Problem 18 - Advanced Engineering Mathematics

Question

In Problems, find \(\mathbf{r}^{\prime}(t)\) and \(\mathbf{r}^{\prime \prime}(t)\) for the given vector function. \(\mathbf{r}(t)=\langle t \cos t-\sin t, t+\cos t\rangle\)

Step-by-Step Solution

Answer

\( \mathbf{r}^{\prime}(t) = \langle -t \sin t, 1 - \sin t \rangle \) and \( \mathbf{r}^{\prime \prime}(t) = \langle -\sin t - t \cos t, -\cos t \rangle \).

Step 1: Understand the Given Vector Function

The vector function given is \( \mathbf{r}(t) = \langle t \cos t - \sin t, t + \cos t \rangle \). This is a two-dimensional vector where \( x(t) = t \cos t - \sin t \) and \( y(t) = t + \cos t \). Our task is to find the first and second derivatives of this vector function with respect to \( t \).

Step 2: Differentiate to Find \( \mathbf{r}^{\prime}(t) \)

To find the first derivative \( \mathbf{r}^{\prime}(t) \), differentiate each component of the vector separately with respect to \( t \). - For \( x(t) = t \cos t - \sin t \), use the product rule for differentiation: \( \frac{d}{dt}[t \cos t] = \cos t - t \sin t \). Then differentiate \( -\sin t \) to get \( -\cos t \). Combining these, the derivative is \( \cos t - t \sin t - \cos t = -t \sin t \).- For \( y(t) = t + \cos t \), differentiate to get \( 1 - \sin t \).Therefore, \( \mathbf{r}^{\prime}(t) = \langle -t \sin t, 1 - \sin t \rangle \).

Step 3: Differentiate Again to Find \( \mathbf{r}^{\prime \, \prime}(t) \)

Now, differentiate \( \mathbf{r}^{\prime}(t) = \langle -t \sin t, 1 - \sin t \rangle \) again to find the second derivative \( \mathbf{r}^{\prime \, \prime}(t) \).- For \( -t \sin t \), use the product rule again: \( \frac{d}{dt}[-t \sin t] = -\sin t - t \cos t \).- For \( 1 - \sin t \), differentiate to get \( -\cos t \).Thus, \( \mathbf{r}^{\prime \, \prime}(t) = \langle -\sin t - t \cos t, -\cos t \rangle \).

Step 4: Compile the Results

We've found both derivatives: First derivative: \( \mathbf{r}^{\prime}(t) = \langle -t \sin t, 1 - \sin t \rangle \) Second derivative: \( \mathbf{r}^{\prime \, \prime}(t) = \langle -\sin t - t \cos t, -\cos t \rangle \).

Key Concepts

Differentiation of Vector Functions

Vector functions are essential when studying motion along a path in space. These functions often involve calculations that include differentiating them with respect to a variable like time (often denoted as \( t \)). Differentiating a vector function generally means finding the derivatives of its individual components.In our example, we explore this with the vector function \( \mathbf{r}(t) = \langle t \cos t - \sin t, t + \cos t \rangle \). Differentiation here involves:

• Identifying each component of the vector function, primarily \( x(t) \) and \( y(t) \).
• Using differentiation rules like the product rule and chain rule as needed for each component.

Understanding these steps helps in calculating the rate of change of the vector with respect to \( t \), making it crucial in fields like physics and engineering, where vector functions model real-world phenomena such as velocity and acceleration.

Product Rule in Calculus

The product rule in calculus is a fundamental principle used when differentiating expressions that are products of two functions. When dealing with vector functions, the product rule becomes indispensable, especially for components involving multiplied terms.For instance, consider the function within the vector \( \, x(t) = t \cos t - \sin t \.\) The term \( t \cos t \) is a product of \( t \) and \( \cos t \), so we use the product rule to differentiate it:

• Derive \( t \) with respect to \( t \) to get 1, multiply that by \( \cos t \)
• Then, derive \( \cos t \) to get \( -\sin t \), and multiply by \( t \)

Putting it together, the derivative becomes \( \cos t - t \sin t \). Remembering rules like these make differentiation more manageable and prevent mistakes that could arise from oversight.

First and Second Derivatives

Differentiation gives rise to two main results for vector functions - the first and second derivatives. Each has distinct roles and interpretations, especially when studying dynamic systems.The first derivative, \( \mathbf{r}^{\prime}(t) \), involves differentiating each component separately. In our function, it yields:

• \( -t \sin t \) for the x-component
• \( 1 - \sin t \) for the y-component

This gives us a vector that can represent the velocity of a moving object if \( \mathbf{r}(t) \) symbolizes position over time.

The second derivative, \( \mathbf{r}^{\prime \prime}(t) \), is the derivative of \( \mathbf{r}^{\prime}(t) \). It results in:

• \( -\sin t - t \cos t \) for the x-component
• \( -\cos t \) for the y-component

Second derivatives are essential in physics because they often represent acceleration. They provide insights into how the velocity of a vector function, such as motion, is changing over time.