Slippery Slope Worksheet Slippery Slope
Investigation: The Sliding Ladder Problem
PROJECT: PYTHAGOREAN MOTION
DOC ID: L1-WS-INV
Student Name
Date
The Big Question
A 10-foot ladder is leaning against a vertical wall. If you pull the base of the ladder away from the wall at a constant rate of 2 feet per second, does the top of the ladder slide down the wall at a constant rate too?
Part 1: Intuition Check
Before doing any math, make a prediction. As the ladder gets closer to the ground, what happens to the speed of the top of the ladder?
It stays constant.
It speeds up.
It slows down.
Part 2: The Model
Sketch the scenario. Label the distance from the wall as \(x\), the height on the wall as \(y\), and the ladder length as \(L\).
Variables
\(x\): dist. from wall
\(y\): height on wall
\(L\): 10 ft (constant)
The Constraint
x² + y² = L²
Part 3: The Calculus
Differentiate the equation \(x^2 + y^2 = 10^2\) with respect to time \(t\).
Solve for \(\frac{dy}{dt}\) (the rate the top is moving):
\(\frac{dy}{dt} = \)
Part 4: Data Analysis
Use your formula to calculate the velocity of the top of the ladder (\(\frac{dy}{dt}\)) at different positions, given the base is moving out at 2 ft/s (\(\frac{dx}{dt} = 2\)).
Distance from wall (\(x\)) Height on wall (\(y\)) dy/dt 6 ft 8 ft 8 ft 6 ft 9.9 ft ~1.41 ft
Conclusion
As the ladder gets closer to the floor (as \(y \rightarrow 0\)), what happens to its downward speed? Why might this be physically dangerous or impossible?
The Speed Limit Challenge
If the ladder is at an angle where \(x = y\), what is the relationship between \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)? Prove it using your equation.
Ladder Slip Slides Ladder Slip
Lesson 1: Related Rates & The Pythagorean Theorem
The Scenario
A 10ft ladder is leaning against a vertical wall.
The base is pulled away at a constant rate of 2 ft/s.
"Does the top slide down at 2 ft/s too?"
10 ft dx/dt = 2
Geometric Constraint
Static Components
Ladder Length (L) = 10
Dynamic Variables
x = Distance from wall
y = Height on wall
The Linkage
x² + y² = 10²
This relationship must hold true at every moment in time (t).
Connecting the Rates
Take the derivative with respect to time (t) :
Step 1: Differentiate
2x \(\frac{dx}{dt}\) + 2y \(\frac{dy}{dt}\) = 0
Step 2: Isolate the unknown rate
\(\frac{dy}{dt} = - \frac{x}{y} \cdot \frac{dx}{dt}\)
Physical Interpretation
The rate \(\frac{dy}{dt}\) depends on the ratio of position \(-x/y\).
As the ladder slides down, x increases and y decreases .
The ratio \(x/y\) grows larger and larger.
The speed must increase.
The Breaking Point
As \(y \rightarrow 0\), the term \(x/y \rightarrow \infty\).
Mathematically, the top of the ladder hits infinite speed just before it hits the ground.
(Note: In the real world, the ladder would lose contact with the wall before this happens!)
Ladder Lead Guide Ladder Lead Guide
Lesson 1: The Sliding Ladder Inquiry
Pacing
50 MIN
Instructional Goals
Students will model physical motion using the Pythagorean Theorem.
Students will differentiate geometric constraints with respect to time.
Students will interpret the significance of the ratio \(x/y\) in the resulting rate equation.
Prerequisites
• Chain Rule mastery
• Implicit Differentiation
• Pythagorean Theorem
Pacing & Facilitation
0-10 min
The Hook & Prediction
Display the first two slides. Pose the "ladder question." Have students complete Part 1 of the worksheet. Resist the urge to give the answer.
Pro Tip: Ask students to visualize pulling the bottom of the ladder slowly, then very fast. Does the top feel like it has a "speed limit"?
10-25 min
Modeling & Calculus
Guide students through Part 2 and 3 . Focus on why we use implicit differentiation. Every variable ($x$, $y$) is a function of time $t$.
Common pitfall: Students often try to plug in specific values (like $x=6$) before differentiating. Emphasize: Differentiate first, substitute second.
25-45 min
Data & Physicality
Students work in pairs to complete the table in Part 4 . They should notice the speed increases rapidly as the ladder falls.
Discussion: At $x=9.9$, $dy/dt \approx 14$ ft/s. Compare this to the constant $dx/dt = 2$ ft/s.
