Problem 9 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

a. Use the double-angle formulas along with the Pythagorean identity \(\sin ^{2} A+\cos ^{2} A=1\) to show that \(\cos ^{2} \theta=\frac{1}{2}(1+\cos 2 \theta) \quad\) and \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) b. Use the identities in part (a) to show that $$ \begin{aligned} &\int \cos ^{2} x d x=\frac{1}{2} x+\frac{1}{4} \sin (2 x)+C \\ &\int \sin ^{2} x d x=\frac{1}{2} x-\frac{1}{4} \sin (2 x)+C \end{aligned} $$ and c. An object moves along a straight line in such a way that after \(t\) seconds, its velocity is given by $$ v(t)=2 t+\sin ^{2}\left(\frac{\pi t}{6}\right) $$ meters per second. Find the average velocity of the object over the time period \(0 \leq t \leq 3\).

Step-by-Step Solution

Answer

The average velocity of the object over the interval \([0,3]\) is 3.5 meters per second.

Step 1: Use the double-angle formula for cosine

Recall the double-angle formula for cosine: \[ \cos 2\theta = 2\cos^{2}\theta - 1\]We can rearrange this equation to solve for \(\cos^2\theta\).

Step 2: Solve for \(\cos^2\theta\)

Rearrange the equation to isolate \(\cos^2 \theta\): \[ \cos 2\theta = 2\cos^{2}\theta - 1 \implies 2\cos^{2}\theta = 1 + \cos 2\theta \implies \cos^{2} \theta = \frac{1}{2}(1 + \cos 2\theta)\]

Step 3: Use the Pythagorean identity

Recall the Pythagorean identity: \[ \sin^{2}\theta + \cos^{2}\theta = 1\]Using the result from Step 2, substitute \(\cos^{2}\theta = \frac{1}{2}(1 + \cos 2\theta)\).

Step 4: Solve for \(\sin^2\theta\)

From the Pythagorean identity: \[ \sin^{2}\theta = 1 - \cos^{2}\theta = 1 - \frac{1}{2}(1 + \cos 2\theta) \implies \sin^{2}\theta = \frac{1}{2}(1 - \cos 2\theta)\]

Step 5: Show the integral for \(\cos^2 x\)

Use the identity for \(\cos^2 x\) in the integral: \[ \int \cos^{2} x \, dx = \int \frac{1}{2}(1 + \cos 2x) \, dx = \frac{1}{2}\int 1 \, dx + \frac{1}{2}\int \cos 2x \, dx\]Evaluate the integrals: \[ \frac{1}{2}x + \frac{1}{4}\sin 2x + C\]

Step 6: Show the integral for \(\sin^2 x\)

Use the identity for \(\sin^2 x\) in the integral: \[ \int \sin^{2} x \, dx = \int \frac{1}{2}(1 - \cos 2x) \, dx = \frac{1}{2}\int 1 \, dx - \frac{1}{2}\int \cos 2x \, dx\]Evaluate the integrals: \[ \frac{1}{2}x - \frac{1}{4}\sin 2x + C\]

Step 7: Find the average velocity of the object

Use the formula for average velocity: \[ \text{Average velocity} = \frac{1}{b-a} \int_{a}^{b} v(t) \, dt\]where \(v(t) = 2t + \sin^{2}\left(\frac{\pi t}{6}\right)\) and the interval is \([0, 3]\).

Step 8: Integrate the velocity function

First, break the integral into two parts: \[ \int_{0}^{3} \left(2t + \sin^{2}\left(\frac{\pi t}{6}\right) \right) dt = \int_{0}^{3} 2t \, dt + \int_{0}^{3} \sin^{2}\left(\frac{\pi t}{6}\right) \,, dt\]Evaluate each part separately using the identities from Step 5 and Step 6.

