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Differentiating Parametric Functions

  1. Mar 30, 2007 #1

    danago

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    Gold Member


    Given:
    [tex]
    \begin{array}{l}
    x = 2\cos t \\
    y = 2\sin t \\
    \end{array}
    [/tex]

    find [tex]
    \frac{{dy}}{{dx}}
    [/tex]



    I started by finding dy/dt and dx/dt

    [tex]
    \begin{array}{l}
    \frac{{dy}}{{dt}} = 2\cos t \\
    \frac{{dx}}{{dt}} = - 2\sin t \\
    \end{array}
    [/tex]

    Now, dy/dx = (dy/dt) / (dx/dt)

    so:

    [tex]
    \frac{{dy}}{{dx}} = \frac{{2\cos t}}{{ - 2\sin t}} = - \cot t
    [/tex]

    [tex]
    \begin{array}{l}
    x = 2\cos t \\
    \therefore t = \cos ^{ - 1} \frac{x}{2} \\
    \end{array}
    [/tex]

    Substituting that into my dy/dx which is currently in terms of t:

    [tex]
    \frac{{dy}}{{dx}} = - \cot t = - \cot \cos ^{ - 1} \frac{x}{2}
    [/tex]

    Now, my book gives a completely different answer, which isnt even in terms of any trigonometric function. Where have i gone wrong?

    Thanks,
    Dan.
     
  2. jcsd
  3. Mar 30, 2007 #2
    Do you know that sin2t + cos2t = 1? ;)
     
  4. Mar 30, 2007 #3

    danago

    User Avatar
    Gold Member

    Yep i do. Not quite seeing where it comes into it though :cry:

    EDIT: ahhh now ive got it. Square my x and y functions, and then add them, which eliminates the parameter :) Thanks for that.
     
    Last edited: Mar 30, 2007
  5. Mar 30, 2007 #4

    Dick

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    Science Advisor
    Homework Helper

    You could also continue on your original course. cot(cos^(-1)(x/2)) can be written without use of trig functions. Draw a right triangle containing an angle cos^(-1)(x/2).
     
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