# Find the linear approximation of the function f(x,y,z) = \sqrt{x^2 + y^2 + z^2} at (2,3,6) and...

## Question:

Find the linear approximation of the function {eq}f(x,y,z) = \sqrt{x^2 + y^2 + z^2} {/eq} at (2,3,6) and use it to approximate the number {eq}\sqrt{2.03^2 + 2.99^2 + 5.98^2} {/eq}

## Linear Approximation:

We can define the linear approximation function as {eq}\displaystyle L(x) {/eq}. A function of three variables at the point is used for finding the gradient with the help of partial derivatives.

The basic formula is {eq}\ \displaystyle L(x, y, z) =f(A, B, C)+f_{x}(A, B, C)(x-A)+f_{y}(A, B, C)(y-B)+f_{z}(A, B, C)(z-C) {/eq}.

## Answer and Explanation:

The given function is {eq}\displaystyle f(x,y,z) = \sqrt{x^2 + y^2 + z^2} {/eq} at the point {eq}\displaystyle (2,3,6) {/eq}.

Finding the linear approximation:

{eq}\begin{align*} \\ \displaystyle f(x,y,z) &= \sqrt{x^2 + y^2 + z^2} \\ \displaystyle f(2,3,6) &= \sqrt{(2)^2+(3)^2+(6)^2} \\ \displaystyle f(2,3,6) &= 7 \\ \\ \displaystyle f_{x}(x,y,z) &=\frac{x}{\sqrt{x^2+y^2+z^2}} \\ \displaystyle f_{x}(2,3,6) &=\frac{2}{\sqrt{(2)^2+(3)^2+(6)^2}} \\ \displaystyle f_{x}(2,3,6) &=\frac{2}{7} \\ \\ \displaystyle f_{y}(x,y,z) &=\frac{y}{\sqrt{x^2+y^2+z^2}} \\ \displaystyle f_{y}(2,3,6) &=\frac{3}{\sqrt{(2)^2+(3)^2+(6)^2}} \\ \displaystyle f_{y}(2,3,6) &=\frac{3}{7} \\ \\ \displaystyle f_{z}(x,y,z) &=\frac{z}{\sqrt{x^2+y^2+z^2}} \\ \displaystyle f_{z}(2,3,6) &=\frac{6}{\sqrt{(2)^2+(3)^2+(6)^2}} \\ \displaystyle f_{z}(2,3,6) &=\frac{6}{7} \\ \\ \\ \displaystyle L(x, y, z) &=f(A, B, C)+f_{x}(A, B, C)(x-A)+f_{y}(A, B, C)(y-B)+f_{z}(A, B, C)(z-C) \\ \displaystyle &=7+\frac{2}{7}(x-2)+\frac{3}{7}(y-3)+\frac{6}{7}(z-6) \\ \displaystyle &=7+\frac{2x}{7}-\frac{4}{7}+\frac{3y}{7}-\frac{9}{7}+\frac{6z}{7}-\frac{36}{7} \\ \displaystyle &=\frac{2x}{7}+\frac{3y}{7}+\frac{6z}{7}-\frac{36}{7}-\frac{4}{7}-\frac{9}{7}+7 \\ \displaystyle L(x, y, z) &=\frac{2x}{7}+\frac{3y}{7}+\frac{6z}{7} \end{align*} {/eq}

Therefore, the linear approximation is{eq}\ \displaystyle \mathbf{\color{blue}{ L(x, y, z)=\frac{2x}{7}+\frac{3y}{7}+\frac{6z}{7} }} {/eq}.

Approximating the number {eq}\displaystyle \sqrt{2.03^2 + 2.99^2 + 5.98^2} {/eq} using the linear approximation:

{eq}\begin{align*} \\ \displaystyle f(2.03+2.99+5.98) & \approx L(x, y, z)=\frac{2x}{7}+\frac{3y}{7}+\frac{6z}{7} \\ \displaystyle f(2.03+2.99+5.98) & \approx \frac{2(2.03)}{7}+\frac{3(2.99)}{7}+\frac{6(5.98)}{7} \\ \displaystyle f(2.03+2.99+5.98) & \approx \frac{48.91}{7} \\ \displaystyle f(2.03+2.99+5.98) & \approx 6.98714 \end{align*} {/eq}

Therefore, the approximated value is {eq}\ \displaystyle \mathbf{\color{blue}{ f(2.03+2.99+5.98) \approx 6.987 }} {/eq}.

#### Learn more about this topic:

How to Estimate Function Values Using Linearization

from Math 104: Calculus

Chapter 10 / Lesson 2
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