Linear Approximation

  • #1

Homework Statement



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

Homework Equations


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?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Since .98 is reasonably close to 1 and 0.01 close to 0, I think x= 1, y= 0 would be a good try.
 
  • #3
Was leaning towards that, just wanted to make sure.

That all worked out nicely, thanks for your advice =D
 

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