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!

Linear Approximation

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the linear approximation of (xy)/z at the point (-3,2,1)



    3. The attempt at a solution
    So the example my book gives has 2 variables so i'm struggling a bit with this, But I started off by taking the partial derivative with respect to each variable and solving for it.

    d/dx = (y/z) = 2 = A
    d/dy = (x/z) = -3 = B
    d/dz = -(xy)/z^2 =6 = C

    now... find the equation of a plane passing through that point maybe?

    A(x-xo)+B(y-yo)+C(z-zo) = 0
    2(x+3) - 3(y-2) + 6(z-1) = 0
    2x+6 - 3y+6 +6z-6 = 0
    2x-3y+6z+6=0

    am I on the right track here..?
     
  2. jcsd
  3. Oct 13, 2013 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Good try, but not quite. Here, you are NOT asked about the equation for some tangent plane, but how the linearized function Lf looks like, with f=xy/z

    That is essentially to determine the first 4 terms of the Taylor series about (-3,2,1), that is:
    [tex]Lf=f(-3,2,1)+\frac{\partial{f}}{\partial{x}}(x-(-3))+ \frac{\partial{f}}{\partial{y}}(y-2) + \frac{\partial{f}}{\partial{z}}(z-3)[/tex]
    where the partial derivatives are evaluated at (-3,2,1)
     
  4. Oct 13, 2013 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    arildno: That last term should be ##\frac{\partial f}{\partial z}(z-1)##. (You had (z-3) instead of (z-1)).
     
  5. Oct 13, 2013 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Close, but no.

    Look at the simpler problem finding a linear approximation to f(x) at some point x0. What you did is the equivalent of saying A=df/dx at x=x0, and thus A(x-x0) = 0. That's obviously *not* a linear approximation of f(x). What one needs to do is to replace that 0 on the right hand side with fapprox(x)-f(x0), yielding fapprox(x)=f(x0)+A(x-x0).

    You need to do the same with your function of three variables.
     
  6. Oct 13, 2013 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    :frown:, :cry:
     
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: Linear Approximation
  1. Linear Approximations (Replies: 4)

  2. Linear approximation (Replies: 4)

  3. Linear approximation (Replies: 1)

  4. Linear Approximations (Replies: 3)

Loading...