Spill Spread Slides Unit: Applied Analysis
Oil Spill Dynamics
Modeling environmental disasters through the lens of related rates.
The Reality of the Spill
In April 2010, the Deepwater Horizon disaster released millions of barrels of oil into the Gulf of Mexico.
"How fast is it spreading?"
This is the critical question for containment teams. The rate of surface expansion dictates the length of boom needed and the impact on coastal ecosystems.
Satellite imagery of the slick expansion (2010)
The Geometric Model
We model the spill as a thin cylinder expanding on the ocean surface.
Variables:
V: Total volume (cm³)
r: Radius (cm)
h: Thickness (cm)
A: Surface Area (cm²)
Volume of a cylinder:
\[ V = \pi r^2 h \]
Assume the oil is released at a constant rate \( \frac{dV}{dt} \).
Coupling the Rates
Using the product rule, we differentiate with respect to time \( t \):
\[ \frac{d}{dt}[V] = \frac{d}{dt}[\pi r^2 h] \]
\[ \frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]
Expansion Rate
\( \frac{dr}{dt} \): How fast the radius is growing outward.
Thinning Rate
\( \frac{dh}{dt} \): How fast the oil slick is becoming thinner.
Containment Strategy
The Magnitude Problem
As the radius \( r \) increases, even a small \( \frac{dr}{dt} \) causes a massive increase in surface area. This overwhelms containment booms.
Critical Window
We must calculate when the thickness \( h \) reaches the "critical thinning" limit where evaporation accelerates or chemical dispersants become ineffective.
// Case Discussion Question
If the volume of oil entering the system is fluctuating, how does this complicate our prediction of the slick's thickness over the next 48 hours?
Spill Spread Analysis Worksheet Spill Spread Analysis
Applied Analysis Case Study: Lesson 1
Name: ____________________________________
Date: _____________________________________
SITUATION OVERVIEW
A subsea valve failure has resulted in an oil release into the ocean. The oil rises to the surface and forms a circular slick that maintains a cylindrical shape. Data from satellite imaging and flow-meter readings suggest the volume of the spill is increasing at a rate of \( 2,000 \text{ cm}^3/\text{s} \) .
01
The Geometric Relation
Let \( V \) be the volume, \( r \) the radius, and \( h \) the thickness of the oil slick. Write the equation for the volume of the cylinder and use the product rule to find the general expression for \( \frac{dV}{dt} \).
02
Instantaneous Expansion
At a specific moment, the slick has a radius of 100 meters and a uniform thickness of 0.5 centimeters . If the thickness is decreasing at a rate of 0.01 cm/s , find the rate at which the radius of the slick is increasing at this instant. (Watch your units!)
03
Surface Area and Containment
Containment booms are deployed around the perimeter of the slick. The total length of available boom is 1,200 meters .
Part A:
What is the maximum radius the slick can reach before it exceeds the containment capacity?
Part B:
Using your result from Problem 2, how much time does the crew have to double the amount of boom before the current supply is breached?
Sensitivity Analysis
If the measurement of the thinning rate \( \frac{dh}{dt} \) is found to be 20% higher than initially estimated, how does this affect our calculated value of \( \frac{dr}{dt} \)? Qualitatively explain whether the containment window shrinks or expands.
Computational Model: Environmental Dynamics Lab - APPL-101
Spill Spread Teacher Guide Teacher Facilitation Guide
Lesson 1: Environmental Impact Analysis
Answer Key & Discussion
LEARNING OBJECTIVES
Model a real-world physical scenario using cylindrical volume.
Correctly apply the product rule to time-dependent variables.
Interpret the physical significance of rate magnitudes in a crisis context.
Perform unit conversions between meters, centimeters, and liters/seconds.
PROBLEM SOLUTIONS
01. The Geometric Relation
Equation: \( V = \pi r^2 h \)
Differentiation: \( \frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \)
02. Instantaneous Expansion
Key step: Convert all units to cm or m. Let's use cm.
