 8 ft deep at the deep end
    and 3 ft deep at the shallow end. The bottom is rectangular.
    If the pool is filled with by pumping water into it at the rate of 100 ft³/min,
    how fast is the water level rising when it is 3 ft deep at the deep end?
    Code:
                          40
        - * - - - - - - - - - - - - - - - - - *
        : |                                   |
        3 |                                   | 3
        : |               40                  |
        _ * - - - - - - - - - - - - - - - - - *
        : |           x                  *
        : + - - - - - - - - - - - *
        5 |:::::::::::::::::*
        : |h::::::::::*
        : |:::::*
        - *

    The pool is filled to a height of h feet.

    The volume of the water is: . \text{(area of triangle) } \times \text{ (width of pool)}

    So we have: . V \;=\;\left(\tfrac{1}{2}xh\right)(20) \quad\Rightarrow\quad V \;=\;10xh .[1]


    From similar right triangles, we have: . \frac{x}{h} \:=\:\frac{40}{5} \quad\Rightarrow\quad x \:=\:8h .[2]

    Substitute [2] into [1]: . V \:=\:10(8h)h\quad\Rightarrow\quad V \:=\:80h^2


    Differentiate with respect to time: . \frac{dV}{dt} \:=\:160h\cdot \frac{dh}{dt} .[3]

    We are given: . h = 3,\;\frac{dV}{dt} = 100

    Substitute into [3]: . 100 \:=\:160(3)\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} \:=\:\frac{100}{480}\:=\:\frac{5}{24}


    Therefore, the water is rising at: . \tfrac{5}{24}\text{ ft} = 2\tfrac{1}{2}\text{ inches per minute.}

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