Temporal Mechanics Slides Calculus II: Dynamic Systems
Temporal Mechanics
Implicit Differentiation with Respect to Time: The Foundation of Related Rates
Module 01 // Lesson 01
The Dynamic Perspective
Static vs. Dynamic
In standard differentiation, we find how \(y\) changes with respect to \(x\): \(\frac{dy}{dx}\).
In Related Rates, we find how both \(x\) and \(y\) change with respect to time (\(t\)).
Variable \(x\) becomes \(x(t)\)
Variable \(y\) becomes \(y(t)\)
Rates are expressed as \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
// The General Chain Rule
\[ \frac{d}{dt}[f(u)] = f'(u) \cdot \frac{du}{dt} \]
Every variable is a function of time. Every differentiation requires a "tail" rate.
A Ripple in a Pond
A pebble is dropped into a still pond, causing a circular ripple to expand. The radius \(r\) of the circle is increasing at a constant rate.
"As the circle grows, is the area growing at the same rate as the radius?"
— Discuss with a partner.
The Differential Mechanism
1
Identify the Core Equation
Start with the geometric or physical relationship (e.g., \(A = \pi r^2\)).
2
Differentiate Both Sides w.r.t. Time (\(t\))
Apply the Power Rule and Chain Rule to every term containing a variable.
\[ \frac{d}{dt}[A] = \frac{d}{dt}[\pi r^2] \implies \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \]
3
Solve for the Unknown Rate
Isolate the rate you are looking for before substituting values.
Guided Practice: Volumetric Expansion
Find the rate of change of the volume of a sphere with respect to time if the radius is changing at rate \(\frac{dr}{dt}\).
Step 1: Formula
\[ V = \frac{4}{3}\pi r^3 \]
The Derivative:
Differentiate: \(\frac{d}{dt}[V]\)
Power Rule: \(\frac{4}{3}\pi \cdot 3r^2\)
Chain Rule: \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
Danger Zones
Forgetting the Tail
Writing \(2x\) instead of \(2x \frac{dx}{dt}\). In Related Rates, there are no "silent" variables.
Premature Substitution
Plugging in numbers before differentiating. Constants stay, but values at a "specific instant" must wait.
Product Rule Neglect
Differentiating \(xy\) as \(\frac{dx}{dt} \cdot \frac{dy}{dt}\) instead of using \(x \frac{dy}{dt} + y \frac{dx}{dt}\).
Time to Build
Open your "Chain Rule Mastery" worksheets. We will now apply these mechanics to abstract equations before tackling full word problems.
Focus: Precision
Focus: Notation
Focus: Flow
Chain Rule Mastery Worksheet CALCULUS_MOD_01 // REF_2026
Chain Rule Mastery
Temporal Differentiation & Implicit Mechanics
Student:
Date:
Operational Protocol
Assume all variables (\(x, y, r, h, V, A\)) are functions of time \(t\). For each equation below, differentiate implicitly with respect to \(t\). Clearly show the application of the Chain Rule (the "temporal tail") for every variable term.
Phase I: Geometric Foundations
1. Area of a Circle: \(A = \pi r^2\)
Find \(\frac{dA}{dt}\)
2. Circumference of a Circle: \(C = 2\pi r\)
Find \(\frac{dC}{dt}\)
3. Pythagorean Theorem (Hypotenuse Constant): \(x^2 + y^2 = 10^2\)
Find \(\frac{dy}{dt}\) in terms of \(x, y, \frac{dx}{dt}\)
Phase II: Product & Chain Compositions
4. Volume of a Cylinder: \(V = \pi r^2 h\)
Note: Both \(r\) and \(h\) are changing. Find \(\frac{dV}{dt}\)
5. Ideal Gas Law (Isothermal): \(PV = k\)
where \(k\) is a constant. Find \(\frac{dP}{dt}\) in terms of \(P, V, \frac{dV}{dt}\)
6. Power Rule with Rational Exponent: \(y = \sqrt{x^2 + 1}\)
Find \(\frac{dy}{dt}\) in terms of \(x, \frac{dx}{dt}\)
Technical Checkpoint
Did you apply the Chain Rule to EVERY term?
