Problem 4 - Thomas Calculus

Question

In Exercises \(1-4, \mathbf{r}(t)\) 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 \(t .\) $$ \mathbf{r}(t)=(\cos 2 t) \mathbf{i}+(3 \sin 2 t) \mathbf{j}, \quad t=0 $$

Step-by-Step Solution

Answer

The path is an ellipse: \( x^2 + \frac{y^2}{9} = 1 \). Velocity at \( t=0 \) is \( 6\mathbf{j} \); acceleration is \( -4\mathbf{i} \).

Step 1: Parametric Equation to Cartesian Form

Given the parametric equations, \( x = \cos(2t) \) and \( y = 3\sin(2t) \). To find an equation in terms of \( x \) and \( y \), we use trigonometric identities: \( \cos^2(\theta) + \sin^2(\theta) = 1 \). \[ x = \cos(2t) \Rightarrow \cos(2t) = x \] \[ y = 3\sin(2t) \Rightarrow \sin(2t) = \frac{y}{3} \] \( \cos^2(2t) + \sin^2(2t) = 1 \) implies: \[ x^2 + \left(\frac{y}{3}\right)^2 = 1 \] This simplifies to the ellipse equation: \[ x^2 + \frac{y^2}{9} = 1 \]

Step 2: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Differentiate each component with respect to \( t \): \[ \frac{d}{dt}(\cos(2t)) = -2\sin(2t) \] \[ \frac{d}{dt}(3\sin(2t)) = 6\cos(2t) \] Therefore, the velocity vector is \( \mathbf{v}(t) = (-2\sin(2t)) \mathbf{i} + (6\cos(2t)) \mathbf{j} \).

Step 3: Evaluating Velocity at \( t=0 \)

Substitute \( t=0 \) into the velocity vector \( \mathbf{v}(t) = (-2\sin(2t)) \mathbf{i} + (6\cos(2t)) \mathbf{j} \): \[ \mathbf{v}(0) = (-2\sin(0)) \mathbf{i} + (6\cos(0)) \mathbf{j} = 0\mathbf{i} + 6\mathbf{j} \] Thus, the velocity at \( t=0 \) is \( 6\mathbf{j} \).

Step 4: Find the Acceleration Vector

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

Step 5: Evaluating Acceleration at \( t=0 \)

Substitute \( t=0 \) into the acceleration vector \( \mathbf{a}(t) = (-4\cos(2t)) \mathbf{i} - (12\sin(2t)) \mathbf{j} \): \[ \mathbf{a}(0) = (-4\cos(0)) \mathbf{i} - (12\sin(0)) \mathbf{j} = -4\mathbf{i} + 0\mathbf{j} \] Thus, the acceleration at \( t=0 \) is \( -4\mathbf{i} \).

Key Concepts

Velocity Vector

The velocity vector is a crucial concept when discussing the motion of particles. It represents how fast and in what direction the particle is moving at any given moment. In the context of parametric equations, the velocity vector is essentially the derivative of the position vector with respect to time.
To find the velocity vector \(\mathbf{v}(t)\) from our given position vector \(\mathbf{r}(t) = (\cos(2t)) \mathbf{i} + (3\sin(2t)) \mathbf{j}\), we need to differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\)\:

• For \(\frac{d}{dt}(\cos(2t)) = -2\sin(2t)\)
• For \(\frac{d}{dt}(3\sin(2t)) = 6\cos(2t)\)

Thus, the velocity vector \(\mathbf{v}(t)\) is \((-2\sin(2t)) \mathbf{i} + (6\cos(2t)) \mathbf{j}\). This tells us how the particle's position changes over time.
When evaluating velocity at a specific time like \(t=0\), plug in the value into the velocity vector to get \((0)i + 6j\), meaning the particle is moving straight up with a speed of 6 units per time interval.

Acceleration Vector

The acceleration vector illustrates how the velocity of a particle changes over time. It is found by differentiating the velocity vector with respect to time \(t\). In essence, it shows how the speed and direction of the particle’s motion are changing.
With our example \(\mathbf{v}(t) = (-2\sin(2t)) \mathbf{i} + (6\cos(2t)) \mathbf{j}\), we derive the acceleration vector \(\mathbf{a}(t)\) by differentiating each component:

• \(\frac{d}{dt}(-2\sin(2t)) = -4\cos(2t)\)
• \(\frac{d}{dt}(6\cos(2t)) = -12\sin(2t)\)

So, \(\mathbf{a}(t)\) becomes \((-4\cos(2t)) \mathbf{i} - (12\sin(2t)) \mathbf{j}\). This indicates the manner in which the velocity's direction and magnitude are changing at any time.
At \(t=0\), the acceleration vector evaluates to \(-4\mathbf{i} + 0\mathbf{j}\)\, demonstrating a deceleration purely in the horizontal direction.

Cartesian Equation

Converting parametric equations into Cartesian form is often necessary to analyze the path of a particle in more straightforward algebraic terms. Given parametric equations \(x = \cos(2t)\) and \(y = 3\sin(2t)\), the goal is to eliminate the parameter \(t\) and express \(x\) and \(y\) in a single equation.
By applying the trigonometric identity \(\cos^2(\theta) + \sin^2(\theta) = 1\), connect \(x\) and \(y\):

• Set \(x = \cos(2t)\)\, then \(\cos(2t) = x\).
• Set \(y = 3\sin(2t)\)\, then \(\sin(2t) = \frac{y}{3}\).

Plug these into the identity:
\(\cos^2(2t) + \sin^2(2t) = 1 \Rightarrow x^2 + \left(\frac{y}{3}\right)^2 = 1\).
Finally, simplify to the Cartesian equation of an ellipse: \(x^2 + \frac{y^2}{9} = 1\). This equation now describes the elliptical path of the particle in the xy-plane.

Trigonometric Identities

Trigonometric identities are fundamental in solving parametric equations, especially when converting to Cartesian form. They relate different trigonometric functions to each other and help simplify complex expressions.
In this exercise:\

• The identity \(\cos^2(\theta) + \sin^2(\theta) = 1\) was essential in connecting x and y for eliminating the parameter \(t\)\.
• Recognizing that \(\cos(2t) = x\) and \(\sin(2t) = \frac{y}{3}\) allowed for substitution into the identity to create a single equation.

By utilizing these identities, you can transform equations from their parametric forms into Cartesian equations easily, which then can be used for further operations like finding slopes or intercepts.
Understanding how these identities work allows for significant simplification of problems involving circles, ellipses, and other trigonometric forms.