Please help with integration attachment.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Please help with integration attachment.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

here we go again, eh?
Lol!
1 Attachment
I've never seen a problem like this before...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i dont even know where to start. The professor says its supoosed to be challenging.
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...
Ok, i'll see what i can work out with that idea.
Thank you!
I summoned someone smarter than I. Hopefully he will show up and save us.
Actually, I just noticed that my battery is about to die, so i have to go. :( Thank you all the same! :)
well help has arrived, so perhaps it will be solved when you return
I've seen this problem before, give me a few minutes and I think I'll have an answer for you Xylienda
ok
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.
ok
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 \[f'(m)=-\frac{\sqrt{L^2-m^2}}{m}\]
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 :)
Ok. Thank you both for your help. I'll definitely study this, but my laptop is threatening to hibernate.
Bye!

Not the answer you are looking for?

Search for more explanations.

Ask your own question