Problem 49 - Calculus 3

Question

Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=3 \mathbf{i}+4 \sin (3 t) \mathbf{j}+t \cos (t) \mathbf{k}$$

Step-by-Step Solution

Answer

\( \mathbf{r}'(t) = 12 \cos (3t) \mathbf{j} + (\cos(t) - t \sin(t)) \mathbf{k} \).

Step 1: Identify Components

The vector-valued function is given as \( \mathbf{r}(t) = 3 \mathbf{i} + 4 \sin (3t) \mathbf{j} + t \cos (t) \mathbf{k} \). It consists of three components: a constant component in the \( \mathbf{i} \) direction, a sinusoidal component in the \( \mathbf{j} \) direction, and a function of product in the \( \mathbf{k} \) direction.

Step 2: Differentiate Each Component

To find the derivative, differentiate each component of \( \mathbf{r}(t) \) separately:1. The derivative of the constant term \( 3 \mathbf{i} \) is \( 0 \).2. For \( 4 \sin (3t) \mathbf{j} \), use the chain rule: the derivative is \( 4 \cdot 3 \cos (3t) = 12 \cos (3t) \mathbf{j} \).3. For \( t \cos (t) \mathbf{k} \), use the product rule: \((t)'\cos(t) + t(-\sin(t)) = \cos(t) - t \sin(t)\). So, the derivative is \(( \cos(t) - t \sin(t) ) \mathbf{k}\).

Step 3: Combine the Derivatives

Combine the derivatives of the components to find the derivative of the vector-valued function:\[\mathbf{r}'(t) = 0 \mathbf{i} + 12 \cos (3t) \mathbf{j} + (\cos(t) - t \sin(t)) \mathbf{k}\].

Key Concepts

Chain Rule

In vector calculus, the chain rule is a fundamental tool used to differentiate composite functions. When dealing with vector-valued functions, applying the chain rule allows us to differentiate components that involve nested functions. For instance, consider the function component \(4 \sin(3t) \mathbf{j}\) from the original exercise.
Here, we have a sine function wrapped around \(3t\). To differentiate this, we recognize the "chain" of functions: the outer function is sine, and the inner function is \(3t\).

• The derivative of the outer function \(\sin(u)\) is \(\cos(u)\) where \(u\) is the inner function.
• The derivative of the inner function \(3t\) is simply \(3\).

Using the chain rule, multiply the two derivatives: \(4 \cdot 3 \cos(3t) \mathbf{j} = 12 \cos(3t) \mathbf{j}\).
This process shows how the chain rule helps in smoothly and effectively finding the derivative of complex, nested expressions within vector-valued functions.

Product Rule

The product rule is particularly important when dealing with vector calculus, especially when differentiating components that involve the product of two functions. In our exercise, this is illustrated by the component \(t \cos(t) \mathbf{k}\).
The product rule states that if you have a product of two functions, say \(u(t)\) and \(v(t)\), the derivative is given by: \(u'v + uv'\).
Applying this to \(t \cos(t)\):

• Let \(u(t) = t\) and \(v(t) = \cos(t)\).
• The derivative of \(u(t) = t\) is \(1\).
• The derivative of \(v(t) = \cos(t)\) is \(-\sin(t)\).

Plug these into the product rule to get: \(1 \cdot \cos(t) + t \cdot (-\sin(t)) = \cos(t) - t \sin(t)\).
Thus, the derivative of the component is \((\cos(t) - t \sin(t)) \mathbf{k}\). This method illustrates how the product rule efficiently handles the differentiation of products within vector-valued functions.

Vector Calculus

Vector calculus provides the tools necessary for analyzing and manipulating vector-valued functions, which are crucial in fields like physics and engineering. When working with vector-valued functions, each component is typically treated separately to simplify the differentiation process. In our initial problem, the vector-valued function \(\mathbf{r}(t) = 3 \mathbf{i} + 4 \sin(3t) \mathbf{j} + t \cos(t) \mathbf{k}\) is a combination of independent functions along the \(\mathbf{i}, \mathbf{j},\) and \(\mathbf{k}\) directions.

• Each component function can be handled using standard differentiation techniques, such as the chain and product rules, as seen in the steps of the original solution.
• The result is the derivative vector \(\mathbf{r}'(t) = 0 \mathbf{i} + 12 \cos(3t) \mathbf{j} + (\cos(t) - t \sin(t)) \mathbf{k}\).

By breaking down the vector function into manageable parts, vector calculus offers an organized approach to find derivatives, enhancing the understanding of how these functions behave and interact over time.