Path and Pace Slides Path and Pace
An Introduction to Parametric Curves
Beyond the Function
Standard functions \( y = f(x) \) tell us the shape of a path, but they are limited:
They cannot model curves that double back (Vertical Line Test).
They don't tell us when an object is at a specific point.
They don't show direction or speed.
y-axis
x-axis
Is this a function? No!
The Power of \( t \)
A parameter is a third variable (often time) that determines both \( x \) and \( y \).
\( x = f(t) \)
\( y = g(t) \)
"As time moves forward, where am I horizontally and where am I vertically?"
Example:
\( x = 2t + 1 \)
\( y = t^2 - 4 \)
t
x
y
0
1
-4
1
3
-3
2
5
0
Orientation: The Arrow of Time
t = 0 t = 4
The orientation is the direction in which the curve is traced as \( t \) increases.
Why it matters:
Two sets of parametric equations might produce the same shape but represent different motion (faster speed, reverse direction).
Graphing Strategies
1. Table of Values
Calculate \( x \) and \( y \) for several values of \( t \) in the given interval.
2. Analyze Components
Look at the behavior of \( x(t) \) and \( y(t) \) separately. Are they periodic? Growing?
3. Technology
Use Desmos or GeoGebra to visualize dynamic changes as you adjust parameters.
Path and Pace Worksheet Path and Pace
Introductory Parametric Exploration
Name:
Date:
1
Component Breakdown
Consider the following parametric equations for a particle's position at time \( t \): \( x = t^2 - 1 \quad \text{and} \quad y = 2t + 1 \) Complete the table for the interval \( -2 \leq t \leq 2 \).
\( t \) \( x = t^2 - 1 \) \( y = 2t + 1 \) \( (x, y) \) -2 -1 0 1 2
2
Tracing the Path
Plot the coordinates from Part 1 on the grid below. Connect the dots smoothly and draw arrows to indicate the orientation of the curve.
x
y
Observations:
Does this curve represent a function \( y = f(x) \)? Why or why not?
3
Same Path, Different Motion
Compare Curve A (from Part 1) to Curve B defined by: \( x = (t/2)^2 - 1 \quad \text{and} \quad y = 2(t/2) + 1 \quad \text{for } -4 \leq t \leq 4 \)
Calculations for Curve B:
Find \( (x, y) \) at \( t = 0 \) and \( t = 4 \):
Analysis:
How does the particle in Curve B move compared to Curve A? Consider speed and time taken to cover the same distance.
Path and Pace Teacher Guide Teacher Guide: Path and Pace
Instructional strategies for introducing parametric equations through dynamic motion.
Lesson Overview
Common Pitfalls
Students forgetting that \( t \) is NOT on the x or y axis.
Confusing "orientation" with "slope".
Treating \( x(t) \) and \( y(t) \) as unrelated functions rather than coordinates.
Instructional Flow
0-10 min: The Hook
Show a video or simulation of a drone following a curved path. Ask: "If I tell you the height as a function of horizontal distance, do I know where the drone is at 10 seconds?" Lead them to the need for a time parameter.
10-25 min: Manual Tracing
Guide students through Part 1 and 2 of the worksheet. Emphasize that the "orientation arrows" represent the direction of time increasing. Discuss why this curve fails the vertical line test but is still "defined".
25-45 min: The Speed Paradox
Introduce Part 3. Use Desmos to demonstrate. Show that while the shape is identical, the particle in Curve B moves slower. This is a crucial distinction: same path, different motion.
Discussion Prompts
Real-world examples?
A racing car vs. a jogger on a track. The track (path) is the same; the parametrization (speed/time) is different.
Backwards time?
What happens if we replace \( t \) with \( -t \)? How does the graph change? (The shape remains, orientation reverses).
Decoding the Curve Slides Algebraic Reconstruction
Decoding the Curve
Eliminating the Parameter to Reveal the Path
The Vanishing Act
Why eliminate \( t \)?
1
Identify the Cartesian equation (the \( x, y \) relationship) of the curve.
2
Recognize standard geometric shapes (lines, parabolas, circles, ellipses).
Current State:
\( x = f(t) \)
Target State:
\( y = F(x) \)
Method 1: Substitution
The Algorithm
1. Pick one equation (usually \( x \)).
2. Solve for \( t \).
3. Substitute into the other equation.
4. Simplify.
Example
\( x = t - 3 \implies t = x + 3 \)
\( y = t^2 + 1 \)
\( y = (x + 3)^2 + 1 \)
Result: A parabola with vertex at \((-3, 1)\).
