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Chapter 13: Problem 27

Solve the initial value problems in Exercises \(27-32\) for \(r\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=-t \mathbf{i}-t \mathbf{j}-t \mathbf{k}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}\end{array} $$

Short Answer

Expert verified
\( \mathbf{r}(t) = \left(-\frac{1}{2}t^2 + 1\right) \mathbf{i} + \left(-\frac{1}{2}t^2 + 2\right) \mathbf{j} + \left(-\frac{1}{2}t^2 + 3\right) \mathbf{k} \).

Step by step solution

01

Integrate the differential equation

The differential equation given is \( \frac{d \mathbf{r}}{dt} = -t \mathbf{i} - t \mathbf{j} - t \mathbf{k} \). To solve for \( \mathbf{r}(t) \), we need to integrate this equation with respect to \( t \). By integrating the components separately, we have:\[\mathbf{r}(t) = \int (-t \mathbf{i} - t \mathbf{j} - t \mathbf{k}) \, dt = \int -t \, dt \mathbf{i} + \int -t \, dt \mathbf{j} + \int -t \, dt \mathbf{k}\]Each integral, \( \int -t \, dt \), results in \( -\frac{1}{2}t^2 \). Thus, the integrated vector function is:\[\mathbf{r}(t) = -\frac{1}{2}t^2 \mathbf{i} - \frac{1}{2}t^2 \mathbf{j} - \frac{1}{2}t^2 \mathbf{k} + \mathbf{C}\]where \( \mathbf{C} \) is the integration constant vector.
02

Apply the initial condition

We apply the initial condition \( \mathbf{r}(0) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \) to determine the constant vector \( \mathbf{C} \).Substitute \( t = 0 \) into the integrated expression for \( \mathbf{r}(t) \):\[\mathbf{r}(0) = -\frac{1}{2}(0)^2 \mathbf{i} - \frac{1}{2}(0)^2 \mathbf{j} - \frac{1}{2}(0)^2 \mathbf{k} + \mathbf{C} = \mathbf{C}\]Since \( \mathbf{r}(0) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \), we have\[\mathbf{C} = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}\]Substituting \( \mathbf{C} \) back into the integrated vector function gives us:\[\mathbf{r}(t) = -\frac{1}{2}t^2 \mathbf{i} - \frac{1}{2}t^2 \mathbf{j} - \frac{1}{2}t^2 \mathbf{k} + \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}\]
03

Combine terms to express \( \mathbf{r}(t) \) fully

Finally, we combine like terms in the vector to express \( \mathbf{r}(t) \) more neatly:\[\mathbf{r}(t) = \left(-\frac{1}{2}t^2 + 1\right) \mathbf{i} + \left(-\frac{1}{2}t^2 + 2\right) \mathbf{j} + \left(-\frac{1}{2}t^2 + 3\right) \mathbf{k}\]This expression represents the vector function \( \mathbf{r}(t) \) that satisfies both the differential equation and the initial condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
In mathematics, a differential equation is an equation that relates a function to its derivatives. It serves as a fundamental tool for modeling dynamic systems in multiple fields such as physics, engineering, and economics. A typical differential equation will describe how a rate of change of a quantity is related to the quantity itself and other variables.
  • Order and Degree: The order of a differential equation is the highest order of derivative present in the equation. For instance, the given differential equation \( \frac{d \mathbf{r}}{dt} = -t \mathbf{i} - t \mathbf{j} - t \mathbf{k} \) is a first-order differential equation.
  • Linear vs Non-linear: Differential equations can be linear or non-linear. The presented equation is linear, involving terms that are proportional to the function or its derivatives.
Differential equations are solved using various techniques such as separation of variables, integrating factors, or numerical methods. In this context, we are dealing with a vector differential equation, which requires integrating the equation to find the function \( \mathbf{r}(t) \).
Initial Value Problems
An initial value problem (IVP) is a type of differential equation accompanied by a specific condition, called the initial condition, provided at the start. The initial condition is used to find particular solutions from a general solution.
  • Initial Conditions: These are values given for the solution at a specific point, usually at \( t = 0 \). For example, the initial condition for our problem is \( \mathbf{r}(0) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \).
  • Uniqueness of Solution: Usually, an IVP has a unique solution under specific conditions, meaning the initial condition will determine a unique function \( \mathbf{r}(t) \).
Solving an IVP involves integrating the differential equation first, then applying the initial condition to obtain a single, unique solution.
Integration of Vector Functions
Integration is a fundamental concept in calculus that allows us to find functions when their rates of change are known. In vector calculus, integrating a vector function involves integrating each of its components separately.
  • Integration of Components: For a vector function like \( \mathbf{r}(t) = -t \mathbf{i} - t \mathbf{j} - t \mathbf{k} \), integration involves handling each of the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) components individually.
  • Definite vs Indefinite Integration: While definite integration finds the area under a curve within limits, indefinite integration as used in this problem finds a family of solutions, incorporating an arbitrary constant.
The outcome of integration gives a general solution which then, through initial conditions, allows finding the specific solution to the problem.
Initial Conditions
Initial conditions are given values that specify the state of a solution at a particular point, commonly at the starting point of the problem. They are crucial because they allow the determination of the constants that arise during the integration process.
  • Role of Initial Conditions: When the vector function \( \mathbf{r}(t) = -\frac{1}{2}t^2 \mathbf{i} - \frac{1}{2}t^2 \mathbf{j} - \frac{1}{2}t^2 \mathbf{k} + \mathbf{C} \) is integrated, the constant vector \( \mathbf{C} \) arises.
  • Application of Initial Conditions: By applying \( \mathbf{r}(0) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \), we can solve for \( \mathbf{C} \), finding it to be \( \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \).
These specific values allow the conversion of a general solution to a particular solution by specifying the exact status of a system at an initial time.

