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Help understanding equation involving a partial derivative

  1. Sep 14, 2015 #1
    Mod note: Moved from a homework section
    1. The problem statement, all variables and given/known data

    N/A

    2. Relevant equations
    f(x + Δx,y) = f(x,y) + ∂f(x,y)/∂x*Δx

    3. The attempt at a solution
    Sorry this isn't really homework. We were given this equation today in order to derive the Taylor expansion formula in two variables and I'm not sure where it came from.It seems similar to the mean value theorem.

    any help would be appreciated
     
    Last edited by a moderator: Sep 14, 2015
  2. jcsd
  3. Sep 14, 2015 #2

    Mark44

    Staff: Mentor

    It's helpful to sketch the surface z = f(x, y), showing the points ##(x_0, y_0, f(x_0, y_0))## and ##(x_0 + \Delta x, y_0, f(x_0 + \Delta x, y_0))##. The formula above should really be ##f(x_0 + \Delta a, y_0) \approx f(x_0, y_0) + \frac{\partial f(x, y)}{\partial x}|_{(x_0, y_0)} \Delta x##, since the right side is only an approximation to the left side.

    What's happening here is that the expression on the right side gives the approximate function value using the line that is tangent to the surface z = f(x,y) at ##(x_0, y_0, z_0)## (with ##z_0 = f(x_0, y_0)##), along a direction parallel to the x-axis. The value that is produced might be smaller than the actual function value on the surface, or it might be larger, depending on whether the surface is concave up or concave down, respectively, near the point ##(x_0, y_0, z_0)##.
     
  4. Sep 14, 2015 #3
    Thanks for the reply. I'm still unsure about how the line to the tangent of the surface can be used to approximate the function in the example given.
     
  5. Sep 14, 2015 #4

    Mark44

    Staff: Mentor

    In pretty much the same way that ##f(x_0 + \Delta x)## can be approximated by ##f(x_0) + f'(x_0) \Delta x##. It's really the same idea. Here's an example using a function of one variable.
    Let ##f(x) = \sqrt{x}##. Approximate ##\sqrt{4.1}##
    Here x0 = 4, f(x0) = 2, and ##\Delta x = .1##
    ##\sqrt{4.1} = f(x_0 + \Delta x) \approx f(x_0) + f'(x_0) \Delta x = 2 + \frac{1}{2\sqrt{4}} .1 = 2.025##
    So ##\sqrt{4.1} \approx 2.025##.
    Compare this answer with what I get from a calculator, approximately 2.02484567.

    My answer overestimates the actual answer because the tangent line approximation gives me a number that is above the curve ##y = \sqrt{x}##.
     
  6. Sep 14, 2015 #5

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Last term is Δf (lim as → 0 bla bla). Take first RHS term to LHS and you see nothing but your first ever formula in beginning calculus.

    They're assuming here only x varies and they must soon get to x and y both vary in which case they'd have to add a + ∂f/∂y*Δy
     
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