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Paramentric eq. of tangent line

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the tangent line to f(t)=<cot(t),csc(t)> at the point (1/sq.rt of 3, 2/sq.rt of 3)


    2. Relevant equations
    n/a


    3. The attempt at a solution
    I started by finding the slope, y'/x', so I got csc(t)cot(t)/csc^2(t). I then used the equation of a line, y=mx+b. Doesn't a tangent line have the opposite slope? So I think i'd plug in 2/sq.rt of 3 for y, csc^2(t)/csc(t)cot(t) for the slope, 1/sq.rt of 3 for x, and then solve for b. Do I need to plug in my x and y points though into the slope? Then that gets tough because I dont know what csc^2(1/sq. rt of 3) is. Any suggestions?
     
  2. jcsd
  3. Dec 6, 2009 #2

    LCKurtz

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    Any law against simplifying that?

    Why use that form instead of the form y - y0 = m(x - x0)? After all, you are given a point on the curve through which the line must pass.
    Opposite? Isn't the slope of the tangent line defined to be the slope of the curve?

    Can't you figure out what t gives the point in question and just use it in your calculations?
     
  4. Dec 6, 2009 #3
    I say you should try to figure out what value of t makes that point. Once you do that you can derive that vector equation, then with the derivative you can find the slope at any given point. With that slope you can find the equation of the line.
     
  5. Dec 7, 2009 #4
    Thank you! I took the advice from the first post, and Ive ended up with y-csc(t)=cos(t) (x-csc(t)). Now i'm just lost as to what to do with the point ive been given, (1/sq. rt of 3, 2/sq. rt of 3). I could plug in those numbers into the equation i arrived at for x and y, but then what would i solve for?
     
  6. Dec 7, 2009 #5
    do what i said on my post after that, because if you have the derived vector equation, you need to plug in a t value
     
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