Step-by-Step Guide To Finding The Taylor Polynomial Of Order 1 For F(x, Y) = Ln(x² + Y²) At P(1, 1)

Hey guys! Ever wondered how to approximate a function using a simpler polynomial, especially when dealing with functions of multiple variables? Well, the Taylor polynomial is your answer! In this article, we're going to dive deep into finding the Taylor polynomial of order 1 for the function f(x, y) = ln(x² + y²) around the point P(1, 1). Trust me, it's not as scary as it sounds. We'll break it down step-by-step, making it super easy to follow. So, grab your favorite beverage, and let's get started!

Understanding Taylor Polynomials

Before we jump into the specifics, let's quickly recap what Taylor polynomials are all about. In essence, Taylor polynomials are used to approximate the value of a function at a specific point using its derivatives at another point. Think of it as creating a local approximation of a function using its behavior at a known location. For functions of multiple variables, like our f(x, y), the concept extends naturally, but we need to consider partial derivatives. The Taylor polynomial essentially gives us a linear approximation of a possibly complex function near a given point. This is incredibly useful in various fields, including physics, engineering, and computer science, where we often deal with functions that are difficult to compute directly. The beauty of the Taylor polynomial lies in its ability to replace a complex function with a simpler polynomial, making calculations much more manageable. For instance, instead of directly evaluating the logarithm of a complicated expression, we can use the Taylor polynomial to get a close approximation with simple arithmetic operations. Understanding this fundamental concept is key to appreciating the power and utility of Taylor polynomials in real-world applications. When working with Taylor polynomials, keep in mind that the accuracy of the approximation generally improves as we include higher-order terms. However, for many practical applications, a first-order or second-order Taylor polynomial provides a sufficiently accurate approximation while keeping the calculations relatively simple. So, let's roll up our sleeves and get into the nitty-gritty of how to construct these polynomials!

Formula for the First-Order Taylor Polynomial

The first-order Taylor polynomial for a function of two variables, f(x, y), around a point (a, b), is given by a specific formula. Let's break it down: P₁(x, y) = f(a, b) + fx(a, b)(x - a) + fy(a, b)(y - b). Here, f(a, b) represents the value of the function at the point (a, b). The terms fx(a, b) and fy(a, b) are the partial derivatives of f with respect to x and y, respectively, evaluated at the point (a, b). These partial derivatives tell us how the function changes with respect to each variable at that specific point. The terms (x - a) and (y - b) represent the differences between the variables x and y and the coordinates of the point (a, b). This formula essentially constructs a linear approximation of the function near the point (a, b) using the function's value and its rate of change in the x and y directions. Now, let's dissect each component of this formula to ensure we grasp its essence. The first term, f(a, b), is simply the function's value at the point of approximation. It serves as the anchor for our polynomial. The second term, fx(a, b)(x - a), accounts for the change in the function's value as we move along the x-axis away from the point (a, b). The partial derivative fx(a, b) acts as the slope in the x-direction, and (x - a) represents the distance we've moved along the x-axis. Similarly, the third term, fy(a, b)(y - b), captures the change in the function's value as we move along the y-axis away from the point (a, b). The partial derivative fy(a, b) is the slope in the y-direction, and (y - b) represents the distance we've moved along the y-axis. Combining these three terms gives us a linear approximation that closely resembles the function's behavior near the point (a, b). Understanding the role of each component in the formula is crucial for applying it effectively. So, keep this breakdown in mind as we move forward and put this formula into action!

Step 1: Evaluate f(x, y) at P(1, 1)

Okay, let's get our hands dirty with our specific function: f(x, y) = ln(x² + y²). The first step in finding the Taylor polynomial is to evaluate the function at our given point, P(1, 1). This means we need to plug in x = 1 and y = 1 into our function. So, f(1, 1) = ln(1² + 1²) = ln(1 + 1) = ln(2). That's it! We've found the first piece of our puzzle. This value, ln(2), will be the constant term in our Taylor polynomial, anchoring our approximation at the point (1, 1). Now, you might be wondering, why do we even need to evaluate the function at this point? Well, the Taylor polynomial is essentially an approximation that's most accurate near the point we're expanding around. By evaluating the function at P(1, 1), we're ensuring that our approximation starts with the correct value at that specific location. It's like setting the starting point for our approximation journey. Think of it as calibrating our GPS before we start driving. We need to know our current location (the function's value at P(1, 1)) before we can accurately estimate where we'll be after moving a certain distance (using the partial derivatives). This initial evaluation sets the stage for the rest of our calculations, so it's crucial to get it right. And in our case, we've nailed it: f(1, 1) = ln(2). Now, with this value in hand, we're ready to move on to the next step: finding the partial derivatives of our function. So, let's keep the momentum going and tackle those derivatives!

