Register to reply

Conversion from Parametric to Cartesian

by W3bbo
Tags: cartesian, conversion, parametric
Share this thread:
W3bbo
#1
Feb19-08, 02:11 PM
P: 31
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
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
cristo
#2
Feb19-08, 03:01 PM
Mentor
cristo's Avatar
P: 8,308
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.
W3bbo
#3
Feb24-08, 08:13 AM
P: 31
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

Tedjn
#4
Feb24-08, 09:53 AM
P: 738
Conversion from Parametric to Cartesian

Well, when you divide by 2, you should have x/2, not 2/x
W3bbo
#5
Feb24-08, 10:05 AM
P: 31
Quote Quote by Tedjn View Post
Well, when you divide by 2, you should have x/2, not 2/x
Note it's Cot, not Tan. So I inverted the argument for Tan^-1. Otherwise it would have been ArcCot(x/2)
Tedjn
#6
Feb24-08, 10:37 AM
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.

[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]


Register to reply

Related Discussions
Cartesian and Parametric General Math 9
Converting from Cartesian to Parametric form Precalculus Mathematics Homework 5
Parametric and cartesian equations? HELP Calculus & Beyond Homework 17
Parametric and Cartesian Equations Calculus & Beyond Homework 6
Line Integrals - Cartesian and Parametric Calculus & Beyond Homework 5