Problem 19 - Calculus Early Transcendentals

Question

Use a graphing utility to determine the number of times the curves intersect and then apply Newton’s Method, where needed, to approximate the \(x\)-coordinates of all intersections. \(y=1\) and \(y=e^{x} \sin x ; 0

Step-by-Step Solution

Answer

The curves intersect once; Newton's Method gives the intersection around \(x \approx 0.589\).

Step 1: Graph the Equations

Use a graphing utility to graph the equations \(y = 1\) and \(y = e^x \sin x\) over the interval \(0 < x < \pi\). This will help to visually determine where the curves intersect within the given range. Look for points where the curves physically cross each other on the graph.

Step 2: Identify Intersection Points

From the graph, identify the number of intersection points within the interval \((0, \pi)\). Count how many times the curve \(y = e^x \sin x\) crosses the line \(y = 1\). According to the visual inspection, the curves should intersect once in this interval.

Step 3: Apply Newton's Method Setup

Now, set up Newton's Method to approximate the x-coordinate of the intersection. The objective is to find the \(x\) value for which \(e^x \sin x = 1\). Define \(f(x) = e^x \sin x - 1\). Newton's Method iteration formula is \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\].

Step 4: Compute Derivative

Calculate the derivative \(f'(x) = \frac{d}{dx}(e^x \sin x - 1)\). Using the product rule, \(f'(x) = e^x \sin x + e^x \cos x\).

Step 5: Choose Initial Approximation

From the graph, estimate an initial approximation \(x_0\) close to the intersection point. Suppose the intersection visually occurs around \(x = 1\). Use \(x_0 = 1\) as the starting point for iteration.

Step 6: Perform Iterations Using Newton's Method

Perform iterations using the equation \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). Substitute \(x_0 = 1\) into the formula and continue the iterations until the difference between consecutive \(x\) values is less than a specified tolerance, such as \(0.00001\). Calculate for: 1. \(f(x_0) = e^1 \sin 1 - 1\)2. \(f'(x_0) = e^1 \sin 1 + e^1 \cos 1\) 3. Update \(x_1 = 1 - \frac{f(1)}{f'(1)}\)4. Continue iterations until convergence.

Key Concepts

Intersection Points

In mathematical problems involving curves, intersection points are where two curves cross or meet each other. To find the intersections between the curves of two functions, we visually inspect a graph to see where they overlap. In our scenario, you are looking at the two equations, \( y = 1 \) and \( y = e^x \sin x \), to determine their intersections. The interval to consider is from 0 to \( \pi \). By plotting these equations, you identify that these curves intersect at specific points in this range.

• Visually identify where curves cross.
• Focus on the interval \( 0 < x < \pi \).
• Usually involves comparing curve shapes and positions.
• Intersection points are solutions to the equation \( e^x \sin x = 1 \).

Graphing Utility

Graphing utilities are powerful tools for visualizing mathematical functions. They help in plotting equations to see their behavior across different domains. When you have equations like \( y=1 \) and \( y=e^x \sin x \), sketching them manually could be challenging. A graphing utility, however, allows you to visualize these curves much more easily.
By inputting these functions into a graphing utility, you can identify their forms and any intersections within an interval. This visual representation supports the analytical process of solving function equations and finding intersections.

• Use a utility to plot functions visually.
• Adjust the view to focus on specific intervals.
• Check the intersection visually as a lead-in to calculations.

Derivative Calculation

In calculus, finding derivatives is pivotal for understanding the rate of change of a function. For Newton’s Method, derivative calculation is crucial to approximate solutions to equations. When you have the function \( f(x) = e^x \sin x - 1 \), calculating its derivative is a major step. Using the product rule, which states that the derivative of a product of two functions is given by \((uv)' = u'v + uv'\), you find the derivative.

• Derivatives help assess function behavior.
• Product rule aids in calculating the derivative of \( e^x \sin x \).
• Compute \( f'(x) = e^x \sin x + e^x \cos x \).

This derivative is essential for Newton's Method, as it allows the calculation of the function's slope, which informs each iteration towards finding an intersection closer to the true value.

Product Rule

The product rule is an essential technique in differentiation used for finding the derivative of products of two functions. When faced with the function \( e^x \sin x \), it becomes necessary to apply the product rule. The product rule is defined as

• \((uv)' = u'v + uv' \)

where \( u=e^x \) and \( v=\sin x \). So, the derivative of \( e^x \sin x \) using the product rule is \( e^x \sin x + e^x \cos x \).
This principle simplifies working with complex expressions, breaking them into manageable parts, leading smoothly into numerical methods such as Newton's Method for finding specific solutions like intersection points. With the derivative complete, you are well equipped to iterate towards an approximation of the x-value of any intersections.