Problem 9 - Advanced Engineering Mathematics

Question

\(\mathbf{r}(t)\) is the position vector of a moving particle. Find the tangential and normal components of the acceleration at any \(t\). $$ \mathbf{r}(t)=t^{2} \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}+2 t^{2} \mathbf{k} $$

Step-by-Step Solution

Answer

The tangential component is \(2\sqrt{6}\); the normal component is 0.

Step 1: Find the Velocity Vector

To find the velocity vector, differentiate the position vector \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t^2 \mathbf{k} \) with respect to time \( t \).ewline\( \mathbf{v}(t) = \frac{d}{dt}(t^2 \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t^2 \mathbf{k}) = 2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \).

Step 2: Find the Acceleration Vector

Differentiate the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \) to find the acceleration vector.ewline\( \mathbf{a}(t) = \frac{d}{dt}(2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k}) = 2 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k} \).

Step 3: Calculate the Tangential Component of Acceleration

The tangential component of acceleration \( a_T \) is found by projecting \( \mathbf{a}(t) \) onto \( \mathbf{v}(t) \).ewlineFirst, find the magnitude of \( \mathbf{v}(t) \): \( \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + (2t)^2 + (4t)^2} = \sqrt{4t^2 + 4t^2 + 16t^2} = \sqrt{24t^2} = 2\sqrt{6}t \).ewlineThe unit tangent vector \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\| \mathbf{v}(t) \|} = \frac{1}{2\sqrt{6}t} (2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k}) = \frac{1}{\sqrt{6}}(\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \).ewlineFinally, project \( \mathbf{a}(t) \) onto \( \mathbf{T}(t) \):ewline\( a_T = \mathbf{a} \cdot \mathbf{T} = (2 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k}) \cdot \frac{1}{\sqrt{6}}(\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \).ewlineCalculating the dot product: \( a_T = \frac{1}{\sqrt{6}}(2 \times 1 + 2 \times 1 + 4 \times 2) = \frac{1}{\sqrt{6}}(2 + 2 + 8) = \frac{12}{\sqrt{6}} = 2\sqrt{6} \).

Step 4: Calculate the Normal Component of Acceleration

The normal component of acceleration \( a_N \) can be found using the formula \( a_N = \sqrt{\| \mathbf{a} \|^2 - a_T^2} \).ewlineFirst, compute \( \| \mathbf{a} \| = \sqrt{2^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \).ewline\( a_N = \sqrt{\| \mathbf{a} \|^2 - a_T^2} = \sqrt{(2\sqrt{6})^2 - (2\sqrt{6})^2} = \sqrt{24 - 24} = \sqrt{0} = 0 \).

Key Concepts

Position Vector

The position vector is a representation of a particle's location in space at a given time, typically denoted as \( \mathbf{r}(t) \). This vector provides a way to trace the path of the particle over time. For instance, in our exercise, the position vector is given by \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t^2 \mathbf{k} \).
This means that the position of the particle at any time \( t \) can be found by substituting the value of \( t \) into each component of the vector.
The position vector gives us a complete picture of where the particle is in a three-dimensional space:

• \( t^2 \mathbf{i} \) shows the position along the x-axis.
• \( (t^2 - 1) \mathbf{j} \) gives the position along the y-axis, showing a slight offset due to -1.
• \( 2t^2 \mathbf{k} \) represents the z-axis position, indicating the particle's elevation over time.

This representation is crucial because it forms the basis for derivatives that lead to velocity and acceleration vectors, illustrating the dynamic aspects of motion.

Velocity Vector

The velocity vector indicates the speed and direction of a moving particle at any given time. It's derived by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \).
By calculating the velocity vector in our example, we found \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \).
This gives us a clear picture of how fast and in what direction each component of the particle's movement is changing:

• \( 2t \mathbf{i} \) shows the rate of change along the x-axis.
• \( 2t \mathbf{j} \) indicates the speed in the y-direction.
• \( 4t \mathbf{k} \) suggests a quicker rise along the z-axis compared to x and y.

The velocity vector is fundamental because it reflects the particle's motion dynamics, providing insights into both instantaneous speed and trajectory direction.

Acceleration Vector

The acceleration vector is crucial for understanding how the velocity of a particle changes over time. It is found by differentiating the velocity vector \( \mathbf{v}(t) \).
In the example provided, the acceleration vector is calculated as \( \mathbf{a}(t) = 2 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k} \).
Each element of this vector provides insight into the particle's change in speed and direction:

\( 2 \mathbf{i} \) dictates the change in velocity in the x-direction.

\( 2 \mathbf{j} \) represents the acceleration along the y-axis.

\( 4 \mathbf{k} \) shows a larger acceleration along the z-axis, indicating swift vertical movement.

The acceleration vector allows us to delve into the forces and influences acting upon a particle, aiding in the analysis of motion under various conditions.

Unit Tangent Vector

The unit tangent vector is a normalized vector that points in the direction of the velocity vector. It is an essential part of understanding the geometry of a particle’s path.
To find it, we normalize the velocity vector \( \mathbf{v}(t) \) by dividing it by its magnitude \( \| \mathbf{v}(t) \| \).
From our exercise, the unit tangent vector is computed as \( \mathbf{T}(t) = \frac{1}{\sqrt{6}}(\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \).
This vector gives information on the path’s direction without being influenced by the particle’s speed:

• It illustrates the direction of motion, which is crucial for calculating the tangential component of acceleration.
• The tangent vector remains unchanged regardless of how fast the particle is moving.

By normalization, we ensure that the length of our tangent vector is always 1, making it easier to compare directions of motion at different points along the path.