1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    [itex]f(x,y)=xy
    \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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding a tangent line to a level curve - Don't understand solution
Loading...