Is your \(\frac{d(\text{var})}{dt}\) notation clear?
Did you treat constants correctly (derivative = 0)?
PAGE_01_OF_02
Secured Document // System Calculus
Chain Rule Mastery Answer Key TEACHER_RESOURCE // SOLUTIONS_KEY
Chain Rule Mastery
Answer Key & Solution Paths
Internal Use Only
Phase I: Geometric Foundations
1. Area of a Circle: \(A = \pi r^2\)
\(\frac{d}{dt}[A] = \frac{d}{dt}[\pi r^2]\)
2. Circumference: \(C = 2\pi r\)
\(\frac{d}{dt}[C] = \frac{d}{dt}[2\pi r]\)
3. Pythagorean Theorem: \(x^2 + y^2 = 10^2\)
\(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)
Phase II: Product & Chain Compositions
4. Volume of a Cylinder: \(V = \pi r^2 h\)
\(\frac{dV}{dt} = \pi \left[ \frac{d}{dt}(r^2) \cdot h + r^2 \cdot \frac{d}{dt}(h) \right]\) (Product Rule)
5. Ideal Gas Law: \(PV = k\)
\(P\frac{dV}{dt} + V\frac{dP}{dt} = 0\)
6. Rational Exponent: \(y = \sqrt{x^2 + 1}\)
\(y = (x^2 + 1)^{1/2}\)
Common Student Error: Forgetting to differentiate the right side of equations like \(x^2 + y^2 = 100\), resulting in a non-zero constant.
Answer Key // Related Rates Module 01
Linear Motion Slides Calculus II: Dynamic Systems
Linear Motion
Modeling Orthogonal Rates with the Pythagorean Theorem
Module 01 // Lesson 02
The S.E.D.S. Framework
01
Sketch
Draw the diagram. Label variables (changing) and constants (fixed).
02
Equation
Relate the variables using geometry (e.g., \(a^2 + b^2 = c^2\)).
03
Derivative
Differentiate both sides with respect to time (t).
04
Substitute
Plug in the given values for the specific "snapshot" in time.
Case Study: The Leaning Ladder
A 13-foot ladder leans against a wall. The bottom of the ladder slides away from the wall at 2 ft/sec.
How fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?
x = 5
y = ?
Length = 13
The Solution Path
1. Relationship
\[ x^2 + y^2 = 13^2 \]
Note: 13 is a constant, so its derivative will be zero.
2. Differentiation
\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \]
3. Snapshot Analysis
When \(x = 5\), solve for \(y\):
\(5^2 + y^2 = 13^2 \implies y = 12\)
Given: \(\frac{dx}{dt} = 2\)
Substitute and solve:
\(2(5)(2) + 2(12)\frac{dy}{dt} = 0\)
\(20 + 24\frac{dy}{dt} = 0\)
\(\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6}\text{ ft/sec}\)
Direction Matters
Positive Rate
The distance is increasing. The objects are moving apart.
Negative Rate
The distance is decreasing. The objects are moving together.
SIMULATION ACTIVE
Triangle Motion Workshop
Today's challenge involves intersection problems and receding objects. Remember: Sketch first, differentiate second, substitute last.
Mission Parameters:
Model vehicle intersections
Calculate rates of separation
Interpret sign values for motion direction
Triangle Motion Workshop Worksheet CALCULUS_MOD_01 // TRIANGLE_DYNAMICS
Triangle Motion Workshop
Modeling Linear Rates in Orthogonal Systems
Student:
UNIT: 01 // PROBLEM_SET: 02
Sketch
Equation
Derivative
Substitute
Scenario 01: Structural Failure
1. The Leaning Ladder
A 25-foot ladder is leaning against a vertical wall. The base of the ladder is pulled away from the wall at a rate of 3 ft/sec. How fast is the top of the ladder sliding down the wall at the instant when the base is 7 feet from the wall?
