# Find the linearization of the function f(x,y)=\sqrt{157-5x^2-y^2} at the point (5, -4). Use the...

## Question:

Find the linearization of the function {eq}f(x,y)=\sqrt{157-5x^2-y^2} {/eq} at the point {eq}(5, -4) {/eq}. Use the linear approximation to estimate the value of {eq}f(4.9,-3.9) {/eq}.

## Linearization of Two-Variable Function

The linearization of a two-variable function {eq}f(x,y) {/eq} at a point {eq}P(x_0, y_0) {/eq} is computed by

{eq}z(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) {/eq}

where {eq}f_x(x_0,y_0) \ and \ f_y(x_0,y_0) {/eq} are the first partial derivatives of the function with respect to {eq}x \ and \ y {/eq} evaluated at the given point.

## Answer and Explanation: 1

Let's begin by evaluating the function at the given point.

{eq}f(x,y)=\sqrt{157-5x^2-y^2} f(5,-4)=\sqrt{157-5(5)^2-(-4)^2} \\ f(5,-4)=\sqrt{157-5(25)-(16)} \\ f(5,-4)=\sqrt{157-125-16} \\ f(5,-4)=\sqrt{16} \\ f(5,-4)= 4 {/eq}

Next, let's get the first partial derivatives then evaluate the the same point.

{eq}f_x = \dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x}(\sqrt{157-5x^2-y^2}) \\ f_x = \dfrac{\partial}{\partial x}(({157-5x^2-y^2})^{\frac{1}{2}}) \\ \displaystyle f_x = \frac{1}{2}(({157-5x^2-y^2})^{\frac{1}{2}-1})\cdot (-(2)5x^{2-1}) \\ \displaystyle f_x = \frac{1}{2}(({157-5x^2-y^2})^{-\frac{1}{2}})\cdot (-10x^{1}) \\ f_x = -5x ({157-5x^2-y^2})^{-\frac{1}{2}} \\ f_x = -\dfrac{5x}{({157-5x^2-y^2})^{\frac{1}{2}} } \\ f_x = -\dfrac{5x}{\sqrt{157-5x^2-y^2}} {/eq}

Evaluate at .

{eq}f_x(5, -4) = -\dfrac{5(5)}{\sqrt{157-5(5)^2-(-4)^2}} \\ f_x(5, -4) = -\dfrac{25}{\sqrt{157-5(25)-(16)}} \\ f_x(5, -4) = -\dfrac{25}{\sqrt{157-125-16}} \\ f_x(5, -4) = -\dfrac{25}{\sqrt{16}} \\ f_x(5, -4) = -\dfrac{25}{4} {/eq}

The other partial derivative is

{eq}f_y = \dfrac{\partial f}{\partial y} = \dfrac{\partial}{\partial y}(\sqrt{157-5x^2-y^2}) \\ f_y = \dfrac{\partial}{\partial y}(({157-5x^2-y^2})^{\frac{1}{2}}) \\ \displaystyle f_y = \frac{1}{2}(({157-5x^2-y^2})^{\frac{1}{2}-1})\cdot (-(2)y^{2-1}) \\ \displaystyle f_y = \frac{1}{2}(({157-5x^2-y^2})^{-\frac{1}{2}})\cdot (-2y^{1}) \\ \displaystyle f_y = -y(({157-5x^2-y^2})^{-\frac{1}{2}}) \\ f_y = -\dfrac{y}{({157-5x^2-y^2})^{\frac{1}{2}}} \\ f_y = -\dfrac{y}{\sqrt{157-5x^2-y^2}} {/eq}

Evaluate at .

{eq}f_y(5,-4) = -\dfrac{(-4)}{\sqrt{157-5(5)^2-(-4)^2}} \\ f_y(5,-4) = \dfrac{4}{\sqrt{157-5(25)-(16)}} \\ f_y(5,-4) = \dfrac{4}{\sqrt{157-125-16}} \\ f_y(5,-4) = \dfrac{4}{\sqrt{16}} \\ f_y(5,-4) = \dfrac{4}{4} \\ f_y(5,-4) = 1 {/eq}

We will substitute these values to the linear approximation equation where {eq}(x_0, y_0) = (5,-4) {/eq}.

{eq}z(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) \\ z(x,y) = f(5,-4) + f_x(5,-4)(x-5) + f_y(5,-4)(y-(-4)) \\ z(x,y) = 4 + \left (-\dfrac{25}{4} \right )(x-5) + (1)(y+4) \\ z(x,y) = 4 -\dfrac{25}{4}x + \dfrac{25(5)}{4} + y + 4 \\ z(x,y) = -\dfrac{25}{4}x + \dfrac{125}{4} + y + 8 \\ z(x,y) = -\dfrac{25}{4}x + y + 8 + \dfrac{125}{4} \\ z(x,y) = -\dfrac{25}{4}x + y + \dfrac{32}{4} + \dfrac{125}{4} \\ z(x,y) = -\dfrac{25}{4}x + y + \dfrac{157}{4} {/eq}

The linearization of the function at {eq}(5,-4) {/eq} is

{eq}\boldsymbol{z(x,y) = -\dfrac{25}{4}x + y + 39.25} {/eq}

We will use this is estimate the value of {eq}f(4.9,-3.9) {/eq}.

{eq}f(4.9,-3.9) \approx z(4.9,-3.9) = -\dfrac{25}{4}(4.9) + (-3.9) + 39.25 \\ f(4.9,-3.9) \approx -\dfrac{122.5}{4} - 3.9 + 39.25 \\ f(4.9,-3.9) \approx - 30.625 - 3.9 + 39.25 \\ \boldsymbol{f(4.9,-3.9) \approx 4.725} {/eq}

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