Problem 17 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) as a vector function of \(t .\) Differential equation: $$\quad \frac{d^{2} \mathbf{r}}{d t^{2}}=-32 \mathbf{k}$$ Initial conditions:$$ \begin{array}{l}{\mathbf{r}(0)=100 \mathbf{k} \text { and }} \\\ {\left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=8 \mathbf{i}+8 \mathbf{j}}\end{array} $$

Step-by-Step Solution

Answer

\(\mathbf{r}(t) = -16 \mathbf{k} t^2 + 8 \mathbf{i} t + 8 \mathbf{j} t + 100 \mathbf{k}\)

Step 1: Understand the Differential Equation

The differential equation given is \( \frac{d^{2} \mathbf{r}}{d t^{2}} = -32 \mathbf{k} \). This tells us that the second derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \) is a constant vector pointing in the negative \( \mathbf{k} \) direction.

Step 2: Integrate to Find Velocity Function

To find the velocity \( \frac{d \mathbf{r}}{d t} \), integrate the second derivative with respect to \( t \):\[\frac{d \mathbf{r}}{d t} = \int (-32 \mathbf{k}) \, dt = -32 \mathbf{k} t + \mathbf{C_1}\]where \( \mathbf{C_1} \) is a constant vector from integration.

Step 3: Apply Initial Condition for Velocity

Use the initial condition \( \left.\frac{d \mathbf{r}}{d t}\right|_{t=0} = 8 \mathbf{i} + 8 \mathbf{j} \) to find \( \mathbf{C_1} \):\[8 \mathbf{i} + 8 \mathbf{j} = -32 \mathbf{k}(0) + \mathbf{C_1} \rightarrow \mathbf{C_1} = 8 \mathbf{i} + 8 \mathbf{j}\]

Step 4: Write the Velocity Function

Substitute \( \mathbf{C_1} \) into the velocity function:\[\frac{d \mathbf{r}}{d t} = -32 \mathbf{k} t + 8 \mathbf{i} + 8 \mathbf{j}\]

Step 5: Integrate to Find the Position Function

Integrate the velocity function to find \( \mathbf{r}(t) \):\[\mathbf{r}(t) = \int (-32 \mathbf{k} t + 8 \mathbf{i} + 8 \mathbf{j}) \, dt = -16 \mathbf{k} t^2 + 8 \mathbf{i} t + 8 \mathbf{j} t + \mathbf{C_2}\]where \( \mathbf{C_2} \) is a constant vector from integration.

Step 6: Apply Initial Condition for Position

Use the initial condition \( \mathbf{r}(0) = 100 \mathbf{k} \) to find \( \mathbf{C_2} \):\[100 \mathbf{k} = -16 \mathbf{k}(0)^2 + 8 \mathbf{i}(0) + 8 \mathbf{j}(0) + \mathbf{C_2} \rightarrow \mathbf{C_2} = 100 \mathbf{k}\]

Step 7: Write the Position Function

Substitute \( \mathbf{C_2} \) into the position function:\[\mathbf{r}(t) = -16 \mathbf{k} t^2 + 8 \mathbf{i} t + 8 \mathbf{j} t + 100 \mathbf{k}\]

Key Concepts

Vector Calculus

Vector calculus is a fundamental field of mathematics that deals with vector fields and operations on them. Vectors are entities that have both magnitude and direction, such as displacement or force in physics, and they are typically represented with arrows in diagrams. In this context, the vector

• \(\mathbf{r}(t)\) refers to a position vector function, which means it describes the location of a point in space as it changes over time \(t\).

The exercise involves the vector function of time \(\mathbf{r}(t)\), which describes how an object's position changes with time in three-dimensional space. In vector calculus, you often deal with operations such as differentiation and integration of these vector functions to explore their behavior. For example, you might examine how velocity and acceleration relate to the position vector through derivatives.Understanding these concepts in vector calculus is essential, as they form the groundwork for solving complex physical problems that involve multi-dimensional changes in direction and magnitude over time.

Differential Equation

Differential equations are mathematical equations that relate some function with its derivatives. In this particular exercise, we are dealing with a differential equation \[\frac{d^{2} \mathbf{r}}{dt^{2}} = -32 \mathbf{k}\]This equation is second-order because it involves the second derivative of \( \mathbf{r} \), which represents acceleration in physical terms. The given equation describes a constant acceleration pointing in the negative \( \mathbf{k} \) direction, such as the force of gravity pointing downwards.These equations are solved by finding functions that satisfy them. They are crucial for modeling the behavior of dynamic systems across physics, engineering, and other fields.In this problem, understanding the mathematical meaning of the differential equation helps us comprehend how an object's acceleration impacts its velocity and position over time.

Integration

Integration is the mathematical process used to find the original function given its derivative, a critical step when solving differential equations. In this exercise, we perform integration to transition from acceleration to velocity, and then from velocity to position.

• The first integration helps us find the velocity function: \[ \frac{d \mathbf{r}}{d t} = \int (-32 \mathbf{k}) \, dt = -32 \mathbf{k} t + \mathbf{C_1} \]Here, \(\mathbf{C_1}\) is a constant vector representing the initial velocity.
• The second integration yields the position function:\[ \mathbf{r}(t) = \int (-32 \mathbf{k} t + 8 \mathbf{i} + 8 \mathbf{j}) \, dt = -16 \mathbf{k} t^2 + 8 \mathbf{i} t + 8 \mathbf{j} t + \mathbf{C_2} \]

This step-by-step integration process is essential to translating the effects of acceleration on positioning over time. It allows us to reconstruct the history of motion from its derivatives, thus offering a full picture of the object's movement.

Constant of Integration

When solving differential equations through integration, you often encounter what's termed as a "constant of integration." This constant arises because integration can only determine a family of functions rather than a specific one without additional information.

• For the velocity function, the constant of integration appearing as \( \mathbf{C_1} \) reflects this ambiguity and can be resolved using the initial condition \( \left. \frac{d \mathbf{r}}{d t} \right|_{t=0} = 8 \mathbf{i} + 8 \mathbf{j} \).
• Similarly, in the position function, \( \mathbf{C_2} \) is another constant of integration resolved by applying the initial condition \( \mathbf{r}(0) = 100 \mathbf{k} \).

Understanding and determining these constants using given initial conditions are crucial steps in solving initial value problems. They ensure that the solution not only obeys the differential equation but also fits any specific, real-world situation or experiment described by those initial conditions.