Math Problem Statement

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

Solution

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

Given:

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

Step 1: Find the Velocity Vector v(t)\mathbf{v}(t)

  1. Integrate each component of a(t)\mathbf{a}(t) with respect to tt: vi(t)=17tdt=172t2+Civ_i(t) = \int 17t \, dt = \frac{17}{2}t^2 + C_i vj(t)=etdt=et+Cjv_j(t) = \int e^t \, dt = e^t + C_j vk(t)=etdt=et+Ckv_k(t) = \int e^{-t} \, dt = -e^{-t} + C_k

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

    • vi(0)=0v_i(0) = 0
    • vj(0)=0v_j(0) = 0
    • vk(0)=1v_k(0) = 1

    Solving for CiC_i, CjC_j, and CkC_k:

    • For vi(0)=0v_i(0) = 0: Ci=0C_i = 0.
    • For vj(0)=0v_j(0) = 0: Cj=1C_j = -1.
    • For vk(0)=1v_k(0) = 1: Ck=2C_k = 2.

    So, the velocity vector is: v(t)=(172t2)i+(et1)j+(et+2)k\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 r(t)\mathbf{r}(t)

  1. Integrate each component of v(t)\mathbf{v}(t) with respect to tt: ri(t)=172t2dt=176t3+Dir_i(t) = \int \frac{17}{2}t^2 \, dt = \frac{17}{6}t^3 + D_i rj(t)=(et1)dt=ett+Djr_j(t) = \int (e^t - 1) \, dt = e^t - t + D_j rk(t)=(et+2)dt=et+2t+Dkr_k(t) = \int (-e^{-t} + 2) \, dt = -e^{-t} + 2t + D_k

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

    • ri(0)=0r_i(0) = 0
    • rj(0)=1r_j(0) = 1
    • rk(0)=1r_k(0) = 1

    Solving for DiD_i, DjD_j, and DkD_k:

    • For ri(0)=0r_i(0) = 0: Di=0D_i = 0.
    • For rj(0)=1r_j(0) = 1: Dj=1D_j = 1.
    • For rk(0)=1r_k(0) = 1: Dk=2D_k = 2.

    Thus, the position vector is: r(t)=176t3i+(ett+1)j+(et+2t+2)k\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 tt 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?


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.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Kinematics
Integration

Formulas

v(t) = ∫ a(t) dt
r(t) = ∫ v(t) dt

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate (Calculus level)