Results 1 to 3 of 3

Math Help - differentiation, parametric equation help

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    607

    differentiation, parametric equation help

    Find  \frac{dy}{dx} for the following, leaving your answer in terms of the parameter t.

      x = 2t-sin2t

     y = 1-cos2t

     \frac{dx}{dt} = 2-2cos2t

     \frac{dy}{dt} = 2sin2t

      \frac{dy}{dx} = \frac{2sin2t}{2-2cos2t}

    I can factor and cancel the two's but that still does give me the correct answer. Can someone please show me what I am doing wrong?


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

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,690
    Thanks
    617
    Hello, Tweety!

    It would help if you told us the correct answer.
    I see nothing wrong with your work.


    Find  \frac{dy}{dx} for the following, leaving your answer in terms of the parameter t.

    . . \begin{array}{ccc}x &=& 2t-\sin2t \\<br />
y & =& 1-\cos2t \end{array}


    My work:

    . . \begin{array}{ccc}\dfrac{dx}{dt} &=& 2-2\cos2t \\ \\[-3mm]<br />
\dfrac{dy}{dt} &=& 2\sin2t \end{array}

    . .   \dfrac{dy}{dx} \:=\: \frac{2\sin2t}{2-2\cos2t}

    I can factor and cancel the two's but that still does give me the correct answer.
    Your answer can be simplified . . .

    We have: . \frac{dy}{dx} \:=\:\frac{\sin2t}{1 - \cos2t} \;=\;\frac{2\sin t\cos t}{2\sin^2\!t} \;=\;\frac{\cos t}{\sin t} \;=\;\cot t


    Is that it?

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2008
    Posts
    607
    Quote Originally Posted by Soroban View Post
    Hello, Tweety!

    It would help if you told us the correct answer.
    I see nothing wrong with your work.


    Your answer can be simplified . . .

    We have: . \frac{dy}{dx} \:=\:\frac{\sin2t}{1 - \cos2t} \;=\;\frac{2\sin t\cos t}{2\sin^2\!t} \;=\;\frac{\cos t}{\sin t} \;=\;\cot t


    Is that it?

    yes that is the correct answer cott, but how did you manage to get that?

    I see you have replace sin2t with 2sin2tcost, but what happened ti the other number two? And the bottom bit, how did you get it?

    thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parametric Differentiation
    Posted in the Calculus Forum
    Replies: 8
    Last Post: February 16th 2011, 01:23 PM
  2. Parametric Differentiation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 29th 2009, 03:39 PM
  3. Parametric Differentiation help....
    Posted in the Calculus Forum
    Replies: 9
    Last Post: April 15th 2009, 01:34 PM
  4. Differentiation of parametric equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 15th 2008, 06:34 AM
  5. Differentiation ( Parametric)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 2nd 2008, 06:33 AM

Search Tags


/mathhelpforum @mathhelpforum