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anonymous
 5 years ago
Please help with integration attachment.
anonymous
 5 years ago
Please help with integration attachment.

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TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1here we go again, eh?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1I've never seen a problem like this before...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont even know where to start. The professor says its supoosed to be challenging.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1I 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:dwThe figure lets me get Lx 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...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, i'll see what i can work out with that idea.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1I summoned someone smarter than I. Hopefully he will show up and save us.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, I just noticed that my battery is about to die, so i have to go. :( Thank you all the same! :)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1well help has arrived, so perhaps it will be solved when you return

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've seen this problem before, give me a few minutes and I think I'll have an answer for you Xylienda

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have that \[f'(x)=\frac{\sqrt{L^2x^2}}{x}\] This is timeconsuming, though not too hard, function to integrate, but wolfram gives the integrated function as \[f(x)=L \ln \frac{L+\sqrt{L^2x^2}}{x}\sqrt{L^2x^2}\] I'll type up the solution in a moment.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry 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 yaxis 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 \[L=\sqrt{(pf(m))^2+m^2}\] Rearranging a bit gives \[pf(m)=\sqrt{L^2m^2}\] 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 yintercept 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 \[f'(m)m=\sqrt{L^2m^2}\] or \[f'(m)=\frac{\sqrt{L^2m^2}}{m}\]

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1So I guess I wasn't too far off. In my pic I should have called Lx 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 :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok. Thank you both for your help. I'll definitely study this, but my laptop is threatening to hibernate.
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