Problem 10 - Essential Calculus Early Transcendentals

Question

Find the most general antiderivative of the function (Check your answer by differentiation $$f(t)=\sin t+2 \sinh t$$

Step-by-Step Solution

Answer

The antiderivative is \(-\cos t + 2\cosh t + C\).

Step 1: Understand the function components

The given function is composed of two separate parts: \( \sin t \), which is the sine function, and \( 2 \sinh t \), which is twice the hyperbolic sine function. We need to find antiderivatives for each of these functions separately.

Step 2: Find the antiderivative of \( \sin t \)

The antiderivative of \( \sin t \) is \( -\cos t \), because the derivative of \( -\cos t \) is \( \sin t \).

Step 3: Find the antiderivative of \( 2 \sinh t \)

The antiderivative of \( \sinh t \) is \( \cosh t \), since the derivative of \( \cosh t \) is \( \sinh t \). Thus, the antiderivative of \( 2 \sinh t \) is \( 2 \cosh t \).

Step 4: Combine the antiderivatives

The most general antiderivative of the function \( f(t) = \sin t + 2 \sinh t \) is obtained by combining the two antiderivatives found in Steps 2 and 3, which is \( -\cos t + 2 \cosh t + C \), where \( C \) is the constant of integration.

Step 5: Check the result by differentiation

Differentiate the result \( -\cos t + 2 \cosh t + C \) to verify that we return to \( f(t) = \sin t + 2 \sinh t \). The derivative of \( -\cos t \) is \( \sin t \), the derivative of \( 2 \cosh t \) is \( 2 \sinh t \), and the derivative of any constant \( C \) is 0. Thus, the differentiation confirms that \( \sin t + 2 \sinh t \) is correct.

Key Concepts

Sine Function

The sine function, denoted as \( \sin t \), is one of the fundamental trigonometric functions in mathematics. It is most commonly used to describe the shape of waves and oscillations, which is why it's essential in fields like physics and engineering.

The sine function takes an angle as input (usually measured in radians) and returns a value between -1 and 1. For example:

• \( \sin(0) = 0 \)
• \( \sin\left(\frac{\pi}{2}\right) = 1 \)
• \( \sin(\pi) = 0 \)

The graph of the sine function is a smooth, periodic wave that repeats every \(2\pi\) radians.

When finding its antiderivative, we reverse the differentiation process. Since the derivative of \( -\cos t \) is \( \sin t \), the antiderivative of \( \sin t \) is \( -\cos t \) plus a constant of integration.

Hyperbolic Sine Function

The hyperbolic sine function, written as \( \sinh t \), is part of the family of hyperbolic functions similar to trigonometric functions but based on hyperbolas rather than circles. The hyperbolic sine function is defined as:
\[ \sinh t = \frac{e^t - e^{-t}}{2} \]
Where \( e \) is the base of the natural logarithm.

Similar to the sine function, \( \sinh t \) is odd and has a graph that passes through the origin. Unlike \( \sin t \), it increases exponentially for large positive or negative values of \( t \).

• \( \sinh(0) = 0 \)
• \( \sinh(1) \approx 1.175 \)

With hyperbolic functions, the process for finding antiderivatives also involves finding a function whose derivative results in the original function. The derivative of \( \cosh t \) is \( \sinh t \), thus the antiderivative of \( \sinh t \) is \( \cosh t \). For the function \( 2 \sinh t \), the antiderivative becomes \( 2\cosh t \).

Hyperbolic functions serve to model growth phenomena and often appear in contexts involving calculus and analysis.

Constant of Integration

The constant of integration \( C \) is an essential concept when calculating antiderivatives. When we integrate a function, we determine a family of functions that differ by a constant. This is because the derivative of a constant is zero, and it doesn’t show up in the differentiation process.

In every antiderivative result, you'll see the constant of integration added, indicating all possible functions that could differentiate into the given function.

• If \( F(x) = f(x) + C \), then \( F'(x) = f(x) \)

This means our original function could have been shifted vertically by any constant amount, and differentiation would still yield the same result.

For instance, in the solution to our problem regarding \( \sin t + 2 \sinh t \), adding \( C \) results in \( -\cos t + 2 \cosh t + C \). This fully encompasses all possible functions that could differ by a constant amount and yet have the same derivative as the given function.