Method 2: Trigonometric Identities
When \( x \) and \( y \) involve \(\sin\) and \(\cos\), solving for \( t \) is messy. Instead, use:
\(\sin^2 t + \cos^2 t = 1\)
Decoding a Circle
\( x = 4 \cos t \) \( \implies \cos t = x/4 \)
\( y = 4 \sin t \) \( \implies \sin t = y/4 \)
\( (x/4)^2 + (y/4)^2 = 1 \)
\( x^2 + y^2 = 16 \)
The Hidden Catch
Eliminating the parameter can hide domain restrictions.
"The Cartesian equation represents the whole path, but the parametric equation might only cover a portion of it."
Watch Out:
\( x = \cos t \)
\( y = \sin^2 t \)
Eliminating yields: \( y = 1 - x^2 \)
BUT:
Since \( x = \cos t \), we must have \( -1 \leq x \leq 1 \).
The graph is only a segment of the parabola!
Decoding the Curve Worksheet Decoding the Curve
Workshop: Eliminating the Parameter
Name:
Date:
To eliminate the parameter \( t \), express \( t \) in terms of \( x \) (or vice-versa) and substitute, or use trigonometric identities like \(\sin^2 t + \cos^2 t = 1\). Always check if the original parametric equations impose restrictions on \( x \) or \( y \).
01
Linear Path Exploration
Eliminate the parameter and identify the resulting curve:
\( x = 2t + 3 \quad \text{and} \quad y = 4t - 1 \)
Show Work Here
02
Parabolic Trajectory
Express \( y \) as a function of \( x \) and state the domain:
\( x = \sqrt{t} \quad \text{and} \quad y = t - 4 \quad \text{for } t \geq 0 \)
Show Work Here
03
Periodic Motion / Conic Sections
Use a trigonometric identity to eliminate \( t \) and identify the shape:
\( x = 3 \cos t + 1 \quad \text{and} \quad y = 5 \sin t - 2 \)
Show Work Here
The Reflection Challenge
Consider the curve \( x = t^2 \) and \( y = t^4 \).
Decoding the Curve Answer Key Answer Key
Decoding the Curve: Parameter Elimination
Teacher Reference
01
Linear Path
Process:
\( x = 2t + 3 \implies t = \frac{x-3}{2} \)
Final Answer:
\( y = 2x - 7 \)
Shape: A Line
02
Parabolic Trajectory
Process:
\( x = \sqrt{t} \implies t = x^2 \)
Final Answer & Domain:
\( y = x^2 - 4 \)
Domain: \( x \geq 0 \)
Note: Since \( \sqrt{t} \) is always non-negative.
03
Periodic Motion
Process:
\( \cos t = \frac{x-1}{3} \)
Final Answer:
\( \frac{(x-1)^2}{9} + \frac{(y+2)^2}{25} = 1 \)
Shape: An Ellipse
Center: (1, -2)
Challenge Analysis
1. \( x = t^2 \), \( y = t^4 \implies y = (t^2)^2 \implies y = x^2 \).
Key takeaway for students:
The entire parabola is NOT traced. Since \( x = t^2 \), the x-values are restricted to \( x \geq 0 \). This results in only the right half of the parabola \( y = x^2 \). This highlights how parameters control the scope of the path.
Flight Dynamics Slides PHYSICS x CALCULUS
Flight Dynamics
Modeling Projectile Motion with Parametric Equations
The Physics Model
Initial Conditions
Initial Velocity \( v_0 \)
Launch Angle \( \theta \)
Gravity (approx.) \( g = 9.8 \, m/s^2 \)
Gravity only acts vertically. Horizontal velocity remains constant (ignoring air resistance).
Horizontal Position
\( x(t) = (v_0 \cos \theta)t \)
Vertical Position
\( y(t) = -\frac{1}{2}gt^2 + (v_0 \sin \theta)t + h_0 \)
Anatomy of a Flight
Flight Time
The total duration until the object hits the ground.
Set \( y(t) = 0 \)
Max Height
The peak altitude of the trajectory.
\( v_y(t) = 0 \)
Total Range
The total horizontal distance covered during the flight.
\( x(\text{time of flight}) \)
Case Study: The 3-Pointer
Release Rim (10ft)
How do we find the exact launch velocity \( v_0 \) needed to hit a rim 10ft high from 23.75ft away?
