91Ó°ÊÓ

In multivariable calculus, partial derivatives help us understand how a function changes when one variable is altered while the other variables remain constant. For the function given in the exercise, which is defined as \(f(x, y) = \ln x + \ln y\), we compute its partial derivatives to examine changes in the \(x\) and \(y\) directions.

• The partial derivative with respect to \(x\), denoted as \(f_x(x, y)\), describes how \(f\) changes as \(x\) varies. For our function, \(f_x(x, y) = \frac{1}{x}\).
• Similarly, the partial derivative with respect to \(y\), denoted as \(f_y(x, y)\), is given by \(f_y(x, y) = \frac{1}{y}\). This indicates how \(f\) changes with variations in \(y\).

Evaluating these derivatives at the point \(P_0 = (1, 1)\) gives us \(f_x(1, 1) = 1\) and \(f_y(1, 1) = 1\). This evaluation helps us in building the linearization or the first-order approximation of the function around that point.

Linear approximation, also known as linearization, simplifies a function to a linear form near a specific point. It's similar to finding the tangent line to a function at one point, but in the case of multivariable functions, we consider tangent planes.
The linearization \(L(x, y)\) for a function \( f(x, y) \) at a point \( P_0 \) uses the function's value and its partial derivatives at that point. Using the given formula:\[L(x, y) = f(1, 1) + f_x(1, 1)(x - 1) + f_y(1, 1)(y - 1)\]Substitute the values we already determined:\[L(x, y) = 0 + 1(x - 1) + 1(y - 1) = x + y - 2\] This linear equation approximates \(f(x, y)\) around the point \(P_0 = (1,1)\), making it easier to work with locally, especially in small neighborhoods around \(P_0\).

Whenever we approximate a function using a linearization, there's inevitably some error. Error estimation allows us to quantify this difference between the actual function and its linear approximation.
To approximate the error, we rely on the second partial derivatives. For \(f(x, y) = \ln x + \ln y\), the second partial derivatives are:

• \(f_{xx}(x, y) = -\frac{1}{x^2}\)
• \(f_{yy}(x, y) = -\frac{1}{y^2}\)

Evaluating these over the rectangle \(R\), where \(|x-1| \leq 0.2\) and \(|y-1| \leq 0.2\), we find the maximum magnitude of the second derivatives in the interval \([0.8, 1.2]\). This gives a maximum of \(\frac{25}{16}\) at \(x = 0.8\) and \(y = 0.8\).
The error bound uses the formula:\[|E| \leq \frac{1}{2}M((\Delta x)^2 + (\Delta y)^2)\] Substituting \(\Delta x = 0.2\), \(\Delta y = 0.2\), and \(M = \frac{25}{16}\), we find:\[|E| \leq \frac{1}{2} \cdot \frac{25}{16} \cdot 0.08 = 0.0625\] This provides an upper bound, ensuring the linear approximation remains accurate within the defined region.