Results 1 to 9 of 9

Math Help - Parametric equations

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    107

    Parametric equations

    A curve has parametric equations :

    x=t^3-\mu t , where \mu is a constant

    y=\frac{t^2}{1+t^2}

    (1) Show that the curve is symmetrical about the y-axis

    (2) For \mu>0 ,find in terms of \mu the coordinates of the point where the curve intersects itself.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Ok, what ideas have you had so far?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by hooke View Post
    A curve has parametric equations :

    x=t^3-\mu t , where \mu is a constant

    y=\frac{t^2}{1+t^2}

    (1) Show that the curve is symmetrical about the y-axis

    (2) For \mu>0 ,find in terms of \mu the coordinates of the point where the curve intersects itself.
    (1) x --> -x => t --> -t and t --> -t => y --> y therefore the function is even (symmetric about y-axis).

    (2) You are trying to find a crunode: Crunode - Wikipedia, the free encyclopedia.

    Note that crunodes occur at points where dy/dx has the indeterminant form 0/0. So get dy/dx, find the value of t for which dy/dx is indeterminant and then substitute this value of t into x = x(t) and y = y(t). (You know in advance that the point is a crunode but you should test it anyway).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2009
    Posts
    107
    Quote Originally Posted by mr fantastic View Post
    (1) x --> -x => t --> -t and t --> -t => y --> y therefore the function is even (symmetric about y-axis).

    (2) You are trying to find a crunode: Crunode - Wikipedia, the free encyclopedia.

    Note that crunodes occur at points where dy/dx has the indeterminant form 0/0. So get dy/dx, find the value of t for which dy/dx is indeterminant and then substitute this value of t into x = x(t) and y = y(t). (You know in advance that the point is a crunode but you should test it anyway).
    thank you sir,why is the gradient of tangent 0/0 at the crunode? I only see two different tangents but not a vertical tangent.


    just out of curiousity, is there an algebraic or non-calculus solution to part (2)?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by hooke View Post
    thank you sir,why is the gradient of tangent 0/0 at the crunode? I only see two different tangents but not a vertical tangent.


    just out of curiousity, is there an algebraic or non-calculus solution to part (2)?
    0/0 does not mean vertical tangent. It means indeterminant (which makes perfect sense if you go back and read the link again).

    Not that I'm aware of.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2009
    Posts
    107
    Quote Originally Posted by mr fantastic View Post
    0/0 does not mean vertical tangent. It means indeterminant (which makes perfect sense if you go back and read the link again).

    Not that I'm aware of.
    i have read it a few times but i do not understand how can a point with 2 different tangents ended up with an indeterminant.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,545
    Thanks
    1394
    Well, what else can it be? If the derivative were not indeterminant, then it would have one "determinant" value and that would be the slope of the tangent line.

    But mr fantastic was answering a different question, "why is 0/0 not "infinite?"(giving a vertical tangent). The answer to that is that limits such as \lim_{x\to 0}\frac{x}{x}= 1, \lim_{x\to 0} \frac{2x}{x}= 2 all involve fractions of the form "0/0" but different limits and so different "slopes". To have an infinite limit your fraction must be of the form "a/0" where a is not 0.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2009
    Posts
    107
    Quote Originally Posted by hooke View Post
    A curve has parametric equations :

    x=t^3-\mu t , where \mu is a constant

    y=\frac{t^2}{1+t^2}

    (1) Show that the curve is symmetrical about the y-axis

    (2) For \mu>0 ,find in terms of \mu the coordinates of the point where the curve intersects itself.
    This is my work,

    \frac{dy}{dx}=\frac{2t}{(1+t^2)^2(3t^2-\mu)}

    so i am supposed to find a value of t which gives 0/0 but that seems impossible..

    If t=\pm\sqrt{\frac{\mu}{3}}, it only makes the denominator 0 and if t=0, only the numerator is 0

    ??
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by hooke View Post
    This is my work,

    \frac{dy}{dx}=\frac{2t}{(1+t^2)^2(3t^2-\mu)}

    so i am supposed to find a value of t which gives 0/0 but that seems impossible..

    If t=\pm\sqrt{\frac{\mu}{3}}, it only makes the denominator 0 and if t=0, only the numerator is 0

    ??
    Perhaps there's a misprint in the question .... (I had half-suspected as much earlier because if the curve is symmetric about the y-axis I would have expected the crunode to lie on the y-axis ....)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with parametric equations?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 13th 2009, 07:20 PM
  2. parametric equations, equations of plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 10th 2009, 02:58 AM
  3. Parametric equations to rectangular equations.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 5th 2009, 10:39 PM
  4. Replies: 3
    Last Post: December 2nd 2008, 10:54 AM
  5. Replies: 1
    Last Post: September 1st 2007, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum