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Tangent line

  1. Oct 10, 2004 #1
    Write the equation of the tangent line to the cure y=cosx at a=pi/4

    (y-y1)=m(x-x1)

    cos(pi/4)=sqrt(2)/2
    y'=-sinx=-sin(pi/4)=-sqrt(2)/2

    (y-sqrt(2)/2)=m(x-pi/4)
    y'=-sqrt(2)/2(x-pi/4)+sqrt(2)/2

    My teacher circled the y' and took a point off. I know that y'=-sinx but what should be in its place?
     
  2. jcsd
  3. Oct 10, 2004 #2

    Zurtex

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    [itex]y = mx + c[/itex] is the tangent line.

    [tex]y' = \frac{dy}{dx}[/tex] which is the gradient of your line. So actually at [itex]x = a[/itex] you will find that [itex]m = y'[/itex]
     
  4. Oct 10, 2004 #3
    The answer for the tangent line is
    -sqrt(2)/2(x-pi/4)+sqrt(2)/2

    However, I wrote y'=-sqrt(2)/2(x-pi/4)+sqrt(2)/2

    which my teacher marked off a point for the y'
    I know that -sqrt(2)/2(x-pi/4)+sqrt(2)/2 is the tangent line, but what does it equal to?
     
  5. Oct 10, 2004 #4
    Because the tangent line is a function on its own, it could just be f(x) or y; you are looking for the equation of the tangent line, not just its slope (which is y').
     
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