S: Sketch & Variables
[Diagram Area]
E: Equation & D: Derivative
S: Final Substitution & Result
Scenario 02: Traffic Reconstruction
2. The Approaching Intersection
Car A is traveling west toward an intersection at 50 mph. Car B is traveling north away from the same intersection at 60 mph. At the instant Car A is 0.3 miles east of the intersection and Car B is 0.4 miles north of the intersection, at what rate is the distance between the two cars changing?
S: Sketch & Variables
[Diagram Area]
E: Equation & D: Derivative
S: Final Substitution & Result
Scenario 03: Marine Mechanics
3. The Winch & The Boat
A boat is being pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 12 feet higher than the bow. If the rope is being pulled in at a constant rate of 4 ft/sec, how fast is the boat approaching the dock when it is 5 feet from the dock?
S: Sketch & Variables
[Diagram Area]
Work Area: Model Construction
Final Calculation & Unit Analysis
Analysis Checkpoint:
Ensure that rates of approach (distance decreasing) are treated as negative derivatives, and rates of separation (distance increasing) are treated as positive derivatives.
DYNAMICS_DOC_V2
Triangle Motion Workshop Answer Key TEACHER_RESOURCE // SOLUTIONS_KEY
Triangle Motion Workshop
Model Solutions & Grading Rubric
Instructor View
1. The Leaning Ladder Solution
Setup:
Eq: \(x^2 + y^2 = 25^2\) (Ladder length is constant)
Diff: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0\)
Values at Snapshot:
\(x = 7\)
\(7^2 + y^2 = 25^2 \implies 49 + y^2 = 625 \implies y = 24\)
\(\frac{dx}{dt} = 3\)
\(7(3) + 24\frac{dy}{dt} = 0 \implies 21 + 24\frac{dy}{dt} = 0\)
\(\frac{dy}{dt} = -\frac{21}{24} = -\frac{7}{8}\text{ ft/sec} \approx -0.875\text{ ft/sec}\)
(Interpretation: The top is sliding DOWN at 7/8 ft/sec)
2. The Approaching Intersection Solution
Setup:
Eq: \(a^2 + b^2 = z^2\) (where \(z\) is separation distance)
Diff: \(2a\frac{da}{dt} + 2b\frac{db}{dt} = 2z\frac{dz}{dt}\)
Values at Snapshot:
\(a = 0.3, \frac{da}{dt} = -50\) (distance decreasing)
\(b = 0.4, \frac{db}{dt} = 60\) (distance increasing)
\(z = \sqrt{0.3^2 + 0.4^2} = 0.5\)
\(0.3(-50) + 0.4(60) = 0.5\frac{dz}{dt} \implies -15 + 24 = 0.5\frac{dz}{dt}\)
\(9 = 0.5\frac{dz}{dt}\)
\(\frac{dz}{dt} = 18\text{ mph}\)
(Interpretation: The cars are moving apart at 18 mph)
3. The Winch & The Boat Solution
Setup:
Eq: \(x^2 + 12^2 = L^2\) (where \(x\) is horizontal distance, \(L\) is rope length)
Diff: \(2x\frac{dx}{dt} + 0 = 2L\frac{dL}{dt} \implies x\frac{dx}{dt} = L\frac{dL}{dt}\)
Values at Snapshot:
\(x = 5\)
\(5^2 + 12^2 = L^2 \implies 25 + 144 = 169 \implies L = 13\)
\(\frac{dL}{dt} = -4\) (rope is being pulled IN)
\(5\frac{dx}{dt} = 13(-4) \implies 5\frac{dx}{dt} = -52\)
\(\frac{dx}{dt} = -10.4\text{ ft/sec}\)
(Interpretation: The boat is approaching the dock at 10.4 ft/sec)
Grading Note:
Award 2 points for correct diagram/setup, 3 points for differentiation (check for dt terms), and 5 points for snapshot substitution and correct units.
Answer Key // Related Rates Module 01
Geometric Solids Slides Calculus II: Dynamic Systems
Expanding Solids
Modeling Volumetric Rates and Surface Area Dynamics
Module 01 // Lesson 03
3D Expansion Mechanics
Linear vs. Volumetric
If the radius of a sphere grows at a constant rate, does the volume grow at a constant rate?
