Sphere Mechanics Slides Sphere Mechanics
Related Rates & Spherical Expansion
The Balloon Paradox
Imagine you are blowing air into a balloon at a constant rate (e.g., 10 cubic inches per second).
Does the balloon's radius grow at a constant speed?
Think: Does it get bigger "faster" at the beginning or at the end?
Geometry Review
Volume
\[ V = \frac{4}{3} \pi r^3 \]
Relates volume (\(V\)) to radius (\(r\)).
Surface Area
\[ A = 4 \pi r^2 \]
Relates surface area (\(A\)) to radius (\(r\)).
Differentiating wrt Time (\(t\))
Step 1: Write Formula
\[ V = \frac{4}{3} \pi r^3 \]
Step 2: Differentiate
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot (3r^2) \cdot \frac{dr}{dt} \]
Step 3: Simplify
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Note: \(4\pi r^2\) is the Surface Area!
Sample Problem
A spherical balloon is inflated at a rate of \(100 \text{ cm}^3/\text{sec}\).
How fast is the radius increasing when the radius is \(5 \text{ cm}\)?
Knowns
\[ \frac{dV}{dt} = 100 \]
\[ r = 5 \]
Find
\[ \frac{dr}{dt} = ? \]
Crucial Observation
\[ \frac{dr}{dt} = \frac{1}{4\pi r^2} \cdot \frac{dV}{dt} \]
As the radius \(r\) increases, the rate of growth \(\frac{dr}{dt}\) decreases.
Start
Small \(r\)
Fast Expansion
End
Large \(r\)
Slow Expansion
Balloon Bursting Worksheet Balloon Bursting
Related Rates: Spherical Expansion & Contraction
Name:
Date:
Sphere Volume
\[ V = \frac{4}{3}\pi r^3 \]
Surface Area
\[ A = 4\pi r^2 \]
Differentiated Volume
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
1
The Party Balloon
A balloon is being inflated at a constant rate of \(15 \text{ in}^3/\text{min}\). At the instant when the radius is \(3 \text{ inches}\), how fast is the radius of the balloon increasing? Provide your answer in terms of \(\pi\).
Given Values
What to Find
Show your differentiation and substitution steps.
2
The Melting Snowball
A spherical snowball melts so that its surface area decreases at a constant rate of \(2 \text{ cm}^2/\text{min}\). Find the rate at which the radius is changing when the radius is \(4 \text{ cm}\).
Note: Use the Surface Area derivative for this problem.
Follow-up Challenge:
Based on your answer above, how fast is the volume of the snowball changing at that same instant?
3
Conceptual Investigation
Suppose a balloon expands at a constant rate \( \frac{dV}{dt} = C \). Explain, using the derivative equation, why the rate of the radius change \( \frac{dr}{dt} \) gets smaller as the balloon gets bigger.
Sphere Logic Teacher Guide Instructional Guide
Sphere Logic
LESSON 01
Related Rates Intro
Learning Objectives
Apply implicit differentiation to geometric formulas with respect to time.
Relate volume, surface area, and radius in a dynamic spherical system.
Interpret the physical meaning of a decreasing rate of change.
The "Big Idea"
In a sphere, growth is distributed across a 3D volume. As the object grows, a "unit of volume" is spread thinner over a larger surface area, meaning the linear dimensions (radius) grow progressively slower.
Common Misconceptions
Missing the Chain Rule
Students often differentiate \(r^3\) as \(3r^2\) but forget the \(\frac{dr}{dt}\). Emphasize that every variable is a function of \(t\).
Confusing Units
Remind students to check units: \( \text{in}^3/\text{min} \) for volume, \( \text{in}^2/\text{min} \) for surface area, and \( \text{in}/\text{min} \) for radius.
Class Discussion Starters
The Snowball Effect
"If a snowball is melting at a constant volume rate, does it disappear 'faster' or 'slower' as it gets smaller?"
Answer: Faster. Since \(dr/dt = (dV/dt) / (4\pi r^2)\), as \(r \to 0\), the rate of radius change approaches infinity.
Thinning Rubber
"How does the surface area rate relate to the volume rate?"
Exploration: Show how \(dA/dt = 8\pi r (dr/dt)\). Use substitution to link \(dA/dt\) to \(dV/dt\).
Worksheet Quick-Key
1. \(\frac{5}{12\pi} \text{ in/min}\)
2. \(\frac{-1}{16\pi} \text{ cm/min}\)
3. Inverse Square Law
Steady Stream Slides Steady Streams
Cylindrical Dynamics & Linear Change
The Pool Problem
You are filling a rectangular swimming pool with a garden hose.
As the pool fills, does the water level rise at a constant speed?
