Problem 40 - Advanced Engineering Mathematics

Question

Find a vector function \(\mathbf{r}\) that satisfies the indicated conditions. $$ \begin{aligned} &\mathbf{r}^{\prime \prime}(t)=\sec ^{2} t \mathbf{i}+\cos t \mathbf{j}-\sin t \mathbf{k} \\ &\mathbf{r}^{\prime}(0)=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{r}(0)=-\mathbf{j}+5 \mathbf{k} \end{aligned} $$

Step-by-Step Solution

Answer

\( \mathbf{r}(t) = (-\ln |\cos t| + t) \mathbf{i} + (-\cos t + t) \mathbf{j} + (-\sin t + 2t + 5) \mathbf{k} \).

Step 1: Integrate the Acceleration to Find Velocity

The given second derivative of the vector function is \( \mathbf{r}''(t) = \sec^2 t \mathbf{i} + \cos t \mathbf{j} - \sin t \mathbf{k} \). Integrate each component separately to find \( \mathbf{r}'(t) \).- The integral of \( \sec^2 t \) is \( \tan t \), so the \( \mathbf{i} \) component becomes \( \tan t \mathbf{i} \).- The integral of \( \cos t \) is \( \sin t \), so the \( \mathbf{j} \) component becomes \( \sin t \mathbf{j} \).- The integral of \( -\sin t \) is \( \cos t \), so the \( \mathbf{k} \) component becomes \( -\cos t \mathbf{k} \).Combine these results to get: \( \mathbf{r}'(t) = \tan t \mathbf{i} + \sin t \mathbf{j} - \cos t \mathbf{k} + \mathbf{C} \), where \( \mathbf{C} \) is a constant vector.

Step 2: Determine the Constant Vector for Velocity

We know \( \mathbf{r}'(0) = \mathbf{i} + \mathbf{j} + \mathbf{k} \). Substitute \( t = 0 \) into \( \mathbf{r}'(t) = \tan t \mathbf{i} + \sin t \mathbf{j} - \cos t \mathbf{k} + \mathbf{C} \) and set it equal to \( \mathbf{i} + \mathbf{j} + \mathbf{k} \):\[ \tan 0 \mathbf{i} + \sin 0 \mathbf{j} - \cos 0 \mathbf{k} + \mathbf{C} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]Simplifying, we have:\[ 0 \mathbf{i} + 0 \mathbf{j} - 1 \mathbf{k} + \mathbf{C} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]Thus, \( \mathbf{C} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \). Substitute back to get \( \mathbf{r}'(t) = \tan t \mathbf{i} + \sin t \mathbf{j} - \cos t \mathbf{k} + \mathbf{i} + \mathbf{j} + 2\mathbf{k} \).

Step 3: Integrate the Velocity to Find Position Vector

Integrate \( \mathbf{r}'(t) = (\tan t + 1) \mathbf{i} + (\sin t + 1) \mathbf{j} + (-\cos t + 2) \mathbf{k} \) to find \( \mathbf{r}(t) \).- The integral of \( \tan t + 1 \) is \( -\ln |\cos t| + t \), so the \( \mathbf{i} \) component is \( (-\ln |\cos t| + t) \mathbf{i} \).- The integral of \( \sin t + 1 \) is \( -\cos t + t \), so the \( \mathbf{j} \) component is \( (-\cos t + t) \mathbf{j} \).- The integral of \( -\cos t + 2 \) is \( -\sin t + 2t \), so the \( \mathbf{k} \) component is \( (-\sin t + 2t) \mathbf{k} \).Combine these results to get: \( \mathbf{r}(t) = (-\ln |\cos t| + t) \mathbf{i} + (-\cos t + t) \mathbf{j} + (-\sin t + 2t) \mathbf{k} + \mathbf{D} \), where \( \mathbf{D} \) is another constant vector.

