1. The problem statement, all variables and given/known data Reduce these parametric functions to a single cartesian equation: [tex] $\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} $ [/tex] 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: [tex] $\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} $ [/tex] For Q2 I thought I had it solved with a trigonometric identity, but the plots look different: [tex] $\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} $ [/tex] Finally, Q3 seems deceptivly simple, but again, the plot doesn't match the original: [tex] $\displaylines{ x = t^2 + 1 \cr y = t^2 + t \cr \cr t^2 = x - 1 \cr y = x - 1 + \sqrt {x - 1} \cr} $ [/tex] I'm not looking for answers, just to find out where I've gone wrong. Thanks
Q2. You've made an error when multiplying through by 25.. you should have a -25 on the RHS of the penultimate line. Q3. I'm not sure what you've done, but I'd try solving one of the parametric equations for t, but the quadratic formula, then subbing into the other equation.
Thanks. I got those questions correct. Further on I'm faced with converting this into cartesian: [tex]$\displaylines{ x = 2{\mathop{\rm Cot}\nolimits} \left( t \right) \cr y = 2{\mathop{\rm Sin}\nolimits} ^2 \left( t \right) \cr} $[/tex] I tried having a go using this: [tex]$\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} $[/tex] ...but the plot is totally different. I was able to solve it eventually using [tex]$\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[/tex] But can you explain why my initial attempt failed? Thanks
Note it's Cot, not Tan. So I inverted the argument for Tan^-1. Otherwise it would have been ArcCot(x/2)
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. [tex]y = 2\sin^2\left[\tan^{-1}\left(\frac{2}{x}\right)\right][/tex] From a picture of the right triangle, we know that [tex]\sin t = \pm\frac{2}{\sqrt{4+x^2}} \implies \sin^2 t = \frac{4}{4 + x^2}[/tex] Then, [tex]\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[/tex]