Contrast this with the balloon: In a sphere, the growth slowed down. In a cylinder, what happens?
dh/dt
Constant Cross-Section
The Cylindrical Advantage
\[ V = \pi r^2 h \]
Crucial Realization:
In a rigid tank, the radius \(r\) is a constant. It does not change over time.
Differentiating
\[ \frac{d}{dt}[V] = \frac{d}{dt}[\pi r^2 h] \]
\[ \frac{dV}{dt} = \pi r^2 \cdot \frac{dh}{dt} \]
\[ \frac{dh}{dt} = \frac{1}{\text{Base Area}} \cdot \frac{dV}{dt} \]
Sphere vs. Cylinder
The Sphere
\[ \frac{dr}{dt} = \frac{1}{4\pi \mathbf{r^2}} \cdot \frac{dV}{dt} \]
Rate depends on the current size. The more you have, the slower it grows.
The Cylinder
\[ \frac{dh}{dt} = \frac{1}{\mathbf{\pi R^2}} \cdot \frac{dV}{dt} \]
Rate is constant for a constant \(dV/dt\). Shape doesn't change with height.
Example: Oil Barrel Leak
An oil barrel (radius \(2 \text{ ft}\)) is leaking at a rate of \(0.5 \text{ ft}^3/\text{min}\). How fast is the oil level dropping?
Step 1 \[ \frac{dV}{dt} = -0.5 \]
Step 2 \[ \frac{dV}{dt} = \pi (2)^2 \frac{dh}{dt} \]
Step 3 \[ -0.5 = 4\pi \frac{dh}{dt} \]
Result \[ \frac{dh}{dt} \approx -0.04 \text{ ft/min} \]
Pool Filling Worksheet Steady Streams
Cylindrical and Constant Cross-Section Tanks
Name:
Class:
CORE THEOREM
In a tank where the cross-sectional area \( A \) is constant (cylinders, prisms, troughs), the rate of volume change relates to the rate of height change as:
\[ \frac{dV}{dt} = A \cdot \frac{dh}{dt} \]
Where \( A \) is the constant area of the base.
1
The Industrial Reservoir
A cylindrical tank with a radius of \( 10 \text{ meters} \) is being filled at a rate of \( 50 \text{ m}^3/\text{min} \).
A. Calculate the rate at which the water level is rising.
B. Does the water level rise faster when the tank is half full compared to when it is nearly full? Explain.
2
The Backyard Pool
A rectangular pool is \( 20 \text{ ft} \) long and \( 12 \text{ ft} \) wide. If water is being pumped in at \( 15 \text{ ft}^3/\text{min} \), find the rate \( \frac{dh}{dt} \).
SCHEMATIC
CALCULATIONS
3
The Growing Oil Spill
An oil spill on the ocean surface takes the shape of a cylinder with a constant thickness of \( 0.01 \text{ meters} \). If the oil is being spilled at a rate of \( 2 \text{ m}^3/\text{hr} \), how fast is the radius of the spill increasing when the radius is \( 100 \text{ meters} \)?
Think carefully: Is the radius constant in this problem? Or is the thickness constant?
Constant Flow Answer Key Answer Key: Steady Streams
1. The Industrial Reservoir
Part A: \( \frac{dV}{dt} = 50, \quad r = 10 \)
\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \implies 50 = \pi(10)^2 \frac{dh}{dt} \]
\[ \frac{dh}{dt} = \frac{50}{100\pi} = \frac{1}{2\pi} \approx 0.159 \text{ m/min} \]
Part B: No. Since the radius (and thus the base area) is constant, the rate of height change remains constant regardless of the current volume of water.
2. The Backyard Pool
Base Area \( A = 20 \times 12 = 240 \text{ ft}^2 \)
\[ \frac{dV}{dt} = A \cdot \frac{dh}{dt} \implies 15 = 240 \cdot \frac{dh}{dt} \]
\[ \frac{dh}{dt} = \frac{15}{240} = \frac{1}{16} = 0.0625 \text{ ft/min} \]
3. The Growing Oil Spill
Thickness \( h = 0.01 \) (constant). Volume formula: \( V = \pi r^2 h \)
Differentiating with respect to \( t \): \[ \frac{dV}{dt} = \pi h \cdot (2r \frac{dr}{dt}) \]
Substitute: \( 2 = \pi (0.01) \cdot (2 \cdot 100 \frac{dr}{dt}) \)
\[ 2 = 2\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi} \approx 0.318 \text{ m/hr} \]
CONFIDENTIAL TEACHER RESOURCE • VOLUME FLOW DYNAMICS
Ratio Rigger Slides Ratio Rigger
Solving the Multi-Variable Conic Constraint
The Geometry Trap
CRITICAL STEP
The volume of a cone is:
\[ V = \frac{1}{3}\pi r^2 h \]
The Problem:
As water fills the cone, BOTH \(r\) and \(h\) change. We can't differentiate yet!
r h
The Solution: Similar Triangles
The water inside the tank forms a smaller cone that is geometrically similar to the tank itself.
\[ \frac{r}{h} = \frac{R_{\text{tank}}}{H_{\text{tank}}} \]
This ratio is fixed. We can use it to replace \(r\) with a function of \(h\).