Step 4: Determine the Constant Vector for Position

We know \( \mathbf{r}(0) = -\mathbf{j} + 5\mathbf{k} \). Substitute \( t = 0 \) into \( \mathbf{r}(t) = (-\ln |\cos t| + t) \mathbf{i} + (-\cos t + t) \mathbf{j} + (-\sin t + 2t) \mathbf{k} + \mathbf{D} \):\[ (-\ln |\cos 0| + 0) \mathbf{i} + (-\cos 0 + 0) \mathbf{j} + (-\sin 0 + 2 \cdot 0) \mathbf{k} + \mathbf{D} = -\mathbf{j} + 5 \mathbf{k} \]This simplifies to:\[ 0 \mathbf{i} - 1 \mathbf{j} + 0 \mathbf{k} + \mathbf{D} = -\mathbf{j} + 5 \mathbf{k} \]Thus, \( \mathbf{D} = 5\mathbf{k} \). Substitute back to get: \( \mathbf{r}(t) = (-\ln |\cos t| + t) \mathbf{i} + (-\cos t + t) \mathbf{j} + (-\sin t + 2t + 5) \mathbf{k} \).

Key Concepts

Integration

Integration is a fundamental concept in calculus, often used to find functions when given their derivatives. In the context of vector calculus, it allows us to discover velocity from acceleration and position from velocity. The process involves reversing the derivation to obtain the original function, along with any constants of integration that may arise.

For example, when given a vector acceleration

• such as \[\mathbf{r}''(t) = \sec^2 t \mathbf{i} + \cos t \mathbf{j} - \sin t \mathbf{k},\]

we integrate each component separately:

• \[\int \sec^2 t \, dt = \tan t + C_1\]
• \[\int \cos t \, dt = \sin t + C_2\]
• \[\int -\sin t \, dt = \cos t + C_3\]

This procedure generates a general solution for the velocity, \(\mathbf{r}'(t) = \tan t \mathbf{i} + \sin t \mathbf{j} - \cos t \mathbf{k} + \mathbf{C},\)where \( \mathbf{C} \)is a constant vector. The constant vector needs to be determined using additional conditions given in the problem.

Constant Vector

After integrating, often the process yields a undefined result due to integration constants which are involved. A constant vector encompasses these integration constants arising from the indefinite integration of vector components.

• The initial integration of the acceleration function gives us\[\mathbf{r}'(t) = \tan t \mathbf{i} + \sin t \mathbf{j} - \cos t \mathbf{k} + \mathbf{C}.\]

Here, \( \mathbf{C} \)is not fully known yet and present because each integrated component might have its integration constant.

• This undetermined constant vector will be later defined using initial conditions.

By plugging in the available initial conditions, we can solve for this constant vector. In our exercise, we determined \( \mathbf{C} \)to be \( \mathbf{i} + \mathbf{j} + 2\mathbf{k} \) through substitution techniques; determining this vector is essential to further our understanding and provide exact solutions.

Initial Conditions

Initial conditions are crucial in dynamics, greatly assisting in fully defining the motion of objects. They are specific known values at a given initial point (often time zero) used to determine unknown constants in integrated functions.

In the problem, initial conditions include the given:

• velocity at time \(t=0\),\[\mathbf{r}'(0) = \mathbf{i} + \mathbf{j} + \mathbf{k},\]
• and the position at time\(t=0\),\[\mathbf{r}(0) = -\mathbf{j} + 5 \mathbf{k}.\]

These values allow us to solve for the constant vector components found during integration. By substituting these initial values back into the general integrated function, the exact constants are found through algebraic manipulation. Thus, enabling the development of a precise description of the system’s motion.

Acceleration and Velocity

The concepts of acceleration and velocity are foundational in physics and calculus equations describing motion through differentials.

• Acceleration describes the rate of change of velocity with respect to time.
• Velocity is the rate at which an object's position changes, being the derivative of position with respect to time.

In our vector exercise, specific derivatives were given, \(\mathbf{r}''(t) \) to represent acceleration, and our objective was to find the initial velocity and position.

• We began with acceleration\[\mathbf{r}''(t) = \sec^2 t \mathbf{i} + \cos t \mathbf{j} - \sin t \mathbf{k},\]
• and applied integration to obtain velocity, ending with\[\mathbf{r}'(t) = \tan t \mathbf{i} + \sin t \mathbf{j} - \cos t \mathbf{k} + \mathbf{i} + \mathbf{j} + 2\mathbf{k}\]

as a combined result of the integration and initial conditions. Acceleration and velocity together provide a dynamic picture of the system’s evolution over time, making them key components in constructing precise models of real-world dynamics.