Consider the Sphere:
\[ V = \frac{4}{3}\pi r^3 \]
Its derivative reveals the truth:
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
// Analysis Insight
The rate of change of volume depends not only on how fast the radius is growing, but also on how large the radius already is.
"As the sphere gets bigger, the same increase in radius results in a much larger increase in volume."
Clinical Application: Tumor Growth
A medical researcher observes that a spherical tumor's radius is increasing at a rate of 0.5 mm per month.
Calculate the rate of increase of the tumor's volume when the radius is 10 mm.
Why is this specific calculation critical for treatment planning?
r
Cube Dynamics
Volume & Surface Area
// Volume
\(V = s^3 \implies \frac{dV}{dt} = 3s^2 \frac{ds}{dt}\)
// Surface Area
\(A = 6s^2 \implies \frac{dA}{dt} = 12s \frac{ds}{dt}\)
Interconnected Rates
Often, we are given the rate of change of surface area and asked to find the rate of change of volume.
Master Key Strategy:
Use the given rate to find \(\frac{ds}{dt}\) first, then substitute it into the second rate equation.
Unit Verification Protocol
📏
Linear Rate
\(\frac{ds}{dt}\) or \(\frac{dr}{dt}\)
units / time (e.g., cm/s)
⬜
Area Rate
\(\frac{dA}{dt}\)
units² / time (e.g., cm²/s)
🧊
Volume Rate
\(\frac{dV}{dt}\)
units³ / time (e.g., cm³/s)
If your derivative's units don't match the dimension, check your Power Rule application.
Volumetric Growth Activity
From inflating balloons to melting ice cubes, we will now model 3D systems in transition. Watch your exponents.
Spheres
Cubes
Surface Area
Volumetric Growth Activity Worksheet CALCULUS_MOD_01 // VOLUMETRIC_DYNAMICS
Volumetric Growth
Modeling 3D Rate Interdependencies
Student:
FORM_ID: EXP_SOLIDS_03
Sphere Protocol:
\(V = \frac{4}{3}\pi r^3 \quad | \quad A = 4\pi r^2\)
Cube Protocol:
\(V = s^3 \quad | \quad A = 6s^2\)
1
The Inflating Sphere
Air is being pumped into a spherical balloon at a rate of 100 cm³/sec. How fast is the radius of the balloon increasing when the diameter is 50 cm?
Setup (Differentiate & Identify)
Snapshot Calculation
2
The Melting Ice Cube
A cube of ice melts such that its side length \(s\) decreases at a constant rate of 0.2 cm/min. At what rate is the volume of the cube decreasing when the side length is 10 cm?
Differential Model
Rate Analysis & Units
3
Interconnected Rates: Surface & Volume
The surface area of a spherical soap bubble is increasing at a rate of 2 cm²/sec. Find the rate at which the volume is increasing at the instant when the radius is 4 cm.
