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Thread: One more homework problem

  1. #1
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    One more homework problem

    This word problem is really confusing me.

    One more homework problem-capture.jpg

    How would I start this problem?
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  2. #2
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    Re: One more homework problem

    What is "the area in the first quadrant bounded above by the curve from (0, 0) to (x, y)= (x, f(x)) and below by y the x-axis? What is "the area of the rectangle with opposite vertices at (0, 0) and (x, y)= (x, f(x))? Set the first equal to 1/3 the second and solve for f(x).
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  3. #3
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    Re: One more homework problem

    I'm still having a little trouble picturing what the question is asking.
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    Re: One more homework problem

    One more homework problem-sketch.png

    The shaded blue area is one third of the area of the rectangle.

    Thus
    $\displaystyle \int \limits_0^x f(t,c) \, \mathrm d t = \tfrac13xf(x,c)$

    If you differentiate that and write $\displaystyle y=f(x,c)$, you should have a differential equation that you can solve for $\displaystyle y$
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  5. #5
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    Re: One more homework problem

    Thank you for the illustration. I was only imagining (x,y) as a point and didn't see it made a box. I wrote down (1/3)xf(x,c) but didn't understand where it came from, now I do.
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