Finding The Tangent Line Equation For H(x) = 18x - 2ln(x)

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In the realm of calculus, finding the equation of a tangent line to a curve at a specific point is a fundamental concept with numerous applications. This article delves into the process of determining the tangent line equation for the function h(x) = 18x - 2ln(x). We will explore the necessary steps, including finding the derivative, evaluating it at a given point, and utilizing the point-slope form of a linear equation. By the end of this guide, you will have a comprehensive understanding of how to tackle such problems and a solid foundation for further exploration of calculus concepts.

Understanding Tangent Lines

Before diving into the specifics of our problem, let's establish a clear understanding of what a tangent line represents. A tangent line to a curve at a point is a straight line that touches the curve at that point and has the same slope as the curve at that point. Imagine zooming in on a curve at a specific location; the tangent line is the straight line that best approximates the curve's direction at that point. This concept is crucial in calculus as it allows us to analyze the instantaneous rate of change of a function.

The slope of the tangent line is given by the derivative of the function at that point. The derivative, denoted as h'(x), represents the instantaneous rate of change of h(x) with respect to x. It tells us how much the function's value changes for an infinitesimally small change in the input variable. To find the equation of the tangent line, we need both the slope and a point on the line. The point will be the specific point on the curve where we want to find the tangent, and the slope will be the value of the derivative at that point.

Steps to Find the Tangent Line Equation

To find the equation of the tangent line to the curve h(x) = 18x - 2ln(x) at a given point, we follow a structured approach involving these key steps:

  1. Find the Derivative: The first crucial step is to determine the derivative of the function h(x). This derivative, h'(x), will provide us with the slope of the tangent line at any point on the curve. We will utilize the rules of differentiation, including the power rule and the derivative of the natural logarithm.
  2. Evaluate the Derivative: Once we have the derivative, we need to evaluate it at the specific x-coordinate of the point where we want to find the tangent line. This evaluation will give us the numerical value of the slope of the tangent line at that particular point.
  3. Find the Point of Tangency: To define a line, we require a point. We'll use the given x-coordinate and plug it into the original function h(x) to find the corresponding y-coordinate. This (x, y) pair represents the point where the tangent line touches the curve.
  4. Use the Point-Slope Form: With the slope (from step 2) and the point of tangency (from step 3), we can utilize the point-slope form of a linear equation to construct the equation of the tangent line. The point-slope form is a convenient way to express the equation of a line when we know its slope and a point it passes through.

By systematically following these steps, we can confidently find the equation of the tangent line for the given function at any specified point. Let's delve into each step in detail, applying them to our function h(x) = 18x - 2ln(x).

1. Finding the Derivative h'(x)

To find the equation of the tangent line, we must first determine the derivative of the function h(x) = 18x - 2ln(x). The derivative, denoted as h'(x), represents the instantaneous rate of change of the function and will give us the slope of the tangent line at any point on the curve. To find the derivative, we will use the rules of differentiation, including the power rule and the derivative of the natural logarithm.

Recall the following differentiation rules:

  • Power Rule: The derivative of x^n is nx^(n-1).
  • Derivative of ln(x): The derivative of ln(x) is 1/x.
  • Constant Multiple Rule: The derivative of cf(x) is cf'(x), where c is a constant.
  • Sum/Difference Rule: The derivative of f(x) ± g(x) is f'(x) ± g'(x).

Applying these rules to our function h(x) = 18x - 2ln(x):

  • The derivative of 18x is 18 (using the power rule and constant multiple rule).
  • The derivative of -2ln(x) is -2 * (1/x) = -2/x (using the derivative of ln(x) and the constant multiple rule).

Therefore, the derivative of h(x) is:

h'(x) = 18 - 2/x

This expression, h'(x) = 18 - 2/x, is the derivative of our original function. It gives us a formula to calculate the slope of the tangent line at any point x on the curve. The next step is to evaluate this derivative at the specific point of interest.

2. Evaluating the Derivative at a Specific Point

Now that we have found the derivative, h'(x) = 18 - 2/x, the next step is to evaluate it at a specific x-coordinate. This evaluation will give us the numerical value of the slope of the tangent line at that point. Let's assume we want to find the tangent line at the point where x = 1. This means we will substitute x = 1 into our derivative expression:

h'(1) = 18 - 2/1

h'(1) = 18 - 2

h'(1) = 16

Thus, the slope of the tangent line at x = 1 is 16. This value, 16, is a crucial piece of information for constructing the equation of the tangent line. It tells us the steepness and direction of the line that touches the curve at the point where x = 1. To fully define the tangent line, we also need a point on the line, which we will find in the next step.

