Problem 10 - Thomas Calculus

Question

Exercises \(9-12\) 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 \(x^{2}+y^{2}=16\) $$\mathbf{r}(t)=\left(4 \cos \frac{t}{2}\right) \mathbf{i}+\left(4 \sin \frac{t}{2}\right) \mathbf{j} ; \quad t=\pi\( and 3\)\pi / 2$$

Step-by-Step Solution

Answer

The velocity and acceleration at \( t=\pi \) are \(-2\mathbf{i}\) and \(-\mathbf{j}\); at \( t=\frac{3\pi}{2} \), they are \(-2\sqrt{2}\mathbf{i}-\sqrt{2}\mathbf{j}\) and \(\frac{\sqrt{2}}{2}\mathbf{i}-\frac{\sqrt{2}}{2}\mathbf{j}\).

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Given \( \mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j} \), differentiate each component:\[ \frac{d}{dt} \left(4 \cos \frac{t}{2}\right) = -2 \sin \frac{t}{2} \]\[ \frac{d}{dt} \left(4 \sin \frac{t}{2}\right) = 2 \cos \frac{t}{2} \]Thus, the velocity vector \( \mathbf{v}(t) \) is \( \mathbf{v}(t) = -2 \sin \frac{t}{2} \ \mathbf{i} + 2 \cos \frac{t}{2} \ \mathbf{j} \).

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time.Differentiate each component of \( \mathbf{v}(t) = -2 \sin \frac{t}{2} \ \mathbf{i} + 2 \cos \frac{t}{2} \ \mathbf{j} \):\[ \frac{d}{dt} (-2 \sin \frac{t}{2}) = - \cos \frac{t}{2} \]\[ \frac{d}{dt} (2 \cos \frac{t}{2}) = - \sin \frac{t}{2} \]So, the acceleration vector \( \mathbf{a}(t) \) is \( \mathbf{a}(t) = - \cos \frac{t}{2} \ \mathbf{i} - \sin \frac{t}{2} \ \mathbf{j} \).

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

First, evaluate the velocity and acceleration vectors at \( t = \pi \):\[ \mathbf{v}(\pi) = -2 \sin \frac{\pi}{2} \ \mathbf{i} + 2 \cos \frac{\pi}{2} \ \mathbf{j} = -2 \ \mathbf{i} \]\[ \mathbf{a}(\pi) = - \cos \frac{\pi}{2} \ \mathbf{i} - \sin \frac{\pi}{2} \ \mathbf{j} = - \mathbf{j} \]Now, evaluate at \( t = \frac{3\pi}{2} \):\[ \mathbf{v}\left(\frac{3\pi}{2}\right) = -2 \sin \frac{3\pi}{4} \ \mathbf{i} + 2 \cos \frac{3\pi}{4} \ \mathbf{j} = -\sqrt{2} \ \mathbf{i} - \sqrt{2} \ \mathbf{j} \]\[ \mathbf{a}\left(\frac{3\pi}{2}\right) = - \cos \frac{3\pi}{4} \ \mathbf{i} - \sin \frac{3\pi}{4} \ \mathbf{j} = \frac{\sqrt{2}}{2} \ \mathbf{i} - \frac{\sqrt{2}}{2} \ \mathbf{j} \]

Step 4: Sketch the Vectors

To sketch the vectors, plot them on the unit circle defined by \( x^2 + y^2 = 16 \).1. At \( t = \pi \), the point is \((-4, 0)\). The velocity vector \( \mathbf{v}(\pi) = -2 \ \mathbf{i} \) points directly left, and the acceleration vector \( \mathbf{a}(\pi) = - \mathbf{j} \) points down.2. At \( t = \frac{3\pi}{2} \), the point is \(\left(-2\sqrt{2}, -2\sqrt{2}\right)\). The velocity vector \( \mathbf{v}\left(\frac{3\pi}{2}\right) = -\sqrt{2} \ \mathbf{i} - \sqrt{2} \ \mathbf{j} \) points diagonally down-left, and the acceleration vector \( \mathbf{a}\left(\frac{3\pi}{2}\right) = \frac{\sqrt{2}}{2} \ \mathbf{i} - \frac{\sqrt{2}}{2} \ \mathbf{j} \) points diagonally up-left.

Key Concepts

Position Vectors

A position vector describes the location of a point in space relative to an origin. In this case, we're dealing with a particle's movement along a circle, defined mathematically by the equation \(x^2 + y^2 = 16\). This equation tells us the particle moves in a circle of radius 4.
The position vector \(\mathbf{r}(t)\) is given by \((4 \cos \frac{t}{2}) \mathbf{i} + (4 \sin \frac{t}{2}) \mathbf{j}\), indicating that it moves along a trajectory formed by combining cosine and sine functions. These functions ensure that the particle's path loops through all possible points on the circle.
Think of the position vector as an active pointer showing where our particle is at any given time \(t\). The coefficients 4 represent the circle's radius and ensure each component stays within the circle's bounds, taking the particle around at different angles as \(t\) changes.

Velocity Vectors

Velocity vectors illustrate how position changes over time. In mathematical terms, the velocity vector \(\mathbf{v}(t)\) is derived by differentiating the position vector with respect to time, \(t\).
For our particle, if we differentiate \(\mathbf{r}(t)\), the velocity vector becomes \(-2 \sin \frac{t}{2} \ \mathbf{i} + 2 \cos \frac{t}{2} \ \mathbf{j}\).
This vector gives both the direction and speed of our particle’s movement around the circle. It’s like an arrow pointing in the direction our particle is heading.
Importantly, velocity vectors change over time, both in magnitude and direction, as the particle moves along its path. Evaluating these at specific times, like \(t = \pi\) or \(t = \frac{3\pi}{2}\), helps us know exactly where the particle wants to go and how fast it's moving at that moment.

Acceleration Vectors

Acceleration vectors represent how velocity changes over time. Much like velocity is the derivative of position, acceleration is the derivative of velocity.
By finding the derivative of our velocity vector \(-2 \sin \frac{t}{2} \ \mathbf{i} + 2 \cos \frac{t}{2} \ \mathbf{j}\), we obtain the acceleration vector\(\mathbf{a}(t)\), which simplifies to \(-\cos \frac{t}{2} \ \mathbf{i} - \sin \frac{t}{2} \ \mathbf{j}\).
This vector tells us how the speed and direction of our particle are evolving as it travels along the curve and can point in a different direction than the velocity vector.
For instance, at given times like \(t = \pi\), the acceleration shows whether the particle is speeding up, slowing down, or changing direction.

Derivatives in Calculus

Derivatives are foundational in calculus and help us understand how things change.
In the context of motion, derivatives allow us to transition from position (\(\mathbf{r}(t)\)) to velocity (\(\mathbf{v}(t)\)) to acceleration (\(\mathbf{a}(t)\)).
Each derivative represents a deeper level of change: the first derivative shows the rate of change of position, and the second indicates the rate at which velocity itself changes.
Calculus uses derivatives to provide insights into dynamic systems, here helping us track how a particle rotates around a circle over time. With derivatives, we can predict not just where a particle is, but also where it will go and how its journey evolves, mathematically navigating through change in an accurate, logical manner.