Problem 8 - Applied Calculus

Question

Differentiate the functions in Problems 1-20. Assume that \(A\) and \(B\) are constants. $$ R=\sin (5 t) $$

Step-by-Step Solution

Answer

The derivative of \( R = \sin(5t) \) is \( R'(t) = 5\cos(5t) \).

Step 1: Identify the Derivative Formula

To differentiate the function, we identify that it is a trigonometric function. The derivative of \( \sin(x) \) is \( \cos(x) \). However, since we have a composite function \( \sin(5t) \), we'll use the chain rule.

Step 2: Apply the Chain Rule

The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). Here, the outer function is \( \sin(u) \) and the inner function \( u = 5t \). The derivative of \( \sin(u) \) is \( \cos(u) \) and of \( 5t \) is \( 5 \).

Step 3: Compute the Derivative

Apply the chain rule: differentiate the outer function \( \sin(5t) \) to get \( \cos(5t) \), and multiply by the derivative of the inner function \( 5t \), which is \( 5 \). Thus, the derivative \( R'(t) = 5 \cdot \cos(5t) \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used for differentiation. It helps us find the derivative of composite functions, where one function is nested inside another. If you have a function in the form of \( f(g(x)) \), think of it as one function \( f \) working on the inside function \( g(x) \).

• Outer function: This is the main function you look at, such as \( f \) in \( f(g(x)) \).
• Inner function: This is the function inside, such as \( g(x) \).

The chain rule can be mathematically expressed as:\[ \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \]This formula tells us to take the derivative of the outer function \( f \) evaluated at \( g(x) \), and multiply it by the derivative of the inner function \( g(x) \). This method is crucial when differentiating trigonometric and other complex functions that are composed of inner parts.

Trigonometric Functions

Trigonometric functions are based on the angles of a right triangle and are fundamental in calculus for modeling periodic phenomena. These functions include sine, cosine, and tangent, among others. For differentiation purposes, it’s important to remember some basic derivatives:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \(-\sin(x) \).

These derivatives help us understand how a small change in the angle affects the value of the function. For example, knowing that the derivative of \( \sin(x) \) is \( \cos(x) \) means that the rate of change of the sine function is determined by the cosine function.In the context of composite functions like \( \sin(5t) \), you can see that knowing these derivatives is essential when applying the chain rule.

Composite Function

Composite functions are functions that contain another function as part of them. Think of them as combining two functions where the output of one function becomes the input for the other. The standard notation is \( f(g(x)) \), showing that function \( f \) is applied to \( g(x) \).When it comes to derivative calculus, composite functions are especially interesting, as they require the chain rule for differentiation. For example, in \( \sin(5t) \):

• The outer function is \( \sin(u) \).
• The inner function is \( g(t) = 5t \).

This creates a situation where you must understand how the inner function’s derivative influences the result of the outer function’s derivative. This is why the chain rule is vital—without it, analyzing how changes in \( t \) impact the overall composite function would be much more challenging. By smoothly applying these concepts, you can manage and find derivatives of more elaborate functions.