Step A: Find \(\frac{dr}{dt}\) from \(\frac{dA}{dt}\)
Step B: Use \(\frac{dr}{dt}\) to find \(\frac{dV}{dt}\)
Dimension Verification:
Length: cm/s
Area: cm²/s
Volume: cm³/s
VOL_SET_03
Volumetric Growth Answer Key TEACHER_RESOURCE // SOLUTIONS_KEY
Volumetric Growth
Model Solutions & Unit Verification
Instructor View
1. The Inflating Sphere Solution
Setup:
Eq: \(V = \frac{4}{3}\pi r^3\)
Diff: \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
Snapshot:
\(d = 50 \implies r = 25\)
\(\frac{dV}{dt} = 100\)
\(100 = 4\pi(25)^2 \frac{dr}{dt} \implies 100 = 2500\pi \frac{dr}{dt}\)
\(\frac{dr}{dt} = \frac{100}{2500\pi} = \frac{1}{25\pi}\text{ cm/sec} \approx 0.0127\text{ cm/sec}\)
2. The Melting Ice Cube Solution
Setup:
Eq: \(V = s^3\)
Diff: \(\frac{dV}{dt} = 3s^2 \frac{ds}{dt}\)
Snapshot:
\(s = 10\)
\(\frac{ds}{dt} = -0.2\) (decreasing)
\(\frac{dV}{dt} = 3(10)^2(-0.2) = 300(-0.2)\)
\(\frac{dV}{dt} = -60\text{ cm}^3\text{/min}\)
(Note: The volume is decreasing at 60 cm³/min)
3. Interconnected Rates Solution
Step A: Find \(\frac{dr}{dt}\) from Surface Area Rate
\(A = 4\pi r^2 \implies \frac{dA}{dt} = 8\pi r \frac{dr}{dt}\)
Given: \(\frac{dA}{dt} = 2, r = 4\)
\(2 = 8\pi(4) \frac{dr}{dt} \implies 2 = 32\pi \frac{dr}{dt}\)
\(\frac{dr}{dt} = \frac{1}{16\pi}\text{ cm/sec}\)
Step B: Find \(\frac{dV}{dt}\) using \(\frac{dr}{dt}\)
\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
Substitute \(r = 4\) and \(\frac{dr}{dt} = \frac{1}{16\pi}\):
\(\frac{dV}{dt} = 4\pi(4)^2 \left(\frac{1}{16\pi}\right)\)
\(\frac{dV}{dt} = 4\pi(16) \cdot \frac{1}{16\pi}\)
\(\frac{dV}{dt} = 4\text{ cm}^3\text{/sec}\)
Pedagogical Insight:
Problem 3 demonstrates the "linkage" between area and volume. Students often try to find a direct formula relating V and A, but using the radius as the intermediary variable is more robust for general calculus application.
Answer Key // Related Rates Module 01
Angular Velocity Slides Calculus II: Dynamic Systems
Angular Velocity
Trigonometric Rates and Rotational Dynamics
Module 01 // Lesson 04
Angles in Motion
The Angular Rate
In problems involving tracking or rotation, the variable is an angle \(\theta\).
The rate of change is \(\frac{d\theta}{dt}\) (Angular Velocity).
// Differentiation Recall
\(\frac{d}{dt}[\tan\theta] = \sec^2\theta \cdot \frac{d\theta}{dt}\)
\(\frac{d}{dt}[\sin\theta] = \cos\theta \cdot \frac{d\theta}{dt}\)
\(\frac{d}{dt}[\cos\theta] = -\sin\theta \cdot \frac{d\theta}{dt}\)
θ
Distance (Fixed)
Object (y)
Case Study: Tracking a Rocket
A rocket is launched vertically from a site 3000 meters from a camera. When the rocket's height is 4000 meters, it is rising at 800 m/s.
How fast is the camera's angle of elevation changing at that instant?
\[ \tan\theta = \frac{h}{3000} \]
The Angular Solution Path
1
Relate the Variables
Identify which trig ratio fits. Usually \(\tan\theta\) for elevation or \(\cos\theta\) for horizontal distance.
2
Differentiate
Remember: \(\sec^2\theta\) will appear. You'll need the hypotenuse to find \(\sec\theta\).
// The Secret: The Hypotenuse
To evaluate \(\sec^2\theta\), you don't actually need the angle \(\theta\). Just find the lengths of the triangle's sides at that instant.
\(\sec\theta = \frac{\text{hypotenuse}}{\text{adjacent}}\)
\(\sec^2\theta = \left( \frac{\text{hyp}}{\text{adj}} \right)^2\)
The Radian Directive
Calculus derivatives for trig functions ONLY work if \(\theta\) is in RADIANS.
If your final answer is \(\frac{d\theta}{dt}\), its units are always radians per unit of time.
Angular Motion Problem Set
We will now model rotating searchlights, tracking cameras, and swinging pendulums. Focus on identifying the fixed distances versus the changing angles.
