Problem 64 - Calculus of a Single Variable

Question

Using a tangent Line In Exercises \(61-64,\) set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function and the tangent line to the graph at the given point. $$ y=\frac{2}{1+4 x^{2}}, \quad\left(\frac{1}{2}, 1\right) $$

Step-by-Step Solution

Answer

Due to the complexity of the integral expression, there might not be any simple forms of \(w\) and \(z\), or the values of the definite integrals. Therefore, numerical methods or a calculator might be needed for their computation.

Step 1: Find the Derivative

Firstly, we need to find the derivative of the function \(y = \frac{2}{1+4x^2}\). Using the quotient rule, we get that the derivative \(y' = \frac{-16x}{(1+4x^2)^2}\).

Step 2: Find the Tangent Line

Now, to find the equation of the tangent line at the point \((\frac{1}{2}, 1)\), we first need the slope, which is the value of the derivative at this point. Substituting \(x = \frac{1}{2}\) gives \(y' = -2\), so the equation of the tangent line using the point-slope form \(y – y_1 = m(x – x_1)\) is \(y - 1 = -2(x - \frac{1}{2})\), or \(y = -2x +2\).

Step 3: Set Up the Definite Integral

In this step, the definite integral is set up to give the area between the curve and the tangent line. This is achieved by integrating the absolute value of the difference between the functions over an interval. For instance, since the tangent line starts to intersect the curve \(y = \frac{2}{1+4x^2}\) when \(x < \frac{1}{2}\), then the equation for the area \(A\) is \(A = \int_{w}^{\frac{1}{2}} ((\frac{2}{1+4x^2}) - (-2x+2)) dx + \int_{\frac{1}{2}}^{z} ((-2x+2) - (\frac{2}{1+4x^2})) dx\), where \(w\) and \(z\) are the x-values for the points where the tangent line intersects the curve. These intersection points occur when \(y = \frac{2}{1+4x^2}\) equals \(y = -2x +2\), i.e., when \(\frac{2}{1+4x^2} = -2x +2\). Solving this equation for \(x\) will yield \(w\) and \(z\).

Step 4: Evaluate the Definite Integral

After finding the limits of integration, the definite integral can then be evaluated. Note that the exact computation depends on the values of \(w\) and \(z\), as well as some non-trivial integral evaluations involved.

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus, and it's crucial for understanding areas under curves and between curves. In our exercise, the definite integral will help us find the area between a curve and its tangent line. The process involves evaluating the integral of a function over a specified interval. When we set up a definite integral, we're essentially summing up an infinite number of infinitesimally small elements over the range.

• In our case, the definite integral will span from the point where the tangent line intersects the curve, on both sides of a given point.
• We integrate the difference between the function of our curve and the equation of the tangent line.

For the region bounded by the curve and its tangent, this difference helps calculate the area accurately, ensuring we're summing the vertical distances between two lines over a range. Evaluating this gives us the area we're looking for.

Derivative

The derivative measures how a function changes as its input changes. In simple terms, it gives us the slope of a line tangent to the function at any point.
In our exercise, using the function's derivative is essential to finding the tangent line at a specific point.

• The derivative of the function \(y = \frac{2}{1+4x^2}\) using the quotient rule is \(y' = \frac{-16x}{(1+4x^2)^2}\).
• This derivative tells us how steep the curve is at any point \(x\).

To find the tangent line's slope at a specified point, we evaluate the derivative at that point. For instance, at \(x = \frac{1}{2}\), the slope is \(-2\). This critical value helps us determine the tangent line equation.

Tangent Line Equation

The tangent line is a straight line that touches a curve at a single point, matching the curve's slope there.
Finding the tangent line equation involves using the slope derived from the function's derivative and the point where they meet.

• First, use the point-slope form: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope from the derivative.
• For our curve \(y = \frac{2}{1+4x^2}\) at point \((\frac{1}{2}, 1)\), we found the slope \(-2\).
• This gives us: \(y - 1 = -2(x - \frac{1}{2})\), or simplified to \(y = -2x + 2\).

This equation tells us precisely how the tangent line behaves around our point and allows us to set boundaries for the area calculation.

Area Between Curves

When tasked with finding the area between curves, we often use definite integrals. This area finds practical use in various fields, such as engineering and sciences, to determine quantities or optimize functions.

• First, we set up integrals that represent the top and bottom curves over specific intervals.
• In our problem, these curves are defined by the original function and its tangent line.

To find this intersection area:

• We integrate the absolute difference between the functions, over intervals \([w, \frac{1}{2}]\) and \([\frac{1}{2}, z]\).
• The intersection points can be found by equating \(\frac{2}{1+4x^2} = -2x + 2\) and solving for \(x\).

By evaluating these integrals, we capture the exact area between the curve and the tangent line. This precise area calculation is essential for many practical applications involving curves.