# Parameterization

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• Dec 7th 2011, 11:48 AM
icelated
Parameterization
I have a problem that seems easy enough but i cant seem to find x and y

Problem: find a Parametrization for the graph of the function $y = x^2$
Starting at (3,9) and moving towards the origin,using angle $\theta$
in the accompanying figure as a parameter.

I drew out the graph and took a picture of it and i will include it.

Attachment 23034

I think i use $Tan \theta = \frac{y}{x} = \frac{9}{3} = 3$http://www.mathhelpforum.com/math-he...isc/pencil.png

Im not sure how to finish solving for x then to solve for y?

Im not sure if i need to find $y = mx + b$ since the slop comes out as zero?

Thank you..
• Dec 7th 2011, 01:17 PM
TheEmptySet
Re: Parameterization
Quote:

Originally Posted by icelated
I have a problem that seems easy enough but i cant seem to find x and y

Problem: find a Parametrization for the graph of the function $y = x^2$
Starting at (3,9) and moving towards the origin,using angle $\theta$
in the accompanying figure as a parameter.

I drew out the graph and took a picture of it and i will include it.

Attachment 23034

I think i use $Tan \theta = \frac{y}{x} = \frac{9}{3} = 3$http://www.mathhelpforum.com/math-he...isc/pencil.png

Im not sure how to finish solving for x then to solve for y?

Im not sure if i need to find $y = mx + b$ since the slop comes out as zero?

Thank you..

Just use the defintion of polar coordinates

$x=r\cos(\theta), \quad y=r\sin(\theta)$

and that fact that $y=x^2$

This gives that

$x=r\cos(\theta), y=x^2=r^2\cos^2(\theta)$

Now you just need to find r and you are done.

Solve

$3=r\cos(\theta), 9=r\sin(\theta)$ for r.
• Dec 7th 2011, 01:40 PM
icelated
Re: Parameterization
Quote:

Originally Posted by TheEmptySet
Solve

$3=r\cos(\theta), 9=r\sin(\theta)$ for r.

if i do that i get $r = 3 sec(\theta)$
and

$r = 9csc(\theta)$

would one be x and one by y?

also, how would i solve

$y = r^2 sin^2 (\theta)$

would i fill in for y and divide?
$r sin (\theta) = r^2 sin^2 (\theta)$

oh, would i just have to r for the final answer like r of 1 and r of 2?
• Dec 7th 2011, 02:04 PM
TheEmptySet
Re: Parameterization
Quote:

Originally Posted by icelated
if i do that i get $r = 3 sec(\theta)$
and

$r = 9csc(\theta)$

would one be x and one by y?

also, how would i solve

$y = r^2 sin^2 (\theta)$

would i fill in for y and divide?
$r sin (\theta) = r^2 sin^2 (\theta)$

I think you may be confusing the idea of a polar function with the parameteric representation of a function.

Given any function $y=f(x)$ we can always parameterize it with the trivial parmatereization

$x=t, y=f(t) \implies (t,f(t))$ where t is the parameter.

They asked you to parameterize with respect to the angle so we know that

$t=x=r\cos(\theta)$ this gives

$(r\cos(\theta),r^2\cos(\theta))$ where theta is the parameter.

You can use the pythagorean theorem to solve the above relationship (or just draw a triangle).

$9=r^2\cos^2(\theta)$ and $81=r^2\sin^2(\theta)$ solving this system gives

$9+81=r^2\cos^2(\theta)+r^2(\sin^2(\theta)$

$90=r^2 \implies r=3\sqrt{10}$

$(3\sqrt{10}\cos(\theta),90\cos^2(\theta))$

Here is a plot of the parameterization

plot&#91; x&#61;3sqrt&#40;10&#41;cos&#40;t&#41;, y&#61;90cos&#94;2&#40;t&#41;&#93; - Wolfram|Alpha
• Dec 7th 2011, 02:09 PM
icelated
Re: Parameterization
this was the example i was given. So, i assumed this is how i do it?
see attached for a drawing.
• Dec 7th 2011, 02:24 PM
icelated
Re: Parameterization
Quote:

Originally Posted by TheEmptySet

$9+81=r^2\cos^2(\theta)+r^2(\sin^2(\theta)$

$90=r^2 \implies r=3\sqrt{10}$

$(3\sqrt{10}\cos(\theta),90\cos^2(\theta))$

How does the sin fall out?

$(\sin^2(\theta)$
• Dec 7th 2011, 02:30 PM
TheEmptySet
Re: Parameterization
Okay, notice we are doing the exact same thing!

$x=r\cos(\theta), y=r\sin(\theta)$

Now if I divide equation 2 by equation 1 I get

$\frac{y}{x}=\frac{r\sin(\theta)}{r\cos(\theta)}=\t an(\theta) \iff \tan(\theta)=\frac{y}{x}$

The two starting points are the same.

If you use the 2nd defintion then

$\tan(\theta)=\frac{y}{x}=\frac{x^2}{x}=x$ and

$\tan(\theta)=\frac{y}{\sqrt{y}}=\sqrt{y} \implies y=\tan^2(\theta)$

This gives another paramterization of the parabola.

Remember that there are many different ways to parameterize a curve.
• Dec 7th 2011, 10:11 PM
icelated
Re: Parameterization
Quote:

Originally Posted by TheEmptySet
Okay, notice we are doing the exact same thing!

If you use the 2nd defintion then

$\tan(\theta)=\frac{y}{x}=\frac{x^2}{x}=x$ and

$\tan(\theta)=\frac{y}{\sqrt{y}}=\sqrt{y} \implies y=\tan^2(\theta)$

Ok, in the first equation you get an = x i can see how you got that but what is that value? thats not merely the x coordinate is it?

Then, in the second equation has me confused. Where did you get $\frac{y}{\sqrt y} = \sqrt y$

Not sure how you got the denominator or the answer square root y

Can you explain a little better. Sorry =(
• Dec 8th 2011, 07:21 AM
TheEmptySet
Re: Parameterization
Quote:

Originally Posted by icelated
Ok, in the first equation you get an = x i can see how you got that but what is that value? thats not merely the x coordinate is it?

Then, in the second equation has me confused. Where did you get $\frac{y}{\sqrt y} = \sqrt y$

Not sure how you got the denominator or the answer square root y

Can you explain a little better. Sorry =(

You are told that $y=x^2 \implies x=\sqrt{y}$

Just sub out y and x.

So $\frac{y}{x}=\frac{x^2}{x}=x$

and

So $\frac{y}{x}=\frac{y}{\sqrt{y}}=\sqrt{y}$
• Dec 8th 2011, 11:38 AM
icelated
Re: Parameterization
So, the final answer is?

$( tan(\theta) , tan^2(\theta) )$