\( r = 100 \text{ m} = 10,000 \text{ cm} \)
\( h = 0.5 \text{ cm} \)
\( \frac{dh}{dt} = -0.01 \text{ cm/s} \) (negative because it's thinning)
\( \frac{dV}{dt} = 2,000 \text{ cm}^3/\text{s} \)
\( 2000 = \pi [ 2(10000)(0.5)\frac{dr}{dt} + (10000)^2(-0.01) ] \)
03. Surface Area and Containment
Part A: \( 2\pi r = 1200 \implies r = \frac{1200}{2\pi} \approx 191 \text{ meters} \).
Part B: Current radius is 100m. Radius capacity is 191m. Distance to breach = 91m. If \( \frac{dr}{dt} \approx 1 \text{ m/s} \), the crew has roughly 91 seconds before the boom is breached. (Discussion point: The urgency of the math!)
Socratic Discussion Prompts
Misconception Alert:
Students often forget the product rule and assume \( \frac{dr}{dt} \) is the only changing variable. Remind them that for an oil slick, thinning (\( dh/dt \)) is a physical reality due to spreading.
The Reality Check:
Ask: "Is \( dr/dt \) actually constant?" In reality, the rate likely slows as thickness reaches a minimum or as volume release slows. How would that change the model?
Shadow Dynamics Slides Simulation & Inquiry
Shadow Dynamics
Visualizing geometric rates through the 'Streetlight Problem'.
The Midnight Sprint
Have you ever noticed your shadow when walking past a streetlight at night?
"As you walk away at a constant speed, your shadow doesn't just follow you—it seems to race ahead of you."
Today we investigate why the tip of your shadow moves significantly faster than you do.
Similar Triangles
H h x s
H: Height of light source (constant)
h: Height of person (constant)
x: Distance from pole to person
s: Length of the shadow
By Similarity:
\[ \frac{H}{x + s} = \frac{h}{s} \]
"The ratio of the full height to the full distance (base) equals the ratio of the person's height to the shadow's length."
Relating the Speeds
Rearranging the similarity equation for \( s \):
\[ Hs = h(x+s) \implies Hs = hx + hs \implies (H-h)s = hx \implies s = \frac{h}{H-h}x \]
Differentiating with respect to time \( t \):
\[ \frac{ds}{dt} = \frac{h}{H-h} \frac{dx}{dt} \]
Walking Speed
\( \frac{dx}{dt} \)
Shadow Growth Speed
\( \frac{ds}{dt} \)
The Velocity of the Tip
The tip of the shadow is at position \( P = x + s \).
\[ v_{tip} = \frac{dx}{dt} + \frac{ds}{dt} \]
Since \( \frac{ds}{dt} > 0 \), the tip always moves faster than the person.
Critical: Height Ratios Matter!
Shadow Sprint Inquiry Worksheet Shadow Sprint Inquiry
Inquiry Activity: Lesson 2
Observation Log
SID-2026-B
Control Variables
Light Height (\(H\))
6 meters
Person Height (\(h\))
1.8 meters
Walking Speed (\(v_x\))
1.2 m/s
Geometry Reference
"Observe how the ratio \( h / (H-h) \) acts as a constant scaling factor for the shadow growth rate."
Task 1: Deriving the Growth Constant
Based on the control variables provided, calculate the exact constant \( k \) such that \( \frac{ds}{dt} = k \frac{dx}{dt} \). Show your work clearly below.
Task 2: Instantaneous Velocities
Complete the following data points for the given system:
Shadow Growth Rate (\( \frac{ds}{dt} \))
Velocity of Shadow Tip (\( v_{tip} \))
Pattern Generalization
Scenario A: The "Shrinking" Light
Suppose the streetlight is on a hydraulic pole that is slowly descending at a rate of 0.1 m/s while the person is walking away. How does this change the relationship? Identify which variables are no longer constant and how many terms would appear in your derivative.
Scenario B: Non-Linear Movement
A student claims that if you walk towards the light, your shadow tip moves at the same speed as when you walk away from the light. Is this true? Justify your answer using the sign of your derivatives.
© 2026 Applied Analysis Curriculum | Dynamics & Optic Interaction
Geometric Rate Relations Reference Geometric Rate Relations
Fundamental Models for Related Rates
Spherical Volume
\[ V = \frac{4}{3}\pi r^3 \]
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Note: The derivative of volume is surface area times the rate of radius change.
