Problem 11 - 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 cycloid \(x=t-\sin t, \quad y=1-\cos t\) $$\mathbf{r}(t)=(t-\sin t) \mathbf{i}+(1-\cos t) \mathbf{j} ; \quad t=\pi \operatorname{and} 3 \pi / 2$$

Step-by-Step Solution

Answer

Velocity at \(t = \pi\) is \(2 \textbf{i}\), acceleration is \(-\textbf{j}\). At \(t = \frac{3\pi}{2}\), velocity is \(\textbf{i} - \textbf{j}\), acceleration is \(-\textbf{i}\).

Step 1: Understand the problem

We need to find the velocity and acceleration vectors for a particle moving on a cycloid given by \(\mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j}\). We will calculate these vectors at \(t = \pi\) and \(t = \frac{3\pi}{2}\). Additionally, we'll sketch these vectors on the cycloid.

Step 2: Find the velocity vector

The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\). \[\mathbf{v}(t) = \frac{d}{dt}(t - \sin t) \mathbf{i} + \frac{d}{dt}(1 - \cos t) \mathbf{j} = (1 - \cos t) \mathbf{i} + (\sin t) \mathbf{j}.\]

Step 3: Calculate the velocity at specific times

Substitute \(t = \pi\) into \(\mathbf{v}(t)\):\[\mathbf{v}(\pi) = (1 - \cos \pi) \mathbf{i} + \sin \pi \mathbf{j} = 2 \mathbf{i} + 0 \mathbf{j} = 2\mathbf{i}.\] Substitute \(t = \frac{3\pi}{2}\) into \(\mathbf{v}(t)\):\[\mathbf{v}\left(\frac{3\pi}{2}\right) = (1 - \cos \frac{3\pi}{2}) \mathbf{i} + \sin \frac{3\pi}{2} \mathbf{j} = 1\mathbf{i} - 1\mathbf{j}.\]

Step 4: Find the acceleration vector

The acceleration vector \(\mathbf{a}(t)\) is the derivative of the velocity vector \(\mathbf{v}(t)\).\[\mathbf{a}(t) = \frac{d}{dt}(1 - \cos t) \mathbf{i} + \frac{d}{dt}(\sin t) \mathbf{j} = (\sin t) \mathbf{i} + (\cos t) \mathbf{j}.\]

Step 5: Calculate the acceleration at specific times

Substitute \(t = \pi\) into \(\mathbf{a}(t)\):\[\mathbf{a}(\pi) = \sin \pi \mathbf{i} + \cos \pi \mathbf{j} = 0 \mathbf{i} - 1 \mathbf{j} = -\mathbf{j}.\] Substitute \(t = \frac{3\pi}{2}\) into \(\mathbf{a}(t)\):\[\mathbf{a}\left(\frac{3\pi}{2}\right) = \sin \frac{3\pi}{2} \mathbf{i} + \cos \frac{3\pi}{2} \mathbf{j} = -1 \mathbf{i} + 0 \mathbf{j} = -\mathbf{i}.\]

Step 6: Sketch the vectors on the curve

Draw the cycloid curve defined by the equations \(x = t - \sin t\) and \(y = 1 - \cos t\). Indicate the position of the particle at \(t = \pi\) and \(t = \frac{3\pi}{2}\) on the curve.At \(t = \pi\), sketch the velocity vector \(2 \mathbf{i}\) as a horizontal arrow pointing right and the acceleration vector \(-\mathbf{j}\) as a vertical arrow pointing down.At \(t = \frac{3\pi}{2}\), sketch the velocity vector \(1 \mathbf{i} - 1 \mathbf{j}\) as a diagonal arrow pointing down-right and the acceleration vector \(-\mathbf{i}\) as a horizontal arrow pointing left.

Key Concepts

Velocity Vector

In calculus, the velocity vector is an essential concept that describes how fast and in what direction a particle is moving. It is derived from the position vector \(\mathbf{r}(t) = (x(t), y(t))\) of a particle, by taking the derivative with respect to time, yielding the velocity vector \(\mathbf{v}(t)\). This tells us the rate of change of the position vector over time.
For example, in the cycloid problem above, the velocity vector \(\mathbf{v}(t)\) is found by differentiating each component of the position vector: \(\mathbf{v}(t) = (1 - \cos t) \mathbf{i} + (\sin t) \mathbf{j}\). This expression provides a comprehensive description of the particle's movement along the curve.

Acceleration Vector

The acceleration vector explains how the velocity of a particle changes over time. It is calculated by taking the derivative of the velocity vector. This helps us understand any changes in the speed or direction of the particle.
In our cycloid motion case, after finding the velocity vector, we further differentiate to get the acceleration vector: \(\mathbf{a}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j}\). This provides the instantaneous acceleration of the particle at any given time. Acceleration tells us whether the particle is speeding up, slowing down, or changing direction.

Derivatives

Derivatives play a crucial role in calculating both velocity and acceleration vectors. When we take the derivative of a function, we're essentially finding the rate at which its value changes.
For position vectors, taking the derivative gives us the rate of change of position, otherwise known as the velocity. Further differentiating the velocity vector gives us the rate of change of velocity, known as acceleration.

• The derivative of the position vector, \(\mathbf{r}(t)\), provides the velocity vector, \(\mathbf{v}(t)\).
• The derivative of the velocity vector provides the acceleration vector, \(\mathbf{a}(t)\).

These calculations are fundamental in the study of motion dynamics.

Position Vectors

Position vectors are used to describe the location of a particle in the plane. In this exercise, the position vector is given as \(\mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j}\). This vector depicts the particle's path as a function of time, tracing out a specific curve, in this case, a cycloid.
The key point is that the position vector provides the exact location of a point at any time \(t\). When we evaluate this position vector at specific times, such as \(t = \pi\) or \(t = \frac{3\pi}{2}\), we determine precise points on the cycloid path, useful for sketching or analysis of motion.

Sketching Vectors

Sketching vectors is an integral skill in understanding the geometric representation of vectors, like the position, velocity, and acceleration vectors, on curves.
At each specified instance of time, such as \(t = \pi\) or \(t = \frac{3\pi}{2}\), we plot the position on the curve. From there, the velocity vector at that time shows the direction and speed of the movement, while the acceleration vector shows changes in that movement.

• At \(t = \pi\), the velocity vector \(2 \mathbf{i}\) is a horizontal arrow pointing right.
• The acceleration vector \(-\mathbf{j}\) points downward.

This visual representation helps in intuitively grasping how the particle transitions along its path.