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

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I think that a possible parametrization can be this: \[\Large \left\{ \begin{gathered} x = \cosh \theta \hfill \\ y = \sinh \theta \hfill \\ \end{gathered} \right.,\quad \theta \in {\mathbf{R}}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1how about x= sec t y = tan t?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please keep in mind that the subsequent condition, holds: \[\Large {\left( {\cosh \theta } \right)^2}  {\left( {\sinh \theta } \right)^2} = 1\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I know hyper trigs are work. However, if I don't know, how to derive to it?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I prefer hyperbolic function, since, your function is an hyperbola, and it is their natural application

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1There 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
 one year ago
Best ResponseYou've already chosen the best response.0I think that your parametrization also works, nevertheless it is the first time that I see it

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1One more question: I know the form is a hyperbola. What is the link between hyperbolic function with a hyperbola?

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1note that : \[x \geq 1\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@amoodarya how do you know?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Oh, I got it @amoodarya

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the requested link is: suppose to have your hyperbola: x^2y^2=1 then we want to compute the area of the hyperbolic sector: dw:1436448106611:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.0then 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
 one year ago
Best ResponseYou've already chosen the best response.1because : \[x^2y^2=1 \\x^2=y^2+1\\y^2 \ge 0\\y^2+1 \geq 1\rightarrow x^2 \geq 1\rightarrow x>1\] 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
 one year ago
Best ResponseYou've already chosen the best response.0yes! Using hyperbolic function we have no restriction conditions on the parameter

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now 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
 one year ago
Best ResponseYou've already chosen the best response.1Got you @amoodarya dw:1436448494788:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@Michele_Laino I did!! I got the hyperbolic function by deriving from the definition of cosh and sinh .

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0so, solving those formulas for x, and y, we get: \[\Large \left\{ \begin{gathered} x = \frac{{{e^A} + {e^{  A}}}}{2} = \cosh A \hfill \\ \hfill \\ y = \frac{{{e^A}  {e^{  A}}}}{2} = \sinh A \hfill \\ \end{gathered} \right.\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1\[x=\cosh(t) \\y=\sinh(t)\\ x \geq 1 \\y \in R\] dw:1436448711471:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1@amoodarya we have x>=1, still have the opposite part!! dw:1436448847889:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0sorry I have made a typo, here is the right formula: \[\Large A = xy  2\int_1^x {\sqrt {{\xi ^2}  1} d\xi } = \log \left( {x + \sqrt {{x^2}  1} } \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0because we have to compute the two half areas: dw:1436448957814:dw

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1do you want to parametrize x^2y^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
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.0since we are computing the area of the shaded region: dw:1436449141018:dw the area of triangle OPQ is (x* 2y)/2=xy

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1because 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
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1@amoodarya how can we know the range of y >=1?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0whereas the area of the nonshaded 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\cosh(t)=\frac{e^t+e^{t}}{2}\\e^t +e^{t}=e^t+\frac{1}{e^t} \geq 2\\so\\ \cosh t \geq \frac{2}{2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0It is the link between the hyperbola and the hyperbolic functions

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@Michele_Laino Got you, thanks a lot

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1hint : 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
 one year ago
Best ResponseYou've already chosen the best response.1@amoodarya when I graph it, I know you are right, but how to prove it algebraically?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1:) , 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436449649722:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thank you very much. I got it.

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1\[weknow\\a+\frac{1}{a} \geq 2\\e^{something} \geq 0\\e^{t}+\frac{1}{e^{t}} \geq 2\\so\\cosh(t) \geq 1\]
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