Loser66
  • Loser66
Find a parametrization of the curve \(x^2-y^2=1\) indicating the range of allowed values of your parametrization. Please, help
Mathematics
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SOLVED
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chestercat
  • chestercat
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Loser66
  • Loser66
@amoodarya
Loser66
  • Loser66
@Michele_Laino
Michele_Laino
  • Michele_Laino
I think that a possible parametrization can be this: \[\Large \left\{ \begin{gathered} x = \cosh \theta \hfill \\ y = \sinh \theta \hfill \\ \end{gathered} \right.,\quad \theta \in {\mathbf{R}}\]

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Loser66
  • Loser66
how about x= sec t y = tan t?
Michele_Laino
  • Michele_Laino
please keep in mind that the subsequent condition, holds: \[\Large {\left( {\cosh \theta } \right)^2} - {\left( {\sinh \theta } \right)^2} = 1\]
Loser66
  • Loser66
I know hyper trigs are work. However, if I don't know, how to derive to it?
Michele_Laino
  • Michele_Laino
I prefer hyperbolic function, since, your function is an hyperbola, and it is their natural application
Loser66
  • Loser66
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
Michele_Laino
  • Michele_Laino
I think that your parametrization also works, nevertheless it is the first time that I see it
Loser66
  • Loser66
One more question: I know the form is a hyperbola. What is the link between hyperbolic function with a hyperbola?
amoodarya
  • amoodarya
note that : \[|x| \geq 1\]
Loser66
  • Loser66
@amoodarya how do you know?
Loser66
  • Loser66
Oh, I got it @amoodarya
Michele_Laino
  • Michele_Laino
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|
Loser66
  • Loser66
@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?
Michele_Laino
  • Michele_Laino
then the requested area is: \[\Large A = xy - \int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)\]
amoodarya
  • amoodarya
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
Michele_Laino
  • Michele_Laino
yes! Using hyperbolic function we have no restriction conditions on the parameter
Michele_Laino
  • Michele_Laino
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
Loser66
  • Loser66
Got you @amoodarya |dw:1436448494788:dw|
Loser66
  • Loser66
@Michele_Laino I did!! I got the hyperbolic function by deriving from the definition of cosh and sinh .
Michele_Laino
  • Michele_Laino
so, solving those formulas for x, and y, we get: \[\Large \left\{ \begin{gathered} x = \frac{{{e^A} + {e^{ - A}}}}{2} = \cosh A \hfill \\ \hfill \\ y = \frac{{{e^A} - {e^{ - A}}}}{2} = \sinh A \hfill \\ \end{gathered} \right.\]
Loser66
  • Loser66
Yes!
amoodarya
  • amoodarya
\[x=\cosh(t) \\y=\sinh(t)\\ x \geq 1 \\y \in R\] |dw:1436448711471:dw|
Loser66
  • Loser66
@Michele_Laino I don't get this \(\Large A = xy - \int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)\)
Loser66
  • Loser66
@amoodarya we have |x|>=1, still have the opposite part!! |dw:1436448847889:dw|
Michele_Laino
  • Michele_Laino
sorry I have made a typo, here is the right formula: \[\Large A = xy - 2\int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)\]
Loser66
  • Loser66
why??
Michele_Laino
  • Michele_Laino
because we have to compute the two half areas: |dw:1436448957814:dw|
amoodarya
  • amoodarya
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
Loser66
  • Loser66
@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?
Michele_Laino
  • Michele_Laino
since we are computing the area of the shaded region: |dw:1436449141018:dw| the area of triangle OPQ is (x* 2y)/2=xy
amoodarya
  • amoodarya
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 , \[y \in R\]
Loser66
  • Loser66
@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?
Loser66
  • Loser66
@amoodarya how can we know the range of y >=1?
Michele_Laino
  • Michele_Laino
whereas the area of the non-shaded region is: |dw:1436449211449:dw| \[\begin{gathered} \int_1^x {dx{\text{ }}\left( {{\text{measure of the segment AB}}} \right)} \hfill \\ \hfill \\ {\text{measure of the segment AB}} = \sqrt {{x^2} - 1} - \left( { - \sqrt {{x^2} - 1} } \right) = \hfill \\ \hfill \\ = 2\sqrt {{x^2} - 1} \hfill \\ \end{gathered} \]
amoodarya
  • amoodarya
\[\cosh(t)=\frac{e^t+e^{-t}}{2}\\e^t +e^{-t}=e^t+\frac{1}{e^t} \geq 2\\so\\ \cosh t \geq \frac{2}{2}\]
Michele_Laino
  • Michele_Laino
It is the link between the hyperbola and the hyperbolic functions
Loser66
  • Loser66
@Michele_Laino Got you, thanks a lot
amoodarya
  • amoodarya
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 "
Loser66
  • Loser66
@amoodarya when I graph it, I know you are right, but how to prove it algebraically?
Loser66
  • Loser66
:) , 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. :)
amoodarya
  • amoodarya
|dw:1436449649722:dw|
Loser66
  • Loser66
Thank you very much. I got it.
amoodarya
  • amoodarya
\[we-know\\|a+\frac{1}{a}| \geq 2\\e^{something} \geq 0\\e^{t}+\frac{1}{e^{-t}} \geq 2\\so\\cosh(t) \geq 1\]

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