# Find the linearization of the function f(x,y) = \sqrt{39 - 5x^2 - 3y^2} at the point (-2,-1). Use...

## Question:

Find the linearization of the function {eq}f(x,y) = \sqrt{39 - 5x^2 - 3y^2} {/eq} at the point (-2,-1).

Use the linear approximation to estimate the value of f(-2.1,-0.9) =

## Linearization and Estimation:

The linear approximation function with two variables at the point is found by using the basic formula {eq}\displaystyle L(x, y) =f(a, b)+f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b) {/eq}. Then, it should be used for estimating the given value of the function.

The given function is {eq}\displaystyle f(x,y) = \sqrt{39-5x^2-3y^2} {/eq} at the point {eq}\displaystyle (-2,-1) {/eq}.

Finding the linearization of the function:

{eq}\begin{align*} \\ \displaystyle L(x, y) &=f(a, b)+f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b) \\ \\ \displaystyle f(x,y) &= \sqrt{39-5x^2-3y^2} \\ \displaystyle f(-2,-1) &= \sqrt{39-5(-2)^2-3(-1)^2}=4 \\ \\ \displaystyle f_{x}(x, y) &=-\frac{5x}{\sqrt{39-5x^2-3y^2}} \\ \displaystyle f_{x}(-2,-1) &=-\frac{5(-2)}{\sqrt{39-5(-2)^2-3(-1)^2}}=\frac{5}{2} \\ \\ \displaystyle f_{y}(x, y) &=-\frac{3y}{\sqrt{39-5x^2-3y^2}} \\ \displaystyle f_{y}(-2,-1) &=-\frac{3(-1)}{\sqrt{39-5(-2)^2-3(-1)^2}}=\frac{3}{4} \\ \\ \displaystyle L(x, y) &=4+\frac{5}{2}(x-(-2))+\frac{3}{4}(y-(-1)) \\ \displaystyle &=4+\frac{5}{2}(x+2)+\frac{3}{4}(y+1) \\ \displaystyle &=4+\frac{5x}{2}+5+\frac{3y}{4}+\frac{3}{4} \\ \displaystyle &=\frac{5x}{2}+\frac{3y}{4}+\frac{3}{4}+5+4 \\ \displaystyle L(x, y) &=\frac{5x}{2}+\frac{3y}{4}+\frac{39}{4} \\ \end{align*} {/eq}

Therefore, the linearization of the function at the point is {eq}\ \displaystyle \mathbf{\color{blue}{ L(x, y)=\frac{5x}{2}+\frac{3y}{4}+\frac{39}{4} }} {/eq}.

Estimating the value of {eq}\displaystyle f(-2.1,-0.9) {/eq} using the linear approximation:

{eq}\begin{align*} \displaystyle f(-2.1,-0.9) & \approx L(x, y)=\frac{5x}{2}+\frac{3y}{4}+\frac{39}{4} \\ \displaystyle f(-2.1,-0.9) & \approx \frac{5(-2.1)}{2}+\frac{3(-0.9)}{4}+\frac{39}{4} \\ \displaystyle f(-2.1,-0.9) & \approx 3.825 \end{align*} {/eq}

Therefore, the estimated value is {eq}\ \displaystyle \mathbf{\color{blue}{ f(-2.1,-0.9) \approx 3.825 }} {/eq}.