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 $$x=40\sec(t) \land y=40\tan(t)$$
Parametrization of $x^2-y^2=1600$
parametricparametrization
Related Solutions
Subtract the two equations. We get $y=0$
Plug in the first $$x^2-z^2=1$$ A parametrization is $$(x=\cosh t,y=0,z=\sinh t); (x=-\cosh t , y=0, z=-\sinh t)$$ In the image below the two surfaces and their intersection.
$$...$$
Best Answer
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.