Problem 10 - Applied Calculus

Question

Differentiate each function. $$ f(t)=6 \sin 2 t $$

Step-by-Step Solution

Answer

The derivative is \( 12 \cos(2t) \).

Step 1: Identify the basic function

We are given the function \( f(t) = 6 \sin(2t) \). The basic trigonometric function here is \( \sin(2t) \). We need the derivative of \( \sin(x) \), which is \( \cos(x) \).

Step 2: Apply the chain rule

The function inside the sine function is \( 2t \). According to the chain rule, the derivative of \( \sin(2t) \) with respect to \( t \) is \( \cos(2t) \) times the derivative of \( 2t \) with respect to \( t \).

Step 3: Differentiate the inner function

The inner function \( 2t \) has a derivative of \( 2 \). Thus, applying the chain rule gives us \( \frac{d}{dt}[\sin(2t)] = \cos(2t) \times 2 \).

Step 4: Differentiate the full function

Now differentiate the given function \( f(t) = 6 \sin(2t) \). Using the derivative found in the previous step and noting that the constant factor 6 remains outside the differentiation, we have \( \frac{d}{dt}[6 \sin(2t)] = 6 \times 2 \times \cos(2t) \).

Step 5: Simplify the expression

Simplifying \( 6 \times 2 \times \cos(2t) \), we obtain \( 12 \cos(2t) \). This is the final derivative of the function.

Key Concepts

Trigonometric Derivatives

Trigonometric derivatives are an essential part of calculus because they help us understand how trigonometric functions change. When we talk about derivatives in calculus, we are often looking for the rate at which a function changes at any given point. This is crucial when working with trigonometric functions like sine, cosine, and tangent. For example, the basic derivative of \( \sin(x) \) is \( \cos(x) \). Similarly, the derivatives for other trigonometric functions are:

• The derivative of \( \cos(x) \) is \(-\sin(x) \)
• The derivative of \( \tan(x) \) is \( \sec^2(x) \)

Understanding these derivatives allows you to tackle more complex problems involving trigonometric functions. Differentiating these functions lets you know how the sine wave, cosine wave, and their respective rates of change behave in varying contexts. This knowledge is particularly useful in physics and engineering applications where wave functions are prevalent.

Chain Rule in Calculus

The chain rule is a powerful tool in calculus for finding the derivative of composite functions. A composite function is essentially a function within another function, like \( \sin(2t) \). The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \times g'(x) \). This means you first differentiate the outer function and then multiply by the derivative of the inner function.
In our example, the function needed differentiation is \( \sin(2t) \). Here:

• The outer function is \( \sin(u) \) with \( u = 2t \)
• The derivative of \( \sin(u) \) is \( \cos(u) \)

Then, you differentiate the inner function, \( 2t \), giving a result of \( 2 \). So, according to the chain rule, the derivative \( \frac{d}{dt} [\sin(2t)] = \cos(2t) \times 2 \).
Using the chain rule efficiently helps solve complex differentiation problems by breaking them down into simpler steps. This approach can significantly simplify algebra involved in calculus problems, especially with nested functions.

Derivative of Trigonometric Functions

Taking the derivative of trigonometric functions is a common task that benefits greatly from the basics of trigonometric derivatives and the chain rule. In the example function \( f(t) = 6 \sin(2t) \), you need to differentiate a trigonometric function with a constant multiplied to it.
The steps in the solution used:

• Recognizing the trigonometric function and applying initial derivative knowledge: knowing \( \frac{d}{dt}[\sin(2t)] = \cos(2t) \times 2 \)
• Multiplying the resulting expression by the constant 6 outside of the function, resulting in \( 6 \times 2 \times \cos(2t) \)
• Simplifying to obtain the final derivative: \( 12 \cos(2t) \)

Understanding how to differentiate trigonometric functions, especially within more complex expressions, is key for deeper applications in fields involving waves, oscillations, and periodic phenomena. Mastering these basic techniques opens doors to more advanced problem-solving situations in calculus.