Problem 4 - Thomas Calculus

Question

\({r}(t)\) is the position of a particle in the \(x y\) -plane at time \(t .\) Find an equation in \(x\) and \(y\) whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of \(t .\) \begin{equation} \mathbf{r}(t)=(\cos 2 t) \mathbf{i}+(3 \sin 2 t) \mathbf{j}, \quad t=0 \end{equation}

Step-by-Step Solution

Answer

Path: \(x^2 + \frac{y^2}{9} = 1\); \(\mathbf{v}(0) = 6\mathbf{j}\); \(\mathbf{a}(0) = -4\mathbf{i}\).

Step 1: Identify the Parametrized Equation

The position vector is given by \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (3 \sin 2t) \mathbf{j} \). This represents the particle's position in terms of \( t \) on the \( xy \)-plane.

Step 2: Eliminate the Parameter t

To find the path of the particle, eliminate \( t \) by using trigonometric identities. Recall \( x = \cos 2t \) and \( y = 3\sin 2t \). Rewriting these in terms of \( \cos^2 \) and \( \sin^2 \): \( x^2 = \cos^2 2t \) and \( \left(\frac{y}{3}\right)^2 = \sin^2 2t \). Utlize the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) to eliminate \( t \), resulting in: \[ x^2 + \left(\frac{y}{3}\right)^2 = 1. \]

Step 3: Find the Velocity Vector

The velocity \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Thus, compute the derivative: \( \mathbf{v}(t) = \frac{d}{dt}[(\cos 2t) \mathbf{i} + (3 \sin 2t) \mathbf{j}] = (-2\sin 2t) \mathbf{i} + (6\cos 2t) \mathbf{j} \).

Step 4: Find the Acceleration Vector

The acceleration \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate the velocity vector: \( \mathbf{a}(t) = \frac{d}{dt}[(-2\sin 2t) \mathbf{i} + (6\cos 2t) \mathbf{j}] = (-4\cos 2t) \mathbf{i} - (12\sin 2t) \mathbf{j} \).

Step 5: Evaluate at t = 0

Plug \( t = 0 \) into the velocity and acceleration vectors to find their values. \( \mathbf{v}(0) = (-2\sin 0) \mathbf{i} + (6\cos 0) \mathbf{j} = 0\mathbf{i} + 6\mathbf{j} \). Similarly, \( \mathbf{a}(0) = (-4\cos 0) \mathbf{i} - (12\sin 0) \mathbf{j} = -4\mathbf{i} + 0\mathbf{j} \).

Key Concepts

Velocity Vector

A velocity vector provides information about both the speed and direction at which a particle is moving. For our exercise, the velocity vector \( \mathbf{v}(t) \) is derived from the position vector \( \mathbf{r}(t) \). In essence, you obtain the velocity vector by taking the derivative of the position vector with respect to time \( t \). This derivative gives us:

• \( \mathbf{v}(t) = (-2\sin 2t) \mathbf{i} + (6\cos 2t) \mathbf{j} \)

Here, \( (-2\sin 2t) \) indicates the rate of change of the \( x \)-component, and \( (6\cos 2t) \) indicates that of the \( y \)-component.
Notice that the components of the velocity vector show variations due to the trigonometric functions, \( \sin \) and \( \cos \), which oscillate between -1 and 1. This means the direction of the velocity isn't constant, rather it changes as time progresses. These oscillations depict how the direction the particle is moving in the xy-plane constantly shifts, which is crucial for understanding motion in multiple dimensions.
Evaluating the velocity vector at \( t = 0 \), we find:

• \( \mathbf{v}(0) = 0\mathbf{i} + 6\mathbf{j} \)

This shows that initially, there is no motion along the \( x \)-axis, but the particle has a velocity of 6 units in the positive \( y \)-direction.

Acceleration Vector

The acceleration vector describes the rate at which a particle's velocity changes with respect to time. Similar to velocity, acceleration has both magnitude and a direction. For our problem, we find the acceleration vector \( \mathbf{a}(t) \) by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \), resulting in:

• \( \mathbf{a}(t) = (-4\cos 2t) \mathbf{i} - (12\sin 2t) \mathbf{j} \)

The \( (-4\cos 2t) \) term represents the acceleration in the \( x \)-direction, while \( - (12\sin 2t) \) represents that in the \( y \)-direction.
This vector indicates how the particle's velocity changes as time passes, through its trigonometric constituents, \( \cos \) and \( \sin \). Due to these oscillating components, the particle's acceleration also changes continuously; sometimes speeding up in a certain direction, other times slowing down or reversing its direction.
Evaluating the acceleration vector for \( t = 0 \) gives:

• \( \mathbf{a}(0) = -4\mathbf{i} + 0\mathbf{j} \)

This means that initially, the particle is subjected to a deceleration of 4 units per time square in the negative \( x \)-direction, with no influence in the \( y \)-direction at that particular instant.

Trigonometric Identities

Trigonometric identities are vital tools used to simplify and solve problems involving parametric equations. In the given exercise, the role of trigonometric identities is highlighted when transforming parametric equations into standard Cartesian forms. For instance, we use the relationships of \( \sin \) and \( \cos \) functions to eliminate the parameter \( t \).
The primary identity relied upon here is:

• \( \cos^2 \theta + \sin^2 \theta = 1 \)

This identity is used to derive the equation of the path in terms of \( x \) and \( y \) by setting \( x = \cos 2t \) and \( y = 3\sin 2t \). Rearranging gives:

• \( x^2 = \cos^2 2t \)
• \( \left(\frac{y}{3}\right)^2 = \sin^2 2t \)

Substituting these into the identity produces the equation:

• \[ x^2 + \left(\frac{y}{3}\right)^2 = 1 \]

This is essentially the equation of an ellipse centered at the origin, showing the path traced out by the particle as it moves in the xy-plane. Trigonometric identities thus provide a bridge between parametric and non-parametric representations, offering deeper insights into the geometric nature of the particle's path.