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For the life of me I can't see where I'm going wrong with this.

Given the vector field

$$ F(x, y) = \left[ xy, \frac{2y}{x} \right] $$

a particle travels on the hyperbola

$$ \frac{x^2}{4} - y^2 = 1 $$

from $ \left(2\sqrt{10}, -3 \right) $ to $ \left(2, 0 \right) $ how much work is done by the force over the path?

My take on this is to solve for x and parametrize $t = y^2 + 1$ such that

$$ x = 2\sqrt{t} $$ $$ y = \sqrt{t-1} $$ $$ t\in\left[10,1\right]$$

for which

$$ \int_C F(r(t)) \cdot r'(t) dt = \int_{10}^1 2\sqrt{t}\sqrt{t-1}\frac{1}{\sqrt{t}} + \frac{2\sqrt{t-1}}{2\sqrt{t}}\frac{1}{2\sqrt{t-1}} $$ $$ = \left. \left(\frac{4}{3}\left(t-1\right)^{\frac{3}{2}} + \sqrt{t} \right) \right|_{10}^1 = -35 - \sqrt{10} $$

would be grateful for any input.

Simon
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    I would think "force" only applies at a specific point on the trajectory, while the integral over the trajectory (force acting over a distance) would represent "work." – Michael Oct 19 '19 at 22:44
  • @Michael yes indeed, bad translation on my part, have updated my question according to your input. – Simon Oct 19 '19 at 22:48
  • I observe that your trajectory starts at $(2\sqrt{10}, 3)$ instead of $(2\sqrt{10},-3)$, not clear if the question was a typo or if this is the issue. Why do you say your work is going wrong? – Michael Oct 19 '19 at 22:53
  • The trajectory starts at $( 2\sqrt{10}, -3) $. – Simon Oct 19 '19 at 23:01
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    So then that is the issue? If you fix your trajectory I believe the answer comes out to $37 -\sqrt{10}$. Do you have an answer you are working towards? – Michael Oct 19 '19 at 23:02
  • I say it is wrong because my professor says it wrong :), not asking for a numeric answer, instead input of what I'm doing wrong. I suppose there is some trick to it that makes this system behave differently then what I have currently seen.. – Simon Oct 19 '19 at 23:08
  • I think I know that if the particle travels against the vector field the net work is negative.. – Simon Oct 19 '19 at 23:10
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    You are giving unclear responses. Do you acknowledge that the problem asks for one starting point, and your trajectory has a different starting point? Your comment "The trajectory starts at $(2\sqrt{10}, -3)$" is not clear to me because you have made no attempt to fix the problem. – Michael Oct 19 '19 at 23:13
  • Maybe? Is there something about this problem that makes the trajectory start at $(2\sqrt{10}, 3)$ while it is declared to be at $(2\sqrt{10}, -3)$? If that is so, I don't understand it yet.

    Way I understand it now is that the trajectory starts at $(2\sqrt{10}, -3)$ and ends at $(2, 0)$, I'm I missing something?

     – Simon Oct 19 '19 at 23:22
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    You are missing the fact that the trajectory $(2\sqrt{t}, \sqrt{t-1})$ that you have defined does not start at the desired point $(2\sqrt{10}, -3)$. – Michael Oct 19 '19 at 23:24
  • Well sir, it was my belief that any parameterization of $ \frac{x^2}{4} - y^2 = 1 $ would be valid. – Simon Oct 19 '19 at 23:30
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    It sounds like you are saying that, while the problem asks for the trajectory to start at a particular point, your belief is that you are allowed to start at any other point you like (meaning that you believe you can ignore the starting point that the problem asks for). If you can choose any place to start, why not make the starting point equal to the ending point, in which case the integral is 0! – Michael Oct 19 '19 at 23:32
  • Wrapping my head around it, I hear your logic. Is it like one of these properties of squaring where the $\sqrt{y^2} = \pm \sqrt{t-1} $, gives one false equation and I have to test against $(2 \sqrt{10},-3) $ to get the correct one? – Simon Oct 19 '19 at 23:47
  • Hehe, I will be sure to choose the ending point to be my starting point in all my integrals! – Simon Oct 19 '19 at 23:55

1 Answers1

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Since the $y$-coordinate of the trajectory is negative, you should parametrize the path as

$$ x = 2\sqrt{t}, \>\>\> y = -\sqrt{t-1},\>\>\> t\in\left[10,1\right]$$

to allow negative values for the $y$-variable. Then, the work integral becomes

$$ \int_C F(r(t)) \cdot r'(t) dt = \int_{10}^1 \left[2\sqrt{t}(-\sqrt{t-1})\frac{1}{\sqrt{t}} + \frac{2\sqrt{t-1}}{2\sqrt{t}}\frac{1}{2\sqrt{t-1}}\right]dt $$

$$ = \left. \left(-\frac{4}{3}\left(t-1\right)^{\frac{3}{2}} + \sqrt{t} \right) \right|_{10}^1 = 37 - \sqrt{10} $$

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