Problem 28 - Applied Calculus

Question

Use an integral to find the specified area. Under \(y=2 \cos (t / 10)\) for \(1 \leq t \leq 2\)

Step-by-Step Solution

Answer

The area is approximately 1.978 square units.

Step 1: Understand the Problem

We are asked to find the area under the curve of the function \(y = 2 \cos\left(\frac{t}{10}\right)\) within the interval \([1, 2]\). This means integrating the function with respect to \(t\) over the interval \([1, 2]\).

Step 2: Set Up the Integral

The area under the curve can be found by calculating the definite integral of the function \(y = 2 \cos\left(\frac{t}{10}\right)\) over the interval \([1, 2]\). Thus, we set up the integral as:\[\int_{1}^{2} 2 \cos\left(\frac{t}{10}\right) \, dt\].

Step 3: Integrate the Function

To solve the integral \(\int 2 \cos\left(\frac{t}{10}\right) \, dt\), we use the substitution method. Let \(u = \frac{t}{10}\), then \(du = \frac{1}{10} \, dt\), so \(dt = 10 \, du\). Substitute back into the integral:\[\int 2 \cos(u) \times 10 \, du = 20 \int \cos(u) \, du\].The integral of \(\cos(u)\) is \(\sin(u)\), so:\[20 \sin(u) + C\].

Step 4: Substitute Back for t

Replace \(u\) with \(\frac{t}{10}\) in the expression:\[20 \sin\left(\frac{t}{10}\right) + C\].

Step 5: Evaluate the Definite Integral

Now, evaluate the definite integral over the limits from \(t = 1\) to \(t = 2\):\[\left[ 20 \sin\left(\frac{t}{10}\right) \right]_{1}^{2} = 20 \sin\left(\frac{2}{10}\right) - 20 \sin\left(\frac{1}{10}\right)\].

Step 6: Calculate Numerical Value

Calculate the sine values using a calculator:\[\sin\left(\frac{2}{10}\right) \approx \sin(0.2) \approx 0.1987\]\[\sin\left(\frac{1}{10}\right) \approx \sin(0.1) \approx 0.0998\].Substitute these values back into the expression:\[20 \times 0.1987 - 20 \times 0.0998 = 3.974 - 1.996 = 1.978\].

Key Concepts

Definite Integral

The concept of a definite integral is fundamental in calculus, especially when calculating the area under a curve. It essentially represents the accumulation of quantities.
When integrating a function, you work with limits of integration, which specify the interval over which you accumulate the area.
- **Definite Integral Notation:** It's expressed as \( \int_{a}^{b} f(t) \, dt \), where \( a \) and \( b \) are the lower and upper limits of integration.- **Importance:** It computes a precise sum, often described as the 'net area' between the curve \( y = f(t) \) and the t-axis.For instance, in the exercise, to find the area under the curve \( y = 2 \cos\left(\frac{t}{10}\right) \) from \( t = 1 \) to \( t = 2 \), the definite integral accumulates small pieces of area to provide an exact result.
This concept helps in providing real-world solutions where measuring area is necessary, making definite integrals invaluable in fields such as physics and engineering.

Trigonometric Functions

Trigonometric functions are crucial in calculus for solving problems involving periodic behavior. These functions, like sine and cosine, appear frequently in various mathematical contexts.
- **Nature of Cosine:** The cosine function \( \cos(t) \) oscillates between -1 and 1, defining the maximum and minimum values of periodic behavior. When multiplied by a scalar, like 2 in this function \( 2 \cos\left(\frac{t}{10}\right) \), it stretches vertically.- **Application in integrals:** When these functions are integrated, it often involves finding antiderivatives. For example, the antiderivative of \( \cos(u) \) is \( \sin(u) \).
In the example exercise, substitution simplifies the integral \( \int 2 \cos\left(\frac{t}{10}\right) \, dt \). The variable substitution \( u = \frac{t}{10} \) helps manage the derivative during integration.
Understanding these trigonometric functions aids in overcoming many calculus challenges that include natural phenomena, wave equations, and oscillatory systems.

Area Under a Curve

Calculating the area under a curve is a key application of definite integrals. This area provides valuable information about functions, especially in physics and economics.
- **Area Formula:** The area under the curve of a function \( y = f(t) \) over an interval \( [a, b] \) is given by the definite integral \( \int_{a}^{b} f(t) \, dt \).- **Significance:** This measurement tells us how much space is enclosed by the curve and the axis. It's essential in calculating quantities like distance, probability, or even economic measurements such as revenue.In the given exercise, the integral \( \int_{1}^{2} 2 \cos\left(\frac{t}{10}\right) \, dt \) calculates the area between the curve and the t-axis from \( t = 1 \) to \( t = 2 \).
This practical application is why understanding the process is crucial: it transforms abstract mathematical concepts into useful tools for solving real-world problems.