Problem 21 - Thomas Calculus

Question

At time \(t=0,\) a particle is located at the point \((1,2,3) .\) It travels in a straight line to the point \((4,1,4),\) has speed 2 at \((1,2,3)\) and constant acceleration \(3 \mathbf{i}-\mathbf{j}+\mathbf{k} .\) Find an equation for the position vector \(\mathbf{r}(t)\) of the particle at time \(t\) .

Step-by-Step Solution

Answer

\(\mathbf{r}(t) = \begin{pmatrix} 1 + \frac{6}{\sqrt{11}}t + \frac{3}{2}t^2 \\ 2 - \frac{2}{\sqrt{11}}t - \frac{1}{2}t^2 \\ 3 + \frac{2}{\sqrt{11}}t + \frac{1}{2}t^2 \end{pmatrix}\)

Step 1: Initial Information

The particle starts at point \((1, 2, 3)\), heading towards \((4, 1, 4)\). Its initial speed is \(2\) and it has a constant acceleration of \(3 \mathbf{i} - \mathbf{j} + \mathbf{k}\).

Step 2: Determine Initial Velocity Direction

The direction from point \((1, 2, 3)\) to \((4, 1, 4)\) is the vector \((4-1, 1-2, 4-3) = (3, -1, 1)\). Normalize this to get the direction unit vector: \(\frac{(3, -1, 1)}{\sqrt{3^2 + (-1)^2 + 1^2}} = \frac{(3, -1, 1)}{\sqrt{11}}\).

Step 3: Calculate Initial Velocity

With speed \(2\), the initial velocity vector \(\mathbf{v}_0\) is \(2 \times \frac{(3, -1, 1)}{\sqrt{11}} = \left(\frac{6}{\sqrt{11}}, -\frac{2}{\sqrt{11}}, \frac{2}{\sqrt{11}}\right)\).

Step 4: Set Up the Position Vector Equation

The position vector \(\mathbf{r}(t)\) can be calculated with the equation \(\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2\), where \(\mathbf{r}_0 = (1, 2, 3)\) is the initial position, \(\mathbf{v}_0 = \left(\frac{6}{\sqrt{11}}, -\frac{2}{\sqrt{11}}, \frac{2}{\sqrt{11}}\right)\) is the initial velocity, and \(\mathbf{a} = (3, -1, 1)\) is the acceleration.

Step 5: Plug Values into the Equation

Substitute the known values into the position equation:\[\mathbf{r}(t) = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} + t\begin{pmatrix} \frac{6}{\sqrt{11}} \ -\frac{2}{\sqrt{11}} \ \frac{2}{\sqrt{11}} \end{pmatrix} + \frac{1}{2}t^2\begin{pmatrix} 3 \ -1 \ 1 \end{pmatrix}\]

Step 6: Simplified Equation

Combine terms to express \(\mathbf{r}(t)\):\[\mathbf{r}(t) = \begin{pmatrix} 1 + \frac{6}{\sqrt{11}}t + \frac{3}{2}t^2 \ 2 - \frac{2}{\sqrt{11}}t - \frac{1}{2}t^2 \ 3 + \frac{2}{\sqrt{11}}t + \frac{1}{2}t^2 \end{pmatrix}\]

Key Concepts

Motion Along a Line

Motion along a line describes the path of a particle moving in a straight line. In our example, the particle travels from (1, 2, 3) to (4, 1, 4). This means its movement is restricted in three-dimensional space but remains linear. The straightforward nature of motion along a line simplifies complex calculations. The position vector \( \mathbf{r}(t) \) captures this motion over time.

• Movement is linear, from one point to another.
• Position vector provides a clear representation of the particle's location at any time.

By understanding motion along a line, we simplify the challenge of determining a particle's future position. This can be especially useful when combined with other factors like velocity and acceleration.

Constant Acceleration

Constant acceleration implies that the rate of change of velocity of an object is consistent over time. In this problem, we have \( \mathbf{a} = 3 \mathbf{i} - \mathbf{j} + \mathbf{k} \), meaning the acceleration vector has constant components. This consistency makes calculations predictable. When dealing with constant acceleration, the key formula to remember is:
\[ \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2 \]

• \( \mathbf{a} \) remains unchanged no matter \( t \).
• Predictable impacts on velocity and position over time.

The formula highlights how each piece (initial position, initial velocity, and acceleration) affects the position vector. This allows us to calculate the future position of the particle accurately.

Initial Velocity

Initial velocity is the speed and direction at which an object begins its motion. Calculating the initial velocity often involves understanding both the speed and the direction of movement. In our problem, the direction of initial velocity is towards the vector \( (3, -1, 1) \), which is derived from the starting and ending points of motion.
After determining this direction, we normalize it to make it a unit vector and then multiply by the speed to find the \( \mathbf{v}_0 \):
\[ \mathbf{v}_0 = 2 \times \frac{(3, -1, 1)}{\sqrt{11}} \]

• Speed of 2 is multiplied by the direction unit vector.
• Provides foundational piece for calculating position over time.

Initial velocity, therefore, gives the first clue on how the particle will behave immediately after it starts moving.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector, which has a length of one but maintains the original vector's direction. This process is crucial when determining direction, as it allows us to separate direction from magnitude.
In our case, the direction of travel is given by \( (3, -1, 1) \), derived from the change in position.

Steps to Normalize:

• Calculate the magnitude: \( \sqrt{3^2 + (-1)^2 + 1^2} = \sqrt{11} \).
• Divide each component of the vector by this magnitude: \( \frac{(3, -1, 1)}{\sqrt{11}} \).

Normalization ensures the vector describes direction only. This is pivotal for finding accurate initial velocities and subsequent particle paths.