Linear Approximation: Finding Point P

[Mdhiggenz]

Homework Statement

Hey guys I'm having a hard time understanding how the book obtained the solution.

Here is the question

A function f is given along with a local linear approximation of L to f at a point P. Use the information given to determine point P.

f(x,y)= x²+y²; L(x,y)=2y-2x-2

Formula for local linear approximation is

f(x,y)+f_x(x,y)+f_y(x,y)=L(x,y)

So plugging in my known values

x0²+y0²+2x0(x-x0)+2y0(y-y0)=x²+y²

I was able to get up to here, and then when I looked at the solution in the book, it simply said x0=-1 y0 =1 and it isn't obvious to me why.

Any help is appreciated

Homework Equations

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

Hey guys I'm having a hard time understanding how the book obtained the solution.

Here is the question

A function f is given along with a local linear approximation of L to f at a point P. Use the information given to determine point P.

f(x,y)= x²+y²; L(x,y)=2y-2x-2

Formula for local linear approximation is

f(x,y)+f_x(x,y)+f_y(x,y)=L(x,y)

That isn't correct.

So plugging in my known values

x0²+y0²+2x0(x-x0)+2y0(y-y0)=x²+y²

That is the correct linear approximation near $(x_0,y_0)$. Buy why do you set it equal to your original $f(x,y)$? The linear approximation is an approximation to $f(x,y)$ but not equal to it unless $f(x,y)$ is linear itself. But you are given the formula for the linear approximation, which I have highlighted in red. Set it equal to that and see if you can see how they got that answer.