Step 2: Find the Partial Derivatives

Now comes the fun part: calculating the partial derivatives! We need to find fx(x, y) and fy(x, y) for our function f(x, y) = ln(x² + y²). Remember, partial derivatives tell us how the function changes with respect to one variable while keeping the other constant. Let's start with fx(x, y), the partial derivative with respect to x. To find this, we'll treat y as a constant and differentiate ln(x² + y²) with respect to x. Using the chain rule, we get: fx(x, y) = (1 / (x² + y²)) * (2x) = 2x / (x² + y²). Great! We've found our first partial derivative. Now, let's tackle fy(x, y), the partial derivative with respect to y. This time, we'll treat x as a constant and differentiate ln(x² + y²) with respect to y. Again, using the chain rule, we get: fy(x, y) = (1 / (x² + y²)) * (2y) = 2y / (x² + y²). Fantastic! We've successfully calculated both partial derivatives. These derivatives are crucial because they represent the slopes of the function in the x and y directions, respectively. They tell us how the function is changing at any given point (x, y). Think of them as the steering wheels that guide our Taylor polynomial approximation. The partial derivative with respect to x shows us how the function's value changes as we move along the x-axis, while the partial derivative with respect to y shows us how the function's value changes as we move along the y-axis. These rates of change are essential for constructing an accurate linear approximation of the function near the point (1, 1). Without them, our Taylor polynomial would be like a car without steering – it might start in the right place, but it wouldn't know how to follow the road. So, with these partial derivatives in our toolbox, we're well-equipped to build our Taylor polynomial. But before we do that, we need to evaluate these derivatives at our specific point, P(1, 1). Let's move on to the next step and see how that's done!

Step 3: Evaluate the Partial Derivatives at P(1, 1)

Alright, we've found our partial derivatives, fx(x, y) = 2x / (x² + y²) and fy(x, y) = 2y / (x² + y²). Now, we need to evaluate these at our point P(1, 1). This means plugging in x = 1 and y = 1 into both expressions. Let's start with fx(1, 1): fx(1, 1) = 2(1) / (1² + 1²) = 2 / 2 = 1. So, the partial derivative with respect to x at P(1, 1) is 1. Now, let's do fy(1, 1): fy(1, 1) = 2(1) / (1² + 1²) = 2 / 2 = 1. And the partial derivative with respect to y at P(1, 1) is also 1. We now have the slopes of our function in the x and y directions at the point (1, 1). These values are super important because they tell us how the function is changing locally around P(1, 1). Think of it like this: if you were standing on the surface defined by our function at the point (1, 1), fx(1, 1) would tell you how steep the slope is in the x-direction, and fy(1, 1) would tell you how steep the slope is in the y-direction. These slopes are crucial for building our linear approximation. They determine the orientation of the plane that will best approximate our function near P(1, 1). A higher value for fx(1, 1) would mean the function is changing more rapidly in the x-direction, and our Taylor polynomial needs to reflect that. Similarly, a higher value for fy(1, 1) would mean a steeper slope in the y-direction. So, by evaluating these partial derivatives at P(1, 1), we've essentially captured the local behavior of our function in a numerical form. We know how much the function is changing in each direction at our point of interest. With these values in hand, we're ready to assemble our Taylor polynomial. We have all the pieces of the puzzle; now it's time to put them together!

Step 4: Plug the Values into the Taylor Polynomial Formula

Okay, guys, this is where the magic happens! We've got all the pieces, and now we're going to assemble our first-order Taylor polynomial. Remember the formula? P₁(x, y) = f(a, b) + fx(a, b)(x - a) + fy(a, b)(y - b). We know: f(1, 1) = ln(2) fx(1, 1) = 1 fy(1, 1) = 1 And our point P(a, b) is (1, 1). Let's plug these values into the formula: P₁(x, y) = ln(2) + 1(x - 1) + 1(y - 1). Now, let's simplify this a bit: P₁(x, y) = ln(2) + x - 1 + y - 1. Further simplification gives us: P₁(x, y) = ln(2) + x + y - 2. Ta-da! We've found the Taylor polynomial of order 1 for f(x, y) = ln(x² + y²) at P(1, 1). This polynomial, P₁(x, y) = ln(2) + x + y - 2, is a linear approximation of our original function near the point (1, 1). It's a plane that