Problem 6 - Thomas Calculus

Question

Give the position vectors of particles moving along various curves in the \(x y\) -plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve. Motion on the circle \(x^{2}+y^{2}=16\) \begin{equation} \mathbf{r}(t)=\left(4 \cos \frac{t}{2}\right) \mathbf{i}+\left(4 \sin \frac{t}{2}\right) \mathbf{j} ; \quad t=\pi \text { and } 3 \pi / 2 \end{equation}

Step-by-Step Solution

Answer

Velocity: \(-2\mathbf{i}\) at \(t=\pi\), \(-\sqrt{2}\mathbf{i}-\sqrt{2}\mathbf{j}\) at \(t=3\pi/2\); Acceleration: \(-1\mathbf{j}\) at \(t=\pi\), \(\frac{1}{\sqrt{2}}\mathbf{i}-\frac{1}{\sqrt{2}}\mathbf{j}\) at \(t=3\pi/2\).

Step 1: Verify the Curve Equation

First, confirm that the given position vector \( \mathbf{r}(t) = 4 \cos \left( \frac{t}{2} \right) \mathbf{i} + 4 \sin \left( \frac{t}{2} \right) \mathbf{j} \) traces a circle. Substituting \( \mathbf{x}(t) = 4 \cos \left( \frac{t}{2} \right) \) and \( \mathbf{y}(t) = 4 \sin \left( \frac{t}{2} \right) \) into the circle equation gives: \[ x^2 + y^2 = (4 \cos \left( \frac{t}{2} \right))^2 + (4 \sin \left( \frac{t}{2} \right))^2 = 16 \cos^2 \left( \frac{t}{2} \right) + 16 \sin^2 \left( \frac{t}{2} \right) = 16. \] Since this satisfies \( x^2 + y^2 = 16 \), the vector describes motion on the circle.

Step 2: Find the Velocity Vector

Next, find the velocity vector \( \mathbf{v}(t) \) by differentiating the position vector with respect to \( t \): \[ \mathbf{v}(t) = \frac{d}{dt} \left[ 4 \cos \left( \frac{t}{2} \right) \mathbf{i} + 4 \sin \left( \frac{t}{2} \right) \mathbf{j} \right]. \] Applying the chain rule gives: \[ \mathbf{v}(t) = -2 \sin \left( \frac{t}{2} \right) \mathbf{i} + 2 \cos \left( \frac{t}{2} \right) \mathbf{j}. \]

Step 3: Find the Acceleration Vector

Now, find the acceleration vector \( \mathbf{a}(t) \) by differentiating the velocity vector with respect to \( t \): \[ \mathbf{a}(t) = \frac{d}{dt} \left[ -2 \sin \left( \frac{t}{2} \right) \mathbf{i} + 2 \cos \left( \frac{t}{2} \right) \mathbf{j} \right]. \] Again using the chain rule, the acceleration vector is: \[ \mathbf{a}(t) = -\cos \left( \frac{t}{2} \right) \mathbf{i} - \sin \left( \frac{t}{2} \right) \mathbf{j}. \]

Step 4: Evaluate at \( t = \pi \)

Substitute \( t = \pi \) into the velocity and acceleration vectors: \[ \mathbf{v}(\pi) = -2 \sin \left( \frac{\pi}{2} \right) \mathbf{i} + 2 \cos \left( \frac{\pi}{2} \right) \mathbf{j} = -2 \mathbf{i}, \] \[ \mathbf{a}(\pi) = -\cos \left( \frac{\pi}{2} \right) \mathbf{i} - \sin \left( \frac{\pi}{2} \right) \mathbf{j} = -1 \mathbf{j}. \] So, at \( t = \pi \), the velocity is \( -2 \mathbf{i} \) and the acceleration is \( -1 \mathbf{j} \).

Step 5: Evaluate at \( t = \frac{3\pi}{2} \)

Substitute \( t = \frac{3\pi}{2} \) into the velocity and acceleration vectors: \[ \mathbf{v}\left( \frac{3\pi}{2} \right) = -2 \sin \left( \frac{3\pi}{4} \right) \mathbf{i} + 2 \cos \left( \frac{3\pi}{4} \right) \mathbf{j} = -\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j}, \] \[ \mathbf{a}\left( \frac{3\pi}{2} \right) = -\cos \left( \frac{3\pi}{4} \right) \mathbf{i} - \sin \left( \frac{3\pi}{4} \right) \mathbf{j} = \frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j}. \] At \( t = \frac{3\pi}{2} \), the velocity is \( -\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j} \) and the acceleration is \( \frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j} \).

