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Find the slope of the curve

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve, as requested. x^6y^6=64, normal at (2,1)


    2. Relevant equations



    3. The attempt at a solution
    64/y^6
    or 6x6y=0, that's as far as I am getting, totally lost.
     
  2. jcsd
  3. Oct 29, 2011 #2

    Simon Bridge

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    You know the slope of the tangent at a point is the deriverative of the curve at that point right? So you need to find the deriverave of:

    [tex]
    x^6 y^6 = 64
    [/tex]
    at (x,y)=(2,1) and (guessing - you check) y is a function of x.

    you can solve the equation for y, then find y' or find the implicit deriverative:
    example

    [tex]\frac{dy}{dx}:x^2y^2=4[/tex][tex]
    \frac{d}{dx} \left ( x^2y^2=4 \right )[/tex][tex]
    y^2\frac{d}{dx}x^2 + x^2\frac{d}{dx}y^2 = 0[/tex][tex]
    2xy^2 + 2yx^2\frac{dy}{dx} = 0[/tex][tex]
    \frac{dy}{dx} = \frac{-2xy^2}{2yx^2} = -xy[/tex]
     
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