Problem 25 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$s=\frac{4}{3 \pi} \sin 3 t+\frac{4}{5 \pi} \cos 5 t$$

Step-by-Step Solution

Answer

\( s' = \frac{4}{\pi} \cos 3t - \frac{4}{\pi} \sin 5t \)

Step 1: Identify the components of the function

The given function is \( s = \frac{4}{3 \pi} \sin 3t + \frac{4}{5 \pi} \cos 5t \). It is composed of two trigonometric functions: \( \frac{4}{3 \pi} \sin 3t \) and \( \frac{4}{5 \pi} \cos 5t \). Each term needs to be differentiated separately.

Step 2: Differentiate the sine component

Differentiate the term \( \frac{4}{3 \pi} \sin 3t \) with respect to \( t \). Recall that the derivative of \( \sin kt \) is \( k \cos kt \). Thus, the derivative of \( \frac{4}{3 \pi} \sin 3t \) is \( \frac{4}{3 \pi} \times 3 \cos 3t = \frac{4}{\pi} \cos 3t \).

Step 3: Differentiate the cosine component

Differentiate the term \( \frac{4}{5 \pi} \cos 5t \) with respect to \( t \). Recall that the derivative of \( \cos kt \) is \( -k \sin kt \). Thus, the derivative of \( \frac{4}{5 \pi} \cos 5t \) is \( \frac{4}{5 \pi} \times -5 \sin 5t = -\frac{4}{\pi} \sin 5t \).

Step 4: Combine the results

Combine the derivatives found in Step 2 and Step 3. The derivative \( s' \) of the given function \( s \) is:\[ s' = \frac{4}{\pi} \cos 3t - \frac{4}{\pi} \sin 5t \].

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in calculus and appear frequently in real-world applications. These functions—sine, cosine, and tangent—are based on the relationships in a right-angled triangle or the coordinates of a point on a unit circle. When it comes to calculus, particularly for derivatives, understanding their behavior and properties is crucial. These functions help to model periodic phenomena such as sound waves, light waves, or any type of cyclical pattern.

• Sine Function (\(\sin\theta\)): Represents the y-coordinate of a point on the unit circle. It oscillates between -1 and 1 as \(\theta\) varies.
• Cosine Function (\(\cos\theta\)): Represents the x-coordinate on the unit circle. Similar to sine, it varies between -1 and 1, and is often used in conjunction with sine.
• Tangent Function (\(\tan\theta\)): Not a part of our immediate topic but useful to mention. It is the ratio of the sine and cosine functions.

Grasping these fundamental properties helps in understanding how to differentiate these functions, a vital step in analyzing their behavior using calculus.

Sine and Cosine Derivatives

Differentiating trigonometric functions like sine and cosine is a cornerstone concept in calculus. The rules are simple and once understood, can be applied to more complex functions. For sine and cosine, their derivatives are particularly straightforward:

• Derivative of Sine (\(\sin x\)): The derivative is \(\cos x\). This means if you have a function \(\sin x\), its rate of change is described by \(\cos x\)
• Derivative of Cosine (\(\cos x\)): The derivative is \(-\sin x\). This indicates the slope of \(\cos x\) at any point is negative \(\sin x\)

For functions like \(\sin kt\) and \(\cos kt\), where \(k\) is a constant, the derivative involves multiplying by \(k\). For example:

• The derivative of \(\sin kt\) is \(k\cos kt\)
• The derivative of \(\cos kt\) is \(-k\sin kt\)

These derivatives are utilized to solve the exercise problem by applying these rules separate for each component of the given function.

Function Differentiation

Function differentiation refers to the process of finding the derivative of a function. It's a way to determine the rate at which a function is changing at any given point. When you differentiate a function, you are essentially finding the slope of the function's graph at a specific point. This is crucial for analyzing and understanding behaviors of different functions.
To differentiate a function that is a sum of terms, like the one in our exercise, we use a principle called linearity of differentiation. This allows us to differentiate each term individually and then sum up the results:

• For \(\frac{4}{3\pi}\sin 3t\), differentiate as \(\frac{4}{3\pi}\times 3\cos 3t\).
• For \(\frac{4}{5\pi}\cos 5t\), differentiate as \(-\frac{4}{5\pi}\times 5\sin 5t\).

After differentiating each part, you simply combine them to get the complete derivative. Understanding these individual steps is pivotal for mastering calculus problems.