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Finding a tangent line to a level curve - Don't understand solution

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data
    If f(x,y) = xy, find the gradient vector [itex]\nabla f(3,2)[/itex] and use it to find the tangent line to the level curve f(x,y) = 6 at the point (3,2)

    2. Relevant equations

    3. The attempt at a solution
    \Rightarrow\nabla f(x,y)=<y,x>,\nabla f(3,2)=<2,3>[/itex]
    [itex]\nabla f(3,2)[/itex] is perpendicular to the tanget line, so the tangent line has equation
    [itex]\nabla f(3,2)\cdot<x-3,y-2>=0[/itex]... and so on

    I understand that the dot product must be 0 if the two vectors are perpendicular.
    What I don't get is how they pick the vector <x-3, y-2> given that the point were concered with is (3,2)
  2. jcsd
  3. Apr 17, 2013 #2

    Simon Bridge

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    As opposed to doing: <2,3>.<x,y>=0 you mean?
    The set of vectors that are perpendicular to <2,3> lie in a plane don't they?
  4. Apr 17, 2013 #3
    yeah they're in a plane. But <x-3,y-2> is not a plane?
  5. Apr 17, 2013 #4

    Simon Bridge

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    If it were [the plane perpendicular to <2,3>] then the dot product would be zero for every value of x and y so the relation would be useless.

    It follows that just knowing that <2,3> is perpendicular to the tangent you want is not good enough - you need a way to select the one you want out of the whole plane. How would you go about that? What is special about this tangent?
  6. Apr 17, 2013 #5


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    hi mrcleanhands! :smile:
    if (x,y) lies on the tangent plane, then the vector from (3,2) to (x,y) is a tangent at (3,2), and is parallel to <x-3, y-2> :wink:
  7. Apr 17, 2013 #6
    I see, it has to lie on the point (2,3) which you get by taking a line from (3,2) to (x,y)

    thanks to both of you!
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