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Parametric curves

  1. Jan 3, 2006 #1
    1. Im given a curve defined parametrically by [tex] x= 2/t , y=1-2t [/tex] i have found the equation of tangent at t=-2 to be y=4x+9, they have asked whether it cuts the curve again. how do i find that, since i dont know the original equation of the curve and cant solve them simultaneously.

    2. Also they have asked to find an approximation for sec 61. I have used [tex] \frac{\delta y}{\delta x}= \frac{dy}{dx} [/tex], but i did not get the answer, 2.0605. How do i get about doing it?
     
  2. jcsd
  3. Jan 3, 2006 #2

    TD

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    Intersect the line with the curve and see if you find any other intersection points besides t = -2. Since y = 4x+9, using the parametric equation for x, y = 4(2/t) + 9. Now substitute in the parametric equation for y and solve for t.

    For the second question, I'm not really sure what you mean.
     
  4. Jan 3, 2006 #3

    VietDao29

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    You should first convert 610 into radians (most work in calculus use radians, instead of degrees), i.e:
    [tex]61 ^ \circ = \frac{61 \pi}{180} \mbox{ rad}[/tex]
    Since you already have:
    [tex]\sec \left( \frac{60 \pi}{180} \right) = \sec \left( \frac{\pi}{3} \right) = 2[/tex]
    You can use
    f(x0 + h) ≈ f(x0) + h f'(x0) (h ≈ 0) to solve your problem.
    Can you go from here?
     
    Last edited: Jan 3, 2006
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