Probably very easy - Domain/range on parametric->cartesian equation

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SUMMARY

The discussion focuses on converting the parametric equations x = tan(t) and y = sec(t) for the interval π/2 < t < π into a rectangular system, resulting in the hyperbolic equation y² - x² = 1. The graph is confined to the third quadrant, where x is negative and y is less than or equal to -1. The domain is confirmed as (-∞, 0) and the range as (-∞, -1), based on the behavior of the tangent and secant functions within the specified interval.

PREREQUISITES
  • Understanding of parametric equations
  • Knowledge of hyperbolic functions
  • Familiarity with graphing techniques in Cartesian coordinates
  • Basic trigonometric identities, specifically sec²(t) = tan²(t) + 1
NEXT STEPS
  • Study the properties of hyperbolic functions and their graphs
  • Learn how to derive domain and range from parametric equations
  • Explore graphing tools like Desmos or GeoGebra for visualizing parametric equations
  • Investigate the relationship between trigonometric functions and their inverses
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Students studying calculus, particularly those focusing on parametric equations and their graphical representations, as well as educators teaching these concepts in a classroom setting.

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Homework Statement


Convert x=tant, y=sect on ∏/2<t<∏ to rectangular system and graph. Show domain, rnage, and orientation

I use sec2=tan2+1 -> y2-x2=1 which is a hyperbolic function with the graph http://www.wolframalpha.com/input/?i=y^2-x^2%3D1

I know that the graph is only in the 3rd quadrant but I'm not sure why - also, I'm not sure what the domain/range is. I think the domain (valid X's) are -∞ to 0, and range would be -∞ to -1 based on the graph being in the third quadrnat only? How would I get that the graph is only in the 3rd quadrant (probably something to do with the boundaries), and are the domain/range correct
 
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yesiammanu said:

Homework Statement


Convert x=tant, y=sect on ∏/2<t<∏ to rectangular system and graph. Show domain, rnage, and orientation

I use sec2=tan2+1 -> y2-x2=1 which is a hyperbolic function with the graph http://www.wolframalpha.com/input/?i=y^2-x^2%3D1
On the interval [##\pi/2, \pi##] x < 0 and y ≤ -1. Sketch graphs of x = tan(t) and y = sec(t) to see this. These graphs should give you some insight into the values of x and y.
Feodalherren said:
I know that the graph is only in the 3rd quadrant but I'm not sure why - also, I'm not sure what the domain/range is. I think the domain (valid X's) are -∞ to 0, and range would be -∞ to -1 based on the graph being in the third quadrnat only? How would I get that the graph is only in the 3rd quadrant (probably something to do with the boundaries), and are the domain/range correct
 

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