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Most popular questions from this chapter

In Exercises 16 and \(17,\) two planets, planet \(A\) and planet \(B\) , are orbiting their sun in circular orbits with \(A\) being the inner planet and \(B\) being farther away from the sun. Suppose the positions of \(A\) and \(B\) at time \(t\) are $$ \mathbf{r}_{A}(t)=2 \cos (2 \pi t) \mathbf{i}+2 \sin (2 \pi t) \mathbf{j} $$ and $$ \mathbf{r}_{B}(t)=3 \cos (\pi t) \mathbf{i}+3 \sin (\pi t) \mathbf{j} $$ respectively, where the sun is assumed to be located at the origin and distance is measured in astronomical units. (Notice that planet \(A\) moves faster than planet \(B . )\) The people on planet \(A\) regard their planet, not the sun, as the center of their planetary system (their solar system). Using planet \(A\) as the origin, graph the path of planet \(B .\) This exercise illustrates the difficulty that people before Kepler's time, with an earth- centered (planet \(A\) ) view of our solar system, had in understanding the motions of the planets (i.e., planet \(B=\) Mars). See D. G. Saari's article in the American Mathematical Monthly, Vol. 97 (Feb. \(1990 ),\) pp. \(105-119\) .

Use a CAS to perform the following steps in Exercises \(58-61 .\) a. Plot the space curve traced out by the position vector \(\mathbf{r}\) . b. Find the components of the velocity vector \(d \mathbf{r} / d t\) . c. Evaluate \(d \mathbf{r} / d t\) at the given point \(t_{0}\) and determine the equation of the tangent line to the curve at \(\mathbf{r}\left(t_{0}\right) .\) d. Plot the tangent line together with the curve over the given interval. $$ \begin{array}{l}{\mathbf{r}(t)=(\sin t-t \cos t) \mathbf{i}+(\cos t+t \sin t) \mathbf{j}+t^{2} \mathbf{k}} \\ {0 \leq t \leq 6 \pi, \quad t_{0}=3 \pi / 2}\end{array} $$

In Exercises \(11-14\) , find the arc length parameter along the curve from the point where \(t=0\) by evaluating the integral $$ s=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau $$ from Equation ( \(3 ) .\) Then find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(\cos t+t \sin t) \mathbf{i}+(\sin t-t \cos t) \mathbf{j}, \quad \pi / 2 \leq t \leq \pi $$

In Exercises \(11-14,\) write a in the form \(\mathbf{a}=a_{\mathrm{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N}\) at the given value of \(t\) without finding \(\mathbf{T}\) and \(\mathbf{N} .\) $$ \mathbf{r}(t)=(t+1) \mathbf{i}+2 t \mathbf{j}+t^{2} \mathbf{k}, \quad t=1 $$

Use a CAS to perform the following steps in Exercises \(58-61 .\) a. Plot the space curve traced out by the position vector \(\mathbf{r}\) . b. Find the components of the velocity vector \(d \mathbf{r} / d t\) . c. Evaluate \(d \mathbf{r} / d t\) at the given point \(t_{0}\) and determine the equation of the tangent line to the curve at \(\mathbf{r}\left(t_{0}\right) .\) d. Plot the tangent line together with the curve over the given interval. $$ \begin{array}{l}{\mathbf{r}(t)=\left(\ln \left(t^{2}+2\right)\right) \mathbf{i}+\left(\tan ^{-1} 3 t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k}} \\\ {-3 \leq t \leq 5, \quad t_{0}=3}\end{array} $$
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