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Find the level curve through the point on the gradient

  1. Dec 13, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-12-13_18-25-48.png
    2. Relevant equations


    3. The attempt at a solution
    The answer is F. I don't how to get this. I know that it is perpendicular and must have a horizontal tangent. How do I come to this answer?
     
  2. jcsd
  3. Dec 13, 2016 #2

    Mark44

    Staff: Mentor

    Let's look at one choice and see why it isn't an answer.
    D. y = 2/x
    Let f(x, y) = y - 2/x
    Then ##\nabla f = <-2x^{-2}, 1>##, so ##\nabla f(1, 2) = <-2, 1> \ne \vec{j}##
     
  4. Dec 13, 2016 #3
    I understand, but why is f(x,y) = y- 2/x
     
  5. Dec 13, 2016 #4

    Mark44

    Staff: Mentor

    I'm defining that way. With this definition, f(x, y) = 0 is equivalent to y = 2/x.
     
  6. Dec 13, 2016 #5
    Can we always define it as being this way ?
     
  7. Dec 13, 2016 #6

    Mark44

    Staff: Mentor

    When you're talking about level curves, as this problem is, the equation y = 2/x represents the level curve f(x, y) = 0, with f(x, y) = y - 2/x.
     
  8. Dec 13, 2016 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Here's another way to look at it. Each of your examples except (E) is of the form ##y=f(x)## for some ##f(x)##. It's not good form to use ##f(x,y)## in the same problem you have ##f(x)##, so call ##F(x,y) = y - f(x)##. Then ##\nabla F(x,y) = -f'(x)\hat i + 1\hat j## so ##\nabla F(1,2) = -f'(1)\hat i + 1\hat j =0 \hat i + \hat j## means ##f'(1) = 0##. That is quick and easy to check for each ##f(x)## and you will easily will see (F) works.
     
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