Copyright

Find the linearization, L(x,y) of the function f(x,y) = tan^(-1) (x + 8y) at the point (9,-1).

Question:

Find the linearization, {eq}L(x,y) {/eq} of the function

{eq}f(x,y) = tan^{-1}(x + 8y) {/eq} at the point (9,-1).

Linearization:

The linear approximation, linearization, of a function of two variables at one point is defined by three elements:

the value of the function and the value of the two partial derivatives of the function at the point.

Answer and Explanation:

First, calculating the value of the function and the partial derivatives, we have:

{eq}f(x,y) = ta{n^{ - 1}}(x + 8y) \to f\left( {9, - 1} \right) = \frac{\pi }{4}\\ \left\{ {\begin{array}{*{20}{c}} {\frac{{\partial f}}{{\partial x}}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {x + 8y} \right)}^2}}} \cdot 1 \to \frac{{\partial f}}{{\partial x}}\left( {9, - 1} \right) = \frac{1}{2}}\\ {\frac{{\partial f}}{{\partial y}}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {x + 8y} \right)}^2}}} \cdot 8 \to \frac{{\partial f}}{{\partial y}}\left( {9, - 1} \right) = 4} \end{array}} \right. {/eq}

So, the linearization of the function can be written as:

{eq}L\left( {x,y} \right) = f\left( {a,b} \right) + \frac{{\partial f}}{{\partial x}}\left( {a,b} \right)\left( {x - a} \right) + \frac{{\partial f}}{{\partial y}}\left( {a,b} \right)\left( {y - b} \right)\\ L\left( {x,y} \right) = f\left( {9, - 1} \right) + \frac{{\partial f}}{{\partial x}}\left( {9, - 1} \right)\left( {x - 9} \right) + \frac{{\partial f}}{{\partial y}}\left( {9, - 1} \right)\left( {y + 1} \right)\\ L\left( {x,y} \right) = \frac{\pi }{4} + \frac{1}{2} \cdot \left( {x - 9} \right) + 4\left( {y + 1} \right) {/eq}


Learn more about this topic:

Loading...
Linearization of Functions

from Math 104: Calculus

Chapter 10 / Lesson 1
7.6K

Related to this Question

Explore our homework questions and answers library