Problem 17 - Calculus

Question

Find the ares under \(v=\cos t\) from the change in \(f=\sin t:\) $$ \text { from } t=0 \text { to } t=2 \pi $$

Step-by-Step Solution

Answer

The area under the curve from \(t=0\) to \(t=2\pi\) is 0.

Step 1: Identify the Problem

We need to find the area under the curve given by the function \(v = \cos t\) from \(t = 0\) to \(t = 2\pi\). This corresponds to evaluating the definite integral of \(v\) over the given interval.

Step 2: Set Up the Integral

The definite integral to calculate the area under the curve is given by: \[ \int_{0}^{2\pi} v \, dt = \int_{0}^{2\pi} \cos t \, dt \] This integral will give us the desired area under \(v = \cos t\) from \(t = 0\) to \(t = 2\pi\).

Step 3: Find the Antiderivative

The antiderivative of \(v = \cos t\) is \(f(t) = \sin t\). So, we use this antiderivative to evaluate the definite integral.

Step 4: Evaluate the Definite Integral

To find the definite integral, substitute the upper and lower limits into the antiderivative and subtract:\[\int_{0}^{2\pi} \cos t \, dt = \left[ \sin t \right]_{0}^{2\pi} = \sin(2\pi) - \sin(0)\]

Step 5: Calculate the Values

Calculate the sine values:\[ \sin(2\pi) = 0 \quad \text{and} \quad \sin(0) = 0\] Subtract the two values: \[ 0 - 0 = 0\]

Step 6: State the Result

The calculated integral, representing the area under the curve from \(t = 0\) to \(t = 2\pi\), is \(0\).

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in mathematics, especially in the context of periodic and oscillatory phenomena. These functions relate the angles of a triangle to the ratios of its sides in the context of the unit circle.

- **Cosine Function**: The cosine function, denoted as \( \cos t \), represents the horizontal coordinate of a point on the unit circle. This function oscillates between -1 and 1 as the angle \( t \) varies.
- **Sine Function**: The sine function, represented as \( \sin t \), correlates with the vertical coordinate on the unit circle. Like the cosine function, it oscillates between -1 and 1.

When considering the relationship between these functions, they help define the motion or real-world phenomena that follow a periodic rhythm. In the original exercise, the goal is to find the area under the curve described by a trigonometric function, \( v = \cos t \). This requires evaluating definite integrals and finding antiderivatives.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function. In other words, it reverses differentiation. For instance, if \( f(t) \) is an antiderivative of \( v(t) \), then \( f'(t) = v(t) \).

In the context of trigonometric functions, finding the antiderivative helps determine accumulated value, such as area under a curve. For example:

• The antiderivative of \( \cos t \) is \( \sin t \).

This was a key step in the problem solution, where the antiderivative \( \sin t \) was used to evaluate the definite integral from \( t = 0 \) to \( t = 2\pi \). This process involves the **Fundamental Theorem of Calculus**, stating that the principle is to find the antiderivative, then apply the limits to evaluate the integral, which in this case yielded zero.

Area Under Curve

The concept of "Area Under the Curve" refers to the region enclosed between the curve of a function and the horizontal axis. This area is a crucial notion in calculus for quantifying total accumulation or total change over an interval.

Calculating this area requires using a definite integral, which finds the net value over a specific range, such as from \( t = 0 \) to \( t = 2\pi \) in the current example. Here, the definite integral of \( v = \cos t \) represents the total area or the net result of the sine wave across one full period.

For periodic functions like sine and cosine, integration over a full period often results in zero net area due to symmetry. This happens because the positive area above the horizontal axis offsets the negative area below it. Hence, for the function \( \cos t \) over the interval \( [0, 2\pi] \), the integral evaluates to zero, reflecting no net area due to complete cancellation over the interval.