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Math Help - Rate Problem Involving Changing Height..

  1. #1
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    Exclamation Rate Problem Involving Changing Height..

    Hello So first this is the problem:

    A Swimming pool is 20 ft. wide, 40 ft. long, 3 ft. deep at the shallow end and 9 ft. deep at it's deepest point. A cross section is shown in the figure. If the pool is being filled at a rate of 0.8 cu ft. per min., how fast is the water level rising when the depth at the deepest end is 5ft.
    Rate Problem Involving Changing Height..-dumb.pngI understand that dV/dT= .8cu ft/min but what I don't understand is how to set the problem up.. V=hlw where length and width are constant but the height is changing.. 9 is also the critical point since it's the highest.. but how do we set this problem up. with the height changing? dh/dt?
    Thank you Very Much for your help... This is my first post.. I love browsing in this forum. cheers.
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  2. #2
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    Re: Rate Problem Involving Changing Height..

    when h < 6 \, ft ...

    V = 20(h^2 + 12h)

    you are given \frac{dV}{dt} and h ... take the time derivative of the volume equation and solve for \frac{dh}{dt}
    Thanks from kspkido
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  3. #3
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    Re: Rate Problem Involving Changing Height..

    hmmm may I ask how did you derive that equation? thank you
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    Re: Rate Problem Involving Changing Height..

    Do you know how to find the area of a trapezoid?
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  5. #5
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    Re: Rate Problem Involving Changing Height..

    A=(b1+b2)h/2 yep....
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  6. #6
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    Re: Rate Problem Involving Changing Height..

    Quote Originally Posted by kspkido View Post
    hmmm may I ask how did you derive that equation? thank you
    note the sketch ...
    Attached Thumbnails Attached Thumbnails Rate Problem Involving Changing Height..-traparea.jpg  
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