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Homework Help: Finding the line tangent to a curve

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Find an equation for the line tangent to the curve at the point defined by the given value of t.
    x = 2 cos t
    y = 2 sin t
    t = pi/4

    2. Relevant equations
    sin (pi/4) = sqrt(2)/2
    cos (pi/4) = sqrt(2)/2

    3. The attempt at a solution

    Determining the slope:
    [dy/dt]/[dx/dt] = [2 cos t]/[-2 sin t]
    = [2 cos (pi/4)]/[-2 sin (pi/4)]
    = [2 (sqrt(2)/2)]/[-2 (sqrt(2)/2)]
    = [sqrt(2)]/[-sqrt(2)]
    = -1
    slope = -1

    Finding the line:
    y = mx+b
    m = -1
    2 sin t = -1(2 cos t)+ b
    2 sin t = -2 cos t + b
    2 sin (pi/4) = -2 cos(pi/4) + b
    2 [sqrt(2)/2] = -2 [sqrt(2)/2] + b
    sqrt(2) = -sqrt(2) + b
    b = 2 sqrt(2)

    y = -1x + 2 sqrt(2)

    The answer the book has says [y = -x + 2]. I'm not sure what I did wrong with the problem.
     
  2. jcsd
  3. Mar 10, 2010 #2

    Dick

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    Science Advisor
    Homework Helper

    The book is certainly wrong. y=(-x)+2 doesn't work at all.
     
  4. Mar 11, 2010 #3

    HallsofIvy

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    Science Advisor

    For one thing, [itex]\sqrt{2}\ne -\sqrt{2}+ 2[/itex] so that line does not even pass through the point [itex](\sqrt{2},\sqrt{2})[/itex].
     
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