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. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Let f(x,y) = [itex](xe^y)^8[/itex]

    i) Find

    [itex]\frac{∂f}{∂x}[/itex] [itex]\frac{∂f}{∂y}[/itex] [itex]\frac{∂^2f}{∂x^2}[/itex]

    ii) Using a tangent plane of f(x,y) find an approximate value of (0.98e^0.01)^8

    2. Relevant equations
    3. The attempt at a solution

    i)
    [itex]\frac{∂f}{∂x}[/itex] =[itex] 8e^{8y}x^{7}[/itex]

    [itex]\frac{∂f}{∂y}[/itex] = [itex] 8x^{8}e^{8y}[/itex]

    [itex]\frac{∂^2f}{∂x^2}[/itex] = [itex] 56e^{8y}x^{6}[/itex]


    ii) I have done many questions on finding linear approximations but I have always had a function, a point to evaluate the function at and points to approximate it at.

    In this I have the function Let f(x,y) = [itex](xe^y)^8[/itex] and want to use it to approximate f(0.98,0.01) but I'm not sure at what point I should evaluate it at.

    Can anyone help out?
     
  2. jcsd
  3. Nov 3, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since .98 is reasonably close to 1 and 0.01 close to 0, I think x= 1, y= 0 would be a good try.
     
  4. Nov 4, 2011 #3
    Was leaning towards that, just wanted to make sure.

    That all worked out nicely, thanks for your advice =D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Linear Approximation
  1. Linear approximation (Replies: 4)

  2. Linear approximation (Replies: 1)

  3. Linear Approximation (Replies: 4)

  4. Linear Approximations (Replies: 3)

Loading...