Register to reply

Linear Approximation

by KingBigness
Tags: approximation, linear
Share this thread:
Nov3-11, 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

[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?
Phys.Org News Partner Science news on
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Nov3-11, 07:49 PM
Sci Advisor
PF Gold
P: 39,569
Since .98 is reasonably close to 1 and 0.01 close to 0, I think x= 1, y= 0 would be a good try.
Nov4-11, 06:29 AM
P: 96
Was leaning towards that, just wanted to make sure.

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

Register to reply

Related Discussions
Linear approximation Calculus & Beyond Homework 4
Linear approximation... Calculus & Beyond Homework 2
Linear Approximation Calculus & Beyond Homework 12
Linear Approximation Introductory Physics Homework 3