Conical Volume
\[ V = \frac{1}{3}\pi r^2 h \]
\[ \frac{dV}{dt} = \frac{1}{3}\pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]
Often used with similar triangles to relate \( r \) and \( h \).
Pythagorean Theorem
\[ a^2 + b^2 = c^2 \]
\[ 2a \frac{da}{dt} + 2b \frac{db}{dt} = 2c \frac{dc}{dt} \]
Classic for "ladder sliding" or "radar tracking" problems.
Similar Triangles
\[ \frac{y_1}{x_1} = \frac{y_2}{x_2} \]
\[ y_2 = \left(\frac{y_1}{x_1}\right) x_2 \implies \frac{dy_2}{dt} = k \frac{dx_2}{dt} \]
Crucial for eliminating variables in cone and shadow problems.
Reference Card // Department of Applied Analysis
Fluid Dynamics Slides Engineering Lab
Fluid Dynamics
Analyzing non-linear outflow in complex reservoir geometries.
RS-404 RESERVOIR ANALYSIS
The Outflow Risk
When a dam releases water, the vertical drop speed \( dh/dt \) is not constant .
Structural Stress:
As the water level drops, the pressure and horizontal forces on the reservoir walls shift. If the drainage is too fast at certain depths, it can lead to cavitation or wall collapse.
\( V = \int A(h) \, dh \)
Variable Cross-Section
The Hemispherical Tank
For a hemisphere of radius \( R \), the volume of water at height \( h \) is:
\[ V = \pi R h^2 - \frac{1}{3}\pi h^3 \]
Notice that \( V \) is a cubic function of \( h \). This means the relationship between volume and height is non-linear.
Predictive Challenge:
"If the tank is being drained at a constant rate \( dV/dt \), is the water level falling faster when the tank is full or when it is nearly empty ?"
High Volume
Large Surface Area
Low Volume
Small Surface Area
Solving for dh/dt
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \pi R h^2 - \frac{1}{3}\pi h^3 \right) \]
\[ \frac{dV}{dt} = (2\pi Rh - \pi h^2) \frac{dh}{dt} \]
"The term in parentheses represents the Surface Area \( A \) at height \( h \)."
Engineering Formula
\[ \frac{dh}{dt} = \frac{dV/dt}{A(h)} \]
The rate of change of height is inversely proportional to the cross-sectional area.
System Warning
The Critical Threshold
In our lab, you will graph \( dh/dt \) vs \( h \). Identify the height at which the water level drops most rapidly. This is the Structural Danger Zone where turbulent forces are maximized.
If \( |dh/dt| > 0.5 \text{ m/s} \), the auto-vent system must trigger.
|dh/dt| h
Reservoir Outflow Lab Sheet Reservoir Outflow Lab
Civil Engineering Task Force // Lesson 3
Engineer ID: ____________
Date: ____________
Scenario Briefing
A hemispherical reservoir with a radius of 20 meters is being drained for seasonal maintenance. To prevent structural cracking from internal pressure shifts, the rate of water level drop, \( \left| \frac{dh}{dt} \right| \), must not exceed 0.5 m/min . The reservoir is currently being drained at a constant rate of \( 100 \text{ m}^3/\text{min} \) .
1
Static Model Calculation
Recall the volume of a hemispherical water cap: \( V = \pi R h^2 - \frac{1}{3}\pi h^3 \).surface area \( A \) of the water when the height of the water is exactly 15 meters .
2
Rate Evaluation
Determine the rate at which the water level is dropping (\( dh/dt \)) when the height is 15 meters.Check: Is this rate currently safe for the structure? Show your comparison.
3
Predictive Graphing
In the space below, sketch a qualitative graph of the drop speed \( |dh/dt| \) versus the water height \( h \) as the tank empties from \( h=20 \) down to \( h=0 \). Label the axes and identify the point where the speed is greatest.
Water Height (h)
Drop Speed |dh/dt|
Observation Prompt
"As \( h \to 0 \), what happens to the surface area \( A(h) \)? How does this mathematically force the behavior of \( dh/dt \)? Consider what this means for the final moments of drainage."
