Problem 26 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$s=\sin \left(\frac{3 \pi t}{2}\right)+\cos \left(\frac{3 \pi t}{2}\right)$$

Step-by-Step Solution

Answer

\(\frac{ds}{dt} = \frac{3\pi}{2} \cos \left(\frac{3\pi t}{2}\right) - \frac{3\pi}{2} \sin \left(\frac{3\pi t}{2}\right)\)

Step 1: Identify the Problem

We need to find the derivative of the function \(s = \sin \left(\frac{3\pi t}{2}\right) + \cos \left(\frac{3\pi t}{2}\right)\) with respect to \(t\).

Step 2: Apply Derivative Rules

Recall that the derivative of \(\sin u\) with respect to \(u\) is \(\cos u\), and the derivative of \(\cos u\) is \(-\sin u\). Both derivatives must be multiplied by the derivative of the inside function according to the chain rule.

Step 3: Differentiate Each Term Separately

Differentiate the sine term: \[\frac{d}{dt} \sin \left(\frac{3\pi t}{2}\right) = \cos \left(\frac{3\pi t}{2}\right) \cdot \frac{d}{dt}\left(\frac{3\pi t}{2}\right) = \cos \left(\frac{3\pi t}{2}\right) \cdot \frac{3\pi}{2} \]Differentiate the cosine term: \[\frac{d}{dt} \cos \left(\frac{3\pi t}{2}\right) = -\sin \left(\frac{3\pi t}{2}\right) \cdot \frac{d}{dt}\left(\frac{3\pi t}{2}\right) = -\sin \left(\frac{3\pi t}{2}\right) \cdot \frac{3\pi}{2} \]

Step 4: Combine the Differentiated Terms

The derivative of the function \(s\), \(\frac{ds}{dt}\), is the sum of the derivatives of the individual terms:\[ \frac{ds}{dt} = \frac{3\pi}{2} \cos \left(\frac{3\pi t}{2}\right) - \frac{3\pi}{2} \sin \left(\frac{3\pi t}{2}\right) \]

Key Concepts

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus, specifically when dealing with periodic phenomena. These functions map angles to ratios of sides in a right-angled triangle and are extensively used in modeling waves and oscillations.
One essential property of trigonometric functions is their derivatives. For instance, the derivative of the sine function, denoted as \( \sin(x) \), is the cosine function \( \cos(x) \). Similarly, the derivative of the cosine function \( \cos(x) \) is \(-\sin(x) \).
The periodic nature of sine and cosine leads to their derivatives being cyclic and repetitive. This characteristic is crucial for solving various differential calculus problems, including the one in this exercise.

Chain Rule

The chain rule is a key principle in differentiation, especially when dealing with composite functions. A composite function is where one function is nested inside another, like \( f(g(x)) \). The chain rule enables us to differentiate complex functions by addressing the derivative of each nested function sequentially.
In simpler terms, if you can break a function into an outer function and an inner function, the derivative of the whole is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. For example, with \( s = \sin(\frac{3\pi t}{2}) \), the outer function is \( \sin(u) \), and the inner function is \( u = \frac{3\pi t}{2} \).

• The derivative of the outer function \( \sin(u) \) is \( \cos(u) \).
• The derivative of the inner function \( u = \frac{3\pi t}{2} \) is \( \frac{3\pi}{2} \).

Multiply these to apply the chain rule effectively, which helps in finding the correct derivative.

Function Differentiation

Differentiation is the process of finding a function’s derivative, which represents the rate at which the function’s value changes. It is a fundamental tool in calculus, helping us understand how variables are related through their rates of change.
In this exercise, we began by separately differentiating each term of the function \( s = \sin \left(\frac{3\pi t}{2}\right) + \cos \left(\frac{3\pi t}{2}\right) \). This approach ensures each part of the function is appropriately handled based on its specific derivative rule.

• Using derivatives for \( \sin \) and \( \cos \), and applying the chain rule, we compute the differentiation accurately.
• We then combine these results to find the complete derivative of the original function with respect to \( t \).

Differentiation allows us to quantify changes and understand dynamic relationships within functions, serving as a building block for more complex calculus problems.