Problem 12 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = AB \cos(Bt) \)

Step 1: Understand the Problem

We need to differentiate the function \( y = A \sin(Bt) \) with respect to \( t \). This means we will find \( \frac{dy}{dt} \), the derivative of \( y \) with respect to \( t \).

Step 2: Differentiate Using Chain Rule

The function is a composition of the sine function and the linear function \( Bt \). We apply the chain rule. \( \frac{d}{dt} [\sin(Bt)] = \cos(Bt) \cdot \frac{d}{dt}[Bt] \). Since \( \frac{d}{dt}[Bt] = B \), then \( \frac{d}{dt} [\sin(Bt)] = B \cos(Bt) \).

Step 3: Differentiate the Entire Function

Apply the constant multiplier rule. Since \( A \) is a constant factor, it remains in the derivative: \( \frac{dy}{dt} = A \cdot B \cos(Bt) \). The entire differentiated expression is \( AB \cos(Bt) \).

Key Concepts

chain rule

The chain rule is a fundamental concept in calculus for finding derivatives of composite functions. When you have a function composed of two or more functions, like \( y = \sin(Bt) \), you need the chain rule to differentiate it. The idea is to first differentiate the outer function, which in this case is the sine function, and then multiply by the derivative of the inner function, which is the linear function \( Bt \).

• Differentiate the outer function: The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \).
• Find the derivative of the inner function: Here \( Bt \), with its derivative being \( B \) since \( B \) is a constant and the derivative of \( t \) with respect to \( t \) is 1.
• Combine them using the chain rule: So, you get \( \cos(Bt) \times B \).

This method simplifies differentiating complex functions by tackling each component step-by-step.

sine function

The sine function, often denoted as \( \sin(x) \), is one of the primary trigonometric functions. It represents the y-coordinate of a point on the unit circle as it varies with the angle \( x \). In calculus, the sine function is crucial because it's involved in modeling periodic phenomena such as waves. Understanding its properties is key to solving many problems.

• The derivative of \( \sin(x) \) is \( \cos(x) \), which means when you find the rate of change of the sine function, you're actually looking at the cosine function.
• This cyclical pattern continues as \( \cos(x) \)'s derivative is \(-\sin(x) \).

Knowing these derivatives is essential, especially when you're dealing with equations involving trigonometric functions.

derivative of trigonometric functions

Differentiating trigonometric functions is a common task in calculus that deals with how these functions change at any point. Each basic trigonometric function has a straightforward derivative:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \(-\sin(x) \).
• Functions such as \( \tan(x) \), \( \csc(x) \), \( \sec(x) \), and \( \cot(x) \) also have respective derivatives, like the derivative of \( \tan(x) \), which is \( \sec^2(x) \).

For the problem at hand, knowing that the derivative of \( \sin(Bt) \) becomes useful, leading to \( \cos(Bt) \). Combined with the chain rule, this knowledge allows accurate calculation of derivatives for more complex trigonometric expressions.

constant multiplier rule

The constant multiplier rule is an important principle in differentiation. It says that when differentiating a term with a constant multiplied by a function, you can factor out the constant, focusing only on differentiating the function itself.This rule simplifies calculations because you don't need to worry about the constant. Here's how it applies to the exercise:

• Consider \( y = A \cdot \sin(Bt) \), and differentiate \( \sin(Bt) \), as previously found to be \( B \cos(Bt) \).
• The constant \( A \) can be brought outside the differentiation since it doesn't change with respect to \( t \).

Therefore, when you derive \( A \cdot \sin(Bt) \), it becomes \( A \cdot (B \cos(Bt)) = AB \cos(Bt) \).This rule is especially helpful when dealing with real-world problems involving fixed rates or quantities, ensuring calculations are straightforward.