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Find partials for ##f## and the equation of plane

  1. Oct 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Function is ##f(x,y)=((x-1)y)^\frac{2}{3}##,##\space\space(a,b)=(1,0)##

    a) Calculate ##f_{x}(a,b)## and ##f_{y}(a,b)## at point ##(a,b)## and write the equation for the plane.

    2. Relevant equations


    3. The attempt at a solution
    So,

    ##f_{x}(x,y)=\frac{2y^\frac{2}{3}}{3(x-1)^\frac{1}{3}}##

    ##f_{y}(x,y)=\frac{2(x-1)^\frac{2}{3}}{3y^\frac{1}{3}}##

    Therefore, at ##(1,0)##, ##f_{x}(x,y)=0## and ##f_{y}(x,y)=0##

    Equation of plane at ##(1,0)## is ##z=0##.

    What I'm doing seems right, but there is a nagging feeling that I'm wrong.

    So thoughts?

    Thanks.
     
    Last edited: Oct 8, 2016
  2. jcsd
  3. Oct 8, 2016 #2

    PeroK

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    Science Advisor
    Homework Helper
    Gold Member

    Your differentiation of ##x^{\frac23}## is wrong.
     
  4. Oct 8, 2016 #3

    Ray Vickson

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    Yes, indeed, both of your partial derivatives are incorrect. I don't get ##\frac{2}{3} - 1 = 3##, as you seem to have written.
     
  5. Oct 8, 2016 #4
  6. Oct 8, 2016 #5

    Mark44

    Staff: Mentor

    The corrected partials are now correct, but the values you show at (1, 0) aren't. Notice that in both cases, both the numerator and denominator are zero.
     
  7. Oct 8, 2016 #6
    Is it undefined in both cases because they involve a division by zero?
     
  8. Oct 8, 2016 #7

    Mark44

    Staff: Mentor

    Yes. Both partials are undefined at (1, 0).
     
  9. Oct 8, 2016 #8
    So that means there is no equation that exists for the tangent plane at that point, right?
     
  10. Oct 8, 2016 #9

    Mark44

    Staff: Mentor

    Yes
     
  11. Oct 8, 2016 #10
    Thanks!
     
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