Problem 10 - Calculus III

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 proportional to the position vector by a factor of \( \omega^2 \).

Step 1: Differentiate to Find Velocity

To find the velocity vector, we need to differentiate \( \mathbf{r}(t) = r \cosh(\omega t) \mathbf{i} + r \sinh(\omega t) \mathbf{j} \) with respect to \( t \). Apply the chain rule:\( \mathbf{v}(t) = \frac{d}{dt}[r \cosh(\omega t)] \mathbf{i} + \frac{d}{dt}[r \sinh(\omega t)] \mathbf{j} \).Differentiating, we get:\( \mathbf{v}(t) = r \omega \sinh(\omega t) \mathbf{i} + r \omega \cosh(\omega t) \mathbf{j} \).

Step 2: Differentiate to Find Acceleration

Next, differentiate the velocity vector \( \mathbf{v}(t) = r \omega \sinh(\omega t) \mathbf{i} + r \omega \cosh(\omega t) \mathbf{j} \) with respect to \( t \):\( \mathbf{a}(t) = \frac{d}{dt}[r \omega \sinh(\omega t)] \mathbf{i} + \frac{d}{dt}[r \omega \cosh(\omega t)] \mathbf{j} \).Differentiating, we get:\( \mathbf{a}(t) = r \omega^2 \cosh(\omega t) \mathbf{i} + r \omega^2 \sinh(\omega t) \mathbf{j} \).

Step 3: Show Proportionality to \( \mathbf{r}(t) \)

To show the acceleration is proportional to \( \mathbf{r}(t) \), compare \( \mathbf{a}(t) = r \omega^2 \cosh(\omega t) \mathbf{i} + r \omega^2 \sinh(\omega t) \mathbf{j} \) with \( \mathbf{r}(t) = r \cosh(\omega t) \mathbf{i} + r \sinh(\omega t) \mathbf{j} \).Notice that:\( \mathbf{a}(t) = \omega^2 (r \cosh(\omega t) \mathbf{i} + r \sinh(\omega t) \mathbf{j}) = \omega^2 \mathbf{r}(t) \).Thus, the acceleration vector \( \mathbf{a}(t) \) is proportional to \( \mathbf{r}(t) \) by a factor of \( \omega^2 \).

Key Concepts

Velocity Vectors

In calculus, a velocity vector is essential for understanding how a point or object moves through space over time. It is derived from the position vector by differentiating it with respect to time. This gives us the velocity vector, which indicates the direction and speed of the movement at any given moment.

In the exercise, the given position vector is \(\mathbf{r}(t) = r \cosh (\omega t) \mathbf{i} + r \sinh (\omega t) \mathbf{j}\).
To find the velocity vector, you actively differentiate each component with respect to \(t\).

The steps include:

• Differentiating \(r \cosh(\omega t)\) yields \(r \omega \sinh(\omega t)\) because of the chain rule.
• Similarly, differentiating \(r \sinh(\omega t)\) gives \(r \omega \cosh(\omega t)\), again using the chain rule.

This results in the velocity vector:
\(\mathbf{v}(t) = r \omega \sinh(\omega t) \mathbf{i} + r \omega \cosh(\omega t) \mathbf{j}\).

It's crucial to grasp the relationship between the original position vector and this new one, as it lays the foundation for deeper analyses of motion, such as acceleration.

Acceleration Vectors

When studying movement, understanding acceleration vectors is another critical step. The acceleration vector describes how the velocity of an object changes with time.

In the context of our exercise, once you've found the velocity vector \(\mathbf{v}(t)\), you differentiate it with respect to time \(t\) to get the acceleration vector.

The process goes as follows:

• Differentiate \(r \omega \sinh(\omega t)\) to obtain \(r \omega^2 \cosh(\omega t)\).
• Similarly, differentiate \(r \omega \cosh(\omega t)\) to yield \(r \omega^2 \sinh(\omega t)\).

Thus, the acceleration vector is:
\(\mathbf{a}(t) = r \omega^2 \cosh(\omega t) \mathbf{i} + r \omega^2 \sinh(\omega t) \mathbf{j}\).

A notable finding here is that the acceleration vector \(\mathbf{a}(t)\) is directly proportional to the original position vector \(\mathbf{r}(t)\) by a factor of \(\omega^2\). This relationship is key in many fields, such as physics and engineering, where understanding the dynamics of systems is crucial.

Hyperbolic Functions

Hyperbolic functions, although they may sound complex, are analogous to trigonometric functions but with unique properties. They crop up frequently in calculus, especially when dealing with exponential functions and their derivatives.

In this exercise, the hyperbolic sine (\(\sinh\)) and hyperbolic cosine (\(\cosh\)) functions play a pivotal role when analyzing the position vector \(\mathbf{r}(t)\).

Some important aspects of hyperbolic functions include:

• \(\cosh(t)\) is defined as \((e^t + e^{-t})/2\) and is always greater than or equal to one.
• \(\sinh(t)\) is given by \((e^t - e^{-t})/2\) and can take any real number value.
• Differentiating \(\cosh(t)\) yields \(\sinh(t)\), and vice versa, just like differentiation of sine and cosine functions in trigonometry.

In the realm of physics and calculus, these functions are indispensable for describing certain types of growth, decay, and oscillatory motion, including scenarios like chain reactions or oscillations in hyperbolic trajectories. Understanding these functions helps unlock a greater comprehension of complex systems.