Problem 14 - Applied Calculus

Question

Differentiate the functions in Problems 1-20. Assume that \(A\) and \(B\) are constants. $$ y=6 \sin (2 t)+\cos (4 t) $$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = 12 \cos(2t) - 4 \sin(4t) \)

Step 1: Differentiate the first term

The function given is \( y = 6 \sin (2t) + \cos (4t) \). We will differentiate it with respect to \( t \). Start by differentiating the first term, \( 6 \sin(2t) \). Using the chain rule, the derivative of \( \sin(2t) \) is \( 2 \cos(2t) \), and we multiply by 6 to get \( 12 \cos(2t) \).

Step 2: Differentiate the second term

Now, differentiate the second term, \( \cos(4t) \), using the chain rule. The derivative of \( \cos(4t) \) is \( -4 \sin(4t) \).

Step 3: Combine the derivatives

Combine the derivatives from Step 1 and Step 2 to obtain the derivative of the entire function: \( \frac{dy}{dt} = 12 \cos(2t) - 4 \sin(4t) \).

Key Concepts

Chain Rule

The Chain Rule is a vital concept when differentiating composite functions. In simple terms, it helps us find the derivative of a function inside another function. Imagine peeling an onion, layer by layer. That's what the Chain Rule does! It tackles each layer separately.
For instance, in the term \(6 \sin(2t)\), "\(2t\)" is the inner function and "\(\sin\)" is the outer one. We first differentiate the outer function: the derivative of \(\sin(u)\) is \(\cos(u)\). Next, we differentiate the inner function, \(2t\), whose derivative is 2. Then, we multiply these results together, taking care of any constants like 6.

• Differentiate the outer function: \(\cos(2t)\).
• Differentiate the inner function: 2.
• Multiply them together and include any constants: \(12 \cos(2t)\).

This principle similarly applies to the differentiation of \(\cos(4t)\). The Chain Rule simplifies complex differentiations into manageable steps.

Trigonometric Derivatives

Trigonometric derivatives are fundamental in solving calculus problems involving trigonometric functions. These derivatives stem from basic trigonometric functions, like \(\sin\) and \(\cos\), and involve rules you must remember.
Here's a quick guide to the basic trigonometric derivatives:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• You remember these by noting the cyclical nature of sine and cosine functions.

In the given problem, to differentiate \(\sin(2t)\) and \(\cos(4t)\), we use these core derivatives along with the Chain Rule.
For \(\sin(2t)\), the derivative becomes \(2\cos(2t)\), considering \(t\) is multiplied by 2. Similarly, \(\cos(4t)\) yields \(-4\sin(4t)\). Understanding these derivatives makes handling trigonometric expressions in calculus more intuitive.

Calculus Problem Solving

Problem-solving in calculus often requires a systematic approach. It's like solving a mystery, where you're guided by rules and patterns. With functions involving constants and trigonometric terms, make sure to keep these concepts in mind.
Here's a strategy for calculus problem-solving:

• Identify each component of the function: Separate constant terms, variable terms, and operations.
• Choose the right rule: Consider whether to apply the Chain Rule, product rule, or quotient rule.
• Simplify step by step: Don't rush. Take each derivative one part at a time.
• Combine results to form a complete answer: Check your work for errors.

In our sample problem, we started by identifying two separate terms to differentiate: \(6 \sin(2t)\) and \(\cos(4t)\). We used the Chain Rule due to the composite nature of these terms. After differentiating each part, we combined them into one derivative \(\frac{dy}{dt} = 12 \cos(2t) - 4 \sin(4t)\), showing clear and accurate results. Developing these problem-solving skills aids in conquering even the most challenging calculus problems.