91Ó°ÊÓ

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.

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.

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 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.