Problem 2 - Advanced Engineering Mathematics

Question

\( \mathbf{r}(t)\) is the position vector of a moving particle. Graph the curve and the velocity and acceleration vectors at the indicated time. Find the speed at that time. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+\frac{1}{t^{2}} \mathbf{j} ; t=1 $$

Step-by-Step Solution

Answer

The speed at \( t = 1 \) is \( 2\sqrt{2} \).

Step 1: Understand the Position Vector

The given position vector is \( \mathbf{r}(t) = t^{2} \mathbf{i} + \frac{1}{t^{2}} \mathbf{j} \). It represents the position of a moving particle in terms of time \( t \) in the xy-plane. \( \mathbf{i} \) and \( \mathbf{j} \) are the unit vectors along the x and y directions, respectively.

Step 2: Calculate the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). Thus, \( \mathbf{v}(t) = \frac{d}{dt}(t^{2} \mathbf{i} + \frac{1}{t^{2}} \mathbf{j}) = 2t \mathbf{i} - \frac{2}{t^{3}} \mathbf{j} \). For \( t = 1 \), \( \mathbf{v}(1) = 2 \mathbf{i} - 2 \mathbf{j} \).

Step 3: Calculate the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Thus, \( \mathbf{a}(t) = \frac{d}{dt}(2t \mathbf{i} - \frac{2}{t^{3}} \mathbf{j}) = 2 \mathbf{i} + \frac{6}{t^{4}} \mathbf{j} \). For \( t = 1 \), \( \mathbf{a}(1) = 2 \mathbf{i} + 6 \mathbf{j} \).

Step 4: Graph the Position, Velocity, and Acceleration

On a graph with axes labeled x and y, plot the point at \( t=1 \) given by the position vector \( \mathbf{r}(1) = 1 \mathbf{i} + 1 \mathbf{j} \). From this point, draw the velocity vector \( \mathbf{v}(1) = 2 \mathbf{i} - 2 \mathbf{j} \) as an arrow originating from the position, and the acceleration vector \( \mathbf{a}(1) = 2 \mathbf{i} + 6 \mathbf{j} \) similarly as another arrow.

Step 5: Calculate the Speed

The speed of the particle at any time \( t \) is the magnitude of the velocity vector \( \mathbf{v}(t) \). At \( t=1 \), the speed is \( ||\mathbf{v}(1)|| = \sqrt{(2)^{2} + (-2)^{2}} = \sqrt{8} = 2\sqrt{2} \).

Key Concepts

Velocity Vector

The velocity vector is a key concept in understanding motion. It tells us the rate at which the position of an object changes with time. To find the velocity vector for a moving particle described by a position vector, we need to determine the derivative of the position vector with respect to time.
Let's break that down with our example:

• The position vector is given as \( \mathbf{r}(t) = t^{2} \mathbf{i} + \frac{1}{t^{2}} \mathbf{j} \).
• The velocity vector \( \mathbf{v}(t) \) is found by taking the derivative of \( \mathbf{r}(t) \).

This derivative gives us \( \mathbf{v}(t) = 2t \mathbf{i} - \frac{2}{t^{3}} \mathbf{j} \).
For a specific time, say \( t = 1 \), you substitute \( t \) with 1 in the velocity formula to get:

• \( \mathbf{v}(1) = 2 \mathbf{i} - 2 \mathbf{j} \)

Here, the velocity vector helps us visualize the direction and rate of change of the particle's position at any given moment.

Acceleration Vector

Acceleration vector provides information about how the velocity of an object changes over time. It's essentially the derivative of the velocity vector.
Let's continue with our example to see how this works.

• We have the velocity vector: \( \mathbf{v}(t) = 2t \mathbf{i} - \frac{2}{t^{3}} \mathbf{j} \).
• To get the acceleration vector, \( \mathbf{a}(t) \), we differentiate the velocity vector with respect to time \( t \).

The calculation then becomes: \( \mathbf{a}(t) = \frac{d}{dt}(2t \mathbf{i} - \frac{2}{t^{3}} \mathbf{j}) = 2 \mathbf{i} + \frac{6}{t^{4}} \mathbf{j} \).
At \( t = 1 \), the acceleration vector is:

• \( \mathbf{a}(1) = 2 \mathbf{i} + 6 \mathbf{j} \)

This vector shows how the velocity is increasing or decreasing, and in which directions.

Speed Calculation

Calculating speed is one way to quantify how fast an object is moving. Speed is the magnitude of the velocity vector and does not have a direction, unlike the velocity vector.
To find speed, we take the absolute value of the velocity vector, which involves squaring its components, summing them, and then taking the square root of the sum.
For the velocity vector \( \mathbf{v}(t) = 2 \mathbf{i} - 2 \mathbf{j} \) at \( t = 1 \):

• Calculate: \( ||\mathbf{v}(1)|| = \sqrt{(2)^2 + (-2)^2} \).
• This results in: \( \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).

The speed, \( 2\sqrt{2} \), tells us how fast the particle is moving at \( t = 1 \) without concerning its direction.

Derivatives

Understanding derivatives is crucial for working with vectors in physics. They measure how a function changes as its input changes, providing insights into quantities such as velocity and acceleration.
In the context of our exercise:

• The derivative of the position vector gives the velocity vector, indicating how position changes with time.
• The derivative of the velocity vector results in the acceleration vector, showing how velocity itself changes with time.

Differentiation is thus a powerful tool which enables us to describe and predict motion mathematically in a precise way.
Using derivatives, we can convert position data into detailed information about the motion dynamics of particles or objects within various coordinate systems.