Problem 28 - Calculus for Biology and Medicine

Question

In Problems 1-40, find the general antiderivative of the given function. $$ f(x)=\cos \left(\frac{x}{5}\right)-\sin \left(\frac{x}{5}\right) $$

Step-by-Step Solution

Answer

The general antiderivative is \( F(x) = 5 \sin(\frac{x}{5}) + 5 \cos(\frac{x}{5}) + C \).

Step 1: Understanding the Problem

We need to find the general antiderivative of the function \( f(x) = \cos\left(\frac{x}{5}\right) - \sin\left(\frac{x}{5}\right) \). An antiderivative is a function whose derivative is the given function.

Step 2: Identifying the Antiderivatives of Sine and Cosine

Recall that the antiderivative of \( \cos(kx) \) is \( \frac{1}{k} \sin(kx) \) and the antiderivative of \( \sin(kx) \) is \( -\frac{1}{k} \cos(kx) \).

Step 3: Applying the Chain Rule for Antiderivatives

For the term \( \cos(\frac{x}{5}) \), we identify \( k = \frac{1}{5} \), so the antiderivative is \( 5 \sin(\frac{x}{5}) \). For the term \( -\sin(\frac{x}{5}) \), the antiderivative is \( 5 \cos(\frac{x}{5}) \), keeping the sign from the function.

Step 4: Combining the Antiderivatives

The general antiderivative is the sum of the individual antiderivatives: \( F(x) = 5 \sin(\frac{x}{5}) + 5 \cos(\frac{x}{5}) + C \), where \( C \) is the constant of integration.

Key Concepts

Understanding Trigonometric Functions

Trigonometric functions are essential in mathematics, especially in calculus, as they describe relationships between the angles and sides of triangles. The two main functions we encounter are sine and cosine. These functions have specific patterns and properties that make them very useful for solving problems involving periodic and oscillatory behavior.

Sine and cosine functions are periodic with a period of \( 2\pi \), meaning they repeat their values every \( 2\pi \) units. This periodic nature is crucial when dealing with complex calculations in physics and engineering.

• For a function \( \, \cos(x) \, \), its graph starts at its maximum value.
• For \( \, \sin(x) \, \), the graph begins at the origin at zero and rises to a peak.

In our example, the function involves \( \cos\left(\frac{x}{5}\right) \) and \( \sin\left(\frac{x}{5}\right) \), indicating a horizontal stretch due to the factor of \( \frac{1}{5} \). This transformation affects the frequency and lengthens the period of both sine and cosine functions.

The Chain Rule in Antiderivatives

The chain rule is a fundamental tool in calculus for differentiating composite functions, but it also has a counterpart when finding antiderivatives. When you find the antiderivative of a function with a nested structure, similar to applying the chain rule in differentiation, you consider the inner function and its derivative.

In our problem, the chain rule helps identify antiderivatives where the function is dependent on a substitution, such as \( \cos\left(\frac{x}{5}\right) \) or \( \sin\left(\frac{x}{5}\right) \). Consider:

• For \( \cos(kx) \), the antiderivative is \( \frac{1}{k} \sin(kx) \).
• For \( \sin(kx) \), it is \( -\frac{1}{k} \cos(kx) \).

The factor of \( \frac{1}{5} \) in the argument modifies the standard formulas. Therefore, the antiderivatives become \( 5\sin\left(\frac{x}{5}\right) \) for cosine, and \( 5\cos\left(\frac{x}{5}\right) \) for sine, due to the inverse factor of \( \frac{1}{k} = 5 \). This adjustment accounts for the velocity at which values rise along the axes.

Approaching Calculus Problem-solving

Calculus problems can appear daunting at first, but understanding the core concepts can make them manageable. The goal is often to find the derivative or antiderivative of a function. In this exercise, we seek the antiderivative, which is a function whose derivative will give us the original function.

Here’s the typical approach:

• Understand the function given: Break it down into recognizable parts, such as sums or differences of standard functions like sine or cosine.
• Identify relevant rules: Use known derivatives and antiderivatives, like those of trigonometric functions.
• Apply the chain rule if a function has layers, considering the adjustment needed for any coefficients inside the function arguments.
• Combine any separate antiderivatives to form the general solution and don’t forget to add \( C \), the constant of integration. This constant is crucial because it represents any potential constant term that could differentiate to zero.

Let's highlight that calculus problem solving heavily revolves around practice. By working through problems step-by-step, you develop skills to recognize patterns and apply rules effectively. Patience and practice are key.