Problem 20 - Advanced Engineering Mathematics

Question

Find \(\mathbf{r}^{\prime}(t)\) and \(\mathbf{r}^{\prime \prime}(t)\) for the given vector function. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+t^{3} \mathbf{j}+\tan ^{-1} t \mathbf{k} $$

Step-by-Step Solution

Answer

\( \mathbf{r}^{\prime}(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \frac{1}{1+t^2} \mathbf{k} \); \( \mathbf{r}^{\prime\prime}(t) = 2 \mathbf{i} + 6t \mathbf{j} - \frac{2t}{(1+t^2)^2} \mathbf{k} \).

Step 1: Differentiate the first component

The first component of the given vector function is \( t^2 \mathbf{i} \). To find its derivative, we differentiate \( t^2 \) with respect to \( t \). The derivative is \( 2t \), thus the derivative of the first component is \( 2t \mathbf{i} \).

Step 2: Differentiate the second component

The second component of the vector function is \( t^3 \mathbf{j} \). Differentiating \( t^3 \) with respect to \( t \) gives \( 3t^2 \). Therefore, the derivative of the second component is \( 3t^2 \mathbf{j} \).

Step 3: Differentiate the third component

The third component is \( \tan^{-1} t \mathbf{k} \). Its derivative is \( \frac{d}{dt}(\tan^{-1} t) = \frac{1}{1+t^2} \), making the derivative of the third component \( \frac{1}{1+t^2} \mathbf{k} \).

Step 4: Combine derivatives for \( \mathbf{r}^{\prime}(t) \)

Combine the derivatives from the previous steps: \( \mathbf{r}^{\prime}(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \frac{1}{1+t^2} \mathbf{k} \).

Step 5: Differentiate the first derivative's first component

Take the derivative of the first component of \( \mathbf{r}^{\prime}(t) \), which is \( 2t \). Differentiating gives \( 2 \mathbf{i} \).

Step 6: Differentiate the first derivative's second component

The second component, \( 3t^2 \), is differentiated to obtain \( 6t \). This makes the derivative \( 6t \mathbf{j} \).

Step 7: Differentiate the first derivative's third component

To differentiate \( \frac{1}{1+t^2} \), use the chain rule: \(-\frac{2t}{(1+t^2)^2} \). This yields \(-\frac{2t}{(1+t^2)^2} \mathbf{k} \).

Step 8: Combine second derivatives for \( \mathbf{r}^{\prime\prime}(t) \)

Combine the second set of derivatives: \( \mathbf{r}^{\prime\prime}(t) = 2 \mathbf{i} + 6t \mathbf{j} - \frac{2t}{(1+t^2)^2} \mathbf{k} \).

Key Concepts

Derivatives of Vector Functions

When dealing with vector functions, such as \( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} + \tan^{-1} t \mathbf{k} \), calculating derivatives involves breaking the vector into its individual components. Each component is treated as a separate scalar function, which can be differentiated with respect to the parameter \( t \). For this vector function:

• Differentiate the scalar component \( t^2 \mathbf{i} \). This results in \( 2t \mathbf{i} \).
• For \( t^3 \mathbf{j} \), differentiation gives \( 3t^2 \mathbf{j} \).
• The third component \( \tan^{-1} t \mathbf{k} \) is differentiated to yield \( \frac{1}{1+t^2} \mathbf{k} \).

By combining these results, we obtain the derivative of the entire vector function: \( \mathbf{r}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \frac{1}{1+t^2} \mathbf{k} \).
Taking derivatives in this way allows us to explore how each part of a vector function changes over time or with respect to another variable. It's an extension of the familiar process of differentiating scalar functions to handle functions with multiple dimensions.

Chain Rule in Calculus

The chain rule is a powerful tool when differentiating complicated functions involving compositions, especially in vector calculus. When we have a function lined inside another, like \( \tan^{-1} t \) in our vector function, the chain rule comes into play.
To differentiate \( \tan^{-1}(t) \) with respect to \( t \), we use the fact that if \( y = f(u) \) and \( u = g(t) \), then the derivative \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \). Applying this to \( \tan^{-1}(t) \):

• Let \( u = t \). Thus, \( \frac{du}{dt} = 1 \).
• Now, \( \frac{d}{du}(\tan^{-1}(u)) = \frac{1}{1+u^2} \).

Combining these gives \( \frac{d}{dt}(\tan^{-1}(t)) = \frac{1}{1+t^2} \).
This chain rule application enables us to deal with derivatives across compositions smoothly, an essential part of vector calculus operations.

Differentiation in Multivariable Calculus

In multivariable calculus, differentiation is not limited to equations with one variable; it often involves several variables, as seen in vector functions. Here, the focus isn't just on ordinary derivatives, but also on partial derivatives and how these components fit together.
Differentiating vector functions, such as \( \mathbf{r}(t) \), requires recognizing each part operates in its dimension:

• The partial derivative of \( t^2 \mathbf{i} \) is taken with respect to \( t \) alone, leading to \( 2t \mathbf{i} \).
• Similarly, \( t^3 \mathbf{j} \) becomes \( 3t^2 \mathbf{j} \), and \( \tan^{-1} t \mathbf{k} \) needs the chain rule to derive \( \frac{1}{1+t^2} \mathbf{k} \).

This step shows how each variable in a vector expression can be handled individually yet stays part of the entire vector calculation. As more complexity is added to problems, understanding differentiation in this multivariate context becomes crucial, especially for fields like physics and engineering.