Variable Reduction
If \(R=10\) and \(H=20\):
\[ \frac{r}{h} = \frac{10}{20} \]
\[ \frac{r}{h} = \frac{1}{2} \]
\[ r = \frac{1}{2}h \]
The Final Equation
Original:
\[ V = \frac{1}{3}\pi r^2 h \]
Substitute \(r = \frac{1}{2}h\):
\[ V = \frac{1}{3}\pi (\frac{1}{2}h)^2 h \]
Reduced Formula:
\[ V = \frac{\pi}{12}h^3 \]
"One variable, one derivative. Now we're ready for calculus."
The Setup Checklist
1
Draw the diagram and label \(R_{\text{tank}}\) and \(H_{\text{tank}}\).
2
Set up the ratio: \( \frac{r}{h} = \frac{\text{Radius}}{\text{Height}} \).
3
Solve for \(r\) (if you know \(dh/dt\)) or solve for \(h\) (if you know \(dr/dt\)).
4
Plug back into \(V = \frac{1}{3}\pi r^2 h\) and simplify.
Variable Reducer Worksheet Variable Reducer
Mastering the Geometric Substitution for Cones
Project:
The Geometry Goal
In related rates problems for cones, we must eliminate one variable (\(r\) or \(h\)) before we differentiate. Your goal today is 100% preparation: Do not differentiate. Only reduce the equation.
Scenario A: The Coffee Filter R = 4 cm, H = 8 cm
Draw Schematic
1. Set up the Ratio (r/h)
2. Solve for 'r' in terms of 'h'
3. Resulting Volume Equation (V = f(h))
V =
Scenario B: The Gravel Pile H is always 2x the Radius
If we want to find \( \frac{dr}{dt} \), we need to eliminate h.
Show substitution for 'h' here...
Resulting Volume Equation (V = f(r))
V =
Mastery Check
Why is it critical to substitute before you differentiate in these conical problems? What would happen if you used the Product Rule on \( \frac{1}{3}\pi r^2 h \) without substituting?
Geometry Gaps Reference Geometry Gaps
Formula Vault
Cone Volume: \[ V = \frac{1}{3}\pi r^2 h \]
Sphere Volume: \[ V = \frac{4}{3}\pi r^3 \]
Cylinder Volume: \[ V = \pi r^2 h \]
The Golden Ratios
STANDARD CONE
\[ \frac{r}{h} = \frac{R}{H} \implies r = \frac{R}{H}h \]
INVERTED TROUGH
\[ \frac{w}{h} = \frac{W}{H} \]
The Differentiation Map
STEP 1
Reduce to One Variable
Substitute to eliminate the variable you don't have a rate for.
STEP 2
Differentiate (Power Rule)
Keep the constants in front and don't forget d/dt!
STEP 3
Plug and Solve
Only plug in the "at that instant" numbers after differentiating.
Pro-Tip
Always solve the ratio for the variable you want to REPLACE. If you want to find \(dh/dt\), solve for \(r\).
Conic Flow Slides Conic Flow
From Geometry Substitution to Dynamic Rates
The Hourglass Paradox
In an hourglass, sand falls at a constant rate (\(dV/dt\)).
Question:
Does the sand level drop faster when the bulb is nearly full, or when it's almost empty?
Calculus Implementation
THE REDUCED FORMULA
\[ V = \frac{\pi}{12}h^3 \]
DIFFERENTIATE wrt \(t\)
\[ \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} \]
\[ \frac{dV}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt} \]
Why is \(h\) in the derivative?
Because a cone widens as it gets taller. To raise the water level by 1cm at the top requires more volume than at the bottom.
Acceleration of Depth
For a constant \(dV/dt\), as \(h\) decreases, \(\frac{dh}{dt}\) must increase to keep the equation balanced.
Solving the Flow
Water leaks out of a conical funnel at \(2 \text{ cm}^3/\text{sec}\). The funnel has \(R=5\) and \(H=10\). Find the rate the level is dropping when \(h=4\).
KNOWNS
\[ \frac{dV}{dt} = -2 \]
\[ h = 4, \quad r = \frac{1}{2}h \]
CALCULATION
\[ -2 = \frac{\pi (4)^2}{4} \frac{dh}{dt} \]
\[ -2 = 4\pi \frac{dh}{dt} \]
\[ \frac{dh}{dt} = -\frac{1}{2\pi} \text{ cm/s} \]
The Design Insight
"The shallower the water in a cone, the more sensitive the level becomes to changes in volume."
