# Find the linear approximation to the function defined by f(x, y, z) = (y^3}z, xe^{x} + 2 y + z)...

## Question:

Find the linear approximation to the function defined by {eq}f(x, y, z) = (y^{3}z, xe^{x} + 2 y + z) {/eq} at the point {eq}(0, 1, 1) {/eq}.

## Linear Approximation for the Function of three variables:

Let us consider {eq}u=f(x,y,z) {/eq} be any function of three variables.

Then Taylor series Linear approximation of that function about any point {eq}(a,b,c) {/eq} is given by:

{eq}L(x,y,z) = f(a,b,c) + \left( {(x - a){f_x}(a,b,c) + (y - b){f_y}(a,b,c)+ (z - c){f_z}(a,b,c)} \right) {/eq}

Here the Given function is:{eq}f(x, y, z) = (y^{3}z+xe^{x} + 2 y + z) {/eq} and we have to find its Taylor series approximation about the origin. Hence {eq}(a,b,c)=( 0,1,1) {/eq}

According to the calculation, we can say:

{eq}\eqalign{ & f(x, y, z) = (y^{3}z+xe^{x} + 2 y + z)\Rightarrow f\left( {0,1,1} \right) = 4 \cr & {f_x} = xe^x+e^x \Rightarrow {f_x}\left(0,1,1 \right) = 1 \cr & {f_y} = 3y^2z+2 \Rightarrow {f_y}\left( 0,1,1 \right) = 5 \cr & {f_z} = y^3+1 \Rightarrow {f_z}\left( 0,1,1 \right) = 2 \cr} {/eq}

Now the Linear approximation is given by:

{eq}\eqalign{ L(x,y,z)& = f\left( 0,1,1 \right) + \left( {\left( {x - 0} \right){f_x}\left( 0,1,1 \right) + \left( {y -1 } \right){f_y}\left( 0,1,1 \right) + \left( {z - 1} \right){f_z}\left( 0,1,1\right)} \right) \cr & = 4 + \left( {\left( {x - 0} \right)\left( {1} \right) + \left( {y -1} \right)\left( { 5} \right) + \left( {z - 1} \right)\left( {2} \right)} \right) \cr & = -3+x+5y+2z \cr} {/eq}

Linearization of Functions

from

Chapter 10 / Lesson 1
3.7K

Over the river and through the woods to Grandmother's house we go ... Are we there yet? In this lesson, apply linearization to estimate when we will finally get to Grandma's house!