Find the linearization of the function f(x, y)= (68 - 3x^2 - 4y^2)^{1/2} at the point (4, 2). Use...

Question:

Find the linearization of the function {eq}f(x, y)= (68 - 3x^2 - 4y^2)^{1/2} {/eq} at the point (4, 2).

Use the linear approximation to estimate the value of f(3.9, 2.1).

Linearization of Nonlinear Function


Given a nonlinear function of two variables, we use partial derivatives at a point on the function to get the linearization of the function at the point. The linearization of the function is a linear function of two variable which agrees with the original function close to the point at which it is calculated. In order words, the linearization of a nonlinear function at a point on its surface is nothing but the equation of the tangent plane to the surface at that point. Following this we use the linear function found to estimate the value of the function at a point close to the point of linearization.


Answer and Explanation:


From Calculus, the linearlization of the given function at the point (4, 2) is given by

{eq}L(x,y)=f(4,2)+(x-4)f_x(4,2)+(y-2)f_y(4,2) \qquad (1) {/eq}

From the given function {eq}f(4,2)=(68-3(4^2)-4(2^2))^{1/2}=(68-48-16)^{1/2}=4^{1/2}=2 \qquad (2) {/eq}

Further,

{eq}\displaystyle f_x(x,y) = \frac {-3x}{(68-3x^2-4y^2)^{1/2}} \implies f_x(4,2)=\frac {-12}{2}=-6 \qquad (3) {/eq}

and

{eq}\displaystyle f_y(x,y) = \frac {-4y}{(68-3x^2-4y^2)^{1/2}} \implies f_y(4,2)=\frac {-8}{2}=-4 \qquad (4) {/eq}

Substituting from (2), (3) and (4) in (1) yields

{eq}L(x,y)=2-6(x-4)-4(y-2)=-6x-4y+34 \qquad (5) {/eq}

Hence the linearization is {eq}L(x,y)=2-6(x-4)-4(y-2)=-6x-4y+34 {/eq}


We use L(x, y) to approximate f(3.9, 2.1) as follows:

{eq}f(3.9, 2.1) \approx L(3.9, 2.1)=-6(3.9)-4(2.1)+34 = 2.2. {/eq}

Hence the approximation of f(3.9, 4.1) is 2.2.


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Linearization of Functions

from Math 104: Calculus

Chapter 10 / Lesson 1
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