Problem 9 - Advanced Engineering Mathematics

Question

Suppose \(\mathbf{r}(t)=t^{2} \mathbf{i}+\left(t^{3}-2 t\right) \mathbf{j}+\left(t^{2}-5 t\right) \mathbf{k}\) is the position vector of a moving particle. At what points does the particle pass through the \(x y\) -plane? What are its velocity and acceleration at these points?

Step-by-Step Solution

Answer

The particle crosses the xy-plane at (0,0,0) and (25,110,0). The velocities at these points are -2j-5k and 10i+73j+5k; accelerations are 2i+2k and 2i+30j+2k.

Step 1: Determine the condition for crossing the xy-plane

A particle crosses the xy-plane when its z-coordinate is zero. The z-coordinate of the given position vector is \( t^2 - 5t \). So, set it equal to zero and solve for \( t \): \[ t^2 - 5t = 0 \].

Step 2: Solve the equation for t

Factor the equation \( t^2 - 5t = 0 \): \[ t(t - 5) = 0 \]. The solutions are \( t = 0 \) and \( t = 5 \). These are the times when the particle crosses the xy-plane.

Step 3: Find position coordinates at t = 0 and t = 5

Substitute \( t = 0 \) into \( \mathbf{r}(t)\): \( \mathbf{r}(0) = 0^2 \mathbf{i} + (0^3 - 2 \cdot 0) \mathbf{j} + (0^2 - 5 \cdot 0) \mathbf{k} = \mathbf{0} \). Thus, the point is \( (0,0,0) \).For \( t = 5 \): \( \mathbf{r}(5) = 5^2 \mathbf{i} + (5^3 - 2 \cdot 5) \mathbf{j} + (5^2 - 5 \cdot 5) \mathbf{k} = 25 \mathbf{i} + 110 \mathbf{j} + 0 \mathbf{k} \). Thus, the point is \( (25, 110, 0) \).

Step 4: Compute the velocity vector

The velocity vector is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \): \[ \mathbf{v}(t) = \frac{d}{dt}(t^2) \mathbf{i} + \frac{d}{dt}(t^3 - 2t) \mathbf{j} + \frac{d}{dt}(t^2 - 5t) \mathbf{k} \]\[ \mathbf{v}(t) = 2t \mathbf{i} + (3t^2 - 2) \mathbf{j} + (2t - 5) \mathbf{k} \].

Step 5: Evaluate velocity at t = 0 and t = 5

For \( t = 0 \): \( \mathbf{v}(0) = 2 \cdot 0 \mathbf{i} + (3 \cdot 0^2 - 2) \mathbf{j} + (2 \cdot 0 - 5) \mathbf{k} = -2 \mathbf{j} - 5 \mathbf{k} \).For \( t = 5 \): \( \mathbf{v}(5) = 2 \cdot 5 \mathbf{i} + (3 \cdot 5^2 - 2) \mathbf{j} + (2 \cdot 5 - 5) \mathbf{k} = 10 \mathbf{i} + 73 \mathbf{j} + 5 \mathbf{k} \).

Step 6: Compute the acceleration vector

The acceleration vector is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \):\[ \mathbf{a}(t) = \frac{d}{dt}(2t) \mathbf{i} + \frac{d}{dt}(3t^2 - 2) \mathbf{j} + \frac{d}{dt}(2t - 5) \mathbf{k} \]\[ \mathbf{a}(t) = 2 \mathbf{i} + 6t \mathbf{j} + 2 \mathbf{k} \].

Step 7: Evaluate acceleration at t = 0 and t = 5

For \( t = 0 \): \( \mathbf{a}(0) = 2 \mathbf{i} + 6 \cdot 0 \mathbf{j} + 2 \mathbf{k} = 2 \mathbf{i} + 2 \mathbf{k} \).For \( t = 5 \): \( \mathbf{a}(5) = 2 \mathbf{i} + 6 \cdot 5 \mathbf{j} + 2 \mathbf{k} = 2 \mathbf{i} + 30 \mathbf{j} + 2 \mathbf{k} \).

Key Concepts

Position Vector

A position vector describes the position of a particle in three-dimensional space by indicating how far it is from the origin in the direction of each axis. In this problem, the position vector \( \mathbf{r}(t) = t^{2} \mathbf{i} + (t^{3} - 2t) \mathbf{j} + (t^{2} - 5t) \mathbf{k} \) provides:

• The x-coordinate as a function of time: \( t^2 \)
• The y-coordinate: \( (t^{3} - 2t) \)
• The z-coordinate: \( (t^{2} - 5t) \)

At any given time \( t \), plugging in the value of \( t \) will give the respective coordinates in vector form within the space. A thorough understanding of position vectors allows students to analyze particle motion and engage with complex vector calculus concepts.

Velocity Vector

Velocity vectors indicate the rate of change of the position vector with respect to time. To find it, we differentiate the position vector function \( \mathbf{r}(t) \) with respect to time:\[ \mathbf{v}(t) = \frac{d}{dt}(t^2) \mathbf{i} + \frac{d}{dt}(t^3 - 2t) \mathbf{j} + \frac{d}{dt}(t^2 - 5t) \mathbf{k} \]The resulting velocity vector is:\[ \mathbf{v}(t) = 2t \mathbf{i} + (3t^2 - 2) \mathbf{j} + (2t - 5) \mathbf{k} \]This means that the velocity at any time \( t \) is given by the components:

• \( 2t \) in the x-direction
• \( 3t^2 - 2 \) in the y-direction
• \( 2t - 5 \) in the z-direction

The velocity vector helps in understanding how fast the particle is moving and in which direction.

Acceleration Vector

The acceleration vector is derived by differentiating the velocity vector with respect to time. This gives us the change in velocity over time and is crucial in understanding dynamics. Here, after differentiating the velocity vector, we get:\[ \mathbf{a}(t) = \frac{d}{dt}(2t) \mathbf{i} + \frac{d}{dt}(3t^2 - 2) \mathbf{j} + \frac{d}{dt}(2t - 5) \mathbf{k} \]This results in:\[ \mathbf{a}(t) = 2 \mathbf{i} + 6t \mathbf{j} + 2 \mathbf{k} \]

• The x-component is constantly \( 2 \)
• The y-component increases linearly with time: \( 6t \)
• The z-component is also a constant: \( 2 \)

The acceleration vector reveals the forces acting on the particle and how its speed is changing.

Crossing the xy-plane

Crossing the xy-plane happens when the position vector's z-coordinate equals zero. This is indicative of a particle that has shifted from being either above or below the xy-plane to crossing it. The z-coordinate is given by \( (t^2 - 5t) \).To find out when the particle crosses the xy-plane, solve the equation:\[ t^2 - 5t = 0 \]By factoring, we find:\[ t(t - 5) = 0 \]This provides two solutions, \( t = 0 \) and \( t = 5 \), identifying the exact moments in time the particle crosses the plane. At these times, further analysis can be done to investigate the velocity and acceleration at these crossing points.

Derivative with Respect to Time

The derivative with respect to time is essential in vector calculus for finding both velocities and accelerations from position vectors. Differentiation provides the rate of change of a function concerning time, offering insights into how fast or slow something is occurring.To compute the velocity vector, differentiate each component of the position vector with respect to time.

• Position components: \( t^2,\ t^3 - 2t,\ t^2 - 5t \)
• Velocity components: \( 2t,\ 3t^2 - 2,\ 2t - 5 \)

Next, computing the acceleration vector involves differentiating velocity components:

• \( 2,\ 6t,\ 2 \)

Mastering these derivatives improves understanding of motion characteristics, portraying the constant change of position or speed in dynamic systems.