Problem 9 - 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 \(x^{2}+y^{2}=1\) $$\mathbf{r}(t)=(\sin t) \mathbf{i}+(\cos t) \mathbf{j} ; \quad t=\pi / 4\( and \)\pi / 2$$

Step-by-Step Solution

Answer

Velocity at \( t=\pi/4 \): \( \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \); Acceleration: \( -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \). Velocity at \( t=\pi/2 \): \( -\mathbf{j} \); Acceleration: \( -\mathbf{i} \).

Step 1: Understand the Problem

We are given a position vector \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} \) of a particle moving on the unit circle \( x^2 + y^2 = 1 \). Our task is to find the velocity and acceleration vectors at times \( t = \pi/4 \) and \( t = \pi/2 \).

Step 2: Differentiate the Position Vector to Find Velocity

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time. Differentiating, we get:\[ \mathbf{v}(t) = \frac{d}{dt}((\sin t) \mathbf{i} + (\cos t) \mathbf{j} ) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} \]

Step 3: Differentiate the Velocity Vector to Find Acceleration

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiating, we get:\[ \mathbf{a}(t) = \frac{d}{dt}((\cos t) \mathbf{i} - (\sin t) \mathbf{j}) = -(\sin t) \mathbf{i} - (\cos t) \mathbf{j} \]

Step 4: Calculate Vectors at \( t = \pi/4 \)

Substitute \( t = \pi/4 \) into the velocity and acceleration equations:- Velocity: \[ \mathbf{v}(\pi/4) = (\cos \frac{\pi}{4}) \mathbf{i} - (\sin \frac{\pi}{4}) \mathbf{j} = \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \]- Acceleration: \[ \mathbf{a}(\pi/4) = -(\sin \frac{\pi}{4}) \mathbf{i} - (\cos \frac{\pi}{4}) \mathbf{j} = -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \]

Step 5: Calculate Vectors at \( t = \pi/2 \)

Substitute \( t = \pi/2 \) into the velocity and acceleration equations:- Velocity: \[ \mathbf{v}(\pi/2) = (\cos \frac{\pi}{2}) \mathbf{i} - (\sin \frac{\pi}{2}) \mathbf{j} = 0 \mathbf{i} - 1 \mathbf{j} = -\mathbf{j} \]- Acceleration: \[ \mathbf{a}(\pi/2) = -(\sin \frac{\pi}{2}) \mathbf{i} - (\cos \frac{\pi}{2}) \mathbf{j} = -1 \mathbf{i} - 0 \mathbf{j} = -\mathbf{i} \]

Step 6: Sketch the Vectors on the Circle

To visualize, plot the unit circle and label the positions at \( t = \pi/4 \) and \( t = \pi/2 \). For \( t = \pi/4 \), draw \( \mathbf{v} = \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \) and \( \mathbf{a} = -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \); for \( t = \pi/2 \), draw \( \mathbf{v} = -\mathbf{j} \) and \( \mathbf{a} = -\mathbf{i} \). Vectors should be tangent and perpendicular to the circle, respectively.

Key Concepts

Position Vector

A position vector is a vector that represents the position of a point in a specific space, like the \(xy\)-plane. It is crucial in describing *where* an object is at any given time.
For our exercise, the position vector is given by \(\mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j}\). Here, \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors along the \(x\) and \(y\) axes, respectively.
- The function \(\sin t\) determines the horizontal (\(x\)) position.- The function \(\cos t\) determines the vertical (\(y\)) position.Because \(\sin^2 t + \cos^2 t = 1\), this position vector represents a particle moving along a unit circle centered at the origin.
The importance of understanding the position vector lies in its ability to help visualize and understand the trajectory of a particle's motion. Whether you want to follow the path of a satellite or a rotor in a machine, the concept of a position vector applies.

Particle Motion

Particle motion is all about how a particle moves along a path over time. In our example, the particle moves along a unit circle.
The curve of the path is determined by the position vector \(\mathbf{r}(t)\). By observing changes in the position vector over time, you can understand various aspects of the particle's motion:
- **Direction:** Where is the particle heading? The direction of the position vector changes as the particle moves.- **Speed:** How fast is the particle moving along the path? Speed is related to the magnitude of the velocity vector.An essential feature of particle motion is that the velocity and acceleration vectors don't just describe how fast a particle moves; they also reveal how quickly the motion is changing.
The velocity vector \(\mathbf{v}(t)\) tells you how fast and in which direction the particle's position is changing, while the acceleration vector \(\mathbf{a}(t)\) indicates how quickly the velocity itself is changing.

Differentiation

Differentiation is a mathematical process used to find the rate of change of a function. In the context of particle motion, it's how we find velocity and acceleration vectors.
To differentiate the position vector \(\mathbf{r}(t)\) and find the velocity vector \(\mathbf{v}(t)\), you compute the derivative of each component of the position vector:- Derivative of \(\sin t\) is \(\cos t\)- Derivative of \(\cos t\) is \(-\sin t\)
This gives us \(\mathbf{v}(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j}\), a vector tangent to the unit circle indicating speed and direction.
Further differentiating the velocity vector gives the acceleration vector \(\mathbf{a}(t)\):- Derivative of \(\cos t\) is \(-\sin t\)- Derivative of \(-\sin t\) is \(-\cos t\)
Giving \(\mathbf{a}(t) = - (\sin t) \mathbf{i} - (\cos t) \mathbf{j}\), a vector pointing towards the center of the circle signaling circular motion.
Understanding differentiation helps us analyze how particles behave over time and predict future positions and movements.