Findings Summary:
Engineering Recommendation
Based on your graph, at what height (if any) should the outflow valve be partially closed to maintain safety? Justify your answer using the threshold of 0.5 m/min.
FILE: LAB_RES_FLUID_V1 // END OF DOCUMENT
Reservoir Outflow Key Lab Solutions & Rubric
Course: Applied Analysis
Module: Fluid Dynamics
1. Static Model Calculation
Formula: \( A(h) = 2\pi Rh - \pi h^2 \)
Given \( R = 20 \text{ m} \) and \( h = 15 \text{ m} \):
\( A(15) = 2\pi(20)(15) - \pi(15^2) \)
2. Rate Evaluation
Relationship: \( \frac{dV}{dt} = A(h) \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{dV/dt}{A(h)} \)
Note: Outflow is negative, but we evaluate speed (magnitude).
\( \left| \frac{dh}{dt} \right| = \frac{100}{375\pi} \approx \frac{100}{1178.1} \approx 0.0849 \text{ m/min} \)
Evaluation: Safe. 0.0849 m/min is well below the 0.5 m/min threshold.
3. Predictive Graphing Analysis
h (Height) |dh/dt|
Grading Key:
Curve Shape: Should show \( |dh/dt| \) increasing as \( h \) decreases.
Critical Point: Speed is maximized as \( h \to 0 \) because Area \( A \to 0 \).
Threshold: Students must calculate where \( \frac{100}{A(h)} = 0.5 \).
Calculation for Threshold:
Instructor Confidential // Not for Distribution
Piston Motion Slides Mechanical Systems
Piston Motion
Translating angular rotation into linear velocity using the Law of Cosines.
DYNAMIC KINEMATICS
The Heart of the Engine
In a combustion engine, the crankshaft rotates at thousands of RPM. This circular motion must be converted into the back-and-forth motion of a piston.
Critical Inquiry:
At what point in the rotation is the piston moving the fastest? And when does it stop completely? Understanding these moments is essential for timing the fuel injection and spark.
The Mathematical Model
Variables:
r: Crank radius (constant)
L: Connecting rod length (constant)
x: Piston position (variable)
θ: Crank angle (variable)
We relate these using the Law of Cosines on the triangle formed by the crank, rod, and vertical axis.
Governing Equation
\[ L^2 = r^2 + x^2 - 2rx\cos(\theta) \]
This implicitly relates the linear position \( x \) to the angular position \( \theta \).
Coupling Angular & Linear Rates
Differentiate with respect to time \( t \), noting that \( L \) and \( r \) are constant:
\[ 0 = 0 + 2x\frac{dx}{dt} - 2r\left[ \frac{dx}{dt}\cos(\theta) - x\sin(\theta)\frac{d\theta}{dt} \right] \]
\[ \frac{dx}{dt} (x - r\cos\theta) = -rx\sin\theta \frac{d\theta}{dt} \]
Linear Velocity
\( \frac{dx}{dt} \)
Angular Velocity (RPM)
\( \frac{d\theta}{dt} \)
Analyzing the Cycle
Dead Centers
When \( \theta = 0^\circ \) or \( 180^\circ \), \( \sin\theta = 0 \). The velocity \( dx/dt \) becomes zero . The piston stops to reverse direction.
Maximum Speed
Occurs near \( \theta = 90^\circ \), where the leverage of the crank is maximized.
// Design Note
"As engineers, we use this to ensure the piston doesn't exceed the sound barrier or structural fatigue limits during high-speed operation."
Crankshaft Calculus Worksheet Crankshaft Calculus
Mechanical Engineering Division // Applied Analysis
Technician Entry
P-MOD_404
Specification Brief
An internal combustion engine has a crank radius (r) of 5 cm and a connecting rod (L) of 15 cm . The crankshaft is rotating at a constant rate of 3,000 RPM (Revolutions Per Minute).
Input RPM 3,000
Angular Rate (\(\omega\)) 100\(\pi\) rad/s
Law of Cosines
\( L^2 = r^2 + x^2 - 2rx\cos(\theta) \)
Where \( \theta \) is measured from the vertical axis (Top Dead Center).