📷
Tracking
🔦
Rotation
📐
Elevation
Angular Motion Problem Set Worksheet CALCULUS_MOD_01 // ANGULAR_DYNAMICS
Angular Motion
Modeling Trigonometric Rates & Rotational Flux
Student:
RADIANS_PROTOCOL_ACTIVE
\(\frac{d}{dt}(\sin\theta) = \cos\theta \frac{d\theta}{dt}\)
\(\frac{d}{dt}(\cos\theta) = -\sin\theta \frac{d\theta}{dt}\)
\(\frac{d}{dt}(\tan\theta) = \sec^2\theta \frac{d\theta}{dt}\)
1
The Tracking Camera
A rocket is launched vertically at a constant speed of 600 m/s. A camera on the ground, located 4000 meters from the launch pad, tracks the rocket. How fast is the camera's angle of elevation changing when the rocket is 3000 meters high?
Model Setup (Trig Ratio)
Secant Evaluation & Substitution
2
The Rotating Searchlight
A searchlight is 3 miles from a long straight beach and rotates at a constant rate of 2 revolutions per minute. How fast is the beam of light moving along the beach when the beam makes an angle of 45° with the line perpendicular to the beach?
Angular Velocity (\(\frac{d\theta}{dt}\)) Prep
Convert rev/min to rad/min here
Differentiation & Calculation
3
The Tracking Spectator
An observer stands 20 meters from a straight running track. A runner passes the observer's closest point on the track at a speed of 10 m/s. How fast is the observer's head rotating (in rad/s) 2 seconds later?
Note: You must first determine the horizontal distance \(x\) and the hypotenuse \(z\) at the 2-second mark.
Snapshot Logic (t=2)
Model & Derivative
Final Computation (rad/sec)
Critical Constants:
1 rev = 2π rad
\(\sec\theta = 1/\cos\theta\)
ANGULAR_SET_04
Angular Motion Answer Key TEACHER_RESOURCE // SOLUTIONS_KEY
Angular Motion
Model Solutions & Radians Verification
Instructor View
1. The Tracking Camera Solution
Setup:
Eq: \(\tan\theta = \frac{h}{4000} \implies \tan\theta = \frac{1}{4000}h\)
Diff: \(\sec^2\theta \frac{d\theta}{dt} = \frac{1}{4000} \frac{dh}{dt}\)
Snapshot:
\(h = 3000\). Hypotenuse \(z = \sqrt{4000^2 + 3000^2} = 5000\).
\(\sec\theta = \frac{5000}{4000} = \frac{5}{4} \implies \sec^2\theta = \frac{25}{16}\)
\(\frac{dh}{dt} = 600\)
\(\frac{25}{16} \frac{d\theta}{dt} = \frac{1}{4000}(600) \implies \frac{25}{16} \frac{d\theta}{dt} = 0.15\)
\(\frac{d\theta}{dt} = 0.15 \cdot \frac{16}{25} = 0.096\text{ rad/sec}\)
2. The Rotating Searchlight Solution
Setup:
Eq: \(\tan\theta = \frac{x}{3} \implies x = 3\tan\theta\)
Diff: \(\frac{dx}{dt} = 3\sec^2\theta \frac{d\theta}{dt}\)
Constants:
\(\frac{d\theta}{dt} = 2\text{ rev/min} = 4\pi\text{ rad/min}\)
\(\theta = 45^\circ = \pi/4\text{ rad}\)
\(\sec(\pi/4) = \sqrt{2} \implies \sec^2(\pi/4) = 2\)
\(\frac{dx}{dt} = 3(2)(4\pi)\)
\(\frac{dx}{dt} = 24\pi\text{ miles/min} \approx 75.4\text{ miles/min}\)
3. The Tracking Spectator Solution
Snapshot (t=2):
\(x = \text{speed} \cdot \text{time} = 10 \cdot 2 = 20\text{ m}\)
Triangle is 20m (adj) by 20m (opp). It's 45°.
