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?Parametric equations?

  1. Dec 8, 2004 #1
    ??Parametric equations??

    hey whats up,
    ok i have a question on differentiating parametric equations. The question says to find the equation of the two tangent lines to the curve. here's the equations...

    [tex] x = 6 cos t [/tex]
    [tex] y= 2sin2t[/tex]

    Now, after i differentiate them, do i just plug in the corresponding point? , here's what i got after i differentiated them...

    [tex] \frac {dy}{dx} = \frac {2cos2t} {(3)(-sint)} [/tex]

    after plugging in the corresponding point for t, i will have the slope, but is that slope for both of the equations? :confused:
     
  2. jcsd
  3. Dec 8, 2004 #2
    Your reasoning is quite correct except for the fact that you will get not a 3 in the denominator but a 6. Yes after you apply the chain rule, all you have to do is substitute t.
     
  4. Dec 8, 2004 #3
    He'll get a 3 since y'=4cos2t, x'=-6sint and 4/6=2/3.
     
  5. Dec 8, 2004 #4

    ehild

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    It is the slope of the tangent line of the CURVE.

    ehild
     
  6. Dec 8, 2004 #5
    How do i find the corresponding points? lol...
     
  7. Dec 8, 2004 #6
    Hello

    Now that you have the slope, you still need to write down the equation for the tangent:

    y = slope*(x - 6cost) + 2sin2t​
    I guess this will answer this question fully.

    I made a drawing of the curve, a choosen point and its tangent, in MS Excel.
    I attach it to this post, compressed in a zip file.
    By changing the reference point, you can see that the equation for the tangent is ok: it touches the curve smoothly.
    There is a small difficulty to t=0 or pi because the slope becomes infinite (the tangent is vertical then).
     
    Last edited: Nov 22, 2011
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