Problem 43 - Thomas Calculus

Question

Motion along an ellipse A particle moves around the ellipse \((y / 3)^{2}+(z / 2)^{2}=1\) in the \(y z\) -plane in such a way that its position at time \(t\) is $$ \mathbf{r}(t)=(3 \cos t) \mathbf{j}+(2 \sin t) \mathbf{k} $$ Find the maximum and minimum values of \(|\mathbf{v}|\) and \(|\mathbf{a}| .\) (Hint: Find the extreme values of \(|\mathbf{v}|^{2}\) and \(|\mathbf{a}|^{2}\) first and take square roots later.)

Step-by-Step Solution

Answer

The maximum and minimum values of \(|\mathbf{v}|\) and \(|\mathbf{a}|\) are both 3 and 2, respectively.

Step 1: Find the Velocity Vector

The position vector of the particle is given by \( \mathbf{r}(t) = (3 \cos t) \mathbf{j} + (2 \sin t) \mathbf{k} \). The velocity vector, \( \mathbf{v}(t) \), is the time derivative of \( \mathbf{r}(t) \). Compute the derivative to find \( \mathbf{v}(t) = \frac{d}{dt} \left( (3 \cos t) \mathbf{j} + (2 \sin t) \mathbf{k} \right) \). This results in \( \mathbf{v}(t) = (-3 \sin t) \mathbf{j} + (2 \cos t) \mathbf{k} \).

Step 2: Find the Square of the Velocity Magnitude

To find \( |\mathbf{v}|^2 \), we calculate: \( |\mathbf{v}|^2 = ((-3 \sin t)^2 + (2 \cos t)^2) = 9 \sin^2 t + 4 \cos^2 t \).

Step 3: Differentiate \(|\mathbf{v}|^2\) to Find Extreme Values

We need to find when \( |\mathbf{v}|^2 \) is maximized or minimized by taking the derivative with respect to \( t \) and setting it to zero: \( \frac{d}{dt}(19 - 5 \cos^2 t) = 0 \). This simplifies, using trigonometric identities, to: \( 5 \cos t \sin t = 0 \). Thus, \( \cos t = 0 \) or \( \sin t = 0 \).

Step 4: Evaluate \(|\mathbf{v}|^2\) at Critical Points

For \( \cos t = 0 \) (i.e., \( t = \frac{\pi}{2}, \frac{3\pi}{2} \)), \( |\mathbf{v}|^2 = 9 \, (\text{largest}) \). For \( \sin t = 0 \) (i.e., \( t = 0, \pi, 2\pi \)), \( |\mathbf{v}|^2 = 4 \, (\text{smallest}) \).

Step 5: Find the Acceleration Vector

The acceleration vector, \( \mathbf{a}(t) \), is the derivative of \( \mathbf{v}(t) \). Thus, \( \mathbf{a}(t) = \frac{d}{dt}\left((-3 \sin t) \mathbf{j} + (2 \cos t) \mathbf{k}\right) = (-3 \cos t) \mathbf{j} - (2 \sin t) \mathbf{k} \).

Step 6: Find the Square of the Acceleration Magnitude

\( |\mathbf{a}|^2 = ((-3 \cos t)^2 + (-2 \sin t)^2) = 9 \cos^2 t + 4 \sin^2 t \).

Step 7: Differentiate \(|\mathbf{a}|^2\) to Find Extreme Values

As before, to find extreme values of \(|\mathbf{a}|^2\), we take the derivative with respect to \( t \): \( \frac{d}{dt}(13 + 5 \cos^2 t) = 10 \cos t \sin t = 0 \), meaning \( \cos t = 0 \) or \( \sin t = 0 \).

Step 8: Evaluate \(|\mathbf{a}|^2\) at Critical Points

For \( \cos t = 0 \) (i.e., \( t = \frac{\pi}{2}, \frac{3\pi}{2} \)), \( |\mathbf{a}|^2 = 4 \, (\text{smallest}) \). For \( \sin t = 0 \) (i.e., \( t = 0, \pi, 2\pi \)), \( |\mathbf{a}|^2 = 9 \, (\text{largest}) \).