\(z = \sqrt{20^2 + 20^2} = 20\sqrt{2}\)
\(\sec\theta = \frac{20\sqrt{2}}{20} = \sqrt{2} \implies \sec^2\theta = 2\)
Setup:
\(\tan\theta = \frac{x}{20} \implies \sec^2\theta \frac{d\theta}{dt} = \frac{1}{20} \frac{dx}{dt}\)
Given: \(\frac{dx}{dt} = 10\)
\(2 \cdot \frac{d\theta}{dt} = \frac{1}{20}(10) \implies 2 \cdot \frac{d\theta}{dt} = 0.5\)
\(\frac{d\theta}{dt} = 0.25\text{ rad/sec}\)
(Interpretation: The observer's head is rotating at 0.25 rad/s)
Grading Note:
Ensure students convert revolutions to radians (Problem 2). A common mistake is using degrees inside the derivative calculation, which is invalid.
Answer Key // Related Rates Module 01
Conical Flow Slides Calculus II: Dynamic Systems
Conical Flow
Variable Elimination and Multidimensional Constraints
Module 01 // Lesson 05
The Problem of Too Many Variables
The Conical Trap
The volume of a cone is:
\[ V = \frac{1}{3}\pi r^2 h \]
When water enters a cone, both \(r\) and \(h\) change simultaneously.
Usually, we only know \(\frac{dV}{dt}\) and want \(\frac{dh}{dt}\). We must eliminate \(r\) before differentiating.
r_tank h_tank
r
h
The Elimination Key: Similar Triangles
The Geometric Constraint
The water always forms a cone that is similar to the container itself.
\[ \frac{r}{h} = \frac{R_{\text{tank}}}{H_{\text{tank}}} \]
This proportion allows us to write \(r\) in terms of \(h\) (or vice-versa).
Pre-Differentiation Substitution
If \(\frac{r}{h} = \frac{5}{10} \implies r = \frac{1}{2}h\).
New Equation:
\[ V = \frac{1}{3}\pi \left(\frac{1}{2}h\right)^2 h \]
\[ V = \frac{\pi}{12} h^3 \]
Now, only one variable remains on the right side!
Variation: The Growing Sand Pile
Sand is being poured onto a level ground, forming a conical pile whose height is always equal to its diameter.
If sand is added at 10 m³/min, how fast is the height of the pile increasing when the pile is 4 meters high?
Analysis:
Height = Diameter \(\implies h = 2r\)
Solve for \(r\): \(r = h/2\)
Substitute into Volume Formula:
\(V = \frac{1}{3}\pi (h/2)^2 h = \frac{\pi}{12}h^3\)
Now differentiate:
\(\frac{dV}{dt} = \frac{\pi}{4}h^2 \frac{dh}{dt}\)
Conical Mastery Protocol
📐
1. Proportion
Use similar triangles to relate \(r\) and \(h\).
🔄
2. Substitution
Replace the variable you don't want in the volume formula.
⚡
3. Derivative
Differentiate the single-variable equation w.r.t time.
✅
4. Evaluation
Substitute known rates and find the final answer.
The Conical Flow Challenge
This is the capstone of our Related Rates sequence. You will model industrial tank filling and drainage. Watch your fractions, and trust the similar triangle proportion.
VARIABLE_ELIMINATION: ACTIVE GEOMETRIC_CONSTRAINTS: LOCKED
Conical Flow Challenge Worksheet CALCULUS_MOD_01 // CONICAL_SYSTEMS
Conical Flow Challenge
Variable Elimination & Multidimensional Constraints
Student:
SIMILAR_TRIANGLES_REQUIRED
Volume Formula:
\[ V = \frac{1}{3}\pi r^2 h \]
1. Identify Proportion \(\frac{r}{h}\)
2. Substitute to Eliminate \(r\)
3. Differentiate w.r.t Time
1
The Industrial Hopper
A conical tank (vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.
Step A: Similar Triangle Proportion
Step B: Variable Elimination & Eq Prep
Step C: Differentiation & Final Snapshot Evaluation
2
The Leaking Reservoir
A conical water tank with a radius of 6 meters and a height of 18 meters is leaking water at a rate of 2 m³/min. How fast is the water level dropping when the water is 10 meters deep?