If we were given a different x-coordinate, we would simply substitute that value into h'(x) to find the corresponding slope. For example, if we wanted to find the tangent line at x = 2, we would calculate:

h'(2) = 18 - 2/2

h'(2) = 18 - 1

h'(2) = 17

In this case, the slope of the tangent line at x = 2 would be 17. This illustrates how the derivative allows us to determine the slope of the tangent line at any point on the curve. Evaluating the derivative is a key step in finding the tangent line equation.

3. Finding the Point of Tangency

To fully define the tangent line, we need a point that lies on the line. This point will be the point where the tangent line touches the original curve, h(x) = 18x - 2ln(x). We already know the x-coordinate of this point, which we assumed to be x = 1 in the previous step. To find the corresponding y-coordinate, we simply substitute x = 1 into the original function:

h(1) = 18(1) - 2ln(1)

Recall that the natural logarithm of 1, ln(1), is equal to 0. Therefore:

h(1) = 18 - 2(0)

h(1) = 18

Thus, the y-coordinate of the point of tangency is 18. This means the tangent line touches the curve at the point (1, 18). We now have all the information we need to construct the equation of the tangent line: the slope (which we found to be 16) and a point on the line (1, 18).

If we had chosen a different x-coordinate, we would repeat this process to find the corresponding y-coordinate. For instance, if we were considering the point where x = 2, we would calculate:

h(2) = 18(2) - 2ln(2)

h(2) = 36 - 2ln(2)

In this case, the y-coordinate would be 36 - 2ln(2), which is approximately 34.61. The point of tangency would then be (2, 36 - 2ln(2)). Finding the point of tangency by substituting the x-coordinate into the original function is a crucial step in determining the equation of the tangent line.

4. Using the Point-Slope Form

Now that we have the slope of the tangent line (m = 16) and a point on the line (1, 18), we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is a convenient way to express the equation of a line when we know its slope and a point it passes through. The point-slope form is given by:

y - y₁ = m(x - x₁)

where:

  • m is the slope of the line
  • (x₁, y₁) is a point on the line

In our case, m = 16 and (x₁, y₁) = (1, 18). Substituting these values into the point-slope form, we get:

y - 18 = 16(x - 1)

This is the equation of the tangent line in point-slope form. While this form is perfectly valid, we can also simplify it to slope-intercept form (y = mx + b) or standard form (Ax + By = C) if desired.

To convert to slope-intercept form, we distribute the 16 and then isolate y:

y - 18 = 16x - 16

y = 16x - 16 + 18

y = 16x + 2

So, the equation of the tangent line in slope-intercept form is y = 16x + 2. This form clearly shows the slope (16) and the y-intercept (2) of the tangent line.

To convert to standard form, we rearrange the equation so that x and y terms are on the same side and the constant term is on the other side:

y = 16x + 2

-16x + y = 2

Multiplying by -1 to make the coefficient of x positive:

16x - y = -2

Thus, the equation of the tangent line in standard form is 16x - y = -2.

No matter which form we choose, the equation represents the same tangent line to the curve h(x) = 18x - 2ln(x) at the point where x = 1. Using the point-slope form is a direct and efficient way to find this equation once we have the slope and a point on the line.

Summary and Conclusion

In this article, we have explored the process of finding the equation of the tangent line to the curve h(x) = 18x - 2ln(x). We followed a systematic approach involving four key steps:

  1. Finding the derivative h'(x)
  2. Evaluating the derivative at a specific point
  3. Finding the point of tangency
  4. Using the point-slope form

We found that the derivative of h(x) is h'(x) = 18 - 2/x. We then evaluated this derivative at x = 1 to find the slope of the tangent line at that point, which was 16. We also found the point of tangency by substituting x = 1 into the original function, which gave us the point (1, 18). Finally, we used the point-slope form to construct the equation of the tangent line: y - 18 = 16(x - 1), which can be simplified to y = 16x + 2 in slope-intercept form.

Understanding how to find tangent lines is crucial in calculus and has various applications in fields such as physics, engineering, and economics. The ability to determine the instantaneous rate of change of a function allows us to model and analyze real-world phenomena more accurately. By mastering the steps outlined in this article, you can confidently tackle tangent line problems and further expand your knowledge of calculus.

The concepts covered here form the foundation for more advanced topics in calculus, such as optimization, related rates, and curve sketching. The process of finding the tangent line is a building block that will enable you to understand and solve a wide range of calculus problems. Remember to practice these steps with different functions and points to solidify your understanding and develop your problem-solving skills. The journey through calculus is a rewarding one, and finding tangent lines is just one of the many fascinating concepts you will encounter along the way.