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Parametric derivatives

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data
    The parametric equations of a curve are
    ##x=\frac{1}{2}(sint cost+ t), y=\frac{1}{2} t-\frac{1}{4} sin2t##,
    ##-\pi/2<t\leq0##. P is a point on the curve such that the gradient at P is 1. Find the equation of the normal at P. Hence, determine if the normal at P meets the curve again.

    2. Relevant equations
    Parametric differentiation, chain rule


    3. The attempt at a solution

    I have found ##\frac{dy}{dt}## and ##\frac{dt}{dx}##. ..Then ##\frac{dy}{dx}=\frac{1-cos 2t}{1+cos2t}##

    I have also found out that when ##\frac{dy}{dx}=1##, ##t=\frac{-\pi}{4}##. Then the equation of the normal is ##y=-x-\frac{\pi}{4}##. Now I don't know how to determine if the normal at P meets the curve again... :confused:
     
  2. jcsd
  3. Jan 21, 2014 #2

    tiny-tim

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    hi sooyong94! :smile:
    hint: what is x + y ? :wink:
     
  4. Jan 21, 2014 #3
    t ? :P
     
  5. Jan 21, 2014 #4

    HallsofIvy

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    Yes, using the fact that sin(2t)= 2sin(t)cos(t), adding the parametric equations for x and y gives x+ y= t.
    But [itex]y= -x- \frac{\pi}{4}[/itex] tells you that [itex]x+ y= \frac{\pi}{4}[/itex]. Wherever the curve and line intersect, they must both be true.
     
  6. Jan 21, 2014 #5
    But that's ##t=\frac{\pi){4}##, which is not within the domain of t. :/
     
  7. Jan 21, 2014 #6

    tiny-tim

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    no, its -π/4
     
  8. Jan 21, 2014 #7
    :hmm: Wasn't there should another coordinate that intersects with the curve other than x=-pi/4?
     
  9. Jan 21, 2014 #8

    tiny-tim

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    ?? :confused:

    do you mean x+y = -π/4 ?​
     
  10. Jan 21, 2014 #9
    I mean the normal line cuts the curve at other points... is it possible for that? :confused:
     
  11. Jan 21, 2014 #10

    Mark44

    Staff: Mentor

    Questions about derivatives belong in the Calculus & Beyond section...
     
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