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anonymous

  • 5 years ago

Please help with integration attachment.

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  1. TuringTest
    • 5 years ago
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    here we go again, eh?

  2. anonymous
    • 5 years ago
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    Lol!

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  3. TuringTest
    • 5 years ago
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    I've never seen a problem like this before...

  4. anonymous
    • 5 years ago
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    i dont even know where to start. The professor says its supoosed to be challenging.

  5. TuringTest
    • 5 years ago
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    I figure if I can make an expression for x and y at various points I can get the slope of the line L, which is the value of the derivative. |dw:1326841893689:dw|The figure lets me get L-x for the horizontal side of the triangle, but what is the height?I don't know if I'm headed in the right direction, since what I called P is a function of x... But if I could get two coordinate values I suppose I could make a formula for the slope of the line L, which would be f'(x) Toughie...

  6. anonymous
    • 5 years ago
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    Ok, i'll see what i can work out with that idea.

  7. anonymous
    • 5 years ago
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    Thank you!

  8. TuringTest
    • 5 years ago
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    I summoned someone smarter than I. Hopefully he will show up and save us.

  9. anonymous
    • 5 years ago
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    Actually, I just noticed that my battery is about to die, so i have to go. :( Thank you all the same! :)

  10. TuringTest
    • 5 years ago
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    well help has arrived, so perhaps it will be solved when you return

  11. anonymous
    • 5 years ago
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    I've seen this problem before, give me a few minutes and I think I'll have an answer for you Xylienda

  12. anonymous
    • 5 years ago
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    ok

  13. anonymous
    • 5 years ago
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    I have that \[f'(x)=-\frac{\sqrt{L^2-x^2}}{x}\] This is time-consuming, though not too hard, function to integrate, but wolfram gives the integrated function as \[f(x)=L \ln \frac{L+\sqrt{L^2-x^2}}{x}-\sqrt{L^2-x^2}\] I'll type up the solution in a moment.

  14. anonymous
    • 5 years ago
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    ok

  15. anonymous
    • 5 years ago
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    Sorry for the wait, just had to remember how I got the negative sign in there :) Okay, first some notation: let (0,p) be the point on the y-axis that the man is standing at. We are going to let f(m) be the equation of the tractrix itself, so f'(m) will be its derivative. Now, the most important part of the problem is to note that the length of the rope is constant (it is always L). Thus, when the man is at point (0,p) and the boat is at (m,f(m)), we have that \[L=\sqrt{(p-f(m))^2+m^2}\] Rearranging a bit gives \[p-f(m)=\sqrt{L^2-m^2}\] Now, consider the fact that the rope is always the tangent line to the tractrix at a given time. The slope of the line is just f'(m) and the y-intercept is p, giving you \[y=f'(m)x+p\] Since (m,f(m)) is also a point on this line, you then have \[f(m)=f'(m)m+p\] Plugging this in to the above equation gives \[-f'(m)m=\sqrt{L^2-m^2}\] or \[f'(m)=-\frac{\sqrt{L^2-m^2}}{m}\]

  16. TuringTest
    • 5 years ago
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    So I guess I wasn't too far off. In my pic I should have called L-x just x then it's easy to see that the base of the triangle is x, and that it will all be negative because it angles down. By that logic I would have wound up with the right formula for f'(x). I don't feel so bad now :)

  17. anonymous
    • 5 years ago
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    Ok. Thank you both for your help. I'll definitely study this, but my laptop is threatening to hibernate.

  18. anonymous
    • 5 years ago
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    Bye!

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