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Linear Approximation function

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Use the linear approximation to approximate a suitable function f(x,y) and thereby estimate the following

    f(x,y) = [tex]\sqrt{(4.01)^2 + (3.98)^2 + (2.02)^2}[/tex]

    2. Relevant equations

    Not going to type it out, but the formula for f(x,y) http://en.wikipedia.org/wiki/Linear_approximation

    3. The attempt at a solution

    I just need to find the equation first, I can do the estimation.

    My guess is, let x = 4, and y = 2 then

    f(x,y) = [tex]\sqrt{(x)^2 + (x)^2 + (y)^2}[/tex]
    =(2(x)2 + (y)2)1/2

    Is that done correctly? Proceeding this would be to solve the Linear Approximation formula, and check to see that it is infact close to [tex]\sqrt{(4.01)^2 + (3.98)^2 + (2.02)^2}[/tex]

    Any help would be great thanks
     
  2. jcsd
  3. Oct 29, 2009 #2

    Mark44

    Staff: Mentor

    Your function is actually one with three variables, f(x, y, z), and its formula is sqrt(x^2 + y^2 + z^2)

    Here's what you want to use:
    f(x0 + dx, y0 + dy, z0 + dy) [itex]\approx[/itex] f(x0, y0, z0) + df
    where df = fx(x0, y0, z0)*dx + fy(x0, y0, z0)*dy + fz(x0, y0, z0)*dz.

    The notation fx(x0, y0, z0) means the partial deriviative of f, evaluated at (x0, y0, z0), and so on for the other two partials.

    For your problem, x0 = 4, y0 = 4, and z0 = 2
    dx = .01, dy = -.02, and dz = .02
     
  4. Oct 29, 2009 #3
    Ah, That all makes sense, but when I computed df, I got that equal to 0

    df/dx = x/[tex]\sqrt{(x)^2 + (z)^2 + (y)^2}[/tex] = 0.6666...
    df/dy = 0.6666...
    df/dz = 0.3333...

    which means, my approximation is exact, which i dont think is right >.<
     
    Last edited: Oct 29, 2009
  5. Oct 29, 2009 #4

    Mark44

    Staff: Mentor

    It just means that df = 0.
    f(x0 + dx, y0 + dy, z0 + dy) [itex]\approx[/itex]
    f(x0, y0, z0) + df
    The fact that df = 0 doesn't turn the above into an equality.

    In this case, the exact value is sqrt(36.009) [itex]\approx[/itex] 6.000075, which is approximately equal to your estimate, sqrt(36) = 6
     
  6. Oct 29, 2009 #5
    Ohh I see perfect, thank you very much
     
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