Deep Dive Questions
Why is dy/dt negative?
Because the distance $y$ is decreasing . In calculus, negative rates denote shrinking distances.
What happens when x = 0?
The ladder is vertical. $dy/dt = 0$. The top isn't moving yet because the bottom is moving purely horizontally.
Intersection Impact Slides Intersection Impact
Lesson 2: Objects in Orthogonal Motion
The Collision Course
Car A travels West at 60 mph.North at 45 mph.
Both are heading toward a single Intersection (P).
"At what rate is the distance between the cars changing?"
B A
Crucial Rule: The Sign of the Derivative
Approaching
Distance is Decreasing.
Rate < 0 (Negative)
Receding
Distance is Increasing.
Rate > 0 (Positive)
The Distance Function
x² + y² = z²
Differentiate
2x \(\frac{dx}{dt}\) + 2y \(\frac{dy}{dt}\) = 2z \(\frac{dz}{dt}\)
Simplify
x \(\frac{dx}{dt}\) + y \(\frac{dy}{dt}\) = z \(\frac{dz}{dt}\)
Solve for dz/dt
\(\frac{dz}{dt} = \frac{x \dot{x} + y \dot{y}}{z}\)
Don't Get Trapped!
The rate the cars approach each other is NOT just "Speed A + Speed B".
The angle and their relative positions matter significantly.
Warning
If \(x=3\) and \(y=4\), then \(z=5\).
Plug in your negative rates for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
The resulting \(\frac{dz}{dt}\) will be negative, showing the cars are getting closer.
Crossing Paths Practice Crossing Paths
Practice: Directional Related Rates
UNIT: PYTHAGOREAN MOTION
SKILL: ORTHOGONAL RATES
Name
Date
Rule 1: Toward = Negative Rate | Rule 2: Away = Positive Rate | Equation: x(dx/dt) + y(dy/dt) = z(dz/dt)
01
The Junction Approach
Car A is 4 miles East of an intersection and traveling West at 50 mph. Car B is 3 miles North of the same intersection and traveling South at 40 mph. At what rate is the distance between them changing at this moment?
Givens
\(x =\) ________ \(y =\) ________ \(dx/dt =\) ________ \(dy/dt =\) ________
Find
\(dz/dt =\) ? when \(z = \) ________
Work Area
02
The Separation
A ship sails South from a port at 15 knots. Two hours later, another ship sails East from the same port at 20 knots. At what rate is the distance between the ships increasing 3 hours after the first ship left ?
Work Area (Sketch, Identify Rates, Solve)
03
Mixed Directions
Two people are running from a central point. Person A runs North at 6 ft/s. Person B runs East at 8 ft/s. However, Person A started 50 feet North of the starting point. At the instant Person B has run 120 feet, how fast is the distance between them changing?
Work Area
Critical Thinking
If one car passes through the intersection while the other is still approaching, does the sign of its derivative change? Explain why and what that means for the distance \(z\).
Impact Answer Key Impact Answer Key
Crossing Paths Solutions
TEACHER RESOURCE ONLY
REF: L2-WS-KEY
Problem 01: The Junction Approach
Givens: \(x = 4\), \(y = 3\), \(dx/dt = -50\), \(dy/dt = -40\)
Calculations: \(z = \sqrt{4^2 + 3^2} = 5\)
x(dx/dt) + y(dy/dt) = z(dz/dt)
The distance is decreasing at 64 mph.
Problem 02: The Separation
Givens (at t=3 hrs): Ship 1 has traveled 3 hrs (y = 15 * 3 = 45). Ship 2 left at t=2, so it has traveled 1 hr (x = 20 * 1 = 20).
Rates: \(dy/dt = 15\), \(dx/dt = 20\)
Hypotenuse: \(z = \sqrt{20^2 + 45^2} = \sqrt{400 + 2025} = \sqrt{2425} \approx 49.24\)
20(20) + 45(15) = 49.24(dz/dt)
Problem 03: Mixed Directions
Givens: Person B at 120 ft (\(x = 120\)). Time elapsed = \(120/8 = 15\) sec. Person A's distance \(y = 50 + 6(15) = 140\) ft.
Hypotenuse: \(z = \sqrt{120^2 + 140^2} = \sqrt{14400 + 19600} = \sqrt{34000} \approx 184.39\)
120(8) + 140(6) = 184.39(dz/dt)
Discussion Solution
Yes, the sign of the derivative changes. As the car approaches, \(dx/dt\) is negative (decreasing distance). Once it passes the intersection, it is moving away , so the distance \(x\) begins increasing, making \(dx/dt\) positive. This often results in \(\frac{dz}{dt}\) switching from negative to positive at the point where the distance between the objects is at its absolute minimum.
