Problem 34 - Applied Calculus

Question

Use an integral to find the specified area. Between \(y=\cos t\) and \(y=\sin t\) for \(0 \leq t \leq \pi\)

Step-by-Step Solution

Answer

The area between \(y=\cos t\) and \(y=\sin t\) from \(0 \leq t \leq \pi\) is \(2\sqrt{2} - 2\).

Step 1: Identify Intersection Points

First, we need to find the values of \(t\) where \(\cos t = \sin t\) as these points will be the bounds of integration for finding the area between the curves. Set \(\cos t = \sin t\), and solve for \(t\). This leads to \(t = \frac{\pi}{4}\) within the given range \(0 \leq t \leq \pi\).

Step 2: Determine Which Function is Greater

Within the interval \([0, \frac{\pi}{4}]\), \(\cos t\) is greater than \(\sin t\). Similarly, between \(\frac{\pi}{4}\) and \(\pi\), \(\sin t\) is greater than \(\cos t\). Thus, we will compute the area where one function is above the other in each segment.

Step 3: Set Up the Integral for Each Section

Calculate the area where \(\cos t\) is above \(\sin t\), and vice versa:1. \(A_1 = \int_{0}^{\frac{\pi}{4}} (\cos t - \sin t) \, dt\)2. \(A_2 = \int_{\frac{\pi}{4}}^{\pi} (\sin t - \cos t) \, dt\)

Step 4: Solve the Integrals

Compute each integral separately.- For \(A_1\), use the antiderivatives: \[ \int \cos t \, dt = \sin t \quad \text{and} \quad \int \sin t \, dt = -\cos t \] \[ A_1 = \left[ \sin t + \cos t \right]_{0}^{\frac{\pi}{4}} = \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - \left( \sin 0 + \cos 0 \right) \] Simplifying gives \(A_1 = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - (0 + 1) = \sqrt{2} - 1\).- For \(A_2\): \[ A_2 = \left[-\cos t - \sin t \right]_{\frac{\pi}{4}}^{\pi} = \left(-\cos \pi - \sin \pi \right) - \left(-\cos \frac{\pi}{4} - \sin \frac{\pi}{4} \right) \] Simplifying gives \(A_2 = -(0 + 1) + (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) = \sqrt{2} - 1\).

Step 5: Calculate Total Area

Sum the areas from the two segments: \[ A = A_1 + A_2 = \sqrt{2} - 1 + \sqrt{2} - 1 = 2\sqrt{2} - 2 \]

Key Concepts

Area between curves

When you want to find the area between two curves, you need to identify where they intersect. These intersection points are crucial as they determine the limits of integration. Imagine two wiggly lines on a graph—your task is to find the space between them.
To do this, follow these steps:

• Determine where the two curves meet. This is done by equating their equations, such as setting the functions equal to each other to solve for the variable in question.
• Decide which curve is on top between these intersection points. This aspect is important because you will compute the integral of the top function minus the bottom one.
• Perform integration over the intervals determined, from the first intersection point to the last.

This way of breaking the problem down helps keep things simple. You can then find the area by integrating the vertical distance between these two curves.

Trigonometric functions

Trigonometric functions, like sine and cosine, are highly periodic and oscillate between certain values. In this exercise, the focus is mainly on \(y = \cos t\) and \(y = \sin t\). These functions:

• Repeat values every \(2\pi\).
• Oscillate between -1 and 1.
• Intersect at several key points, \(t = \rac{\pi}{4}\), for instance, is one of them.

Understanding the nature of these curves not only helps in finding intersection points but also in determining which function is larger over a given interval.
Between the start and end of the given range, remember that:

• Cosine starts at 1 when \(t = 0\) while sine starts at 0.
• They meet at \(t = \rac{">\pi}{4}\), both having a value of \(\frac{\sqrt{2}}{2}\).
• Beyond this point, sine overtakes cosine until they finish their cycle.

Noticing these characteristics allows you to appropriately set boundaries for your integral.

Definite integrals

Definite integrals are a fundamental concept for finding the area between curves. They take the antiderivative of a function and evaluate it over a specific interval. Here's how to use them in exercises like these:

• Set limits: Use the intersection points of curves as the bounds of your integral.
• Evaluate: Substitute the bounds into the antiderivative function.
• Subtract: Find the difference between the upper and lower evaluated values.

In the provided exercise, the antiderivatives were calculated separately:

• For \((\int \cos t \ dt = \sin t)\) and \((\int \sin t \ dt = -\cos t)\), evaluate these from 0 to \(\pi\).

By following these signs and steps carefully, you get the exact area between the trigonometric curves, revealing how integral calculus helps to solve such problems effectively.