Problem 5 - 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. $$ \mathbf{r}(t)=2 \mathbf{i}+(t-1)^{2} \mathbf{j}+t \mathbf{k} ; t=2 $$

Step-by-Step Solution

Answer

The speed at \( t = 2 \) is \( \sqrt{5} \).

Step 1: Write Down the Position Vector

The position vector of the moving particle is given by \( \mathbf{r}(t) = 2 \mathbf{i} + (t-1)^{2} \mathbf{j} + t \mathbf{k} \). This represents the location of the particle in 3D space at any time \( t \).

Step 2: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Thus, \( \mathbf{v}(t) = \frac{d}{dt}[2 \mathbf{i} + (t-1)^{2} \mathbf{j} + t \mathbf{k}] = 0 \mathbf{i} + 2(t-1) \mathbf{j} + 1 \mathbf{k} \).

Step 3: Evaluate the Velocity Vector at t=2

Substitute \( t = 2 \) into the velocity vector function: \( \mathbf{v}(2) = 0 \mathbf{i} + 2(2-1) \mathbf{j} + 1 \mathbf{k} = 0 \mathbf{i} + 2 \mathbf{j} + 1 \mathbf{k} \). Thus, the velocity vector at \( t = 2 \) is \( 2 \mathbf{j} + \mathbf{k} \).

Step 4: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Thus, \( \mathbf{a}(t) = \frac{d}{dt}[0 \mathbf{i} + 2(t-1) \mathbf{j} + 1 \mathbf{k}] = 0 \mathbf{i} + 2 \mathbf{j} + 0 \mathbf{k} = 2 \mathbf{j} \).

Step 5: Evaluate the Acceleration Vector at t=2

Since the acceleration vector is constant, at \( t = 2 \), it is simply \( 2 \mathbf{j} \).

Step 6: Calculate the Speed at t=2

Speed is the magnitude of the velocity vector. At \( t = 2 \), \( \mathbf{v}(2) = 2 \mathbf{j} + \mathbf{k} \). The speed is \( \| \mathbf{v}(2) \| = \sqrt{(0)^2 + (2)^2 + (1)^2} = \sqrt{5} \).

Step 7: Graph the Curve and Vectors

Graph the 3D parametric curve from \( \mathbf{r}(t) = 2 \mathbf{i} + (t-1)^{2} \mathbf{j} + t \mathbf{k} \) over a range including \( t=2 \). At that point on the curve, draw the velocity vector \( 2\mathbf{j} + \mathbf{k} \) and the acceleration vector \( 2 \mathbf{j} \). Ensure these vectors are emanating from the point \( \mathbf{r}(2) \).

Key Concepts

Position Vector

A position vector is a foundational concept in vector calculus and physics. It represents the location of a particle in space relative to a fixed origin point. In this exercise, the position vector given is \( \mathbf{r}(t) = 2 \mathbf{i} + (t-1)^{2} \mathbf{j} + t \mathbf{k} \). This notation signifies a 3D space where:

• \( 2 \mathbf{i} \) represents a constant position along the x-axis.
• \( (t-1)^2 \mathbf{j} \) describes a parabolic motion component along the y-axis.
• \( t \mathbf{k} \) indicates a linear motion along the z-axis.

When analyzing the motion at a specific time, such as \( t = 2 \), we substitute this value into \( \mathbf{r}(t) \) to pinpoint the exact location in space where the particle is situated.

Velocity Vector

The velocity vector is derived by taking the derivative of the position vector with respect to time. This gives us a vector that represents the rate of change of the position over time, or simply how fast and in which direction the particle is moving. The equation for the velocity vector in this exercise is:\( \mathbf{v}(t) = 0 \mathbf{i} + 2(t-1) \mathbf{j} + 1 \mathbf{k} \), which simplifies to identify the velocity at any given moment:

• \( 0 \mathbf{i} \) indicates no movement along the x-axis.
• \( 2(t-1) \mathbf{j} \) suggests that the particle's velocity along the y-axis depends on time, t.
• \( 1 \mathbf{k} \) shows a constant velocity along the z-axis.

At \( t = 2 \), we have \( \mathbf{v}(2) = 2 \mathbf{j} + \mathbf{k} \), revealing the particle's directional movement in space at that time.

Acceleration Vector

Acceleration is the rate of change of velocity over time, and it's represented as a vector as well. This vector gives insight into how the velocity of the particle is accelerating or decelerating within the coordinate system. In this context:\( \mathbf{a}(t) = 0 \mathbf{i} + 2 \mathbf{j} + 0 \mathbf{k} \),which tells us that:

• There is no change in velocity along the x-axis (\( 0 \mathbf{i} \)).
• An acceleration of \( 2 \mathbf{j} \) along the y-axis, indicating a steady increase or decrease in velocity in this direction.
• No change in velocity along the z-axis (\( 0 \mathbf{k} \)).

Since the acceleration is constant, it remains the same regardless of the value of \( t \) and for \( t = 2 \), it stays as \( 2 \mathbf{j} \). This constant acceleration simplifies calculations for dynamic systems.

Speed Calculation

Speed, unlike velocity, is a scalar quantity, meaning it only has magnitude without a directional component. To find the speed, we compute the magnitude of the velocity vector. In this exercise, the velocity vector at \( t = 2 \) is \( \mathbf{v}(2) = 2 \mathbf{j} + \mathbf{k} \).The formula for speed is:\[ \| \mathbf{v} \| = \sqrt{(v_x)^2 + (v_y)^2 + (v_z)^2} \]Substituting the given components:

• No x-component: \( (0)^2 \)
• y-component: \( (2)^2 = 4 \)
• z-component: \( (1)^2 = 1 \)

Thus, the speed is:\[ \sqrt{4 + 1} = \sqrt{5} \]This value, \( \sqrt{5} \), represents the speed of the particle, showing how fast it moves through space at the moment considered. This step is crucial for understanding movement without worrying about directionality.