Problem 16 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\theta^{3} e^{-2 \theta} \cos 5 \theta$$

Step-by-Step Solution

Answer

\( \frac{dy}{d\theta} = e^{-2\theta} (3\theta^2 \cos 5\theta - 2\theta^3 \cos 5\theta - 5\theta^3 \sin 5\theta) \).

Step 1: Identify the Differentiation Rule

The function is given as \( y = \theta^3 e^{-2\theta} \cos 5\theta \). We observe that this is a product of three functions: \( u(\theta) = \theta^3 \), \( v(\theta) = e^{-2\theta} \), and \( w(\theta) = \cos 5\theta \). We will use the product rule for three functions: if \( y = f \cdot g \cdot h \), then \( \frac{dy}{d\theta} = f'gh + fg'h + fgh' \).

Step 2: Differentiate Each Function

Calculate the derivatives of the individual functions:- The derivative of \( u(\theta) = \theta^3 \) is \( u'(\theta) = 3\theta^2 \).- The derivative of \( v(\theta) = e^{-2\theta} \) is \( v'(\theta) = -2e^{-2\theta} \) by the chain rule.- The derivative of \( w(\theta) = \cos 5\theta \) is \( w'(\theta) = -5\sin 5\theta \) by the chain rule.

Step 3: Apply the Product Rule

Substitute these derivatives back into the product rule formula. The derivative is given by:\[ \frac{dy}{d\theta} = (3 \theta^2)(e^{-2\theta})(\cos 5\theta) + (\theta^3)(-2e^{-2\theta})(\cos 5\theta) + (\theta^3)(e^{-2\theta})(-5\sin 5\theta) \].

Step 4: Simplify the Expression

Combine all terms to simplify the expression:\[ \frac{dy}{d\theta} = 3\theta^2 e^{-2\theta} \cos 5\theta - 2\theta^3 e^{-2\theta} \cos 5\theta - 5\theta^3 e^{-2\theta} \sin 5\theta \].Group the common factor \( e^{-2\theta} \) to further simplify:\[ \frac{dy}{d\theta} = e^{-2\theta} \left( 3\theta^2 \cos 5\theta - 2\theta^3 \cos 5\theta - 5\theta^3 \sin 5\theta \right) \].

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus that comes into play when we need to find the derivative of a product of two or more functions. In a simple form, if you have two functions, say \( u(x) \) and \( v(x) \), and need to differentiate their product, the product rule states:

• \( \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

When applied to more than two functions, as in our example \( y = \theta^3 e^{-2\theta} \cos 5\theta \), the product rule expands a bit. For three functions, \( f\cdot g\cdot h \), it becomes:

• \( \frac{d}{d\theta} [f(\theta) \cdot g(\theta) \cdot h(\theta)] = f'gh + fg'h + fgh' \).

Understanding this rule is crucial because it lets you break down the differentiation process into simpler parts. Each segment is differentiated individually while treating the others as constants for that moment. This approach minimizes mistakes especially in complex products.

Differentiation

Differentiation is a cornerstone of calculus, involving the process of finding the rate at which a function is changing at any given point. Specifically, the derivative of a function gives us this rate of change.To differentiate a function means to apply algebraic rules to find its derivative. The primary step is identifying the type of function or combination of functions, as in our exercise, which combines polynomial, exponential, and trigonometric functions. The product rule was the appropriate differentiation technique because we had a product of three functions to differentiate.In our problem, you first differentiate each individual function:

• The polynomial function \( \theta^3 \) becomes \( 3\theta^2 \).
• The exponential function \( e^{-2\theta} \) differentiates into \( -2e^{-2\theta} \).
• The trigonometric function \( \cos 5\theta \) becomes \( -5\sin 5\theta \).

Each derivative reflects how its corresponding function changes as \( \theta \) changes. Knowing how to handle each type of function simplifies solving complex derivatives.

Chain Rule

The chain rule is an essential rule for differentiation in calculus, particularly useful when dealing with composite functions. A composite function is essentially a function within another function, and the chain rule helps differentiate these stacked functions.Consider the chain rule formula: if you have a function \( z = g(f(x)) \), the derivative is given by:

• \( \frac{dz}{dx} = g'(f(x)) \cdot f'(x) \).

In our exercise, the chain rule was necessary for differentiating both the exponential function \( e^{-2\theta} \) and the trigonometric function \( \cos 5\theta \). For example:

• For \( e^{-2\theta} \): The outer function is an exponential, and the inner function is \( -2\theta \). Differentiating gives \( -2e^{-2\theta} \).
• For \( \cos 5\theta \): The outer function is cosine, and the inner is \( 5\theta \). The derivative becomes \( -5\sin 5\theta \).

By understanding and implementing the chain rule, we simplify complex operations by focusing on the derivatives of the inner and outer functions separately. This makes conquering complicated differentiation tasks much more manageable.