Problem 17 - Advanced Engineering Mathematics

Question

In Problems, 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

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

Step 1: Differentiate the First Component

To find the derivative of a vector function \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} \), we differentiate each component separately. The first component is \( f(t) = \ln t \). The derivative of \( \ln t \) with respect to \( t \) is \( \frac{1}{t} \).

Step 2: Differentiate the Second Component

The second component of the vector function is \( g(t) = 1 \), since \( \mathbf{j} \) represents a constant vector. The derivative of a constant is 0.

Step 3: Combining Derivatives for \( \mathbf{r}^{\prime}(t) \)

Combine the derivatives found for each component to find the first derivative of the vector function: \( \mathbf{r}^{\prime}(t) = \frac{1}{t} \mathbf{i} + 0 \mathbf{j} = \frac{1}{t} \mathbf{i} \).

Step 4: Differentiate \( \mathbf{r}^{\prime}(t) \)

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

Step 5: Combine for \( \mathbf{r}^{\prime \prime}(t) \)

The derivative of a vector with only one component becomes \( \mathbf{r}^{\prime \prime}(t) = -\frac{1}{t^2} \mathbf{i} + 0 \mathbf{j} = -\frac{1}{t^2} \mathbf{i} \).

Key Concepts

Vector Derivatives

In vector calculus, understanding how to differentiate vector functions is crucial. A vector function can be thought of as a function that takes in a scalar (such as time, in this case) and outputs a vector. Differentiating a vector function is analogous to differentiating scalar functions, but it's done component-wise. For instance, consider the vector function \( \mathbf{r}(t) = \ln t \mathbf{i} + \mathbf{j} \). Here, \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors pointing in the "x" and "y" directions respectively.

• The derivative \( \mathbf{r}^{\prime}(t) \) is found by taking the derivative of each component separately. This means that if \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} \), then \( \mathbf{r}^{\prime}(t) = f^{\prime}(t) \mathbf{i} + g^{\prime}(t) \mathbf{j} \).
• It's important to remember that this derivative respects both the magnitude and direction changes according to each component.

Learning to differentiate vector functions like this helps in various applications, including physics and engineering, where vector quantities like velocity and acceleration are considered.

Component Differentiation

Component differentiation is a process where we differentiate each element of a vector function independently. In the given problem, we dealt with a vector function \( \mathbf{r}(t) = \ln t \mathbf{i} + 1 \mathbf{j} \).

• The first component \( f(t) = \ln t \) involves calculating its derivative using known calculus rules. The derivative, \( f^{\prime}(t) \), is \( \frac{1}{t} \).
• The second component, which is \( g(t) = 1 \), is already a constant, so its derivative, \( g^{\prime}(t) \), is 0, as the derivative of a constant is always zero.

For further ease, remember that each component is treated independently, and the results are combined to form the new vector derivative. This concept simplifies the sometimes daunting task of working with vector functions by breaking them into more manageable parts.

Power Rule

The power rule is a fundamental tool in calculus used to differentiate functions of the form \( x^n \). It states that the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \). This rule is quite versatile and not limited to integer powers.In the context of vector calculus, especially when dealing with functions like \( \frac{1}{t} \), which can be rewritten as \( t^{-1} \), the power rule is again applied. Here:

• Differentiating \( \frac{1}{t} \) or \( t^{-1} \) using the power rule gives \( -t^{-2} \) or \( -\frac{1}{t^2} \).
• This rule applies similarly regardless of whether the function forms part of a vector or is standalone, making it vital for finding vector derivatives multiple times efficiently.

Being conversant with the power rule not only eases solving scalar problems but extends to vector calculations, an essential skillset for students advancing in vector calculus.