Step 6: Sketch Vectors on the Circle

Finally, sketch the velocity and acceleration vectors for both times on the circle \( x^2 + y^2 = 16 \). At \( t = \pi \), plot the vector \( -2 \mathbf{i} \) (3 o'clock direction) and \( -1 \mathbf{j} \) (downward) at the point \( (0, -4) \). At \( t = \frac{3\pi}{2} \), plot \( -\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j} \) at \((-\sqrt{2}, -\sqrt{2}) \) and \( \frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j} \) at the same point, ensuring the directions are clear.

Key Concepts

Velocity Vectors

In the context of motion along a circular path, velocity vectors play a significant role in determining the direction and speed of an object at any given moment. A velocity vector is the derivative of a position vector with respect to time. This means it represents how an object's position is changing over time.

For the given position vector on a circle, \[\mathbf{r}(t) = 4 \cos \left( \frac{t}{2} \right) \mathbf{i} + 4 \sin \left( \frac{t}{2} \right) \mathbf{j},\]we found the velocity vector by differentiating this position vector.

• Using differentiation, the velocity vector \(\mathbf{v}(t)\) is \[ \mathbf{v}(t) = -2 \sin \left( \frac{t}{2} \right) \mathbf{i} + 2 \cos \left( \frac{t}{2} \right) \mathbf{j}. \]
• At specific times, such as \( t = \pi \), the velocity vector is \( -2 \mathbf{i} \), which indicates movement horizontally to the left.
• For \( t = \frac{3\pi}{2} \), the velocity becomes \( -\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j} \), suggesting a diagonal movement in the vector direction.

These vectors help illustrate not only the speed but the specific pathway direction that the object follows due to these calculated changes in its position.

Acceleration Vectors

Acceleration vectors are crucial in understanding how an object's velocity changes over time. They indicate the rate at which the velocity vector itself changes.

By differentiating the velocity vector, we can find the acceleration vector. For the circular motion example:

• The velocity vector was given by \[ \mathbf{v}(t) = -2 \sin \left( \frac{t}{2} \right) \mathbf{i} + 2 \cos \left( \frac{t}{2} \right) \mathbf{j}. \]
• We found the acceleration by differentiating this velocity vector, yielding \[ \mathbf{a}(t) = -\cos \left( \frac{t}{2} \right) \mathbf{i} - \sin \left( \frac{t}{2} \right) \mathbf{j}. \]
• At \( t = \pi \), the acceleration is \( -\mathbf{j} \), indicating a downward change in velocity.
• For \( t = \frac{3\pi}{2} \), the acceleration vector is \( \frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j} \), suggesting a balanced change in the direction of movement.

These calculations show how the direction and magnitude of velocity are altering at specific instances, giving a complete picture of motion dynamics.

Motion on a Circle

The concept of motion on a circle is beautifully represented through vectors, as it provides a geometric understanding of how objects travel along circular paths.

For any circle described by \( x^2 + y^2 = 16 \) — with a radius of 4 — the positions are predictable with trigonometric functions:

• The position vector on a circle is \[ \mathbf{r}(t) = 4 \cos \left( \frac{t}{2} \right) \mathbf{i} + 4 \sin \left( \frac{t}{2} \right) \mathbf{j}. \]
• This vector ensures all points satisfy the equation of a circle, confirming that the object truly travels in circular motion.

The motion pathway provides insight into the constant change in direction despite a consistent path distance, clearly illustrated by the corresponding velocity and acceleration vectors. Motion on a circle emphasizes the perpetual presence of centripetal force, as velocity and acceleration always point in perpendicular directions to one another.

Differentiation of Vectors

Understanding how to differentiate vectors is fundamental in physics and engineering since it allows us to analyze how objects vary in motion over time.

The differentiation of vectors helps in transitioning from position to velocity and then to acceleration. The general steps to differentiate a vector concerning time are as follows:

• Identify the components of the vector, such as \[ \mathbf{r}(t) = 4 \cos \left( \frac{t}{2} \right) \mathbf{i} + 4 \sin \left( \frac{t}{2} \right) \mathbf{j}. \]
• Differentiating each component separately allows us to find derivatives that represent real-world concepts like velocity, \[ \mathbf{v}(t) = -2 \sin \left( \frac{t}{2} \right) \mathbf{i} + 2 \cos \left( \frac{t}{2} \right) \mathbf{j}, \] and acceleration, \[ \mathbf{a}(t) = -\cos \left( \frac{t}{2} \right) \mathbf{i} - \sin \left( \frac{t}{2} \right) \mathbf{j}. \]

Differentiation unravels the dynamic behavior embedded within the vectors, shedding light on the instantaneous changes in movements, providing a structured way of predicting future positions and states.