Question

Solve the given initial value problems. Find \(\vec{r}(t),\) given that \(\vec{r}^{\prime \prime}(t)=\left\langle\cos t, \sin t, e^{t}\right\rangle,\) \(\vec{r}^{\prime}(0)=\langle 0,0,0\rangle\) and \(\vec{r}(0)=\langle 0,0,0\rangle\)

Short answer

The solution is \(\vec{r}(t) = \langle -\cos t + 1, -\sin t + t, e^{t} - t - 1 \rangle\).

Step-by-step solution

Understand the Problem

We are given a second derivative of a vector function \(\vec{r}^{\prime \prime}(t) = \langle \cos t, \sin t, e^{t} \rangle\) and need to find \(\vec{r}(t)\) using the initial conditions \(\vec{r}^{\prime}(0) = \langle 0,0,0 \rangle\) and \(\vec{r}(0) = \langle 0,0,0 \rangle\).

Integrate to Find the First Derivative

Find \(\vec{r}^{\prime}(t)\) by integrating \(\vec{r}^{\prime \prime}(t)\):\[ \vec{r}^{\prime}(t) = \int \langle \cos t, \sin t, e^{t} \rangle \, dt = \langle \sin t, -\cos t, e^{t} \rangle + \vec{C}_1 \]where \(\vec{C}_1\) is the vector of integration constants \(\langle C_{1x}, C_{1y}, C_{1z} \rangle\).

Apply Initial Condition for First Derivative

Use \(\vec{r}^{\prime}(0) = \langle 0,0,0 \rangle\) to find \(\vec{C}_1\):\[ \langle \sin(0), -\cos(0), e^{0} \rangle + \vec{C}_1 = \langle 0,0,0 \rangle \]\[ \langle 0, -1, 1 \rangle + \langle C_{1x}, C_{1y}, C_{1z} \rangle = \langle 0,0,0 \rangle \]Solving, we get \(\vec{C}_1 = \langle 0, 1, -1 \rangle\).

Integrate to Find the Position Vector

Find \(\vec{r}(t)\) by integrating \(\vec{r}^{\prime}(t)\):\[ \vec{r}(t) = \int \langle \sin t, -\cos t + 1, e^{t} - 1 \rangle \, dt = \langle -\cos t, -\sin t + t, e^{t} - t \rangle + \vec{C}_2 \]where \(\vec{C}_2\) is another vector of integration constants \(\langle C_{2x}, C_{2y}, C_{2z} \rangle\).

Apply Initial Condition for Position Vector

Use \(\vec{r}(0) = \langle 0,0,0 \rangle\) to find \(\vec{C}_2\):\[ \langle -\cos(0), -\sin(0) + 0, e^{0} - 0 \rangle + \langle C_{2x}, C_{2y}, C_{2z} \rangle = \langle 0,0,0 \rangle \]\[ \langle -1, 0, 1 \rangle + \langle C_{2x}, C_{2y}, C_{2z} \rangle = \langle 0,0,0 \rangle \]Solving, we get \(\vec{C}_2 = \langle 1, 0, -1 \rangle\).

Write the Final Vector Function

Combine results from the integrations and initial conditions to write:\[ \vec{r}(t) = \langle -\cos t + 1, -\sin t + t, e^{t} - t - 1 \rangle \]

Key concepts

Initial Value Problems

Initial value problems are a key concept in differential equations and vector calculus. They involve finding a function that satisfies the given differential equation and specific values, known as initial conditions, at a certain point. In our exercise, we have to find a vector function \(\vec{r}(t)\) given its second derivative \(\vec{r}^{\prime \prime}(t)\) and initial conditions \(\vec{r}^{\prime}(0) = \langle 0,0,0 \rangle\) and \(\vec{r}(0) = \langle 0,0,0 \rangle\).
These initial conditions are crucial as they determine the specific solution from a family of possible solutions. Without these conditions, we could only determine a general solution, which would contain arbitrary constants. By solving such problems, students learn to find explicit functions that model real-world phenomena given initial states, making these problems immensely practical.

Second Derivative

The second derivative is a measure of how a function's rate of change is itself changing. In terms of vector functions, the second derivative \(\vec{r}^{\prime \prime}(t)\) indicates how the direction or speed of movement changes over time. For the given problem, \(\vec{r}^{\prime \prime}(t) = \langle \cos t, \sin t, e^{t} \rangle\), each component of the vector describes different behaviors in the velocity changes:

• \(\cos t\) and \(\sin t\) indicate oscillatory changes typical of circular or harmonic motion.
• \(e^t\) represents exponential growth, aligning with increasing acceleration in that direction.

Understanding the second derivative helps in predicting future states of the system by understanding its current acceleration trajectory.

Integration

Integration is the inverse operation to differentiation. To solve our initial value problem, we need to integrate the given second derivative to find the first derivative \(\vec{r}^{\prime}(t)\) and then integrate again to find the original function \(\vec{r}(t)\).
Through integration, we sum the changes to recover original quantities that have been differentiated. Each integration step introduces an arbitrary constant, which is where the initial conditions come into play, allowing us to determine these constants. For instance, integrating \( \langle \cos t, \sin t, e^{t} \rangle \) yields the first derivative \(\vec{r}^{\prime}(t)\), and we use the condition \(\vec{r}^{\prime}(0) = \langle 0,0,0 \rangle\) to find the constant vector \(\vec{C}_1\).
This method is essential for reconstructing functions in physics, engineering, and other fields where a known rate of change must be connected back to original conditions.

Vector Functions

Vector functions allow us to describe multi-dimensional systems in a compact and comprehensive manner. They express quantities like position, velocity, and acceleration in terms of vectors, encapsulating multiple dimensions within a single function.
For our specific problem, the vector function \(\vec{r}(t)\) describes the position of a point in three-dimensional space over time. Its components \(\langle x(t), y(t), z(t) \rangle\) are functions of time representing movements along the x, y, and z axes, respectively. By solving the given differential equation with initial conditions, we obtain \[ \vec{r}(t) = \langle -\cos t + 1, -\sin t + t, e^{t} - t - 1 \rangle \].
This function gives us a detailed picture of how the position changes over time, highlighting the unique path the object follows given its specific initial conditions.