A man, a boat, and the tractrix formed...

[Prodigy Girl]

I really need help solving question #2 from the following webpage: Applied Homework 2 (http://www.math.psu.edu/anand/M141/applied2/). I've attempted to solve this, but looking at my answers, I'm not overly confident. In part a, I'm suppose to prove that tractrix somehow, and in part b, I believe I am to just integrate that given equation...

But I'm not 100% sure (especially with part a)...


I'd be eternally grateful to any help given. [:)]

[HallsofIvy]

"Applied Homework"? Now that's a peculiar name. The only good reason for doing homework is to learn something that you can later apply!

In any case, you have a problem about a "tractrix", the curve that has the property that "the rope is always tangent to the curve".

Okay, call the curve "y= f(x)". The man walks along the "y-axis" (I get that from the picture) so let's designate his postion at any time by (0, Y). The boat is at (x,y)= (x,f(x)) and the distance between him and the boat is &sqrt((0-x)2+ (Y-f(x))2)= L. The slope of the line from the boat to the man is (Y-f(x))/(0-x)= f(x)/x- Y/x and, by definition of a "tractrix" that is the tangent of the curve: df/dx= f(x)x- Y/x. "Y" varies from moment to moment but we can solve the "L" equation for Y: x2+(Y-f(x))2= L2 so Y= f(x)+ L2- x2.

From that you should be able to get the equation in part (a).
That looks to me like a straight forward first order, separable differential equation. Try to solve it and if you have trouble with it come back here.

[Prodigy Girl]

Originally posted by HallsofIvy

The boat is at (x,y)= (x,f(x)) and the distance between him and the boat is &sqrt((0-x)2+ (Y-f(x))2)= L. The slope of the line from the boat to the man is (Y-f(x))/(0-x)= f(x)/x- Y/x and, by definition of a "tractrix" that is the tangent of the curve: df/dx= f(x)x- Y/x. "Y" varies from moment to moment but we can solve the "L" equation for Y: x2+(Y-f(x))2= L2 so Y= f(x)+ L2- x2.


Thank you for the help, HallsofIvy.

I was able to follow just about everything you mentioned. I understand that solving for Y would yield Y= f(x)+ sqrt(L2- x2), and then when that Y value is placed into the equation of the slope f(x)/x- Y/x and simplify, I will get that original equation.

There is only one thing I am confused about-- When you said "by definition of a "tractrix" that is the tangent of the curve: df/dx= f(x)x- Y/x," were you referring to df/dx= dY/dx= f'(x)? I just want to make sure I understand perfectly. Thanks.
[:)]

[HallsofIvy]

Yes.