Problem 17 - Advanced Engineering Mathematics

Question

Find \(\mathbf{r}^{\prime}(t)\) and \(\mathbf{r}^{\prime \prime}(t)\) for the given vector function. $$ \mathbf{r}(t)=\ln t \mathbf{i}+\mathbf{j}, t>0 $$

Step-by-Step Solution

Answer

\( \mathbf{r}^{\prime}(t) = \frac{1}{t} \mathbf{i} \); \( \mathbf{r}^{\prime \prime}(t) = -\frac{1}{t^2} \mathbf{i} \).

Step 1: Identify the components of the vector function

The given vector function is \( \mathbf{r}(t) = \ln t \mathbf{i} + \mathbf{j} \). This means it has two components: the \( \mathbf{i} \) component is \( \ln t \) and the \( \mathbf{j} \) component is \( 1 \).

Step 2: Differentiate with respect to t to find \( \mathbf{r}^{\prime}(t) \)

To find \( \mathbf{r}^{\prime}(t) \), differentiate each component of \( \mathbf{r}(t) \) separately. The derivative of \( \ln t \) with respect to \( t \) is \( \frac{1}{t} \), and the derivative of a constant \( 1 \) is \( 0 \). Thus, \( \mathbf{r}^{\prime}(t) = \frac{1}{t} \mathbf{i} + 0 \mathbf{j} \), which simplifies to \( \frac{1}{t} \mathbf{i} \).

Step 3: Differentiate again to find \( \mathbf{r}^{\prime \prime}(t) \)

To find \( \mathbf{r}^{\prime \prime}(t) \), differentiate \( \mathbf{r}^{\prime}(t) = \frac{1}{t} \mathbf{i} \). The derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \). Therefore, \( \mathbf{r}^{\prime \prime}(t) = -\frac{1}{t^2} \mathbf{i} \).

Key Concepts

Derivative of Vector Functions

Vector functions describe curves in a plane or space using a combination of vector components. For the function given in the exercise, \(\mathbf{r}(t) = \ln t \mathbf{i} + \mathbf{j}\), we have two distinct parts corresponding to the vectors \(\mathbf{i}\) and \(\mathbf{j}\).

The process of finding the derivative of a vector function involves dealing with each component separately:

• The \(\ln t\) component for \(\mathbf{i}\)
• The constant component for \(\mathbf{j}\)

Each part is differentiated with respect to \(t\). Let's look at why we handle these separately.
When differentiating a vector function like \(\mathbf{r}(t)\), we aim to describe how the curve changes as \(t\) changes. With this specific function, the change influenced by \(\ln t\) only impacts the \(\mathbf{i}\) direction, while the \(\mathbf{j}\) component remains unchanged, as it's constant.
This is why the derivative, \(\mathbf{r}^{\prime}(t)\), results in \(\frac{1}{t} \mathbf{i}\), effectively showing that only the \(\mathbf{i}\) direction varies with \(t\).

Natural Logarithm Differentiation

Differentiating the natural logarithm function is a fundamental concept in calculus. It applies to vector functions similarly to single-variable functions.

The natural logarithm of \(t\) (\(\ln t\)) is essential in understanding more complex functions. The rules for differentiating \(\ln t\) are straightforward:

• The derivative of \(\ln t\) with respect to \(t\) is simply \(\frac{1}{t}\).
• This rule comes from the chain rule where \(\ln u\) differentiates to \(\frac{1}{u} \cdot \frac{du}{dt}\).

This concise rule helps find how quickly the logarithmic function changes as the input \(t\) changes, an essential operation across calculus fields.
In our exercise, the logarithm part accounts for the entire change in the \(\mathbf{i}\) component's magnitude. Thus, understanding and applying this derivative is key to solving both ordinary and vector functions involving natural logarithms.

Second Derivative Calculation

Calculating the second derivative involves taking the derivative of the first derivative. In our vector function case, this means differentiating \(\mathbf{r}^{\prime}(t) = \frac{1}{t} \mathbf{i}\).

This process often shows the function's concavity or how the rate of change itself changes. Here's how to approach it:

• The first derivative of \(\ln t\) yielded \(\frac{1}{t}\), representing the rate of change from the natural log component.
• The derivative of \(\frac{1}{t}\), using power rule techniques, equates to \(-\frac{1}{t^2}\).

These derivatives show a step-by-step peeling away at how \(\mathbf{r}(t)\) evolves. In the vector context, finding \(\mathbf{r}^{\prime \prime}(t) = -\frac{1}{t^2} \mathbf{i}\) clarifies how the initial function's rate of change itself is altering. This allows deeper insights into the underlying behavior and influences on the curve represented by the vector function.