Finding The Tangent Equation Of Y=2ln(x) At X=1 A Step-by-Step Guide

Introduction: Understanding Tangent Lines and Curve Equations

In the realm of calculus, understanding the relationship between curves and their tangent lines is fundamental. A tangent line to a curve at a specific point is a straight line that touches the curve at that point and has the same slope as the curve at that point. This concept is crucial in various applications, including optimization problems, physics, and engineering. To find the equation of a tangent line, we need two key pieces of information: the point of tangency and the slope of the tangent at that point. This article dives deep into how to find the tangent line when given a curve and a specific point.

In our specific case, we are tasked with finding the equation of the tangent to the curve defined by the equation y = 2ln(x) at the point where the x-coordinate is x = 1. This involves several steps, including finding the y-coordinate of the point, calculating the derivative of the function to find the slope of the tangent, and using the point-slope form of a line to write the equation of the tangent. The natural logarithm function, denoted as ln(x), is the logarithm to the base e, where e is an irrational number approximately equal to 2.71828. The derivative of ln(x) is a fundamental concept in calculus and is equal to 1/x. This knowledge is essential for finding the slope of the tangent to the curve y = 2ln(x).

To find the equation of the tangent, we will leverage the principles of differential calculus. Differential calculus provides us with the tools to determine the instantaneous rate of change of a function, which is represented by the derivative. The derivative of a function at a point gives the slope of the tangent line to the curve at that point. The equation y = 2ln(x) represents a logarithmic curve, and its tangent lines will vary depending on the x-coordinate. When we are given x = 1, we are pinpointing a specific location on the curve where we want to find the tangent. Understanding the behavior of logarithmic functions, such as their domain and range, is also important in this context. The domain of ln(x) is all positive real numbers, and its range is all real numbers. This means that x must be greater than 0 for the function to be defined. In our case, x = 1 is within the domain, so we can proceed with finding the tangent line.

Step 1: Finding the y-coordinate

To begin, we need to find the corresponding y-coordinate for the point on the curve where x = 1. We can do this by substituting x = 1 into the equation of the curve, which is y = 2ln(x). The substitution is a straightforward process: replace every instance of x in the equation with the value 1. This gives us y = 2ln(1). Now, we need to evaluate ln(1). By definition, the natural logarithm of 1 is 0, because e^0 = 1. Therefore, ln(1) = 0. Substituting this back into our equation, we get y = 2 * 0 = 0. Thus, the point on the curve where x = 1 is (1, 0). This point is crucial because it is the point of tangency, where the tangent line will touch the curve.

Understanding the properties of logarithms is essential in this step. The logarithm of 1 to any base is always 0. This is a fundamental property that simplifies many calculations involving logarithmic functions. In the context of our problem, knowing that ln(1) = 0 allows us to quickly find the y-coordinate. The point (1, 0) is also significant because it lies on the x-axis. This gives us a visual understanding of where the tangent line will be located relative to the curve. Furthermore, the point of tangency is not just any point; it is the point where the tangent line best approximates the curve locally. This means that near (1, 0), the tangent line will closely resemble the shape of the curve y = 2ln(x). This approximation is a key concept in calculus and is used in many applications, such as numerical methods and optimization.

In summary, finding the y-coordinate involves a simple substitution and evaluation of the natural logarithm. The result, (1, 0), is the point of tangency and is a critical component in determining the equation of the tangent line. This step highlights the importance of understanding the basic properties of logarithmic functions and their behavior. Without knowing that ln(1) = 0, we would not be able to accurately find the y-coordinate and proceed with the problem. The accuracy of this step is paramount, as an incorrect point of tangency will lead to an incorrect equation for the tangent line. Therefore, it is crucial to carefully perform the substitution and evaluation to ensure the correctness of the result.

Step 2: Finding the Derivative

To find the slope of the tangent line, we need to calculate the derivative of the function y = 2ln(x). The derivative of a function gives us the instantaneous rate of change of the function at any given point. In geometric terms, the derivative at a point represents the slope of the tangent line to the curve at that point. The derivative of ln(x) is a fundamental result in calculus and is equal to 1/x. Using this, we can find the derivative of 2ln(x) by applying the constant multiple rule, which states that the derivative of a constant times a function is the constant times the derivative of the function. In this case, the constant is 2, and the function is ln(x). Therefore, the derivative of y = 2ln(x) is dy/dx = 2 * (1/x) = 2/x.

The derivative dy/dx = 2/x represents the slope of the tangent line to the curve y = 2ln(x) at any point x. To find the slope at the specific point where x = 1, we substitute x = 1 into the derivative. This gives us dy/dx|_(x=1) = 2/1 = 2. Therefore, the slope of the tangent line at the point (1, 0) is 2. This value is crucial for determining the equation of the tangent line, as it provides the direction of the line. A slope of 2 indicates that for every 1 unit increase in x, the y-value increases by 2 units. This steepness of the tangent line is a key characteristic that we will use to define its equation.

