Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(\mathbf{r}^{\prime}(t)\) and \(\mathbf{r}^{\prime \prime}(t)\) for each of the following: (a) \(\mathbf{r}(t)=\left(e^{t}+e^{-t^{2}}\right) \mathbf{i}+2^{t} \mathbf{j}+t \mathbf{k}\) (b) \(\mathbf{r}(t)=\tan 2 t \mathbf{i}+\arctan t \mathbf{j}\)

Step-by-Step Solution

Answer

(a) \( \mathbf{r}^{\prime}(t) = (e^t - 2te^{-t^2}) \mathbf{i} + 2^t \ln 2 \mathbf{j} + \mathbf{k} \). \( \mathbf{r}^{\prime \prime}(t) = (e^t + 4t^2e^{-t^2} - 2e^{-t^2}) \mathbf{i} + 2^t (\ln 2)^2 \mathbf{j} \). (b) \( \mathbf{r}^{\prime}(t) = 2\sec^2 2t \mathbf{i} + \frac{1}{1+t^2} \mathbf{j} \), \( \mathbf{r}^{\prime \prime}(t) = 4\sec^2 2t \tan 2t \mathbf{i} - \frac{2t}{(1+t^2)^2} \mathbf{j} \).

Step 1: Understand the Problem

We are asked to find the first and second derivatives, \( \mathbf{r}^{\prime}(t) \) and \( \mathbf{r}^{\prime \prime}(t) \), of the vector functions provided. This involves differentiating the components of each function with respect to \( t \).

Step 2: Differentiate Part (a)

For the vector function \( \mathbf{r}(t) = \left(e^{t} + e^{-t^{2}}\right) \mathbf{i} + 2^{t} \mathbf{j} + t \mathbf{k} \):- Differentiate the \( \mathbf{i} \)-component: \( \frac{d}{dt}(e^{t} + e^{-t^{2}}) = e^t - 2te^{-t^2} \).- Differentiate the \( \mathbf{j} \)-component: \( \frac{d}{dt}(2^{t}) = 2^{t} \ln 2 \).- Differentiate the \( \mathbf{k} \)-component: \( \frac{d}{dt}(t) = 1 \).Thus, \( \mathbf{r}^{\prime}(t) = \left(e^{t} - 2te^{-t^{2}}\right) \mathbf{i} + 2^{t} \ln 2 \mathbf{j} + 1 \mathbf{k} \).

Step 3: Differentiate Again for Part (a)

Calculate the second derivative \( \mathbf{r}^{\prime \prime}(t) \) by differentiating \( \mathbf{r}^{\prime}(t) \):- Differentiate \( e^{t} - 2te^{-t^{2}} \): use product rule for \(-2te^{-t^{2}}\): \( e^{t} + 4t^2e^{-t^{2}} - 2e^{-t^{2}} \).- Differentiate \( 2^{t} \ln 2 \): \( 2^t (\ln 2)^2 \).- The last component is constant: \( \frac{d}{dt}(1) = 0 \).Therefore, \( \mathbf{r}^{\prime \prime}(t) = \left(e^{t} + 4t^2e^{-t^{2}} - 2e^{-t^{2}}\right) \mathbf{i} + 2^{t} (\ln 2)^2 \mathbf{j} + 0 \mathbf{k} \).

Step 4: Differentiate Part (b)

For \( \mathbf{r}(t) = \tan 2t \mathbf{i} + \arctan t \mathbf{j} \):- Differentiate the \( \mathbf{i} \)-component: \( \frac{d}{dt}(\tan 2t) = 2 \sec^2 2t \).- Differentiate the \( \mathbf{j} \)-component: \( \frac{d}{dt}(\arctan t) = \frac{1}{1 + t^2} \).Thus, \( \mathbf{r}^{\prime}(t) = 2 \sec^2 2t \mathbf{i} + \frac{1}{1 + t^2} \mathbf{j} \).

Step 5: Differentiate Again for Part (b)

Calculate \( \mathbf{r}^{\prime \prime}(t) \) by differentiating the first derivative:- For \( 2 \sec^2 2t \), use chain rule: \( 4 \sec^2 2t \tan 2t \).- For \( \frac{1}{1 + t^2} \), use quotient rule: \( - \frac{2t}{(1 + t^2)^2} \).Thus, \( \mathbf{r}^{\prime \prime}(t) = 4 \sec^2 2t \tan 2t \mathbf{i} - \frac{2t}{(1 + t^2)^2} \mathbf{j} \).

Key Concepts

Vector Differentiation

Vector differentiation is an essential concept in vector calculus, primarily concerning the differentiation of vector-valued functions with respect to a scalar variable. Imagine a vector function \(\mathbf{r}(t)\) that describes the path of a moving object in space with components in the \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) directions. Differentiating a vector function means finding the rate of change of each of these components with respect to the variable \(t\).

• The resulting derivative of a vector function is also a vector, which often represents velocity or the directional change of the original function.
• This process involves applying standard rules of differentiation to each component function individually.
• Understanding vector differentiation lays the foundation for more complex applications such as vector fields and flows in physics and engineering.

Applying vector differentiation accurately and understanding its principles helps in solving real-world problems by interpreting the dynamic behavior of vectors in systems modeled by vector functions.

First Derivative

The first derivative of a vector function, often noted as \(\mathbf{r}^{\prime}(t)\), reveals the instantaneous rate of change of each component of the vector with respect to \(t\). It provides important information such as the velocity of an object if \(\mathbf{r}(t)\) represents its position. To find it, differentiate each component of \(\mathbf{r}(t)\):

• Apply the derivative rules like the chain rule, product rule, or power rule to each component of the vector separately.
• In part (a) of our problem: For the component \(e^{t} + e^{-t^{2}}\), differentiate to get \(e^t - 2te^{-t^2}\).
• Each differentiated component is then reassembled into a new vector \(\mathbf{r}^{\prime}(t)\), giving the velocity vector.

This step is crucial in understanding how a vector changes direction and magnitude over time, providing insight into dynamic systems.

Second Derivative

The second derivative of a vector function, represented by \(\mathbf{r}^{\prime \prime}(t)\), provides the rate of change of the first derivative and often represents acceleration if considering motion. Just like finding the first derivative, the process involves taking the derivative of each already-differentiated component:

• This can involve more advanced rules, such as product or quotient rules, requiring careful handling.
• In our example exercise, for \(e^{t} - 2te^{-t^2}\), using the product rule for the second term gives \(e^{t} + 4t^2e^{-t^2} - 2e^{-t^2}\).
• The second derivative vector gives you an even deeper understanding of how the dynamics of the system evolve over time.

The second derivative is fundamental in physics and engineering when analyzing systems subject to forces, showing how the effect of forces changes the velocity of the system.

Component-wise Differentiation

Component-wise differentiation is the method used when differentiating vector functions and involves treating each component independently. Each component's derivative is calculated as if it were a single-variable function. Here’s how it works:

• While differentiating vectors, each component needs individual attention to ensure precise calculation.
• In example (b), the \(\mathbf{i}\) component \(\tan 2t\) results in \(2 \sec^2 2t\) by applying the chain rule.
• The same rule applies to the \(\mathbf{j}\) component that becomes \(\frac{1}{1 + t^2}\) for \(\arctan t\).

This approach allows for stepwise solving, ensuring each derivative is correctly calculated and contributes to the complete understanding of vectors' roles in various applications.