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Find the linearization of the function z=x sqrt y at the point (-3,49).

Question:

Find the linearization of the function {eq}\ z=x \sqrt y \ {/eq} at the point {eq}(-3,49) {/eq}.

Linearization:

For a function of more than one variables, we can define the linearization function as {eq}\ \displaystyle L(x, y) =f(a, b)+f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b) {/eq} at the specified point. We should consider the given function as {eq}\displaystyle z=f(x, y) {/eq}. And it will be used for finding the partial derivatives of the function at the point.

Answer and Explanation:

The given function is {eq}\displaystyle z=x \sqrt{y} {/eq} at the point {eq}\displaystyle (-3,49) {/eq}.

Finding the partial derivatives at the point:

{eq}\begin{align*} \\ \displaystyle f(x, y) &=x \sqrt{y} \\ \displaystyle f(-3,49) &=(-3) \sqrt{49}=-21 \\ \\ \displaystyle \frac{\partial f}{\partial x} &=\frac{\partial }{\partial x}\left(x\sqrt{y}\right) \\ \displaystyle &=\sqrt{y}\frac{\partial \:}{\partial \:x}\left(x\right) \\ \displaystyle f_{x}(x, y) &=\sqrt{y} \\ \displaystyle f_{x}(-3,49) &=\sqrt{49}=7 \\ \\ \displaystyle \frac{\partial f}{\partial y} &=\frac{\partial }{\partial y}\left(x\sqrt{y}\right) \\ \displaystyle &=x\frac{\partial \:}{\partial \:y}\left(\sqrt{y}\right) \\ \displaystyle &=x\frac{\partial \:}{\partial \:y}\left(y^{\frac{1}{2}}\right) \\ \displaystyle &=x\frac{1}{2}y^{\frac{1}{2}-1} \\ \displaystyle f_{y}(x, y) &=\frac{x}{2\sqrt{y}} \\ \displaystyle f_{y}(-3,49) &=\frac{-3}{2\sqrt{49}}=-\frac{3}{14} \end{align*} {/eq}


Finding the linearization:

{eq}\begin{align*} \\ \displaystyle L(x, y) &=f(a, b)+f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b) \\ \displaystyle &=-21+(7)(x-(-3))+\left( -\frac{3}{14} \right)(y-49) \\ \displaystyle L(x, y) &=7x-\frac{3y}{14}+\frac{21}{2} \end{align*} {/eq}

Therefore, the linearization function at the point is {eq}\ \displaystyle \mathbf{\color{blue}{ L(x, y) =7x-\frac{3y}{14}+\frac{21}{2} }} {/eq}.


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Linearization of Functions

from Math 104: Calculus

Chapter 10 / Lesson 1
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