# One more homework problem

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• Feb 16th 2016, 03:43 PM
marlonbustamante
One more homework problem
This word problem is really confusing me.

Attachment 35351

How would I start this problem?
• Feb 16th 2016, 04:48 PM
HallsofIvy
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).
• Feb 16th 2016, 05:57 PM
marlonbustamante
Re: One more homework problem
I'm still having a little trouble picturing what the question is asking.
• Feb 16th 2016, 06:29 PM
Archie
Re: One more homework problem
Attachment 35353

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 $\displaystyle y$
• Feb 16th 2016, 06:42 PM
marlonbustamante
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.