Model Construction
Rate Substitution (Negative \(\frac{dV}{dt}\))
3
The Conical Sand Pile
Sand is falling off a conveyor belt onto a conical pile at a rate of 15 ft³/min. The geometry of the pile is such that the radius is always three-fourths the height. How fast is the height increasing when the pile is 10 feet high?
Model & Complete Solution Pathway
Geometric Logic Check:
If water is filling a cone, why does \(\frac{dh}{dt}\) decrease as \(h\) increases? Consider the expanding area of the top surface.
CONE_CHALLENGE_05
Conical Flow Answer Key TEACHER_RESOURCE // SOLUTIONS_KEY
Conical Flow Challenge
Model Solutions & Variable Elimination Logic
Instructor View
1. The Industrial Hopper Solution
Setup:
Proportion: Top width = 10 \(\implies\) Radius \(R = 5\). Depth \(H = 12\).
Similar Triangles: \(\frac{r}{h} = \frac{5}{12} \implies r = \frac{5}{12}h\).
Sub: \(V = \frac{1}{3}\pi \left(\frac{5}{12}h\right)^2 h = \frac{25\pi}{432}h^3\).
Snapshot:
\(h = 8, \frac{dV}{dt} = 10\).
\(\frac{dV}{dt} = \frac{25\pi}{144}h^2 \frac{dh}{dt}\)
\(10 = \frac{25\pi}{144}(8^2) \frac{dh}{dt} \implies 10 = \frac{25\pi \cdot 64}{144} \frac{dh}{dt} \implies 10 = \frac{100\pi}{9} \frac{dh}{dt}\)
\(\frac{dh}{dt} = \frac{90}{100\pi} = \frac{9}{10\pi}\text{ ft/min} \approx 0.286\text{ ft/min}\)
2. The Leaking Reservoir Solution
Setup:
Proportion: \(\frac{r}{h} = \frac{6}{18} = \frac{1}{3} \implies r = \frac{1}{3}h\).
Sub: \(V = \frac{1}{3}\pi \left(\frac{1}{3}h\right)^2 h = \frac{\pi}{27}h^3\).
Diff: \(\frac{dV}{dt} = \frac{\pi}{9}h^2 \frac{dh}{dt}\).
Snapshot:
\(h = 10, \frac{dV}{dt} = -2\).
\(-2 = \frac{\pi}{9}(10^2) \frac{dh}{dt} \implies -2 = \frac{100\pi}{9} \frac{dh}{dt}\)
\(\frac{dh}{dt} = -\frac{18}{100\pi} = -\frac{9}{50\pi}\text{ m/min} \approx -0.057\text{ m/min}\)
(Level is dropping at ~5.7 cm/min)
3. The Conical Sand Pile Solution
Setup:
Condition: \(r = \frac{3}{4}h\).
Sub: \(V = \frac{1}{3}\pi \left(\frac{3}{4}h\right)^2 h = \frac{3\pi}{16}h^3\).
Diff: \(\frac{dV}{dt} = \frac{9\pi}{16}h^2 \frac{dh}{dt}\).
Snapshot:
\(h = 10, \frac{dV}{dt} = 15\).
\(15 = \frac{9\pi}{16}(10^2) \frac{dh}{dt} \implies 15 = \frac{900\pi}{16} \frac{dh}{dt} \implies 15 = \frac{225\pi}{4} \frac{dh}{dt}\)
\(\frac{dh}{dt} = \frac{15 \cdot 4}{225\pi} = \frac{60}{225\pi} = \frac{4}{15\pi}\)
\(\frac{dh}{dt} = \frac{4}{15\pi}\text{ ft/min} \approx 0.085\text{ ft/min}\)
Pedagogical Insight:
The "Elimination First" rule is the most critical takeaway for students. Differentiating \(r^2h\) as a product rule is mathematically correct but creates a dead-end if \(\frac{dr}{dt}\) is unknown. Similar triangles bypass this obstacle entirely.
Answer Key // Related Rates Module 01
