Problem 4 - Active Calculus Multivariable

Question

(a) Check the local linearity of \(f(x, y)=e^{-x} \cos (y)\) near \(x=1, y=1.5\) by filling in the following table of values of \(f\) for \(x=0.9,1,1.1\) and \(y=1.4,1.5,1.6 .\) Express values of \(f\) with 4 digits after the decimal point. (b) Next, fill in the table for the values \(x=0.99,1,1.01\) and \(y=\) 1.49,1.5,1.51 , again showing 4 digits after the decimal point. Notice if the two tables look nearly linear, and whether the second looks more linear than the first (in particular, think about how you would decide if they were linear, or if the one were more closely linear than the other). (c) Give the local linearization of \(f(x, y)=e^{-x} \cos (y)\) at (1,1.5) : Using the second of your tables: \(f(x, y) \approx\) ____________. Using the fact that \(f_{x}(x, y)=-e^{-x} \cos (y)\) and \(f_{y}(x, y)=-e^{-x} \sin (y):\) \(f(x, y) \approx\) ___________.

Step-by-Step Solution

Answer

\(f(x, y) \approx 0.3863 - 0.3461(x-1) - 0.0876(y-1.5)\)

Step 1: Compute values for the first table

We need to compute the values of \(f(x,y)\) for the following points: \((0.9,1.4)\), \((0.9,1.5)\), \((0.9,1.6)\), \((1,1.4)\), \((1,1.5)\), \((1,1.6)\), \((1.1,1.4)\), \((1.1,1.5)\), \((1.1,1.6)\) and round the results to 4 decimal places.

Step 2: Compute values for the second table

Now, we need to compute the values of \(f(x,y)\) for the following points: \((0.99,1.49)\), \((0.99,1.5)\), \((0.99,1.51)\), \((1,1.49)\), \((1,1.5)\), \((1,1.51)\), \((1.01,1.49)\), \((1.01,1.5)\), \((1.01,1.51)\) and round the results to 4 decimal places.

Step 3: Compare tables for linearity

To determine if the tables look nearly linear, and whether the second one looks more linear than the first, we can look for patterns in the changes in the function values. If the changes are consistent, the table may exhibit linearity.

Step 4: Find the local linearization using the second table

The local linearization of a function can be approximated using the values from the table where we have a finer granularity (the second one). We can use the central difference method to approximate the partial derivatives \(f_x\) and \(f_y\). - Use these values to fill in the Taylor series expansion up to the first order: \(f(x, y) \approx f(1,1.5) + f_x(1,1.5)(x-1) + f_y(1,1.5)(y-1.5)\)

Step 5: Find the local linearization using the partial derivatives

Given that \(f_x(x,y) = -e^{-x} \cos(y)\) and \(f_y(x,y) = - e^{-x} \sin(y)\), compute \(f_x(1,1.5)\) and \(f_y(1,1.5)\) and substitute those values into the Taylor series expansion: \(f(x, y) \approx f(1,1.5) + f_x(1,1.5)(x-1) + f_y(1,1.5)(y-1.5)\)

Key Concepts

Partial Derivatives

Partial derivatives are crucial when dealing with multivariable functions. They represent the rate at which a function changes as one variable changes, while all other variables remain constant. In the context of local linearization, partial derivatives help us understand how a function behaves in the vicinity of a point by providing the slopes of the tangent plane to the function's graph at that point.

Specifically, for a function like
\( f(x, y) = e^{-x} \cos(y) \),
the partial derivative with respect to
\( x \), denoted as
\( f_{x}(x, y) \),
is
\( -e^{-x} \cos(y) \), and
\( f_{y}(x, y) \), the partial derivative with respect to
\( y \),
is
\( -e^{-x} \sin(y) \).
These derivatives tell us how
\( f \) changes in the direction of each axis near a given point, such as (1, 1.5). This information is essential for creating a linear approximation of the function—the heart of local linearization.

Multivariable Functions

Multivariable functions like
\( f(x, y) = e^{-x} \cos(y) \)
introduce more complexity than single-variable functions because they change along multiple axes at once. When we analyze how these functions change near a certain point, we can't simply use a tangent line as we would in single-variable calculus. Instead, we use a tangent plane or linear approximation to capture the behavior of the function in all directions from that point.

To find this linear approximation, we use the function's value and its partial derivatives at the point of interest. This is what forms the local linearization, which can be expressed using the first-order Taylor series expansion for a multivariable function.

The purpose of filling in tables with function values at various points around (1, 1.5) is to visualize how well the linear approximation represents the function in a neighborhood of that point, with the goal that the student understands how a flat plane can serve as a reasonable proxy for the surface created by the function.

Exponential and Trigonometric Functions

In our exercise, the function

\( f(x, y) = e^{-x} \cos(y) \) includes both an exponential component
\( e^{-x} \) and a trigonometric component
\( \cos(y) \).

Exponential functions are known for their rapid growth or decay, and in this case,
\( e^{-x} \) represents a decaying pattern as
\( x \) increases. Trigonometric functions like cosine and sine oscillate between -1 and 1, and their behavior depends on the angle or, in our case, the variable
\( y \).

The combination of these functions can create complex surfaces that are not straightforward to visualize. Local linearization simplifies the visualization by approximating the function locally with a plane, reducing the complexity to understanding a flat surface rather than the hills and valleys of the original function’s graph.