Problem 164 - Calculus Volume 3

Question

Let \(\mathbf{r}(t)=r \cosh (\omega t) \mathbf{i}+r \sinh (\omega t) \mathbf{j} .\) Find the velocity and acceleration vectors and show that the acceleration is proportional to \(\mathbf{r}(t)\)

Step-by-Step Solution

Answer

The acceleration vector is \( \omega^2 \mathbf{r}(t) \), showing it is proportional to \( \mathbf{r}(t) \).

Step 1: Find the Velocity Vector

The velocity vector is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). We start by differentiating \( \mathbf{r}(t) = r \cosh(\omega t) \mathbf{i} + r \sinh(\omega t) \mathbf{j} \).The derivative of \( \cosh(u) \) with respect to \( u \) is \( \sinh(u) \), and the derivative of \( \sinh(u) \) is \( \cosh(u) \). So, we get:\[ \frac{d}{dt}[r \cosh(\omega t)] = r \omega \sinh(\omega t), \]\[ \frac{d}{dt}[r \sinh(\omega t)] = r \omega \cosh(\omega t). \]Thus, the velocity vector is:\[ \mathbf{v}(t) = r \omega \sinh(\omega t) \mathbf{i} + r \omega \cosh(\omega t) \mathbf{j}. \]

Step 2: Find the Acceleration Vector

The acceleration vector is the second derivative of the position vector \( \mathbf{r}(t) \), which is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiating \( \mathbf{v}(t) = r \omega \sinh(\omega t) \mathbf{i} + r \omega \cosh(\omega t) \mathbf{j} \), we proceed:\[ \frac{d}{dt}[r \omega \sinh(\omega t)] = r \omega^2 \cosh(\omega t), \]\[ \frac{d}{dt}[r \omega \cosh(\omega t)] = r \omega^2 \sinh(\omega t). \]Thus, the acceleration vector is:\[ \mathbf{a}(t) = r \omega^2 \cosh(\omega t) \mathbf{i} + r \omega^2 \sinh(\omega t) \mathbf{j}. \]

Step 3: Show Proportionality of Acceleration to Position

To show that the acceleration is proportional to \( \mathbf{r}(t) \), we compare the acceleration vector \( \mathbf{a}(t) \) and the position vector \( \mathbf{r}(t) \):\( \mathbf{a}(t) = r \omega^2 \cosh(\omega t) \mathbf{i} + r \omega^2 \sinh(\omega t) \mathbf{j} \) is simply \( \omega^2 \) times the position vector:\[ \mathbf{a}(t) = \omega^2 \left( r \cosh(\omega t) \mathbf{i} + r \sinh(\omega t) \mathbf{j} \right) = \omega^2 \mathbf{r}(t). \]Therefore, the acceleration vector is directly proportional to the position vector with proportionality constant \( \omega^2 \).

Key Concepts

Velocity Vector

In vector calculus, a velocity vector represents the rate of change of a position vector with respect to time. It gives both the speed and direction of an object's movement. If you imagine yourself moving along a path, the velocity vector shows exactly where and how fast you're going at any moment.

• Position Vector: This is the initial location of the object, often denoted as \( \mathbf{r}(t) \), where \( t \) stands for time.
• Calculation: To find the velocity vector \( \mathbf{v}(t) \), differentiate the position vector. Differentiation essentially means finding the slope or the rate at which something changes.
• Example: For \( \mathbf{r}(t) = r \cosh(\omega t) \mathbf{i} + r \sinh(\omega t) \mathbf{j} \), the derivative with respect to \( t \) provides the velocity vector \( \mathbf{v}(t) = r \omega \sinh(\omega t) \mathbf{i} + r \omega \cosh(\omega t) \mathbf{j} \).

In summary, the velocity vector reveals the dynamic aspect of motion, capturing how quickly and in what direction things are changing.

Acceleration Vector

An acceleration vector extends the concept of a velocity vector. It represents the rate of change of the velocity vector over time. Think of acceleration as the change in your speed or direction.

• Velocity Vector: This is the vector that gives speed and direction at any time point.
• Calculation: To find the acceleration vector \( \mathbf{a}(t) \), differentiate the velocity vector. This tells us how the velocity is changing.
• Example: Using \( \mathbf{v}(t) = r \omega \sinh(\omega t) \mathbf{i} + r \omega \cosh(\omega t) \mathbf{j} \), differentiating again gives the acceleration vector \( \mathbf{a}(t) = r \omega^2 \cosh(\omega t) \mathbf{i} + r \omega^2 \sinh(\omega t) \mathbf{j} \).

If you look closely, you'll notice that the acceleration vector is proportional to the position vector \( \mathbf{r}(t) \). This means the direction and form of the acceleration match that of the position, changing only by a scaling factor \( \omega^2 \).

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas rather than circles. They can often describe the shape and characteristics of certain curves and physical phenomena.

• Basics: Key hyperbolic functions include the hyperbolic sine \( \sinh \) and hyperbolic cosine \( \cosh \). They are defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \) and \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
• Properties: Just like their trigonometric counterparts, hyperbolic functions exhibit unique properties. For instance, the derivatives exchange roles: the derivative of \( \cosh(x) \) is \( \sinh(x) \), and vice versa.
• Application: In the original exercise, hyperbolic functions describe the motion of the position vector as it evolves over time, giving its characteristic velocities and accelerations.

Understanding hyperbolic functions is crucial in contexts where distances, time, or other continuous factors need a smooth, gradual change, typical in natural processes and certain mathematical models.