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

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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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Bump if I may, my test is in 30 minutes :)
 
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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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