Problem 15 - 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 \(t\). $$ \mathbf{r}(t)=e^{-t}(\mathbf{i}+\mathbf{j}+\mathbf{k}) $$

Step-by-Step Solution

Answer

Tangential component: \(-\sqrt{3} e^{-t}\), Normal component: 0.

Step 1: Differentiate to Find Velocity

The velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector \( \mathbf{r}(t) \). Differentiate \( \mathbf{r}(t) = e^{-t} \mathbf{i} + e^{-t} \mathbf{j} + e^{-t} \mathbf{k} \) with respect to \( t \) to obtain \( \mathbf{v}(t) = -e^{-t} \mathbf{i} - e^{-t} \mathbf{j} - e^{-t} \mathbf{k} \).

Step 2: Differentiate to Find Acceleration

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate \( \mathbf{v}(t) = -e^{-t} \mathbf{i} - e^{-t} \mathbf{j} - e^{-t} \mathbf{k} \) to get \( \mathbf{a}(t) = e^{-t} \mathbf{i} + e^{-t} \mathbf{j} + e^{-t} \mathbf{k} \).

Step 3: Compute Speed for Tangential Component

The magnitude of the velocity \( \lVert \mathbf{v}(t) \rVert \) is the speed of the particle. Calculate this: \( \lVert \mathbf{v}(t) \rVert = \sqrt{(-e^{-t})^2 + (-e^{-t})^2 + (-e^{-t})^2} = \sqrt{3} e^{-t} \).

Step 4: Find Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is \( \frac{\mathbf{v}(t)}{\lVert \mathbf{v}(t) \rVert} = \frac{-e^{-t} \mathbf{i} - e^{-t} \mathbf{j} - e^{-t} \mathbf{k}}{\sqrt{3} e^{-t}} = \frac{-1}{\sqrt{3}} \mathbf{i} + \frac{-1}{\sqrt{3}} \mathbf{j} + \frac{-1}{\sqrt{3}} \mathbf{k} \).

Step 5: Tangential Component of Acceleration

The tangential component of acceleration \( a_T \) is \( \mathbf{a}(t) \cdot \mathbf{T}(t) \). Compute this dot product: \( e^{-t} \mathbf{i} + e^{-t} \mathbf{j} + e^{-t} \mathbf{k} \cdot \frac{-1}{\sqrt{3}} \mathbf{i} + \frac{-1}{\sqrt{3}} \mathbf{j} + \frac{-1}{\sqrt{3}} \mathbf{k} = -\sqrt{3} e^{-t} \).

Step 6: Find Unit Normal Vector

To find \( \mathbf{N}(t) \), compute the derivative of \( \mathbf{T}(t) \) with respect to \( t \), and normalize it: \( \frac{d}{dt}\left(\frac{-1}{\sqrt{3}} \mathbf{i} + \frac{-1}{\sqrt{3}} \mathbf{j} + \frac{-1}{\sqrt{3}} \mathbf{k}\right) = 0 \), hence \( \mathbf{N}(t) \) is not defined as \( \frac{d \mathbf{T}(t)}{dt} \) is zero.

Step 7: Normal Component of Acceleration

Since \( \frac{d \mathbf{T}(t)}{dt} = 0 \), the normal component of acceleration \( a_N \) is zero because \( a_N = \left\lVert \frac{d \mathbf{T}(t)}{dt} \right\rVert = 0 \).

Key Concepts

Position Vector

In physics, particularly in kinematics, a **position vector** is an entity that depicts the position of a point in space concerning the origin of a coordinate system. It tells us the location of a particle at any given time. For the exercise given, the position vector is

• \( \mathbf{r}(t) = e^{-t}(\mathbf{i} + \mathbf{j} + \mathbf{k}) \)

This expression essentially describes the path traced by a moving particle such that as "t" changes, the direction and magnitude of \( \mathbf{r}(t) \) would change according to these exponential decay functions. Here, \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the unit vectors in the direction of the x, y, and z axes respectively. From this vector, we can derive essential characteristics of the particle's motion like velocity and acceleration.

Velocity Vector

The derivative of the position vector with respect to time gives us the **velocity vector**. The velocity vector guides us in understanding a particle's rate of change of position and direction of motion. From the given position vector, we determine:

• \( \mathbf{v}(t) = -e^{-t} \mathbf{i} - e^{-t} \mathbf{j} - e^{-t} \mathbf{k} \)

Each component of this vector tells how the particle's position in each coordinate direction changes over time. The negative sign indicates that as time increases, the position is decreasing exponentially. It essentially means that the particle is retreating faster from the origin initially perhaps but then slowing down as time passes, consistent with an exponential decay.

Tangential Component

The **tangential component** of acceleration refers to the rate at which the speed of the particle changes. It reflects how fast the particle is moving along its direction of motion. To find this, one must understand the particle's speed, which is the magnitude of the velocity vector. Given:

• The speed is \( \lVert \mathbf{v}(t) \rVert = \sqrt{3} e^{-t} \).
• The unit tangent vector \( \mathbf{T}(t) \) is found by normalizing the velocity vector relative to its magnitude.

Afterward, the tangential component of acceleration \( a_T \) is determined as:

• \( a_T = \mathbf{a}(t) \cdot \mathbf{T}(t) = -\sqrt{3} e^{-t} \)

This computation involved a dot product between the acceleration vector and the unit tangent vector, explaining how speed changes along the trajectory.

Normal Component

Unlike the tangential component, the **normal component** of acceleration is concerned with how the direction of the velocity changes, not its magnitude. It typically arises in curved paths where a particle undergoes centripetal acceleration. However, in this situation,

• Given: \( \mathbf{N}(t) \) is undefined as the derivative of the unit tangent vector \( \mathbf{T}(t) \) is zero, \( \frac{d \mathbf{T}(t)}{dt} = 0 \).
• Thus, the normal component \( a_N \) is zero, meaning there is no change in the path's curvature at any point.

This implies that the particle's movement is straightforward, without any directional acceleration, following a straight path even when imagined in a 3D axis.