1. Evaluate: To find a point, plug $t$ into both $x(t)$ and $y(t)$.
2. Derivative: Velocity Vector is $\mathbf{v}(t) = \langle x'(t), y'(t) \rangle$.
3. Slope: $\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$. (Y on Top!).
4. Magnitude: Speed is the length of the Velocity Vector.
A path is given by: $x(t) = 3t + 1$ and $y(t) = t^2$.
The Early Moment: Find the $(x, y)$ coordinate at time $t = 0$ and $t = 2$.
The Velocity: Find the Velocity Vector $\mathbf{v}(t)$ for any time $t$.
The Slope: What is the slope of the path ($ rac{dy}{dx}$) at $t = 5$?
A point moves in a circle: $x(t) = \cos(t)$ and $y(t) = \sin(t)$.
The Quarter-Turn: Find the position at $t = \pi/2$.
The Speed: Calculate the Magnitude of the velocity vector at any time $t$.
In the circle problem, your "Speed" ($|\mathbf{v}|$) should be exactly 1. Does this mean the object is moving at a constant speed even though its $x$ and $y$ values are changing? How does this reflect a heart that is "Stable" even in "Seasons of Change"?
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Convert the path $x(t) = t - 5$ and $y(t) = 2t + 1$ into a rectangular equation ($y = f(x)$).
A path is defined by $x(t) = \cos(t)$ and $y(t) = \sin(t)$ from $t = 0$ to $t = \pi$.
Task: Set up the integral for the Arc Length.
(Hint: We found the magnitude was 1 in Part II).
Objective: Explain "Parameter" to a younger sibling using a toy car and a counting game.
The Activity:
1. Say "One-two-three-GO!"
2. On "One," move the car to spot 1.
3. On "Two," move the car to spot 2.
The Lesson: "The car moves because I am counting. The 'Count' is the Parameter. It's the secret number that makes the car go."
Response: __________________________________________________________