Step 9: Conclusion on Maximum and Minimum Values

The maximum and minimum values of \( |\mathbf{v}| \) are the square roots of the extreme values of \( |\mathbf{v}|^2 \), which are 2 (min) and 3 (max). Similarly, the maximum and minimum values of \( |\mathbf{a}| \) are the square roots of the extreme values of \( |\mathbf{a}|^2 \), which are 2 (min) and 3 (max).

Key Concepts

Motion Along an Ellipse

When studying motion along an ellipse, it's crucial to understand the path a particle takes as it moves. An ellipse is a curve in a plane, encasing two focal points. As a particle moves along the ellipse, its position changes continuously in relation to the shape of the ellipse.

In the given scenario, the particle follows an elliptical path described by the equation \((\frac{y}{3})^2+(\frac{z}{2})^2=1\). This is a standard form of an ellipse equation, indicating the particle moves in a plane with axes lengths determined by the denominators 3 and 2. The particle’s motion can be expressed using a position vector, which gives the coordinates of the particle at any given time \(t\).

Understanding motion along an ellipse is not only about tracing the path but also involves examining how the particle's speed and direction change over time.

Velocity Vector

The velocity vector of a particle describes its speed and direction at any specific moment. To find the velocity vector, you need to take the derivative of the position vector with respect to time.

Given the position vector \( \mathbf{r}(t) = (3 \cos t) \mathbf{j} + (2 \sin t) \mathbf{k} \), the velocity vector, \( \mathbf{v}(t) \), is derived by differentiating \( \mathbf{r}(t) \) with respect to \( t \). This results in \( \mathbf{v}(t) = (-3 \sin t) \mathbf{j} + (2 \cos t) \mathbf{k} \). Here, the coefficients of \( \mathbf{j} \) and \( \mathbf{k} \) define the components of velocity in the respective directions.

In essence, each vector component tells us how fast and in which direction the position is changing along the \(y\) and \(z\) axes at any specific time \(t\). Understanding the velocity vector is crucial because it shows how dynamic the motion is, and becomes a foundational step in solving for acceleration and analyzing overall motion.

Acceleration Vector

The acceleration vector indicates how the velocity of a particle changes over time. It is the derivative of the velocity vector concerning time, showing the rate at which the particle's speed and direction shift.

The previous step provided the velocity vector \( \mathbf{v}(t) = (-3 \sin t) \mathbf{j} + (2 \cos t) \mathbf{k} \). Taking the derivative once more with respect to \( t \), we find the acceleration vector \( \mathbf{a}(t) = (-3 \cos t) \mathbf{j} - (2 \sin t) \mathbf{k} \).

The acceleration vector details in which direction and how strongly the velocity vector is changing. This information helps us understand how forces might be acting on the particle and allows us to explore concepts such as circular motion and dynamics related to elliptical paths.

Extreme Values

Extreme values in the context of motion refer to the maximum and minimum magnitudes of velocity and acceleration. To find these extreme values, we first look at the squares of the magnitudes for simplifying calculations, later taking square roots.

For the given motion, the magnitude squared of the velocity, \(|\mathbf{v}|^2\), is computed as \(9 \sin^2 t + 4 \cos^2 t\). By differentiating this with respect to \(t\) and setting the derivative to zero, the extreme values occur at \(t\) where \( \cos t = 0 \) or \( \sin t = 0 \). Evaluating these points shows that the maximum \(|\mathbf{v}|^2\) is 9 and the minimum is 4, meaning that the magnitude of velocity \(|\mathbf{v}|\) ranges between 2 and 3.

Similarly, for the acceleration, \(|\mathbf{a}|^2 = 9 \cos^2 t + 4 \sin^2 t\), the analysis of critical points gives minimum and maximum magnitudes consistent with those of velocity. This exploration is important as it helps in determining how a system behaves and physically interprets changes in motion conditions.

Derivative

The derivative is a core concept in calculus, serving as a tool to determine the rate at which quantities change. In the context of motion, the derivative helps us find velocity from position and then acceleration from velocity.

Taking a derivative involves applying rules, such as the chain rule and power rule, to break down complex expressions into understandable rates of change. For instance, in our problem, deriving the position vector to obtain the velocity vector and deriving again to get the acceleration is all based on the mechanism provided by derivatives.

Ultimately, derivatives allow us to analyze how physical systems evolve over time, evaluate instantaneous rates, and predict future behavior. They make it possible to simplify complex relationships in motion, such as those involved in elliptical paths, into more manageable computations.