local linear approximations

[sambellamy]

I am stuck. What am I mixing up here? Someone who can help, please take a look.

I am given a function f and LLA L at point P. The problem I'm working on asks me to find the point P.

given: f(x,y) = x² + y² , and L(x,y) = 2y - 2x - 2

I know the LLA formula is L(x,y) ~ f(x₀,y₀) + f_x(x₀,y₀)(x-x₀) + f_y(x₀,y₀)(y-y₀)

I have this so far:

f_x = 2x
f_y = 2y

plugging these in, I get:

-2 = x² + 2x + y² - 2y + 2x₀(x-x₀) + 2y₀(y-y₀)

I am not sure what to do from here, I have four variables. I know I am looking for (x₀,y₀) as the coordinates of P.
I tried completing the square with x² + 2x, for instance, and also distributing 2x₀(x-x₀), but neither of these seemed to make it clearer what the next step should be.

Thanks for any help.

 

[Ishuda]

I am stuck. What am I mixing up here? Someone who can help, please take a look.

I am given a function f and LLA L at point P. The problem I'm working on asks me to find the point P.

given: f(x,y) = x² + y² , and L(x,y) = 2y - 2x - 2

I know the LLA formula is L(x,y) ~ f(x₀,y₀) + f_x(x₀,y₀)(x-x₀) + f_y(x₀,y₀)(y-y₀)

I have this so far:

f_x = 2x
f_y = 2y

plugging these in, I get:

-2 = x² + 2x + y² - 2y + 2x₀(x-x₀) + 2y₀(y-y₀)

I am not sure what to do from here, I have four variables. I know I am looking for (x₀,y₀) as the coordinates of P.
I tried completing the square with x² + 2x, for instance, and also distributing 2x₀(x-x₀), but neither of these seemed to make it clearer what the next step should be.

Thanks for any help.

I'm not sure what LLA L is. Could you explain further please? Do you want to find the "points of intersection" between the functions
f(x,y) = x² + y²
and
L(x,y) = 2y - 2x - 2
using a linear approximation?

 

[sambellamy]

LLA L means local linear approximation, L. So we are using the term L(x,y) instead of f(x₀,y₀) to mean the estimated value of f' at (x,y), using the formula.

 

[Ishuda]

LLA L means local linear approximation, L. So we are using the term L(x,y) instead of f(x₀,y₀) to mean the estimated value of f' at (x,y), using the formula.

O.K. then, line L(x,y)
L(x,y) = 2y - 2x - 2
is the local approximation to f(x,y)
f(x,y) = x² + y²

Since you know the LLA formula and the partial derivatives, just match up terms
L(x,y) ~ f(x₀,y₀) + f_x(x₀,y₀)(x-x₀) + f_y(x₀,y₀)(y-y₀)
So
f_x(x₀,y₀) = 2 x₀ = -2
which is the x coefficient in L(x,y) and x₀ = -1. Now do the same for the y coefficient in L(x,y) as given and in the formula to find y₀.

After that you have x₀ and y₀ but, IMO, you should check your answer. So what is the constant coefficient in L(x,y) as given and in the LLA formula. Those should match up.

Note that you will not aways have the values of x₀ and y₀ after matching up the coefficients of x and y and might need to work with the constant term. For example, suppose
f(x,y) = x³ + y²
and
L(x,y) = 1 + 3 x + 2 y
What is x₀ and y₀? Or maybe if
L(x,y) = -3 + 3 x + 2 y
What is x₀ and y₀?