Results 1 to 10 of 10

Math Help - Parameterization

  1. #1
    Member
    Joined
    Aug 2011
    Posts
    117

    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.

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

    I think i use 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  y = mx + b since the slop comes out as zero?

    Thank you..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    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 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.

    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 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  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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2011
    Posts
    117

    Re: Parameterization

    Quote Originally Posted by TheEmptySet View Post
    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?
    Last edited by icelated; December 7th 2011 at 02:58 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Parameterization

    Quote Originally Posted by icelated View Post
    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[ x=3sqrt(10)cos(t), y=90cos^2(t)] - Wolfram|Alpha
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2011
    Posts
    117

    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  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2011
    Posts
    117

    Re: Parameterization

    Quote Originally Posted by TheEmptySet View Post

    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)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Aug 2011
    Posts
    117

    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

    \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 =(
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    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  \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}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Aug 2011
    Posts
    117

    Re: Parameterization

    So, the final answer is?

    ( tan(\theta) , tan^2(\theta) )
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parameterization
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2011, 05:47 PM
  2. parameterization
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: May 24th 2010, 05:27 AM
  3. Parameterization
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 22nd 2010, 10:02 AM
  4. Parameterization
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 12th 2009, 11:56 AM
  5. Parameterization
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 7th 2008, 07:35 PM

Search Tags


/mathhelpforum @mathhelpforum