# Thread: Linear Approximation problem

1. ## Linear Approximation problem

Hi, I'm given the function f(x,y)= sqrt(x^2+y^2) and the points (3.01, 3.98). I think that once plugging the points in is simple enough but I'm confused as to where to begin. Am I supposed to just take the derivative of the entire function? Which I believe would lead to (1/2(x^2+y^2)^-1/2)+(2x^-1/2+2y^-1/2).^-1/2+2y^-1/2). Thank you in advance

2. ## Re: Linear Approximation problem

Originally Posted by grandmarquis84
Hi, I'm given the function f(x,y)= sqrt(x^2+y^2) and the points (3.01, 3.98). I think that once plugging the points in is simple enough but I'm confused as to where to begin. Am I supposed to just take the derivative of the entire function? Which I believe would lead to (1/2(x^2+y^2)^-1/2)+(2x^-1/2+2y^-1/2).^-1/2+2y^-1/2). Thank you in advance
You are telling us what you are given. What is the question, that is, what you are asked to do with this.

3. ## Re: Linear Approximation problem

This is the linear approximation formula:

$f(x,y) \approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$

So, $f_x(x,y) = \dfrac{x}{\sqrt{x^2+y^2}}, f_y(x,y) = \dfrac{y}{\sqrt{x^2+y^2}}$

Then,
\begin{align*}f(3.01,3.98) & \approx f(3,4) + f_x(3,4)(3.01-3) + f_y(3,4)(3.98-4) \\ & = 5+0.006 -0.016 \\ & = 4.99\end{align*}