# Find the linear approximation of the function f(x,y) = \sqrt{14 - x^2 - 4y^2} at the point...

## Question:

Find the linear approximation of the function {eq}f(x,y) = \sqrt{14 - x^2 - 4y^2} {/eq} at the point (1,1).

Use this approximation to find {eq}f(0.94, 1.09) {/eq}

## Linear Approximation:

The linear approximation function is evaluated by using the actual value of the function and the partial derivatives of the function at the point. The linearization function is used for approximating the given value of the function.

## Answer and Explanation:

The function is {eq}\displaystyle f(x,y) = \sqrt{14 - x^2 - 4y^2} {/eq} at the point {eq}\displaystyle (1,1) {/eq}.

Finding the linear approximation:

{eq}\begin{align*} \displaystyle f(x,y) &= \sqrt{14-x^2-4y^2} \\ \displaystyle f(1,1) &= \sqrt{14-(1)^2-4(1)^2} \\ \displaystyle f(1,1) &= 3 \\ \displaystyle f_{x}(x, y) &=-\frac{x}{\sqrt{14-x^2-4y^2}} \\ \displaystyle f_{x}(1, 1) &=-\frac{1}{\sqrt{14-(1)^2-4(1)^2}} \\ \displaystyle f_{x}(1, 1) &=-\frac{1}{3} \\ \displaystyle f_{y}(x, y) &=-\frac{4y}{\sqrt{14-x^2-4y^2}} \\ \displaystyle f_{y}(1, 1) &=-\frac{4(1)}{\sqrt{14-(1)^2-4(1)^2}} \\ \displaystyle f_{y}(1, 1) &=-\frac{4}{3} \\ \displaystyle L(x, y) &=f(A, B)+f_{x}(A, B)(x-A)+f_{y}(A, B)(y-B) \\ \displaystyle &=3-\frac{1}{3}(x-1)-\frac{4}{3}(y-1) \\ \displaystyle &=3-\frac{x}{3}+\frac{1}{3}-\frac{4y}{3}+\frac{4}{3} \\ \displaystyle &=-\frac{x}{3}-\frac{4y}{3}+\frac{4}{3}+\frac{1}{3}+3 \\ \displaystyle L(x, y) &= -\frac{x}{3}-\frac{4y}{3}+\frac{14}{3} \end{align*} {/eq}

Therefore, the linear approximation is {eq}\ \displaystyle \mathbf{\color{blue}{ L(x, y) = -\frac{x}{3}-\frac{4y}{3}+\frac{14}{3} }} {/eq}.

Using the linear approximation to find {eq}\displaystyle f(0.94, 1.09) {/eq}:

{eq}\begin{align*} \displaystyle f(0.94, 1.09) & \approx L(x, y) = -\frac{x}{3}-\frac{4y}{3}+\frac{14}{3} \\ \displaystyle f(0.94, 1.09) & \approx -\frac{0.94}{3}-\frac{4(1.09)}{3}+\frac{14}{3} \\ \displaystyle f(0.94, 1.09) & \approx 2.9 \end{align*} {/eq}

Therefore, the approximated value is {eq}\ \displaystyle \mathbf{\color{blue}{ f(0.94, 1.09) \approx 2.9 }} {/eq}.

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from Math 104: Calculus

Chapter 10 / Lesson 2