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Finding the equation for a tangent line (Parametric Equation)

  1. Sep 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Find an equation of the tangent to the curve given by
    x=tan(∅)
    y=sec(∅)
    at the point (1,sqrt(2))

    The answer should be in the form y=f(x) without ∅


    3. The attempt at a solution
    Tangent line equation...
    y-y*=m(x-x*)
    m = dy/dx
    m = sec(∅)tan(∅) / (sec^2(∅)
    m = tan∅ / sec(∅)
    m = sin∅

    This is what I did so far, i'm a bit lost obviously
     
  2. jcsd
  3. Sep 5, 2013 #2

    eumyang

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    Hint: use "θ" for theta, not "∅".

    Find out what θ is when x = 1 and y = sqrt(2) (this is from the given point). Plug the value for θ in your expression for dy/dx to get your slope.
     
  4. Sep 5, 2013 #3
    Okay, so i did that and end up with m = sqrt(2)/2.
    If i use the equation
    y-y*=m(x-x*)
    with
    y* = sqrt(2)
    x* = 1 and solve it i get
    y=(sqrt(2)/2)x+2.1
    but i type this into my online course and it says it's wrong?
     
  5. Sep 5, 2013 #4

    eumyang

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    I'm not getting 2.1.
     
  6. Sep 5, 2013 #5
    When i type in
    y=.71x+.71 it gives me the wrong answer as well. ;-(
     
  7. Sep 5, 2013 #6

    eumyang

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    Maybe it is because you are using a decimal approximation? I'm pretty sure the exact form of the last equation you wrote is correct.
     
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