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Thread: Parameterization

  1. #1
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    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.

    Parameterization-2011-12-07_11-48-08_252.jpg

    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..
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    Re: Parameterization

    Quote Originally Posted by icelated View Post
    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.

    Click image for larger version. 

Name:	2011-12-07_11-48-08_252.jpg 
Views:	10 
Size:	392.0 KB 
ID:	23034

    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.
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  3. #3
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    Re: Parameterization

    Quote Originally Posted by TheEmptySet View Post
    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?
    Last edited by icelated; Dec 7th 2011 at 01:58 PM.
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    Re: Parameterization

    Quote Originally Posted by icelated View Post
    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[ x=3sqrt(10)cos(t), y=90cos^2(t)] - Wolfram|Alpha
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    Re: Parameterization

    this was the example i was given. So, i assumed this is how i do it?
    see attached for a drawing.
    Attached Thumbnails Attached Thumbnails Parameterization-img_0243cc.jpg  
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  6. #6
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    Re: Parameterization

    Quote Originally Posted by TheEmptySet View Post

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

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

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

    How does the sin fall out?

    $\displaystyle (\sin^2(\theta)$
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    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.
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    Re: Parameterization

    Quote Originally Posted by TheEmptySet View Post
    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 =(
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    Re: Parameterization

    Quote Originally Posted by icelated View Post
    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 $\displaystyle \frac{y}{x}=\frac{y}{\sqrt{y}}=\sqrt{y}$
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  10. #10
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    Re: Parameterization

    So, the final answer is?

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