1. ## 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 $\displaystyle y = x^2$
Starting at (3,9) and moving towards the origin,using angle$\displaystyle \theta$
in the accompanying figure as a parameter.

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

I think i use $\displaystyle Tan \theta = \frac{y}{x} = \frac{9}{3} = 3$

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

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

Thank you..

2. ## Re: Parameterization

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 $\displaystyle y = x^2$
Starting at (3,9) and moving towards the origin,using angle$\displaystyle \theta$
in the accompanying figure as a parameter.

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

I think i use $\displaystyle Tan \theta = \frac{y}{x} = \frac{9}{3} = 3$

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

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

Thank you..
Just use the defintion of polar coordinates

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

and that fact that $\displaystyle y=x^2$

This gives that

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

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

Solve

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

3. ## Re: Parameterization

Originally Posted by TheEmptySet
Solve

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

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

$\displaystyle r = 9csc(\theta)$

would one be x and one by y?

also, how would i solve

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

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

oh, would i just have to r for the final answer like r of 1 and r of 2?

4. ## Re: Parameterization

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

$\displaystyle r = 9csc(\theta)$

would one be x and one by y?

also, how would i solve

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

would i fill in for y and divide?
$\displaystyle r sin (\theta) = r^2 sin^2 (\theta)$
I think you may be confusing the idea of a polar function with the parameteric representation of a function.

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

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

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

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

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

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

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

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

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

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

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

5. ## Re: Parameterization

this was the example i was given. So, i assumed this is how i do it?
see attached for a drawing.

6. ## Re: Parameterization

Originally Posted by TheEmptySet

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

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

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

How does the sin fall out?

$\displaystyle (\sin^2(\theta)$

7. ## Re: Parameterization

Okay, notice we are doing the exact same thing!

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

Now if I divide equation 2 by equation 1 I get

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

The two starting points are the same.

If you use the 2nd defintion then

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

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

This gives another paramterization of the parabola.

Remember that there are many different ways to parameterize a curve.

8. ## Re: Parameterization

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

If you use the 2nd defintion then

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

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

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

Can you explain a little better. Sorry =(

9. ## Re: Parameterization

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$\displaystyle \frac{y}{\sqrt y} = \sqrt y$

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

Can you explain a little better. Sorry =(
You are told that $\displaystyle y=x^2 \implies x=\sqrt{y}$

Just sub out y and x.

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

and

So $\displaystyle \frac{y}{x}=\frac{y}{\sqrt{y}}=\sqrt{y}$

10. ## Re: Parameterization

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