System of Equations:
1. \( 23.75 = (v_0 \cos \theta)t \)
2. \( 10 = -16t^2 + (v_0 \sin \theta)t + 7 \)
The Power of decoupling
Independent Control
Adjust launch angle without changing initial thrust. Parametrics allow us to tweak horizontal and vertical forces separately.
Temporal Resolution
We don't just know where the ball lands; we know exactly when it is at the peak of its arc or when it crosses the rim.
Flight Dynamics Worksheet Flight Dynamics
Modeling Projectile Motion Lab
Name: __________________________
Date: __________________________
Master Equations
Horizontal Position
\( x(t) = (v_0 \cos \theta)t \)
Vertical Position
\( y(t) = -4.9t^2 + (v_0 \sin \theta)t + h_0 \)
Note: Uses \( g = 9.8 \, m/s^2 \) for calculations.
01 Scenario: The Human Cannonball
A performer is launched from a cannon at an initial velocity of 25 m/s at an angle of 40°. The mouth of the cannon is 3 meters above the ground.
A. Find \( x(t) \)
B. Find \( y(t) \)
C. Cartesian Path
D. Time of Flight:
Calculate the time \( t \) when the performer hits the ground. Round to two decimal places.
\( t = \) ________ s
02 Peak Performance
For the performer in Scenario 1, calculate the following critical values. Show all differentiation or algebraic steps.
1. Maximum Altitude
Find the time \( t \) where vertical velocity is zero, then find \( y(t) \).
2. Horizontal Range
How far from the cannon should the safety net be placed?
Challenge: The Basket Shot
A player shoots from a distance of 7 meters. The basket is 3.05 meters high. The ball is released from a height of 2.2 meters.
If the player launches at a 45° angle, what must the initial velocity \( v_0 \) be for a perfect "swish"?
\( v_0 = \) __________ m/s
Trajectory Designer Slides Trajectory Designer
MISSION: NAVIGATE THE OBSTACLE GRID
The Problem
Constraint Navigation
Your robot must travel from (0, 0) to (10, 0) in exactly 5 seconds.
Obstacle Alert
A vertical wall exists at x = 5 with a small gap between y = 4 and y = 6.
Input Requirements:
\( x(t) = \text{?} \)
\( y(t) = \text{?} \)
Domain: \( 0 \leq t \leq 5 \)
Strategy 1: Piecewise Control
Break the motion into stages using domain restrictions.
Stage 1: Move to the wall.
Stage 2: Rise to clear the gap.
Stage 3: Move to the finish.
Pseudo-code Example:
If \( 0 \leq t < 2 \): \( x=2.5t, y=0 \)
If \( 2 \leq t < 3 \): \( x=5, y=5(t-2) \)
If \( 3 \leq t \leq 5 \): \( \dots \)
"How do we ensure the motion is smooth (continuous)?"
Strategy 2: Continuous Functions
The Polynomial Approach
Use a single high-degree polynomial or trigonometric function to "bend" around the obstacle without stopping.
\( y(t) = A \sin(Bt) \)
Pros:
No jagged corners; easier for robot motors to handle.
Cons:
Much harder to calculate precise clearance of the gap.
Mission Success Criteria
1. Precision
Does the path clear the obstacle gap by at least 0.5 units? No wall clipping allowed.
2. Timing
Does the robot reach the final destination at exactly \( t = 5 \)? Not before, not after.
3. Elegance
Is the math readable? Did you explain your choices for \( x(t) \) and \( y(t) \)?
DESIGN PHASE: INITIATED
Trajectory Designer Lab Report Trajectory Designer
Parametric Motion Engineering Report
Engineer ID:
Squad:
Start Point
(0, 0)
End Point
(10, 0)
Time Limit
\( t = 5 \)
Obstacle Gap
\( x=5, 4 < y < 6 \)
1
Trajectory Model
Provide your final equations for robot navigation:
\( x(t) = \) ___________
\( y(t) = \) ___________
Define Domain/Restrictions
Visual Simulation Sketch:
2
Verification & Analysis
A. Endpoint Check
Show the calculation for \( x(5) \) and \( y(5) \). Do they match the goal?
B. Obstacle Clearance
At what time \( t \) does \( x(t) = 5 \)? What is the value of \( y \) at that moment?
C. Maximum Speed
Identify the time of maximum velocity. (Optional challenge: Find the speed magnitude).
Engineering Summary
Describe the logic behind your choice of functions. Why did you choose polynomials vs. trigonometric vs. piecewise functions? How did you account for the timing constraint?
