Problem 9 - Advanced Engineering Mathematics

Question

In Problems \(7-16, \mathbf{r}(t)\) 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

Tangential acceleration is \(2\sqrt{6}\) and normal acceleration is \(0\).

Step 1: 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 \( t \) to find the velocity vector \( \mathbf{v}(t) \). \[ \mathbf{v}(t) = \frac{d}{dt}(t^2 \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t^2 \mathbf{k}) = 2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \]

Step 2: Find the Acceleration Vector

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

Step 3: Magnitude of the Velocity Vector

Calculate the magnitude of \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \). \[ \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + (2t)^2 + (4t)^2} = \sqrt{4t^2 + 4t^2 + 16t^2} = \sqrt{24t^2} = 2\sqrt{6}t \]

Step 4: Tangential Component of Acceleration

The tangential component of acceleration \( a_T \) is given by the derivative of the magnitude of the velocity vector.\[ a_T = \frac{d}{dt}(2\sqrt{6}t) = 2\sqrt{6} \]

Step 5: Normal Component of Acceleration

The normal component of acceleration \( a_N \) can be calculated using the formula: \[ a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \]First, find \( \| \mathbf{a}(t) \| \):\[ \| \mathbf{a}(t) \| = \sqrt{2^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \]Then compute \( a_N \):\[ a_N = \sqrt{24 - (2\sqrt{6})^2} = \sqrt{24 - 24} = 0 \]

Key Concepts

Position Vector

The position vector, often denoted as \( \mathbf{r}(t) \), is a fundamental concept in physics used to describe the exact position of a particle in space over time. In this specific problem, the position vector is given by:\[ \mathbf{r}(t) = t^2 \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t^2 \mathbf{k} \]This equation tells us the particle's position in three-dimensional space, reflected by its components along the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) axes, which correspond to the x, y, and z coordinates, respectively. Each component is a function of time \( t \), showcasing how the particle moves and changes its position as time progresses. Understanding the position vector helps in visualizing and detailing the path a particle follows.

Velocity Vector

The velocity vector, \( \mathbf{v}(t) \), is derived from the position vector and describes the rate of change of position with respect to time. It essentially tells you how quickly and in what direction the particle is moving at any given moment. By differentiating the position vector with respect to time \( t \), we can find the velocity vector:\[ \mathbf{v}(t) = \frac{d}{dt}(t^2 \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t^2 \mathbf{k}) = 2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \]This vector is key to determining not only the direction of the particle at each moment but also the pace at which it changes its position, making it crucial for analyzing motion dynamics in physics.

Acceleration Vector

The acceleration vector, \( \mathbf{a}(t) \), is obtained by differentiating the velocity vector with respect to time, and it represents how the velocity of the particle changes over time. It's an indication of the particle's change in speed and direction. Using the velocity vector derived earlier, the formula becomes:\[ \mathbf{a}(t) = \frac{d}{dt}(2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k}) = 2 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k} \]This constant acceleration vector shows that the particle's rate of change of velocity is uniform across the x, y, and z dimensions, making it a critical factor in studying and predicting the future motion of the particle.

Magnitude of Velocity

The magnitude of the velocity vector reflects the speed of the particle regardless of its direction. This is important for assessing how fast the particle is actually moving through space. We compute it by taking the square root of the sum of the squares of its components:\[ \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + (2t)^2 + (4t)^2} = \sqrt{24t^2} = 2\sqrt{6}t \]So, the magnitude calculates the actual speed of the particle at any given time \( t \). Knowing this helps in problems where only the magnitude, or speed, is of interest rather than the full directional velocity.

Vector Differentiation

Vector differentiation is essential when studying motion since it helps find rates of change in vector quantities like position and velocity. In this context, differentiating the position vector and the velocity vector provides the velocity and acceleration vectors, respectively.

• The first derivative gives the velocity vector: rate of change of position.
• The second derivative provides the acceleration vector: rate of change of velocity.

Differentiation turns instantaneous positional and velocity changes into a deeper understanding of a particle's motion dynamics. It's used extensively in physics to study trajectories and forces affecting motion.