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While trying to compute the line integral along a path K on a function, I need to parametrize my path K in terms of a single variable, let's say this single variable will be $t$. My path is defined by the following ensemble: $$K=\{(x,y)\in(0,\infty)\times[-42,42]|x^2-y^2=1600\}$$ I know how to calculate the line integral, that is not my issue. My problem is to parametrize $x^2-y^2=1600$. I tried using the identities: $$\sin^2(t)+\cos^2(t)=1$$ $$\sec^2(t)-\tan^2(t)=1$$ But I did not get anywhere with my parametrization (see below for my poor try into parametrizing). I would welcome any help/hints and if you happen to know some good reading to learn more about parametrization, I am also interested. $$r(t)=1600\sec^2(t)-1600\tan^2(t)=1600$$ for $$x=40\sec(t) \land y=40\tan(t)$$

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    $\begingroup$ Are you familiar with the hyperbolic functions? $\endgroup$ – N. F. Taussig Jun 17 '19 at 14:35
  • $\begingroup$ Hello! I am a little bit familiar, yes. I know about the cosh^2-sinh^2=1 but in what degree is it different than using sec^2-tan^2=1? $\endgroup$ – Bilalord Jun 17 '19 at 14:38
  • $\begingroup$ When parametrizing using $\sec$ and $\tan$ it is not particularly clear what $t$ represents geometrically. When parametrizing using $\sinh$ and $\cosh$ however, then $t$ very nicely corresponds to the angle. See this post $\endgroup$ – JMoravitz Jun 17 '19 at 14:45
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I think that using trigonometric function is overcomplicating it in this case. You can let $y$ correspond to a parameter $t$, then, since $x$ is given to be positive, we can say that $x$ is the following positive root $$x = \sqrt{1600 + t^2}.$$ Your parameterised curve is subsequently given by: $$\left\{\left(\sqrt{1600 + t^2},t\right): t \in [-42,42]\right\}.$$


Letting $y = 40\sinh(t)$ is also an option, in which case the parameterisation is given by $$\left(40 \cosh(t), 40 \sinh(t) \right).$$ This perhaps looks more appealing although finding the correct bounds on $t$ now involves inverse hyperbolic functions, which I will leave up to you if you are willing to do it.

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    $\begingroup$ Thank you Pjotr, this was very helpful! I could solve my problem after reading your answer. In fact, I forgot to consider the "simple" way of parametrization... $\endgroup$ – Bilalord Jun 17 '19 at 16:48
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Notice that $x^2-y^2=(x+y)(x-y)=1600$, therefore you are dealing with a hyperbola with asymptotes $x+y=0$ and $x-y=0$ as shown here:

enter image description here

It is natural to try $x+y=t$, therefore $$ x-y= {1600\over t}. $$ That gives $$ x = {t+ {1600\over t} \over 2} \quad y= {t- {1600\over t} \over 2}. $$ As far as bounds go, $-42\leq y \leq 42$ therefore $$ -84 \leq t- {1600\over t} \leq 84 \rightarrow 16\leq t \leq 100. $$

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    $\begingroup$ Thank you for your answer A.G. I would not have seen this way to parametrize and it is interesting to me to consider your idea! $\endgroup$ – Bilalord Jun 17 '19 at 16:50
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    $\begingroup$ “Asymptotes,” not “axes.” $\endgroup$ – amd Jun 17 '19 at 21:18
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$$x^2-y^2=1600$$ $$(\frac{x}{40})^2 - (\frac{y}{40})^2=1$$

Let $x=40\cdot(e^t+e^-t)/2=40\cdot\cosh{t}$.

Let $y=40\cdot (e^t-e^{-t})/2=40 \cdot\sinh{t}$

Use positive values of t if you need positive values of $x$.

$d/dt(\cosh{t})=\sinh{t}$

$d/dt(\sinh{t})=\cosh{t}$

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  • $\begingroup$ Thank you for your answer! Helped me to see it with hyperbolic functions. $\endgroup$ – Bilalord Jun 17 '19 at 16:51

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