Chapter 15: Problem 33
Find \(r(t)\) subject to the given conditions. $$ \begin{aligned} \mathbf{r}^{\prime \prime}(t)=6 t \mathbf{i}-12 t^{2} \mathbf{j}+\mathbf{k} & \\\ \mathbf{r}^{\prime}(0)=\mathbf{i}+2 \mathbf{j}-3 \mathbf{k}, & \mathbf{r}(0)=7 \mathbf{i}+\mathbf{k} \end{aligned} $$
Short Answer
Expert verified
The position function is \(\mathbf{r}(t) = (t^3 + t + 7) \mathbf{i} - (t^4 - 2t) \mathbf{j} + (\frac{1}{2}t^2 - 3t) \mathbf{k}.\)
Step by step solution
01
Integrate Acceleration to Find Velocity
Start by integrating the acceleration vector \( \mathbf{r}^{\prime \prime}(t) = 6t \mathbf{i} - 12t^{2} \mathbf{j} + \mathbf{k} \). The integration of each component gives us:\[\mathbf{r}^{\prime}(t) = \left(\int 6t \, dt\right) \mathbf{i} + \left(\int -12t^2 \, dt\right) \mathbf{j} + \left(\int 1 \, dt\right) \mathbf{k}\]This yields:\[\mathbf{r}^{\prime}(t) = (3t^2 + C_1) \mathbf{i} - \left( 4t^3 + C_2 \right) \mathbf{j} + (t + C_3) \mathbf{k}\]where \(C_1\), \(C_2\), and \(C_3\) are constants of integration.
02
Use Initial Conditions to Find Unknown Constants of Velocity
Apply the initial condition \( \mathbf{r}^{\prime}(0) = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \) to determine the constants:\[(3(0)^2 + C_1) \mathbf{i} - (4(0)^3 + C_2) \mathbf{j} + (0 + C_3) \mathbf{k} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\]Comparing components, we find:\( C_1 = 1 \), \( C_2 = -2 \), and \( C_3 = -3 \). Thus, the velocity function is:\[\mathbf{r}^{\prime}(t) = (3t^2 + 1) \mathbf{i} - (4t^3 - 2) \mathbf{j} + (t - 3) \mathbf{k}\]
03
Integrate Velocity to Find Position
Now, integrate the velocity function to find the position vector:\[\mathbf{r}(t) = \int (3t^2 + 1) \, dt \mathbf{i} + \int (-4t^3 - 2) \, dt \mathbf{j} + \int (t - 3) \, dt \mathbf{k}\]This results in:\[\mathbf{r}(t) = (t^3 + t + D_1) \mathbf{i} - \left(t^4 - 2t + D_2\right) \mathbf{j} + \left( \frac{1}{2}t^2 - 3t + D_3 \right) \mathbf{k}\]where \(D_1\), \(D_2\), and \(D_3\) are constants of integration.
04
Use Initial Conditions to Find Unknown Constants of Position
Apply the initial condition \( \mathbf{r}(0) = 7\mathbf{i} + \mathbf{k} \) to solve for the constants:\[(t^3 + t + D_1) \mathbf{i} - (t^4 - 2t + D_2) \mathbf{j} + (\frac{1}{2}t^2 - 3t + D_3) \mathbf{k} = 7 \mathbf{i} + \mathbf{k}\]At \( t = 0 \), comparing components gives:\( D_1 = 7 \), \( D_2 = 0 \), and \( D_3 = 0 \).Thus, the position function is:\[\mathbf{r}(t) = (t^3 + t + 7) \mathbf{i} - (t^4 - 2t) \mathbf{j} + (\frac{1}{2}t^2 - 3t) \mathbf{k}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
In vector calculus, integration is a fundamental concept used to find antiderivatives or to determine the area under a curve. When dealing with motion, we often integrate acceleration to find velocity, and integrate velocity to find position. Here, we're working with a vector function rather than a simple scalar function. This means we have to perform integration on each component of the vector separately. For example, consider the acceleration vector:
- \( 6t \mathbf{i} \)
- \( -12t^2 \mathbf{j} \)
- \( \mathbf{k} \)
Initial Conditions
Initial conditions are crucial for determining the specific solution to a differential equation. When we integrate, we introduce constants of integration; these constants make the solution of the problem more general. However, to find a particular solution, we use initial conditions. They are essentially starting values for variables at a specific time point. For example, we have initial conditions given for velocity and position:
- \( \mathbf{r}^{\prime}(0) = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \)
- \( \mathbf{r}(0) = 7\mathbf{i} + \mathbf{k} \)
Acceleration and Velocity
Acceleration is the rate of change of velocity over time. In vector terms, it's expressed as a vector function. For example, given:
- \( \mathbf{r}^{\prime\prime}(t) = 6t \mathbf{i} - 12t^2 \mathbf{j} + \mathbf{k} \)
- \( \mathbf{r}^{\prime}(t) = (3t^2 + C_1) \mathbf{i} - (4t^3 + C_2) \mathbf{j} + (t + C_3) \mathbf{k} \)
Constants of Integration
Constants of integration arise during the process of integrating a function. They are essential in determining the particular solution from a general solution. When we integrate a function, we essentially reverse differentiation, meaning any constant that might have been in the original function would have vanished upon differentiation. In the velocity example, when integrating
- \( 6t \mathbf{i} \),
- \( -12t^2 \mathbf{j} \),
- \( \mathbf{k} \)