Problem 16 - Thomas Calculus

Question

Solve the initial value problemsfor \(\mathbf{r}\) as a vector function of \(t .\) $$\text {Differential equation:}\quad \frac{d^{2} \mathbf{r}}{d t^{2}}=-(\mathbf{i}+\mathbf{j}+\mathbf{k})$$ $$\text{Initial conditions:}\begin{array}{l}{\mathbf{r}(0)=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k} \text { and }} \\ {\left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=\mathbf{0}}\end{array}$$

Step-by-Step Solution

Answer

The solution is \(\mathbf{r}(t) = \left(10 - \frac{t^2}{2}\right)\mathbf{i} + \left(10 - \frac{t^2}{2}\right)\mathbf{j} + \left(10 - \frac{t^2}{2}\right)\mathbf{k}\).

Step 1: Solve the second order differential equation

The given differential equation is \(\frac{d^{2} \mathbf{r}}{dt^{2}}=-(\mathbf{i}+\mathbf{j}+\mathbf{k})\). To find \(\mathbf{r}(t)\), we first need to integrate the equation with respect to \(t\) twice. \[\frac{d \mathbf{r}}{dt} = \int -(\mathbf{i}+\mathbf{j}+\mathbf{k}) \, dt = -t(\mathbf{i}+\mathbf{j}+\mathbf{k}) + \mathbf{C_1}\] where \(\mathbf{C_1}\) is the constant of integration.

Step 2: Apply initial condition for velocity

The initial condition given is \(\left.\frac{d \mathbf{r}}{dt}\right|_{t=0} = \mathbf{0}\). Using this condition in the velocity equation: \[-0(\mathbf{i}+\mathbf{j}+\mathbf{k}) + \mathbf{C_1} = \mathbf{0} \] Thus, \(\mathbf{C_1} = \mathbf{0}\). Therefore, \[\frac{d \mathbf{r}}{dt} = -t(\mathbf{i}+\mathbf{j}+\mathbf{k})\]

Step 3: Find the position function \(\mathbf{r}(t)\)

Now integrate \(\frac{d \mathbf{r}}{dt}\) to find \(\mathbf{r}(t)\): \[\mathbf{r}(t) = \int -t(\mathbf{i}+\mathbf{j}+\mathbf{k}) \, dt = -\frac{t^2}{2}(\mathbf{i}+\mathbf{j}+\mathbf{k}) + \mathbf{C_2}\] where \(\mathbf{C_2}\) is another constant of integration.

Step 4: Apply initial position condition

Use the initial condition \(\mathbf{r}(0) = 10 \mathbf{i} + 10 \mathbf{j} + 10 \mathbf{k}\) to solve for \(\mathbf{C_2}\): \[-\frac{0^2}{2}(\mathbf{i}+\mathbf{j}+\mathbf{k}) + \mathbf{C_2} = 10 \mathbf{i} + 10 \mathbf{j} + 10 \mathbf{k}\] Thus, \(\mathbf{C_2} = 10 \mathbf{i} + 10 \mathbf{j} + 10 \mathbf{k}\). Therefore, \[\mathbf{r}(t) = -\frac{t^2}{2}(\mathbf{i}+\mathbf{j}+\mathbf{k}) + 10 \mathbf{i} + 10 \mathbf{j} + 10 \mathbf{k}\]

Step 5: Conclusion: Solution for \(\mathbf{r}(t)\)

The function \(\mathbf{r}(t)\) is given by \[\mathbf{r}(t) = \left(10 - \frac{t^2}{2}\right)\mathbf{i} + \left(10 - \frac{t^2}{2}\right)\mathbf{j} + \left(10 - \frac{t^2}{2}\right)\mathbf{k}\]. This is the solution to the initial value problem.

Key Concepts

Initial Value Problems

An initial value problem in mathematics consists of a differential equation that comes with a set of conditions referred to as initial conditions. The initial conditions are values of the solution and possibly its derivatives at a specific point, which help us determine the constants that appear in the general solution of the differential equation. In this exercise, the initial value problem is set for the vector function \( \mathbf{r}(t) \).

Here, the given differential equation is \( \frac{d^{2} \mathbf{r}}{dt^{2}}=-(\mathbf{i}+\mathbf{j}+\mathbf{k}) \).

The conditions provided are:

• \( \mathbf{r}(0) = 10 \mathbf{i} + 10 \mathbf{j} + 10 \mathbf{k} \)
• The velocity condition \( \left.\frac{d \mathbf{r}}{dt}\right|_{t=0} = \mathbf{0} \)

These initial conditions are crucial as they ensure that the determined function \( \mathbf{r}(t) \) is specific to this problem and not a general one formed out of multiple possibilities. Given these conditions, solving the initial value problem involves integrating the given differential equation twice and using these initial conditions to find any constants of integration.

Second Order Differential Equations

Second order differential equations involve derivatives up to the second degree. In simple terms, they include terms such as the acceleration of a function, as shown by the second derivative \( \frac{d^2 \mathbf{r}}{dt^2} \).

This particular type of equation often arises in contexts where acceleration or curvature is described, such as in physics for motion problems, as seen in this exercise with the differential equation \( \frac{d^{2} \mathbf{r}}{dt^{2}}=-(\mathbf{i}+\mathbf{j}+\mathbf{k}) \).

• The equation confirms that the vector function \( \mathbf{r}(t) \) will be influenced by uniform acceleration in all three components, i.e., \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).
• To solve it, one must integrate the function twice, each time introducing a constant of integration.

This leads to a first derivative (velocity) and a second function (position) that can be modified using the initial values to obtain a complete, unique solution.

Vector Functions

Vector functions describe quantities that have both magnitudes and directions. They establish a relationship where each input, usually a scalar like time \(t\), maps to a vector. In this exercise, \( \mathbf{r}(t) \) is a vector that changes over time and combines three components \( (x(t), y(t), z(t)) \).

The given vector function is expressed in terms of the standard basis vectors:\( \mathbf{i}, \mathbf{j}, \mathbf{k} \), which correspond to the \(x\)-axis, \(y\)-axis, and \(z\)-axis in 3D space. For instance, \( \mathbf{r}(t) = \left(10 - \frac{t^2}{2}\right)\mathbf{i} + \left(10 - \frac{t^2}{2}\right)\mathbf{j} + \left(10 - \frac{t^2}{2}\right)\mathbf{k} \).

• Vector functions like \( \mathbf{r}(t) \) can represent the position of an object in three-dimensional space.
• The evolution of this function over time is typically governed by differential equations, as shown in this problem.

Understanding how vector functions work is essential in analyzing physical systems and phenomena, helping predict movements and changes effectively across different fields such as physics and engineering.