Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. Then on your own using a computer, graph the path of the particle.
a(t) = 17t i + et j + e−t k,    v(0) = k,    r(0) = j + k

To find the position vector \( \mathbf{r}(t) \) of the particle, we start by integrating the acceleration vector \( \mathbf{a}(t) \) twice, using the given initial velocity \( \mathbf{v}(0) \) and initial position \( \mathbf{r}(0) \).

Given:
- Acceleration: \( \mathbf{a}(t) = 17t \, \mathbf{i} + e^t \, \mathbf{j} + e^{-t} \, \mathbf{k} \)
- Initial velocity: \( \mathbf{v}(0) = \mathbf{k} \)
- Initial position: \( \mathbf{r}(0) = \mathbf{j} + \mathbf{k} \)

### Step 1: Find the Velocity Vector \( \mathbf{v}(t) \)
1. Integrate each component of \( \mathbf{a}(t) \) with respect to \( t \):
   \[
   v_i(t) = \int 17t \, dt = \frac{17}{2}t^2 + C_i
   \]
   \[
   v_j(t) = \int e^t \, dt = e^t + C_j
   \]
   \[
   v_k(t) = \int e^{-t} \, dt = -e^{-t} + C_k
   \]

2. Apply the initial condition \( \mathbf{v}(0) = \mathbf{k} \), which means:
   - \( v_i(0) = 0 \)
   - \( v_j(0) = 0 \)
   - \( v_k(0) = 1 \)

   Solving for \( C_i \), \( C_j \), and \( C_k \):
   - For \( v_i(0) = 0 \): \( C_i = 0 \).
   - For \( v_j(0) = 0 \): \( C_j = -1 \).
   - For \( v_k(0) = 1 \): \( C_k = 2 \).

   So, the velocity vector is:
   \[
   \mathbf{v}(t) = \left( \frac{17}{2}t^2 \right) \mathbf{i} + (e^t - 1) \mathbf{j} + (-e^{-t} + 2) \mathbf{k}
   \]

### Step 2: Find the Position Vector \( \mathbf{r}(t) \)
1. Integrate each component of \( \mathbf{v}(t) \) with respect to \( t \):
   \[
   r_i(t) = \int \frac{17}{2}t^2 \, dt = \frac{17}{6}t^3 + D_i
   \]
   \[
   r_j(t) = \int (e^t - 1) \, dt = e^t - t + D_j
   \]
   \[
   r_k(t) = \int (-e^{-t} + 2) \, dt = -e^{-t} + 2t + D_k
   \]

2. Apply the initial condition \( \mathbf{r}(0) = \mathbf{j} + \mathbf{k} \), which gives:
   - \( r_i(0) = 0 \)
   - \( r_j(0) = 1 \)
   - \( r_k(0) = 1 \)

   Solving for \( D_i \), \( D_j \), and \( D_k \):
   - For \( r_i(0) = 0 \): \( D_i = 0 \).
   - For \( r_j(0) = 1 \): \( D_j = 1 \).
   - For \( r_k(0) = 1 \): \( D_k = 2 \).

   Thus, the position vector is:
   \[
   \mathbf{r}(t) = \frac{17}{6}t^3 \, \mathbf{i} + (e^t - t + 1) \, \mathbf{j} + (-e^{-t} + 2t + 2) \, \mathbf{k}
   \]

### Graphing the Path of the Particle
To visualize the path, we would typically plot each component as a function of \( t \) in 3D space, which would show the trajectory of the particle.

Would you like further details, or do you have any questions about this problem?

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**Follow-up Questions for Further Exploration:**
1. How do we interpret each component of the position vector in terms of particle motion?
2. How would the path change if the initial conditions were different?
3. Can we determine the particle’s speed as a function of time?
4. How does the exponential term affect the particle's trajectory over time?
5. What would the particle’s trajectory look like if the acceleration were constant?

**Tip:** When finding position from acceleration, always integrate step-by-step and carefully apply initial conditions at each stage to avoid errors.

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. Then on your own using a computer, graph the path of the particle.
a(t) = 17t i + et j + e−t k, v(0) = k, r(0) = j + k

This problem involves finding the position vector of a particle given its acceleration vector a(t) = 17t i + e^t j + e^-t k and initial conditions v(0) = k and r(0) = j + k. Learn how to solve it using integration and graph the particle's path.

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