#AP Pre-Calculus: Parametric Functions & Planar Motion 🚀

Hey there! Let's get you prepped for the AP Pre-Calculus exam with a deep dive into parametric functions and planar motion. This guide is designed to be your go-to resource, especially the night before the exam. Let's make sure you're feeling confident and ready to ace it! 💪

#4.2 Parametric Functions Modeling Planar Motion

#What are Parametric Functions? 🤔

Parametric functions are your secret weapon for modeling motion in two dimensions. Think of them as a way to track an object's position (both x and y coordinates) as it moves over time. Instead of just having y as a function of x, we have both x and y as functions of a third variable, usually time (t). 🚗

x(t): Represents the horizontal position of the object at time t.
y(t): Represents the vertical position of the object at time t.

Key Concept

These functions are super useful for analyzing the position, velocity, and acceleration of an object moving in a plane. You can even predict its future path! 💫

#Modeling Time ⏱️

The functions x(t) and y(t) tell you the coordinates of a particle's position at any given time t.
The domain of t usually represents a specific time interval during which the motion is being modeled. For example, if t is in the interval [0, 10], you're looking at the motion of the particle over 10 seconds, starting at t = 0.

Exam Tip

Visualizing the path of the particle can be incredibly helpful. If x(t) and y(t) are sinusoidal, you might be looking at a circular or elliptical motion. 👁️

#🔃 Horizontal and Vertical Extrema

#What are Extrema? 👐

Extrema are the maximum and minimum values of a function. For planar motion, we're interested in the horizontal and vertical extrema, which tell us how far the particle moves in the x and y directions.

Horizontal Extrema: The maximum and minimum values of the x(t) function. These represent the farthest left and right points the particle reaches.
Vertical Extrema: The maximum and minimum values of the y(t) function. These represent the highest and lowest points the particle reaches.

Quick Fact

Extrema help you understand the boundaries of the particle's motion. 🤝

#How to Find Them? 🧐

Plugging in Values:
- Choose several values of t within the given domain and evaluate x(t) and y(t).
- Compare the outputs to find the maximum and minimum values for both functions.
Caption: Finding extrema using a table of values.
Algebraic Methods:
- Analyze the shape of the functions x(t) and y(t).
- Look for patterns, symmetry, or periodicity.
- Periodic functions will have extrema at the start and end of their intervals. Symmetric functions will have extrema at their center of symmetry. 🔎

Memory Aid

Think of extrema as the turning points of the motion – where the particle changes direction in either the horizontal or vertical direction. 😁

#📈 Intercepts

#What are Intercepts? 📌

Intercepts are the points where a graph crosses the x or y-axis. For parametric functions, they have a special meaning:

y-intercepts: The points where the particle crosses the y-axis. These occur when x(t) = 0.
x-intercepts: The points where the particle crosses the x-axis. These occur when y(t) = 0.

Key Concept

Real zeros of x(t) correspond to y-intercepts, and real zeros of y(t) correspond to x-intercepts. 💡

Y- and x-intercepts on a graph

Caption: Visualizing x and y intercepts on a graph.

#How to Find Them? ✅

Set x(t) = 0: Solve for t to find the times when the particle crosses the y-axis. These t values will give you the y-intercepts when plugged into y(t).
Set y(t) = 0: Solve for t to find the times when the particle crosses the x-axis. These t values will give you the x-intercepts when plugged into x(t).

Common Mistake

Don't confuse the t-values with the x and y coordinates of the intercepts! Remember to plug the t values back into the respective functions to find the intercept coordinates. ✍️

#Example:

Let's say f(t) = (t^2 - 4, t):

To find x-intercepts, set y(t) = t = 0. Thus, t = 0. The x-intercept is (x(0),0) = (-4,0)
To find y-intercepts, set x(t) = t^2 - 4 = 0. Thus, t = 2 and t = -2. The y-intercepts are (0,2) and (0,-2).

