# Find the linearization, L(x,y), of f(x,y) = \sqrt(7 - x^2 - 2y^2) at the point P(2,-1).

## Question:

Find the linearization, {eq}L(x,y) {/eq}, of {eq}f(x,y) = \sqrt{7 - x^2 - 2y^2} {/eq}

at the point {eq}P(2,-1) {/eq}.

## Linearization

This problem is quite straight forward and it uses the concept of linearization of a function. To linearize a function we need to be familiar with the concept of partial differentiation. We should be careful that our work is not complete just by finding the linearization our last step would be finding the linearization at the given point.

We have,

{eq}\displaystyle f(x,y) = \sqrt{7 - x^2 - 2y^2} {/eq}

Our aim is to find the linearization of the given function at {eq}(2,-1). {/eq}

Linearization of the function is given by,

{eq}\displaystyle L(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) {/eq}

Let us find the individual component:

{eq}\displaystyle f(x_0,y_0)=\sqrt{7 - (2)^2 - 2(-1)^2}=\sqrt{7-4-2}=1 {/eq}

Let us find {eq}f_x {/eq}

{eq}\displaystyle f_x=\frac{d}{dx}( \sqrt{7 - x^2 - 2y^2}) {/eq}

{eq}\displaystyle f_x=\frac{1}{2}( {7 - x^2 - 2y^2})^{\frac{-1}{2}}(-2x) {/eq}

Let us now find {eq}f_x {/eq} at (2,-1)

{eq}\displaystyle f_x(2,-1)=\frac{1}{2}( {7 - (2)^2 - 2(-1)^2})^{\frac{-1}{2}}(-2(2)) {/eq}

{eq}\displaystyle f_x(2,-1)=\frac{1}{2}( {7 - 4 - 2})^{\frac{-1}{2}}(-4) {/eq}

{eq}\displaystyle f_x(2,-1)=(1)^{\frac{-1}{2}}(-2) {/eq}

{eq}\displaystyle f_x(2,-1)=(1)^{\frac{-1}{2}}(-2)=-2 {/eq}

Let us now find {eq}f_y {/eq}

{eq}\displaystyle f_y=\frac{d}{dy}( \sqrt{7 - x^2 - 2y^2}) {/eq}

{eq}\displaystyle f_y=\frac{1}{2}( {7 - x^2 - 2y^2})^{\frac{-1}{2}}(-4y) {/eq}

Let us now find {eq}f_y {/eq} at (2,-1)

{eq}\displaystyle f_y(2,-1)=\frac{1}{2}( {7 - (2)^2 - 2(-1)^2})^{\frac{-1}{2}}(-4(-1)) {/eq}

{eq}\displaystyle f_y(2,-1)=\frac{1}{2}( {7 - 4 - 2})^{\frac{-1}{2}}(4) {/eq}

{eq}\displaystyle f_y(2,-1)=\frac{1}{2}( 1)^{\frac{-1}{2}}(4) {/eq}

{eq}\displaystyle f_y(2,-1)=2 {/eq}

Plugging in the values we get the linearization as,

{eq}\displaystyle L(x,y)=1+(-2)(x-2)+2(y-(-1)) {/eq}

{eq}\displaystyle \boxed{\displaystyle L(x,y)=1-2(x-2)+2(y+1)} {/eq}

And the value of linearization at (2,-1) is,

{eq}\displaystyle \boxed{\displaystyle L(2,-1)=1-2(2-2)+2(-1+1)=1} {/eq}

How to Estimate Function Values Using Linearization

from Math 104: Calculus

Chapter 10 / Lesson 2
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