Problem 14 - 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 \(t\). $$ \mathbf{r}(t)=\cosh t \mathbf{i}+\sinh t \mathbf{j} $$

Step-by-Step Solution

Answer

Tangential: 0, Normal: 1.

Step 1: Compute the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). By differentiating, we have:\[\mathbf{v}(t) = \frac{d}{dt}[\cosh t \mathbf{i} + \sinh t \mathbf{j}] = \sinh t \mathbf{i} + \cosh t \mathbf{j}.\]

Step 2: Compute the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiating the velocity, we find:\[\mathbf{a}(t) = \frac{d}{dt}[\sinh t \mathbf{i} + \cosh t \mathbf{j}] = \cosh t \mathbf{i} + \sinh t \mathbf{j}.\]

Step 3: Compute the Magnitude of Velocity

Calculate the magnitude of the velocity vector \( \mathbf{v}(t) \):\[||\mathbf{v}(t)|| = \sqrt{(\sinh t)^2 + (\cosh t)^2}.\]Using the identity \((\cosh t)^2 - (\sinh t)^2 = 1\), this simplifies to:\[||\mathbf{v}(t)|| = \sqrt{(\sinh^2 t + \cosh^2 t)} = \sqrt{1} = 1.\]

Step 4: Find the Tangential Component of Acceleration

The tangential component of acceleration \( a_T \) is given by the derivative of the magnitude of the velocity:\[a_T = \frac{d}{dt} ||\mathbf{v}(t)|| = \frac{d}{dt}[1] = 0.\]

Step 5: Compute the Normal Component of Acceleration

The normal component of acceleration \( a_N \) is given by:\[a_N = \sqrt{||\mathbf{a}(t)||^2 - a_T^2}.\]First, calculate \(||\mathbf{a}(t)||\):\[||\mathbf{a}(t)|| = \sqrt{(\cosh t)^2 + (\sinh t)^2} = 1.\]Therefore, substituting in:\[a_N = \sqrt{1^2 - 0^2} = 1.\]

Key Concepts

Position Vector

In mathematics, a position vector represents the position of a point in space. It is usually denoted as \( \mathbf{r}(t) \), where \( t \) represents time. This vector helps us track the motion of a particle relative to a specific origin.
For example, consider the position vector given by \( \mathbf{r}(t) = \cosh t \mathbf{i} + \sinh t \mathbf{j} \). Here:

• \( \cosh t \) and \( \sinh t \) are hyperbolic functions.
• However, unlike circular trigonometric functions, these describe hyperbolic geometry.
• \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors in the Cartesian plane, denoting movement along the x-axis and y-axis, respectively.

This vector indicates that the particle moves along a hyperbola as time progresses, and at any instant \( t \), its exact position can be recalled from the vector expression.

Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is paramount in determining how fast and in what direction a particle is moving. It is the derivative of the position vector with respect to time, essentially showing the rate of change of the particle's position.
To determine \( \mathbf{v}(t) \) for our position vector, differentiate \( \mathbf{r}(t) = \cosh t \mathbf{i} + \sinh t \mathbf{j} \) with respect to \( t \):

• This results in \( \mathbf{v}(t) = \sinh t \mathbf{i} + \cosh t \mathbf{j} \).
• The velocities in the \( \mathbf{i} \) and \( \mathbf{j} \) directions use hyperbolic functions, mirroring the change of position components during motion.

This expression dictates not only the speed but also the direction in which the particle is traveling at any time \( t \).

Acceleration Vector

Acceleration vectors describe how the velocity of a particle changes over time, giving insight into how the particle's speed and direction are evolving. It is the second derivative of the position vector, or simply the derivative of the velocity vector.
Take the velocity \( \mathbf{v}(t) = \sinh t \mathbf{i} + \cosh t \mathbf{j} \) and differentiate it:

• This results in the acceleration vector \( \mathbf{a}(t) = \cosh t \mathbf{i} + \sinh t \mathbf{j} \).
• The structure mirrors the position vector, signalizing consistent movement along a hyperbola.

This vector helps in predicting how the path curves or changes, contributing to how forces could be acting upon the particle over time.

Tangential Component

The tangential component of acceleration, denoted as \( a_T \), is directly associated with changes in the speed of the particle alone, ignoring curvature. It is calculated as the derivative of the magnitude of the velocity vector.
To compute \( a_T \) for our exercise:

• Evaluate the magnitude of \( \mathbf{v}(t) \), given as 1, indicating constant speed.
• Then differentiate to get \( a_T = \frac{d}{dt}[1] = 0 \).

This outcome implies that there is no change in the speed of the particle along its path, meaning it moves uniformly in terms of speed.

Normal Component

The normal component of acceleration, expressed as \( a_N \), relates to changes in the direction of the velocity vector and hence indicates the curvature of the particle's path.
For this exercise:

• Calculate the magnitude of acceleration \( ||\mathbf{a}(t)|| \), which simplifies to 1.
• Use the formula \( a_N = \sqrt{||\mathbf{a}(t)||^2 - a_T^2} \).
• Given that \( a_T = 0 \), it results in \( a_N = 1 \).

This component being non-zero suggests that while the speed of the particle doesn't change, the direction continuously shifts along its hyperbolic path, highlighting the inherent curvature.