91Ó°ÊÓ

When working with functions of several variables, partial derivatives play an essential role in understanding how the function behaves around a point. A partial derivative represents the rate at which the function changes with respect to one variable, while all other variables are held constant. In this exercise, we are dealing with a function \( f(x, y) = \frac{1}{\sqrt{x^2 + y^2}} \), which depends on two variables: \( x \) and \( y \).
The partial derivative of \( f \) with respect to \( x \) is calculated by treating \( y \) as a constant and differentiating with respect to \( x \). The result is \( f_x(x, y) = -\frac{x}{(x^2 + y^2)^{3/2}} \). Similarly, the partial derivative with respect to \( y \) is \( f_y(x, y) = -\frac{y}{(x^2 + y^2)^{3/2}} \).
These derivatives help us construct the local linear approximation by assessing how much \( f \) changes near a given point \((a, b) \). By evaluating these derivatives at \( P(4, 3) \), we can approximate the function's value near this point using the changes in \( x \) and \( y \).

Euclidean distance is a measure of the "straight line" distance between two points in Euclidean space. It's like using a ruler to measure how far apart two dots are on a piece of paper. In mathematics, for two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), the Euclidean distance is given by the formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Using the given points \( P(4, 3) \) and \( Q(3.92, 3.01) \), this formula gives us:
\[ \text{Distance} = \sqrt{(3.92 - 4)^2 + (3.01 - 3)^2} \]
This results in a distance of approximately \( 0.0806 \).
This value is crucial because it helps us assess how close the point \( Q \) is to the point \( P \) where the linear approximation is actually centered. The closer the points, the more reliable our linear approximation is likely to be.

Error approximation helps us see how the local linear approximation compares to the actual function value at a particular point. It's like using a map and checking if it points you to the correct location. In this exercise, we approximate \( f \) at the point \( Q(3.92, 3.01) \) using the linear approximation \( L(x, y) \):
\[ L(3.92, 3.01) = 0.2116 \]
The actual function value \( f(3.92, 3.01) \) is about \( 0.2020 \).
The error is the difference between these values, calculated as:
\[ |0.2116 - 0.2020| \approx 0.0096 \]
This small error indicates that the local linear approximation \( L(x, y) \) is quite accurate near point \( P \). Comparing this error to the Euclidean distance \( 0.0806 \), we verify our linear approximation effectiveness. Remember that smaller errors reflect a more faithful approximation, especially when the point of interest \( Q \) is near the approximation point \( P \).
In conclusion, estimating the error helps us understand the reliability of the approximation in real-world contexts.