Loser66 one year ago Find a parametrization of the curve $$x^2-y^2=1$$ indicating the range of allowed values of your parametrization. Please, help

1. Loser66

@amoodarya

2. Loser66

@Michele_Laino

3. Michele_Laino

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

4. Loser66

how about x= sec t y = tan t?

5. Michele_Laino

please keep in mind that the subsequent condition, holds: $\Large {\left( {\cosh \theta } \right)^2} - {\left( {\sinh \theta } \right)^2} = 1$

6. Loser66

I know hyper trigs are work. However, if I don't know, how to derive to it?

7. Michele_Laino

I prefer hyperbolic function, since, your function is an hyperbola, and it is their natural application

8. 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

9. Michele_Laino

I think that your parametrization also works, nevertheless it is the first time that I see it

10. Loser66

One more question: I know the form is a hyperbola. What is the link between hyperbolic function with a hyperbola?

11. amoodarya

note that : $|x| \geq 1$

12. Loser66

@amoodarya how do you know?

13. Loser66

Oh, I got it @amoodarya

14. 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|

15. 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?

16. Michele_Laino

then the requested area is: $\Large A = xy - \int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)$

17. 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

18. Michele_Laino

yes! Using hyperbolic function we have no restriction conditions on the parameter

19. 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

20. Loser66

Got you @amoodarya |dw:1436448494788:dw|

21. Loser66

@Michele_Laino I did!! I got the hyperbolic function by deriving from the definition of cosh and sinh .

22. Michele_Laino

so, solving those formulas for x, and y, we get: $\Large \left\{ \begin{gathered} x = \frac{{{e^A} + {e^{ - A}}}}{2} = \cosh A \hfill \\ \hfill \\ y = \frac{{{e^A} - {e^{ - A}}}}{2} = \sinh A \hfill \\ \end{gathered} \right.$

23. Loser66

Yes!

24. amoodarya

$x=\cosh(t) \\y=\sinh(t)\\ x \geq 1 \\y \in R$ |dw:1436448711471:dw|

25. Loser66

@Michele_Laino I don't get this $$\Large A = xy - \int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)$$

26. Loser66

@amoodarya we have |x|>=1, still have the opposite part!! |dw:1436448847889:dw|

27. Michele_Laino

sorry I have made a typo, here is the right formula: $\Large A = xy - 2\int_1^x {\sqrt {{\xi ^2} - 1} d\xi } = \log \left( {x + \sqrt {{x^2} - 1} } \right)$

28. Loser66

why??

29. Michele_Laino

because we have to compute the two half areas: |dw:1436448957814:dw|

30. 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

31. 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?

32. 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

33. 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 , $y \in R$

34. 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?

35. Loser66

@amoodarya how can we know the range of y >=1?

36. Michele_Laino

whereas the area of the non-shaded region is: |dw:1436449211449:dw| $\begin{gathered} \int_1^x {dx{\text{ }}\left( {{\text{measure of the segment AB}}} \right)} \hfill \\ \hfill \\ {\text{measure of the segment AB}} = \sqrt {{x^2} - 1} - \left( { - \sqrt {{x^2} - 1} } \right) = \hfill \\ \hfill \\ = 2\sqrt {{x^2} - 1} \hfill \\ \end{gathered}$

37. amoodarya

$\cosh(t)=\frac{e^t+e^{-t}}{2}\\e^t +e^{-t}=e^t+\frac{1}{e^t} \geq 2\\so\\ \cosh t \geq \frac{2}{2}$

38. Michele_Laino

It is the link between the hyperbola and the hyperbolic functions

39. Loser66

@Michele_Laino Got you, thanks a lot

40. 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 "

41. Loser66

@amoodarya when I graph it, I know you are right, but how to prove it algebraically?

42. 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. :)

43. amoodarya

|dw:1436449649722:dw|

44. Loser66

Thank you very much. I got it.

45. amoodarya

$we-know\\|a+\frac{1}{a}| \geq 2\\e^{something} \geq 0\\e^{t}+\frac{1}{e^{-t}} \geq 2\\so\\cosh(t) \geq 1$