Finding The Tangent Equation Of Y = Ln(x) At (1, 0)

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Introduction

In the realm of calculus, understanding tangent lines is fundamental. A tangent line to a curve at a specific point represents the instantaneous rate of change of the function at that point. This concept is crucial in various applications, including optimization problems, curve sketching, and approximations. This article will delve into the process of finding the equation of the tangent line to the natural logarithm function, y = ln(x), at the point (1, 0). We will explore the necessary calculus principles, including derivatives and the point-slope form of a linear equation, to arrive at the solution. This exploration is not only a valuable exercise in calculus but also provides a deeper understanding of the behavior of logarithmic functions and their graphical representations.

Understanding the Problem

Before diving into the calculations, it's essential to understand what we're trying to find. We are given the function y = ln(x) and a specific point (1, 0) on its graph. Our goal is to determine the equation of the line that touches the curve at this point and has the same slope as the curve at that point. This line is the tangent line. To find its equation, we need two key pieces of information: the slope of the line and a point on the line. We already have a point (1, 0), so our primary task is to find the slope. The slope of the tangent line at a point is given by the derivative of the function evaluated at that point. The derivative, in essence, gives us the instantaneous rate of change of the function, which is precisely what the slope of the tangent line represents. Therefore, the first step in solving this problem is to find the derivative of y = ln(x).

Finding the Derivative

The derivative of a function gives us the slope of the tangent line at any point on the curve. For the natural logarithm function, y = ln(x), the derivative is a well-known result in calculus. The derivative of ln(x) with respect to x is 1/x. This is a fundamental rule that can be derived using the definition of the derivative or by considering the inverse relationship between the natural logarithm and the exponential function. Understanding this derivative is crucial for many calculus problems involving logarithmic functions. It tells us how the function y = ln(x) changes as x changes. Now that we have the derivative, we can find the slope of the tangent line at the specific point (1, 0) by evaluating the derivative at x = 1. This will give us the numerical value of the slope at that particular point on the curve.

Evaluating the Derivative at x = 1

Now that we know the derivative of y = ln(x) is 1/x, we can find the slope of the tangent line at the point (1, 0). To do this, we substitute x = 1 into the derivative. So, the slope, m, is given by m = 1/1 = 1. This means that at the point (1, 0), the tangent line has a slope of 1. A slope of 1 indicates that for every unit increase in x, the tangent line increases by one unit in y. This is a positive slope, so the tangent line will be increasing as we move from left to right. Now that we have the slope and a point on the line, we have all the necessary information to write the equation of the tangent line. The next step is to use the point-slope form of a linear equation, which is a convenient way to express the equation of a line when we know its slope and a point it passes through.

Using the Point-Slope Form

The point-slope form of a linear equation is a powerful tool for finding the equation of a line when you know a point on the line and its slope. The general form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. In our case, we have the point (1, 0), so x1 = 1 and y1 = 0. We also found that the slope, m, is 1. Substituting these values into the point-slope form gives us y - 0 = 1(x - 1). This equation represents the tangent line to the curve y = ln(x) at the point (1, 0). We can simplify this equation to obtain the slope-intercept form, which is a more familiar way to write the equation of a line. This will give us a clear understanding of the line's y-intercept and its relationship to the x-axis.

Simplifying the Equation

After substituting the point (1, 0) and the slope m = 1 into the point-slope form, we obtained the equation y - 0 = 1(x - 1). To simplify this equation, we can distribute the 1 on the right side, which gives us y = x - 1. This is the equation of the tangent line in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept. In this case, the slope is 1 and the y-intercept is -1. This simplified form makes it easy to visualize the line and its relationship to the coordinate axes. The line crosses the y-axis at -1 and has a positive slope, indicating that it rises as x increases. This final equation, y = x - 1, is the solution to our problem: it is the equation of the tangent line to the curve y = ln(x) at the point (1, 0). Now, let's summarize the steps we took to arrive at this solution.

Summary of Steps

To find the equation of the tangent line to y = ln(x) at the point (1, 0), we followed these steps:

  1. Found the derivative: The derivative of y = ln(x) is 1/x. This is a fundamental rule in calculus and is essential for finding the slope of the tangent line.
  2. Evaluated the derivative: We evaluated the derivative at x = 1 to find the slope of the tangent line at the point (1, 0). This gave us a slope of m = 1.
  3. Used the point-slope form: We used the point-slope form of a linear equation, y - y1 = m(x - x1), with the point (1, 0) and the slope m = 1.
  4. Simplified the equation: We simplified the equation y - 0 = 1(x - 1) to get the equation of the tangent line in slope-intercept form: y = x - 1.

This systematic approach allowed us to determine the equation of the tangent line, which represents the instantaneous rate of change of the function at the specified point. This process is applicable to finding tangent lines for many different functions and is a core concept in differential calculus.

Conclusion

In conclusion, we have successfully found the equation of the tangent line to the curve y = ln(x) at the point (1, 0). By applying the principles of calculus, specifically finding the derivative and using the point-slope form of a linear equation, we determined that the tangent line is given by the equation y = x - 1. This exercise demonstrates the power of calculus in analyzing the behavior of functions and their graphical representations. Understanding tangent lines is crucial for various applications in mathematics, physics, and engineering. The ability to find tangent lines allows us to approximate function values, optimize processes, and gain deeper insights into the relationships between variables. This example serves as a valuable illustration of how calculus concepts can be applied to solve real-world problems and enhance our understanding of the mathematical world.