Step 9: Evaluate the integrals

\[ \int_{0}^{3} 2t \, dt = t^2 \Bigg|_{0}^{3} = 9\]For the second part, use \(u = \frac{\pi t}{6}\): \[ \int_{0}^{3} \sin^{2}\left(\frac{\pi t}{6}\right) \, dt = \int_{0}^{3} \frac{1}{2}\left(1 - \cos \left(\frac{\pi t}{3}\right) \right)\]Evaluate: \[ \frac{1}{2}\left[t - \frac{6}{\pi} \sin \left(\frac{\pi t}{3}\right) \right]_{0}^{3} = \frac{3}{2}\]

Step 10: Calculate average velocity

Add the results of the integrals and divide by the time interval length: \[ \frac{1}{3 - 0} \left(9 + \frac{3}{2}\right) = \frac{1}{3} \cdot \frac{21}{2} = \frac{7}{2} = 3.5\]So, the average velocity over the interval \([0,3]\) seconds is 3.5 meters per second.

Key Concepts

Pythagorean identity

In trigonometry, the Pythagorean identity is a very fundamental equation. It's one of the first identities you learn when you're introduced to the unit circle. The Pythagorean identity states that:
\[ \sin^{2}\theta + \cos^{2}\theta = 1 \] This identity is essential because it connects the sine and cosine functions. Think of it as the equation of a circle with radius 1. Regardless of the angle \(\theta\), the sum of the squares of sine and cosine for that angle will always equal 1.
For example, if \(\cos(\theta)\) is known, you can easily find \(\sin(\theta)\), and vice versa. This relationship is pivotal when working with double-angle formulas and integration as seen later in our solution process.

Double-Angle Identity

Double-angle identities simplify expressions involving angles. Particularly, for an angle \(\theta\), the double-angle identities for sine and cosine are:
\[ \cos(2\theta) = 2\cos^{2}(\theta) - 1 \] \[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \] Using these identities is crucial in rearranging expressions for easier integration. For instance, we saw in the original solution that we needed to express \(\cos^{2}\theta\) in a different form. By manipulating the double-angle identity for cosine, we rearranged it to:
\[ \cos^{2}\theta = \frac{1}{2}(1 + \cos(2\theta)) \] Similarly, using the Pythagorean identity, we derived:
\[ \sin^{2}\theta = \frac{1}{2}(1 - \cos(2\theta)) \] These converted forms are much easier to integrate.

Integration

Integration is a fundamental concept in calculus that's often used to find areas under curves. In our solution, we dealt with the integrals of \(\cos^{2}x\) and \(\sin^{2}x\) by using their double-angle identity forms.
To integrate \(\cos^{2}x\), we used:
\[ \int \cos^{2} x \, dx = \int \frac{1}{2}(1 + \cos(2x)) \, dx = \frac{1}{2}\int 1 \, dx + \frac{1}{2}\int \cos(2x) \, dx \] Evaluating these integrals separately leads us to:
\[ \frac{1}{2}x + \frac{1}{4}\sin(2x) + C \] Similarly, for \(\sin^{2}x\), we followed:
\[ \int \sin^{2} x \, dx = \int \frac{1}{2}(1 - \cos(2x)) \, dx = \frac{1}{2}\int 1 \, dx - \frac{1}{2}\int \cos(2x) \, dx \] Evaluating these gives us:
\[ \frac{1}{2}x - \frac{1}{4}\sin(2x) + C \] These simplified forms are necessary for solving integration problems involving trigonometric functions.

Average Velocity Calculation

Calculating average velocity is a common application of integration in physics. Average velocity over a time interval \([a, b]\) is found by integrating the velocity function \(v(t)\) over that interval and then dividing by the length of the interval. The formula is:
\[ \text{Average velocity} = \frac{1}{b-a} \int_{a}^{b} v(t) \, dt \] In our solution, the given velocity function was:
\[ v(t) = 2t + \sin^{2}\left(\frac{\pi t}{6}\right) \] Over the interval \([0, 3]\), we broke down the integration to:
\[ \int_{0}^{3} \left(2t + \sin^{2}\left(\frac{\pi t}{6}\right)\right) dt = \int_{0}^{3} 2t \, dt + \int_{0}^{3} \sin^{2}\left(\frac{\pi t}{6}\right) \, dt \] Evaluating these parts separately, we found:

• \( \int_{0}^{3} 2t \, dt = 9 \)
• \( \int_{0}^{3} \sin^{2}\left(\frac{\pi t}{6}\right) \, dt = \frac{3}{2} \)

Adding these results and dividing by 3 gave us the average velocity:
\[ \text{Average velocity} = \frac{1}{3}(9+\frac{3}{2}) = 3.5 \text{ meters per second} \].