Problem 29 - 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 is 2.

Step 1: Set up the integral

To find the area between the curves, we start by finding the integral of the difference of the functions over the given interval. Take the integral of (\( \cos t - \sin t \)) from \(t=0\) to \(t=\pi\). This will give us the area between \( y = \cos t \) and \( y = \sin t \).

Step 2: Evaluate the integral

Integrate the function \( \cos t \) to get \( \sin t \), and integrate \( \sin t \) to get \(-\cos t \). So the integral \( \int ( \cos t - \sin t ) \, dt = \int \cos t \, dt - \int \sin t \, dt = \sin t + \cos t \).

Step 3: Apply the limits

Substitute the upper limit (\(\pi\)) and the lower limit (\(0\)) into \( \sin t + \cos t \). Compute \( \sin(\pi) + \cos(\pi) = 0 - 1 = -1 \) and \( \sin(0) + \cos(0) = 0 + 1 = 1 \).

Step 4: Find the area

Subtract the result from the lower limit calculation from the upper limit calculation: \( -1 - 1 = -2 \). Take the absolute value as area cannot be negative, resulting in an area of 2.

Key Concepts

Area Between Curves

Integrating to find the area between curves is a fascinating aspect of calculus that helps in determining how much space lies between two given functions within a specified interval. In this exercise, we look at two trigonometric functions,

• The function where the equation describes a curve.
• The region where one curve is above the other within a particular interval of \( t \).

To find the area between these curves, we perform the following steps:

• Identify the upper and lower curves: In our problem, the area between \( y = \cos t \) and \( y = \sin t \) is evaluated.
• Take the difference between these functions: Subtract \ sin t \ from \ cos t \ to get \( \cos t - \sin t \). This is because the area between involves integrating the difference over the interval \( 0 \leq t \leq \pi \).
• Set limits of integration: Our region of interest is from \( t = 0 \) to \( t = \pi \).
• Calculate definite integral: This integral provides the necessary area, and we take the absolute value to ensure the area is positive.

This approach allows us to find precise geometric space between two curves.

Definite Integral

The definite integral represents the net area under a curve and can be seen as the integral of a continuous function over a closed interval, providing numerical value, rather than a function. In the exercise context, we used a definite integral to identify the area between two trigonometric curves.
Here's a quick breakdown:

• Evaluate at limits: The function is integrated and then evaluated at both the lower and upper limits of our interval.
• The Fundamental Theorem of Calculus: This bridges the concept of derivative and integral, providing a straightforward way to find the area. \[ \int_a^b f(t) \, dt = F(b) - F(a) \] where \ F(t) \ is an antiderivative of \ f(t) \.
• Calculation: Substitute the upper and lower bounds into the integrated function and subtract: \[ \sin(\pi) + \cos(\pi) - (\sin(0) + \cos(0)) = -1 - 1 = -2 \]
• Absolute Value: Since areas cannot be negative, we take the absolute value, which is 2.

Trigonometric Functions

Trigonometric functions, such as \( \cos \) and \( \sin \), are fundamental in representing periodic phenomena and waves. They play a pivotal role not only in trigonometry but also in calculus, modeling cyclic patterns and intervals. Here's how they feature in this problem:

• Cosine Function: Represented by \ y = \cos t \, this function starts at maximum value of 1 when \ t = 0 \ and decreases to -1.
• Sine Function: Represented by \ y = \sin t \, begins at 0, rising to 1 at \ t = \pi/2 \ and returning to 0 at \ t = \pi \.
• Periodicity: Both functions are periodic, with a period of \ 2\pi \, which means they repeat their values every \ 2\pi \.
• Interval Analysis: For \ 0 \leq \ t \leq \pi \, the behavior of these functions gives a symmetric property that's useful for calculating the area where one is greater than the other.

By using these properties, we can effectively analyze how these functions interact over the given interval.