Find a parametrization of the curve \(x^2-y^2=1\) indicating the range of allowed values of your parametrization. Please, 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 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.

Find a parametrization of the curve \(x^2-y^2=1\) indicating the range of allowed values of your parametrization. Please, help

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

I think that a possible parametrization can be this: \[\Large \left\{ \begin{gathered} x = \cosh \theta \hfill \\ y = \sinh \theta \hfill \\ \end{gathered} \right.,\quad \theta \in {\mathbf{R}}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

how about x= sec t y = tan t?
please keep in mind that the subsequent condition, holds: \[\Large {\left( {\cosh \theta } \right)^2} - {\left( {\sinh \theta } \right)^2} = 1\]
I know hyper trigs are work. However, if I don't know, how to derive to it?
I prefer hyperbolic function, since, your function is an hyperbola, and it is their natural application
There must be something to restrict tan and sec, but I don't know it. Since it is nothing wrong with tan^2 -sec^2 =1
I think that your parametrization also works, nevertheless it is the first time that I see it
One more question: I know the form is a hyperbola. What is the link between hyperbolic function with a hyperbola?
note that : \[|x| \geq 1\]
@amoodarya how do you know?
Oh, I got it @amoodarya
the requested link is: suppose to have your hyperbola: x^2-y^2=1 then we want to compute the area of the hyperbolic sector: |dw:1436448106611:dw|
@Michele_Laino if I use sec and tan, the restriction is \(x\neq \pi/2 +k\pi\), right? while using hyperbolic function, there is no restriction, it is defined for all x, right?
then the requested area is: \[\Large A = xy - \int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)\]
because : \[x^2-y^2=1 \\x^2=y^2+1\\y^2 \ge 0\\y^2+1 \geq 1\rightarrow x^2 \geq 1\rightarrow |x|>1\] so \[x=\sec(t)\\y=\tan(t) \] |dw:1436448301983:dw| is true but.... |dw:1436448354062:dw| is not complete because :cosh >=1 only gave on side
yes! Using hyperbolic function we have no restriction conditions on the parameter
now we express x, and y as function of A, so we get: \[\Large \left\{ \begin{gathered} x + y = x + \sqrt {{x^2} - 1} = {e^A} \hfill \\ x - y = x - \sqrt {{x^2} - 1} = {e^{ - A}} \hfill \\ \end{gathered} \right.\] please check those formulas
Got you @amoodarya |dw:1436448494788:dw|
@Michele_Laino I did!! I got the hyperbolic function by deriving from the definition of cosh and sinh .
so, solving those formulas for x, and y, we get: \[\Large \left\{ \begin{gathered} x = \frac{{{e^A} + {e^{ - A}}}}{2} = \cosh A \hfill \\ \hfill \\ y = \frac{{{e^A} - {e^{ - A}}}}{2} = \sinh A \hfill \\ \end{gathered} \right.\]
Yes!
\[x=\cosh(t) \\y=\sinh(t)\\ x \geq 1 \\y \in R\] |dw:1436448711471:dw|
@Michele_Laino I don't get this \(\Large A = xy - \int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)\)
@amoodarya we have |x|>=1, still have the opposite part!! |dw:1436448847889:dw|
sorry I have made a typo, here is the right formula: \[\Large A = xy - 2\int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)\]
why??
because we have to compute the two half areas: |dw:1436448957814:dw|
do you want to parametrize x^2-y^2=1 ? with respect to range of variables ? so x= sect , y=tan t is true but x=cosht , y= sinht give one side of hyperbola
@Michele_Laino I don't get why we have to find the area @amoodarya why cosh t and sinh t give one side of hyperbola?
since we are computing the area of the shaded region: |dw:1436449141018:dw| the area of triangle OPQ is (x* 2y)/2=xy
because range of y=cosh(t) is \[y \geq 1\\\] so you will not number like x=-1 , -2 , -4 ,... points are moving in x>1 , \[y \in R\]
@Michele_Laino why ??? you said "since we are computing the area of the shaded region:" the problem didn't ask we to do that, right?
@amoodarya how can we know the range of y >=1?
whereas the area of the non-shaded region is: |dw:1436449211449:dw| \[\begin{gathered} \int_1^x {dx{\text{ }}\left( {{\text{measure of the segment AB}}} \right)} \hfill \\ \hfill \\ {\text{measure of the segment AB}} = \sqrt {{x^2} - 1} - \left( { - \sqrt {{x^2} - 1} } \right) = \hfill \\ \hfill \\ = 2\sqrt {{x^2} - 1} \hfill \\ \end{gathered} \]
\[\cosh(t)=\frac{e^t+e^{-t}}{2}\\e^t +e^{-t}=e^t+\frac{1}{e^t} \geq 2\\so\\ \cosh t \geq \frac{2}{2}\]
It is the link between the hyperbola and the hyperbolic functions
@Michele_Laino Got you, thanks a lot
hint : make a table with y= cosht , y= sinh t put some t in it and draw points you will get "why this is on side "
@amoodarya when I graph it, I know you are right, but how to prove it algebraically?
:) , it is not my problem, it is the problem from the students I am tutoring. I would like to know the easiest and clearest way to explain them why hyper but not regular trigs. Feel free to help me with your explanation. I don't cheat for myself. :)
|dw:1436449649722:dw|
Thank you very much. I got it.
\[we-know\\|a+\frac{1}{a}| \geq 2\\e^{something} \geq 0\\e^{t}+\frac{1}{e^{-t}} \geq 2\\so\\cosh(t) \geq 1\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question