01
Position Analysis
Determine the position of the piston (\( x \)) when the crank has rotated exactly 90 degrees (\( \pi/2 \) radians).
02
Dynamic Velocity Calculation
Using your answer from Problem 1 and the constant angular rate provided, calculate the linear velocity of the piston (\( dx/dt \)) at this exact moment.
03
Failure Mode Analysis
Suppose the connecting rod length \( L \) is not perfectly constant but expands slightly due to heat. If \( dL/dt = 0.001 \text{ cm/s} \), how would this term affect your differentiation of the Law of Cosines?
Briefly derive the modified rate equation:
Critical Observation Prompt
"At \( \theta = 0^\circ \), the piston velocity is zero regardless of the rod length. Why does the geometric arrangement (collinearity) force this result, and how is it reflected in the sine/cosine terms of your derivative?"
// END OF TASK P-MOD_404 //
Synthesis Seminar Guide Synthesis Seminar Guide
Calculus III: Applied Analysis
Lesson 5: Sensitivity & Error
Objectives
This seminar serves as the capstone for the Related Rates module. Students transition from solving static "textbook" problems to analyzing stochastic (uncertain) systems where measurement error propagates through mathematical models.
Skill A
Error Propagation
Skill B
System Coupling
Skill C
Comm. Strategy
The "Butterfly Effect" Prompts
01
The Measurement Dilemma
In the oil spill model (Lesson 1), suppose our satellite sensor for the thinning rate \( dh/dt \) has a \(\pm 5\%\) margin of error. Does this error result in a \(\pm 5\%\) error in our prediction of radius expansion, or does the non-linear relationship amplify the mistake?
02
The Complexity of Reality
Consider a weather balloon rising (\( dh/dt \)) while being pushed horizontally by wind (\( dx/dt \)). If the wind speed is fluctuating erratically, how do we establish a "confidence interval" for the rate of change of the distance from the launch pad?
Facilitation Framework
Question 1: Sensitivity
Ask: "If the output of a related rate problem is highly sensitive to a specific input, what does that mean for an engineer's design choices?"
Target Answer: It indicates where we need the most expensive, high-precision sensors.
Question 2: Linearization
Ask: "Why do we use derivatives to estimate error propagation instead of just recalculating the formula with a slightly different number?"
Target Answer: Differentials (\( dy = f'(x) dx \)) provide a fast, linear approximation of sensitivity across a whole range of values.
Department of Applied Analysis // Professional Development Series
System Sensitivity Brief System Sensitivity Brief
Final Performance Task // APPL-SENS-05
DEADLINE: END OF SESSION
The Tracking Station Problem
A high-altitude drone is flying horizontally at a constant altitude of 10 km directly away from a radar station. The station measures the line-of-sight distance \( s \) to be 26 km and the rate of change of that distance (\( ds/dt \)) to be 300 m/s .
Calculated Output
Determine the horizontal velocity (\( dx/dt \)) of the drone based on the nominal data.
Sensitivity Analysis
If the radar has a measurement uncertainty of \(\pm 5\) m/s in \( ds/dt \), what is the range of possible horizontal velocities?
System Interdependency
Suppose the drone is also climbing at a rate of 20 m/s . How does adding this second variable change the sensitivity of our calculation? Will our estimate of the horizontal speed become more or less stable?
[INPUT_X: NOMINAL] [INPUT_Y: VAR] [OUTPUT: CONFIRMED]
SENSITIVITY_LOG_05_COMPLETE
Propagation Check Exit Ticket Propagation Check
Exit Ticket // Module Finale
Name
1. In a related rates problem \( y = f(x, t) \), if the rate of change of an input \( dx/dt \) is measured with an error of \( \epsilon \), what mathematical tool do we use to estimate the resulting error in \( dy/dt \)?
A) Integration by Parts
B) Differentials and the Chain Rule
C) The Mean Value Theorem
D) Lagrange Multipliers
2. Explain why a non-linear geometric relationship (like a volume formula \( V = r^3 \)) can make a system highly sensitive to measurement errors as the magnitude of the variables increases.
Synthesis Check: "Real-world math isn't just about finding the derivative; it's about knowing how much you can trust it."
Applied Analysis // End of Series