Understanding the rules of differentiation is essential for finding the derivative of a function. The derivative of ln(x) and the constant multiple rule are fundamental concepts that are widely used in calculus. The derivative 2/x is a function in itself, and it describes how the slope of the tangent line to the curve y = 2ln(x) changes as x varies. This understanding of the derivative as a function is crucial for further applications of calculus, such as optimization and curve sketching. In our case, we are interested in the specific value of the derivative at x = 1, but the derivative function 2/x provides a more general description of the slope of the tangent line at any point on the curve. This highlights the power of differential calculus in providing both specific and general information about the behavior of functions.

Step 3: Finding the Slope at x=1

Now that we have the derivative, dy/dx = 2/x, we can find the slope of the tangent line at the point where x = 1. To do this, we substitute x = 1 into the derivative expression. This gives us the slope, m, at that specific point. So, m = dy/dx|_(x=1) = 2/1 = 2. The slope of the tangent line at x = 1 is 2. This means that the tangent line rises 2 units for every 1 unit it moves to the right. The positive slope indicates that the tangent line is increasing, which aligns with the increasing nature of the logarithmic function y = 2ln(x) in the vicinity of x = 1.

The slope m = 2 is a crucial piece of information for determining the equation of the tangent line. It tells us the steepness and direction of the line. A larger absolute value of the slope indicates a steeper line, while the sign indicates whether the line is increasing (positive slope) or decreasing (negative slope). In this case, the slope of 2 is moderate, suggesting that the tangent line is neither too steep nor too flat. The fact that the slope is positive confirms that the tangent line is increasing as we move from left to right. This is consistent with the graph of the function y = 2ln(x), which is also increasing for x > 0.

This step highlights the importance of evaluating the derivative at a specific point. The derivative function 2/x gives us the slope of the tangent line at any x-value, but to find the slope at a particular point, we need to substitute that x-value into the derivative. This process of evaluation is fundamental in calculus and is used extensively in various applications, such as finding maximum and minimum values of functions, determining the concavity of curves, and analyzing rates of change. In our case, evaluating the derivative at x = 1 allows us to pinpoint the slope of the tangent line at the point of tangency, which is essential for writing the equation of the tangent line.

Step 4: Using the Point-Slope Form

The point-slope form of a line is a convenient way to write the equation of a line when we know a point on the line and its slope. The point-slope form is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope of the line. We have already found the point on the curve where x = 1, which is (1, 0), and the slope of the tangent line at that point, which is m = 2. Now, we can substitute these values into the point-slope form to get the equation of the tangent line.

Substituting x₁ = 1, y₁ = 0, and m = 2 into the point-slope form, we get y - 0 = 2(x - 1). This simplifies to y = 2(x - 1). This is the equation of the tangent line in point-slope form. However, it is often more convenient to write the equation in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. To convert our equation to slope-intercept form, we distribute the 2 on the right side of the equation: y = 2x - 2. This is the equation of the tangent line in slope-intercept form. It tells us that the tangent line has a slope of 2 and a y-intercept of -2.

The point-slope form is a powerful tool for finding the equation of a line because it directly uses the information we have: a point and a slope. It avoids the need to solve for the y-intercept separately, which can be more cumbersome. The conversion to slope-intercept form is useful because it provides a clear understanding of the line's slope and y-intercept, which are important characteristics of the line. The equation y = 2x - 2 tells us that the tangent line crosses the y-axis at -2 and that it increases by 2 units for every 1 unit increase in x. This provides a complete description of the tangent line and its relationship to the curve y = 2ln(x) at the point (1, 0).

Conclusion: The Tangent Line Equation

In conclusion, we have successfully found the equation of the tangent to the curve y = 2ln(x) at the point on the curve with x-coordinate x = 1. We accomplished this by following a series of steps, each building upon the previous one. First, we found the y-coordinate of the point of tangency by substituting x = 1 into the equation of the curve, giving us the point (1, 0). Next, we calculated the derivative of the function y = 2ln(x), which is dy/dx = 2/x. This derivative represents the slope of the tangent line at any point on the curve. We then evaluated the derivative at x = 1 to find the slope of the tangent line at the point of tangency, which was m = 2. Finally, we used the point-slope form of a line, y - y₁ = m(x - x₁), to write the equation of the tangent line. Substituting the point (1, 0) and the slope m = 2 into the point-slope form, we obtained y - 0 = 2(x - 1), which simplifies to y = 2x - 2. This is the equation of the tangent line in slope-intercept form.

The equation y = 2x - 2 is the final answer to our problem. It represents a straight line that touches the curve y = 2ln(x) at the point (1, 0) and has the same slope as the curve at that point. This tangent line provides a linear approximation of the curve near the point (1, 0). This linear approximation is a fundamental concept in calculus and is used in various applications, such as numerical methods, optimization, and approximation of function values. The tangent line gives us a simplified representation of the curve's behavior in the immediate vicinity of the point of tangency.

Finding the equation of a tangent line is a classic problem in calculus that combines several important concepts, including derivatives, slopes, points, and equations of lines. This problem demonstrates the power of differential calculus in providing tools to analyze the behavior of functions and their graphs. The process we followed in this article can be applied to find the equation of the tangent line to any differentiable function at any point on its curve. The key is to understand the relationship between the derivative, the slope of the tangent line, and the point of tangency. By mastering these concepts, one can effectively solve a wide range of problems in calculus and related fields.