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Graph the Cartesian equation

  • #1
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Homework Statement


Identify the particle's path by finding a Cartesian equation for it . Graph the Cartesian equation . indicate the portion of the graph traced by the particle .
##x=2sinh t## , ##y=2cosh t## , ##-\infty < t < \infty##

Homework Equations


##cosh^2 t - sinh^2 t = 1##

The Attempt at a Solution


##sinh t = \frac{x}{2}##
Square both sides :
##sinh^2 t = \frac{x^2}{4}## (1)
##cosh t = \frac{y}{2}##
Square both sides :
##cosh^2 t = \frac{y^2}{4}## (2)
(2) - (1) :
##cosh^2 t - sinh^2 t = 1##
##\frac{y^2}{4} - \frac{x^2}{4} = 1##
put x= 0 → y=±2
Capture.png

Is my answer correct ?
 

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Answers and Replies

  • #2
George Jones
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Is my answer correct ?
What do you know about the possible values of ##\cosh t## for ##-\infty < t < \infty##?
 
  • #3
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What do you know about the possible values of ##\cosh t## for ##-\infty < t < \infty##?
The Cartesian equation forms a parbola opening up . Right ?
 
  • #4
George Jones
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I just noticed something else.
$$\frac{x^2}{4} - \frac{y^2}{4} =1$$
is not correct, but I think this is a "typo".
 
  • #5
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The Cartesian equation forms a parbola opening up . Right ?
No. The graph you show is not a parabola.
 
  • #6
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Identify the particle's path by finding a Cartesian equation for it .
Graph the Cartesian equation . indicate the portion of the graph traced by the particle .
##x=2sinh t## , ##y=2cosh2## , ##-\infty < t < \infty##
Should the second equation be ##y = 2\cosh(t)##?
 
  • #7
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Should the second equation be ##y = 2\cosh(t)##?
yeah
 
  • #8
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I agree with @George Jones's comment in post #4.
(2) - (1) :
##cosh^2 t - sinh^2 t = 1##
##\frac{x^2}{4} - \frac{y^2}{4} = 1##
Take a closer look at the work I've quoted above.
 
  • #9
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##\frac{y^2}{4} - \frac{x^2}{4}## forms a hyperbola and to graph it , we should go 2 units up from center point .
##cosh ( t) ## for ##-\infty < t < \infty## is always positive .
 
  • #10
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I agree with @George Jones's comment in post #4.
Take a closer look at the work I've quoted above.
##\frac{y^2}{4} - \frac{x^2}{4} = 1##
 
  • #11
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  • #12
George Jones
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##\frac{y^2}{4} - \frac{x^2}{4} = 1## forms a hyperbola
Yes, a big hint is given by the names of the functions in the original equations. :wink::biggrin:
 
  • #14
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View attachment 231720
Is it correct ?
No.
The equation ##\frac{y^2}{4} - \frac{x^2}{4} = 1## is NOT a parabola. The graph you showed appears to be the graph of ##y = x^2 + 2##, which is completely unrelated to your equation.
 
  • #15
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The equation ##\frac{y^2}{4} - \frac{x^2}{4} = 1## is NOT a parabola.
Capture.png
 

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  • #17
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Bingo!
(Meaning, yes, that's it.)
Thanks for your help.
 

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