Conversion from Parametric to Cartesian

by W3bbo
Tags: cartesian, conversion, parametric
 P: 31 1. The problem statement, all variables and given/known data Reduce these parametric functions to a single cartesian equation: $$\displaylines{ x = at^2 \cr y = 2at \cr}  \displaylines{ x = 3{\mathop{\rm Sec}\nolimits} \left( \alpha \right) \cr y = 5{\mathop{\rm Tan}\nolimits} \left( \alpha \right) \cr}  \displaylines{ x = t^2 + 1 \cr y = t^2 + t \cr}$$ 2. Relevant equations N/A 3. The attempt at a solution For Q1, I think I got the hang of it... in that the plots of my cartesian and the paremetric are identical: $$\displaylines{ x = at^2 \cr y = 2at \cr \cr t^2 = {x \over a} \cr t = \sqrt {{x \over a}} \cr \cr y = 2a\sqrt {{x \over a}} \cr}$$ For Q2 I thought I had it solved with a trigonometric identity, but the plots look different: $$\displaylines{ x = 3{\mathop{\rm Sec}\nolimits} \left( \alpha \right) \cr y = 5{\mathop{\rm Tan}\nolimits} \left( \alpha \right) \cr \cr {\mathop{\rm Sec}\nolimits} ^2 \left( \alpha \right) = 1 + {\mathop{\rm Tan}\nolimits} ^2 \left( \alpha \right) \cr \left( {{\textstyle{1 \over 3}}x} \right)^2 = 1 + \left( {{\textstyle{1 \over 5}}y} \right)^2 \cr {\textstyle{1 \over 9}}x^2 = 1 + {\textstyle{1 \over {25}}}y^2 \cr {\textstyle{1 \over {25}}}y^2 = {\textstyle{1 \over 9}}x^2 - 1 \cr y^2 = {\textstyle{{25} \over 9}}x^2 - 1 \cr y = \pm \sqrt {{\textstyle{{25} \over 9}}x^2 - 1} \cr}$$ Finally, Q3 seems deceptivly simple, but again, the plot doesn't match the original: $$\displaylines{ x = t^2 + 1 \cr y = t^2 + t \cr \cr t^2 = x - 1 \cr y = x - 1 + \sqrt {x - 1} \cr}$$ I'm not looking for answers, just to find out where I've gone wrong. Thanks
 P: 31 Thanks. I got those questions correct. Further on I'm faced with converting this into cartesian: $$\displaylines{ x = 2{\mathop{\rm Cot}\nolimits} \left( t \right) \cr y = 2{\mathop{\rm Sin}\nolimits} ^2 \left( t \right) \cr}$$ I tried having a go using this: $$\displaylines{ x = 2{\mathop{\rm Cot}\nolimits} \left( t \right) \cr t = {\mathop{\rm Tan}\nolimits} ^{ - 1} \left( {{\textstyle{2 \over x}}} \right) \cr y = 2{\mathop{\rm Sin}\nolimits} ^2 \left( {{\mathop{\rm Tan}\nolimits} ^{ - 1} \left( {{\textstyle{2 \over x}}} \right)} \right) \cr}$$ ...but the plot is totally different. I was able to solve it eventually using $$\displaylines{ 1 + {\mathop{\rm Cot}\nolimits} ^2 \left( t \right) = {\mathop{\rm Cosec}\nolimits} ^2 \left( t \right) \cr 1 + \left( {{\textstyle{1 \over 2}}x} \right)^2 = \left( {{\textstyle{y \over 2}}} \right)^{ - 1} \cr} p$$ But can you explain why my initial attempt failed? Thanks
 P: 738 Sorry, I see that now. It's probably some quirk of the graphing software. The first set should work out to the second set if you get rid of the inverse trig functions. $$y = 2\sin^2\left[\tan^{-1}\left(\frac{2}{x}\right)\right]$$ From a picture of the right triangle, we know that $$\sin t = \pm\frac{2}{\sqrt{4+x^2}} \implies \sin^2 t = \frac{4}{4 + x^2}$$ Then, $$\frac{y}{2} = \frac{4}{4+x^2}\longrightarrow \left(\frac{y}{2}\right)^{-1} = \frac{4+x^2}{4} = 1 + \frac{x^2}{4} = 1 + \left(\frac{1}{2}x\right)^2$$