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Linear Approximation 
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#1
Nov311, 07:14 PM

P: 96

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
Nov311, 07:49 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345

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
Nov411, 06:29 AM

P: 96

Was leaning towards that, just wanted to make sure.
That all worked out nicely, thanks for your advice =D 


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