#Final Exam Focus 🎯

Okay, here's the lowdown on what to focus on for the exam:

Parametric Equations: Understand how to use x(t) and y(t) to represent motion.
Extrema: Be able to find horizontal and vertical extrema using both numerical and algebraic methods.
Intercepts: Master the concept of intercepts and how they relate to the zeros of x(t) and y(t).
Visualizing Motion: Practice visualizing the path of a particle based on its parametric equations.

Exam Tip

Time management is key! Start with the questions you know best, and don't get stuck on a single problem for too long. If you're unsure, make an educated guess and move on. You can always come back later if time allows. ⏱️

#Practice Questions

Practice Question

#Multiple Choice Questions

A particle's position is given by x(t) = 2t - 1 and y(t) = t^2. At what time does the particle cross the x-axis? (A) t = 0 (B) t = 0.5 (C) t = 1 (D) t = 2
The horizontal extrema of a particle moving according to x(t) = 3cos(t) and y(t) = 2sin(t) are: (A) -3 and 3 (B) -2 and 2 (C) -1 and 1 (D) 0 and 3

#Free Response Question

A particle moves in the xy-plane so that its position at any time t, where 0 ≤ t ≤ 4, is given by x(t) = t^2 - 3t + 2 and y(t) = t - 1. (a) Find the x- and y-intercepts of the particle's path. (b) Find the horizontal and vertical extrema of the particle's path. (c) Sketch the path of the particle in the xy-plane.

Scoring Rubric:

(a) 2 points: - 1 point for correctly finding the x-intercept(s). - 1 point for correctly finding the y-intercept(s).

(b) 3 points: - 1 point for identifying the correct method to find the extrema. - 1 point for correctly finding the horizontal extrema. - 1 point for correctly finding the vertical extrema.

(c) 2 points: - 1 point for correctly plotting the intercepts and extrema. - 1 point for sketching a smooth curve that connects these points.

#Answers:

MCQ Answers:

(C) To cross the x-axis, y(t) = 0. So, t^2 = 0, which means t = 0. However, we need to check the x-coordinate at t=0 which is 2(0) - 1 = -1. So, the answer is (A) t = 0.5, because the question is asking when the particle crosses the x-axis, which means y(t) = 0, we set t^2 = 0, t = 0, then we need to check the x-coordinate at t=0, which is 2(0) - 1 = -1. The answer is when y(t) = 0, then we solve for t in y(t) = 0, which is t-1 = 0, t = 1, then we plug t = 1 into x(t), which is 2(1) - 1 = 1. So the answer is (C).
(A) The horizontal extrema occur at the maximum and minimum values of x(t) = 3cos(t), which are -3 and 3. FRQ Answers:

(a) x-intercepts: Set y(t) = t - 1 = 0, so t = 1. Then, x(1) = 1^2 - 3(1) + 2 = 0. So, the x-intercept is (0,0). y-intercepts: Set x(t) = t^2 - 3t + 2 = 0, so (t - 1)(t - 2) = 0. Thus, t = 1 and t = 2. Then, y(1) = 0 and y(2) = 1. So, the y-intercepts are (0,0) and (0,1).

(b) Horizontal Extrema: The x(t) function is a quadratic, which has a minimum at its vertex. The t-coordinate of the vertex is t = -(-3)/(2*1) = 1.5. x(1.5) = (1.5)^2 - 3(1.5) + 2 = -0.25. So the horizontal extrema is -0.25. Vertical Extrema: The y(t) function is linear, and in the given domain, the extrema occur at the endpoints. y(0) = -1 and y(4) = 3. So the vertical extrema are -1 and 3. (c) The path starts at (2,-1) at t=0, passes through (0,0) at t=1, reaches (-0.25, 0.5) at t=1.5, passes through (0,1) at t=2, and ends at (6,3) at t=4. The path is a parabola.

Remember, you've got this! Review these notes, take a deep breath, and go crush that exam! 🎉