Distance Chase Slides Distance Chase
Lesson 3: Hypotenuse Rates & Fixed Altitudes
Plane Overhead
A plane flies at a constant altitude of 5 miles.
Its horizontal speed is 600 mph.
"When the plane is 13 miles from you, how fast is its distance from you changing?"
You x y = 5 (Fixed) z = 13
What Stays Constant?
If altitude \(y\) is constant, then its rate of change is:
\(\frac{dy}{dt} = 0\)
Implicit differentiation becomes:
2x \(\frac{dx}{dt} + 0 = 2z \frac{dz}{dt}\)
The Calculation
1
x² + 5² = 13² \(\rightarrow\) x = 12
2
x \(\dot{x}\) = z \(\dot{z}\)
3
12(600) = 13(\(\dot{z}\))
Result
553.8 mph
Note: \(\dot{z}\) is less than the plane's actual speed. Why?
The Limit
As the plane flies further away, the angle of elevation decreases.
As \(z \rightarrow \infty\), the rate \(\dot{z} \rightarrow \dot{x}\).
At a great distance, the plane moves almost directly away from you.
Inverse Question
What is \(\dot{z}\) at the exact moment the plane is directly overhead ?
0 mph
At x=0, distance z isn't changing yet!
Hypotenuse Hunt Workshop Hypotenuse Hunt
Workshop: Fixed Altitudes & Variable Distances
LAB: CALC-03
TOPIC: z(dz/dt) APPLICATIONS
Name
Date
Strategy
When one dimension is fixed, its derivative is zero . This simplifies the Pythagorean derivative to just two terms!
1
The High-Flying Kite
A child is flying a kite at a constant height of 80 feet. The wind carries the kite horizontally away from the child at 10 ft/s. How fast must the child be letting out the string when the string is 100 feet long?
Sketch Here
Values at Instance
y (Altitude) = _________
dy/dt = _________
z (String) = _________
dx/dt (Wind) = _________
Solution Steps
2
Tracking the Rocket
A television camera at ground level is filming the lift-off of a rocket that is rising vertically according to the position \(s(t) = 50t^2\), where \(s\) is in feet and \(t\) is in seconds. The camera is 2000 feet from the launch pad. Find the rate of change of the distance from the camera to the rocket 10 seconds after lift-off.
Step 1: Find height and vertical velocity at t=10
Step 2: Solve for dz/dt
Related Rates Blueprint Related Rates Blueprint
Standard Reference for Pythagorean Motion
The Procedure
1
Sketch & Label
Identify which sides are Variables and which are Constants.
2
Identify Given Rates
List \(\frac{dx}{dt}, \frac{dy}{dt}\) or \(\frac{dz}{dt}\). WATCH THE SIGNS!
3
The Governing Equation
Write \(x^2 + y^2 = z^2\) based on your model.
4
Implicit Differentiation
x \(\dot{x}\) + y \(\dot{y}\) = z \(\dot{z}\)
5
Substitute & Solve
Plugin your "Snapshot" values and solve for the unknown rate.
Common Models
Case A: Sliding Ladder
\(z\) is constant
x \(\dot{x}\) + y \(\dot{y}\) = 0
Case B: Intersection
\(x, y, z\) are variable
x \(\dot{x}\) + y \(\dot{y}\) = z \(\dot{z}\)
Case C: Fixed Altitude
\(y\) is constant
x \(\dot{x}\) = z \(\dot{z}\)
Sign Convention
Approaching Negative (-)
Receding Positive (+)
Shrinking Side Negative (-)
💡
The "Snapshot" Rule
Never plug in the distance values (like "13 miles") before you take the derivative. If you plug them in first, the derivative will be zero because you've turned a dynamic variable into a static number. Derivative first, snapshot second!
Digital Dynamics Project Digital Dynamics
Lab Guide: Modeling Non-Linear Rates
Software
DESMOS / GEOGEBRA
Objective
In this lab, you will build a dynamic model of the "Sliding Ladder" problem. You will visualize how a constant horizontal velocity (\(dx/dt\)) creates a non-linear vertical velocity (\(dy/dt\)) and identify where the speed becomes physically impossible.
