Problem 35 - Single Variable Calculus

Question

In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ y=\sin 2 x, \quad y=\cos x, \quad x=\frac{\pi}{6}, \quad x=\frac{\pi}{2} $$

Step-by-Step Solution

Answer

The area of the region bounded by the given equations is \(\frac{11}{24}\).

Step 1: Sketch the graphs of the given equations

Plot the following four equations on the same coordinate plane: 1. \(y = \sin 2x\) 2. \(y = \cos x\) 3. \(x = \frac{\pi}{6}\) 4. \(x = \frac{\pi}{2}\) to visualize the region bounded by these graphs.

Step 2: Find the intersection points

To find the intersection points of the curves \(y = \sin 2x\) and \(y = \cos x\), we need to find the \(x\) values where these two functions are equal: \(\sin 2x = \cos x\) Keep in mind that we need to find the intersection points within the bounds \(\frac{\pi}{6} \le x \le \frac{\pi}{2}\).

Step 3: Solve the trigonometric equation

Use the double-angle identity for sine: \(\sin 2x = 2 \sin x \cos x\), then the equation becomes: \(2 \sin x \cos x = \cos x\) Now solve this equation for \(x\) within the given bounds: 1. Divide both sides by \(2 \cos x\): \(\sin x = \frac{1}{2}\). 2. Within the given bounds, there is only one solution for x: \(x = \frac{\pi}{6}\). Therefore, the intersection points are at the given boundaries, namely \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\).

Step 4: Find the area of the bounded region

Since the functions are continuous in the given interval, we can find the area of the bounded region by taking the definite integral of their difference: Area = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} [\cos x - \sin 2x] dx\)

Step 5: Evaluate the definite integral

Now, we need to evaluate the definite integral: Area = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} [\cos x - 2\sin x\cos x] dx\) By applying the power-reduction formula (integration by substitution) with \(u = \sin x\) and \(du = \cos x dx\), we get: Area = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} [\cos x - u^{2}](\frac{du}{\cos x})\) Area = \(\int_{\frac{1}{2}}^{0} [\frac{1}{\cos x} - u^{2}] du\) Now integrate: Area = \(\int_{\frac{1}{2}}^{0} (1 - u^{2}) du = \left[u - \frac{u^{3}}{3}\right]_{\frac{1}{2}}^{0} = \left(\frac{1}{2} - \frac{1}{24}\right) - (0 - 0) = \frac{11}{24}\) Thus, the area of the region bounded by the given equations is \(\frac{11}{24}\).

Key Concepts

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are fundamental in mathematics, particularly in understanding periodic phenomena. In this exercise, we encounter the functions \(y = \sin 2x\) and \(y = \cos x\). Both of these are periodic functions, meaning they repeat their values in a regular interval.

• For \(\sin 2x\), the period is \(\pi\) because the argument of the sine function is multiplied by 2, effectively halving the period of the classic sine function which is \(2\pi\).
• For \(\cos x\), the period remains \(2\pi\).

Understanding these trigonometric functions is crucial for analyzing relationships between angles and their side lengths in right-angled triangles, as well as modeling waves and oscillations. The graphs of these functions are symmetrical and have specific ranges, with sine and cosine functions oscillating between [-1,1]. Recognizing the shape and periodicity of these functions assists greatly when sketching them to determine bounded regions.

Sketching Graphs

Sketching the graphs of trigonometric functions allows visual comprehension of how these functions behave and intersect. The exercise involves plotting the graphs of \(y = \sin 2x\) and \(y = \cos x\) between the vertical lines \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\).
To accomplish this:

• Identify key points such as intersections and extremum points within the given interval.
• Draw horizontal lines to represent the amplitude limits \(-1\) and \(1\).
• Mark vertical lines for \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\) to define the boundaries.

Plot critical points like where \(\sin 2x\) and \(\cos x\) intersect each other. Visualizing these intersections helps identify the bounded region. Good graph sketching includes clear labeling and symmetry awareness, especially for trigonometric functions.

Area Under Curve

The concept of area under the curve is a central idea in calculus, frequently used to find spaces bounded by functions. In this case, the area between \(y = \cos x\) and \(y = \sin 2x\) from \(x = \frac{\pi}{6}\) to \(x = \frac{\pi}{2}\) needs to be calculated.
To find this area, we focus on the definite integral of the difference of functions:\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} [\cos x - \sin 2x] \; dx\]

• Set up the integrals correctly. \(\cos x\) is usually above \(\sin 2x\) within this interval.
• Use properties and identities in trigonometry to simplify calculations, such as the identity \(\sin 2x = 2 \sin x \cos x\).
• Substitute and simplify where applicable to make integration straightforward.

After solving the integral, the resultant value is the area between the two curves, representing the actual physical space under the top graph minus the space under the lower graph.

Trigonometric Equation Solving

Solving trigonometric equations is a process that involves finding angles that satisfy equation conditions. In this exercise, solving \(\sin 2x = \cos x\) involved finding points where the curves intersect within the given bounds.
The double angle identity \(\sin 2x = 2 \sin x \cos x\) is employed to simplify and solve:

• The equation \(2\sin x \cos x = \cos x\) simplifies to \(\sin x = \frac{1}{2}\) after dividing by \(\cos x\), except \(\cos x eq 0\).
• Solve \(\sin x = \frac{1}{2}\) for \(x\) within the interval, using known angle values such as \(x = \frac{\pi}{6}\).

This solution process also reveals intersection points. Besides using algebraic manipulation, make sure to check for possible multiple solutions within given limits due to the periodic nature of trig functions.