Question

find the linear approximation of the function below at the indicated point.

$f(x,y)=\sqrt{14 - x^{2}-4y^{2}}$ at $(1,1)$

$f(x,y)\approx$

use this approximation to find $f(1.09,0.94)$. (round your answer to three decimal places.)

$f(1.09,0.94)\approx$

Explanation:

Step1: Find \( f(1,1) \)

Substitute \( x = 1,y = 1 \) into \( f(x,y)=\sqrt{14 - x^{2}-4y^{2}} \).

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Step2: Find \( f_{x}(x,y) \)

Differentiate \( f(x,y)=\sqrt{14 - x^{2}-4y^{2}}=(14 - x^{2}-4y^{2})^{\frac{1}{2}} \) with respect to \( x \) using the chain rule \( \frac{d}{dx}(u^{n})=nu^{n - 1}\frac{du}{dx} \), where \( u = 14 - x^{2}-4y^{2},n=\frac{1}{2} \).

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Substitute \( x = 1,y = 1 \) into \( f_{x}(x,y) \), \( f_{x}(1,1)=\frac{-1}{3} \)

Step3: Find \( f_{y}(x,y) \)

Differentiate \( f(x,y)=(14 - x^{2}-4y^{2})^{\frac{1}{2}} \) with respect to \( y \) using the chain rule.

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Substitute \( x = 1,y = 1 \) into \( f_{y}(x,y) \), \( f_{y}(1,1)=\frac{-4}{3} \)

Step4: Write the linear approximation formula

The linear approximation formula for \( z = f(x,y) \) at \( (a,b) \) is \( L(x,y)=f(a,b)+f_{x}(a,b)(x - a)+f_{y}(a,b)(y - b) \). Here \( a = 1,b = 1 \), so \( L(x,y)=3-\frac{1}{3}(x - 1)-\frac{4}{3}(y - 1) \)

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Step5: Use the linear approximation to find \( f(1.09,0.94) \)

Substitute \( x = 1.09,y = 0.94 \) into \( L(x,y) \)

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Answer:

The linear approximation \( f(x,y)\approx-\frac{1}{3}x-\frac{4}{3}y+\frac{14}{3} \), and \( f(1.09,0.94)\approx3.05 \)