1
Build the Geometric Model
Set up your workspace with the following sliders and equations:
Sliders
L = 10 (Ladder length)
v = 2 (Horizontal speed dx/dt)
t = 0 (Time)
Points & Lines
x(t) = v * t
y(t) = sqrt(L^2 - (v * t)^2)
Line: (0, 0) to (x(t), y(t))
2
Define the Velocity Function
Create a new function for the vertical velocity based on your derivative work:
f(t) = - (x(t) / y(t)) * v
Graph this function as Velocity vs. Time . What do you observe about the shape of the graph as \(t\) approaches \(L/v\)?
Observations
1. When is the vertical velocity (\(dy/dt\)) equal to the horizontal velocity (\(v\))? Use the graph to find the exact time \(t\).
2. Describe the vertical asymptote of your velocity graph. What does this mean in terms of the "speed limit" of the ladder?
Graphing Velocity Reflection Graphing Velocity
Reflection: Modeling Non-Linear Rates
DOC ID: L4-REF
Student Name
Date
Visualizing the Asymptote
Sketch the shape of the vertical velocity function \(V_y(t)\) from your digital model. Label the axes and show where the ladder hits the ground.
Time (t)
Velocity (dy/dt)
Comparing Rates
Why does the vertical velocity start at zero even though the horizontal velocity is immediately 2 ft/s?
Predictive Modeling
Suppose we changed the constant velocity \(dx/dt\) from 2 ft/s to 4 ft/s. How would the shape of the velocity graph change? Would the vertical asymptote occur sooner or later?
Diamond Duel Slides Diamond Duel
Lesson 5: Complex Pythagorean Scenarios
The Runner's Triangle
A baseball diamond is a square with sides of 90 feet.
A runner on first base heads for second at 25 ft/s.
"At what rate is the runner's distance from Home Plate changing when they are 30 ft from first base?"
Home 1st z y x=90
Advanced Synthesis
Nested Geometry
Problems where a triangle's sides are themselves functions of other geometric shapes.
Multiple Rates
Scenarios where \(dx/dt\) and \(dy/dt\) are not constant (acceleration or variable speed).
Polar Links
Connecting angular rates (\(d\theta/dt\)) to linear distances in the same triangle.
The Searchlight Problem
A searchlight 40 feet from a straight wall rotates at 0.5 rad/s.
We need the rate at which the beam moves along the wall when it is 30 feet from the point closest to the light.
tan(\(\theta\)) = x / 40
This links an angle rate to a linear Pythagorean rate.
Key Takeaway
Sometimes we use Trigonometry instead of \(a^2+b^2=c^2\), but the process of related rates remains identical.
Mastery Challenge
You are now equipped to model any geometric motion in 2D space. The Pythagorean theorem is the link between where things are and how fast they move .
"Calculus is the bridge from the static to the dynamic."
Complex Curves Challenge Complex Curves Challenge
Summative: Advanced Related Rates Scenarios
Complexity
LEVEL 5/5
01
The Diamond Rundown
A baseball diamond is a square with side length 90 ft. A runner is on 1st base and begins running toward 2nd base at 20 ft/s. At the same instant, the catcher throws the ball toward 2nd base at 100 ft/s. How fast is the distance between the ball and the runner changing when the runner is halfway to 2nd base?
Part A: The Geometry
Hint: The ball moves along the diagonal from Home to 2nd. The runner moves from 1st to 2nd. Sketch the triangle connecting the Ball, Runner, and 2nd Base.
Part B: Solving
02
The Accelerated Pursuit
Ship A is 100 km North of Ship B and is sailing South at a constant speed of 30 km/h. Ship B is sailing East at an accelerating rate, where its position is given by \(x(t) = 5t^2\) km. Find the rate at which the distance between the ships is changing at \(t = 2\) hours.
Mastery Work Area (Show all differentiation and substitution)
The Grand Master Problem
A man 6 feet tall walks at 5 ft/s away from a lamppost that is 15 feet high.
A
At what rate is the length of his shadow changing?
B
At what rate is the tip of his shadow moving?
*Hint: Use similar triangles for the shadow problem, then apply your related rates differentiation steps.
Pythagorean Pro Exit Ticket Pythagorean Pro
Exit Ticket: Unit Mastery
Name
Score
1. The Setup
An object moves along the path \(y = \sqrt{x}\). As it passes the point (4, 2), its \(x\)-coordinate is increasing at 3 units/s. How fast is its distance from the origin changing at this moment?
2. Conceptual Logic
In a right triangle where side \(a\) is constant and side \(b\) is increasing, which will be larger: the rate of change of the hypotenuse (\(dc/dt\)) or the rate of change of the variable side (\(db/dt\))? Explain.
Self-Assessment
Still Climbing
Steady Pace
Peak Mastery
