Eliminating the Parameter in a Hyperbolic Curve

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Homework Help Overview

The discussion revolves around eliminating the parameter from a hyperbolic curve defined by the equations x = 2cosh(t) and y = 5sinh(t). Participants are tasked with finding a Cartesian equation and sketching the curve while indicating the direction of tracing as the parameter increases.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the elimination of the parameter and the resulting Cartesian equation, with one participant expressing uncertainty about their approach and the accuracy of their results. There is also a focus on the differences between the parametric and Cartesian graphs, particularly regarding the quadrants represented.

Discussion Status

Some participants have provided guidance on the implications of the parametric equations and the resulting Cartesian form. There is recognition of the potential for multiple solutions arising from squaring terms, and the discussion highlights the need to consider the range of the parameter t to accurately reflect the graph's behavior.

Contextual Notes

Participants note that the parametric form only covers the first quadrant for t in the range (0, ∞), while the Cartesian form appears to cover all quadrants. There is an acknowledgment of the need to clarify the notation used in the equations to avoid confusion.

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Homework Statement



a) Eleminate the parameter to find a Cartesian equation of the curve.

b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.

Homework Equations



x = 2cosh(t) y = 5sinh(t)
cosh^2(x) - sinh^2(x) = 1

The Attempt at a Solution



cosh(t) = x/2 sinh(t) = y/5

(x^(2)/4) - (y^(2)/25) = 1

y = +/- sqrt(25/4(x^2) - 25)

This is the best guess I have here, and the graphs look similar... I just really don't know what to do from here (or even if I'm going about it the right way). Any help would be appreciated.
 
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Sarangalex said:

Homework Statement



a) Eleminate the parameter to find a Cartesian equation of the curve.

b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.

Homework Equations



x = 2cosh(t) y = 5sinh(t)
cosh^2(x) - sinh^2(x) = 1

The Attempt at a Solution



cosh(t) = x/2 sinh(t) = y/5

(x^(2)/4) - (y^(2)/25) = 1

y = +/- sqrt(25/4(x^2) - 25)

This is the best guess I have here, and the graphs look similar... I just really don't know what to do from here (or even if I'm going about it the right way). Any help would be appreciated.

I think you are doing fine so far. What part of this are you having a problem with?
 
Dick said:
I think you are doing fine so far. What part of this are you having a problem with?

I graphed the Cartesian equation(s) I got, and the graphs didn't exactly match up.
 
Sarangalex said:
I graphed the Cartesian equation(s) I got, and the graphs didn't exactly match up.

In what way didn't they match up? I hope in y = +/- sqrt(25/4(x^2) - 25) you meant y = +/- sqrt((25x^2)/4 - 25). You didn't use enough parentheses in the 25/4(x^2) to make it clear what you meant.
 
Dick said:
In what way didn't they match up? I hope in y = +/- sqrt(25/4(x^2) - 25) you meant y = +/- sqrt((25x^2)/4 - 25). You didn't use enough parentheses in the 25/4(x^2) to make it clear what you meant.

I'm sorry; that is what I was meaning to type. Well, the parametric graph only takes up the first quadrant; however, the Cartesian graph takes up all four.

Cartesian graph:

http://www3.wolframalpha.com/Calculate/MSP/MSP419219ih3hd5eb510179000016g700d6idee9f8f?MSPStoreType=image/gif&s=4&w=200&h=204&cdf=RangeControl
 
Last edited by a moderator:
Sarangalex said:
I'm sorry; that is what I was meaning to type. Well, the parametric graph only takes up the first quadrant; however, the Cartesian graph takes up all four.

Cartesian graph:

http://www3.wolframalpha.com/Calculate/MSP/MSP419219ih3hd5eb510179000016g700d6idee9f8f?MSPStoreType=image/gif&s=4&w=200&h=204&cdf=RangeControl

That sort of thing would be normal. It happens when you square things. You wind up getting more solutions than you started with. If you take 0<t<infinity you are only going to get the first quadant of this graph. If you take -infinity<t<infinity then you should get both quadrants with x>0. The parametric form won't give you any of the points where x<0 since cosh(t)>0. If you really want the cartesian form to reflect that then you could solve for x in terms of y and then just take the positive square root.
 
Last edited by a moderator:
Dick said:
That sort of thing would be normal. If you take 0<t<infinity you are only going to get the first quadant of this graph. If you take -infinity<t<infinity then you should get both quadrants with x>0. The parametric form won't give you any of the points where x<0 since cosh(t)>0. If you really want the cartesian form to reflect that then you could solve for x in terms of y and then just take the positive square root.

Oh, I see. Thank you very much for your help.
 

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