This is why the last few drops in a funnel seem to disappear instantly!
Hourglass Hustle Workshop Hourglass Hustle
Solving Volumetric Flow in Conical Containers
Level: ADVANCED
1
The Industrial Funnel
A conical funnel is \( 10 \text{ inches} \) deep and \( 6 \text{ inches} \) across at the top. Liquid is flowing out of the bottom at a constant rate of \( 4 \text{ cubic inches per minute} \).
Step A: Formula Reduction
Find the radius \( r \) in terms of height \( h \). Then write the volume formula \( V = f(h) \).
Step B: Differentiation
Find \( \frac{dV}{dt} \) in terms of \( h \) and \( \frac{dh}{dt} \).
Step C: Calculation
How fast is the depth of the liquid dropping when the depth is \( 5 \text{ inches} \)?
dh/dt =
2
The Sand Conveyor
Sand is falling from a conveyor belt onto a conical pile at a rate of \( 10 \text{ ft}^3/\text{min} \). Due to the friction of the sand, the height of the pile is always equal to its radius.
How fast is the radius of the pile increasing when the pile is \( 4 \text{ feet} \) high?
Space for Full Proof and Calculation
Conceptual Reflection
Compare your results from Problem 1 (emptying a cone) and Problem 2 (filling a cone). In which scenario does the height change accelerate as time goes on, and in which does it decelerate? Why?
Conic Master Answer Key Conic Master Key
1. The Industrial Funnel
Setup: \( H=10, R=3 \) (radius is half the diameter of 6). Ratio: \( \frac{r}{h} = \frac{3}{10} \implies r = 0.3h \).
Volume: \( V = \frac{1}{3}\pi (0.3h)^2 h = \frac{0.09\pi}{3}h^3 = 0.03\pi h^3 \).
Derivative: \( \frac{dV}{dt} = 0.03\pi (3h^2) \frac{dh}{dt} = 0.09\pi h^2 \frac{dh}{dt} \).
Calculation: \( -4 = 0.09\pi (5^2) \frac{dh}{dt} \implies -4 = 2.25\pi \frac{dh}{dt} \).
\[ \frac{dh}{dt} = -\frac{4}{2.25\pi} \approx -0.566 \text{ in/min} \]
2. The Sand Conveyor
Setup: \( h = r \). Since we want \( \frac{dr}{dt} \), substitute \( h=r \).
Volume: \( V = \frac{1}{3}\pi r^2 (r) = \frac{\pi}{3}r^3 \).
Derivative: \( \frac{dV}{dt} = \pi r^2 \frac{dr}{dt} \).
Calculation: \( 10 = \pi (4^2) \frac{dr}{dt} \implies 10 = 16\pi \frac{dr}{dt} \).
\[ \frac{dr}{dt} = \frac{10}{16\pi} = \frac{5}{8\pi} \approx 0.199 \text{ ft/min} \]
Pedagogical Note: Acceleration
Draining Cone (Emptying)
As \( h \) decreases, \( \frac{dh}{dt} \) increases (absolute value). The water appears to "rush" out faster as the funnel gets empty.
Filling Cone (Sand Pile)
As \( r \) increases, \( \frac{dr}{dt} \) decreases. The pile seems to grow slower and slower as it gets larger.
INTERNAL RESOURCE // UNIT: VOLUME DYNAMICS
Plant Pressure Slides Plant Pressure
Calculus for Net Flow Management
The Reactor Logic
In real industrial systems, containers aren't just filling or emptying—they are doing both simultaneously.
THE NET RATE FORMULA
\[ \frac{dV}{dt} = \text{Rate}_{\text{in}} - \text{Rate}_{\text{out}} \]
Inflow: Chemical Reactants
Outflow: Product Extraction
System Equilibrium:
If Rate(In) = Rate(Out), then dV/dt = 0, and the fluid level remains constant.
Emergency Analysis
Alert
A cylindrical cooling tank (\(R=4 \text{ ft}\)) has a stuck input valve pumping in water at \(20 \text{ ft}^3/\text{min}\).
How fast must the emergency drain pump (\(\text{Rate}_{\text{out}}\)) operate so that the water level drops at \(0.5 \text{ ft}/\text{min}\)?
Step 1: Relate dV and dh
\[ \frac{dV}{dt} = \pi(4)^2(-0.5) = -8\pi \]
Step 2: Solve for Outflow
\[ -8\pi = 20 - \text{Rate}_{\text{out}} \]
\[ \text{Rate}_{\text{out}} \approx 45.1 \text{ ft}^3/\text{min} \]
Net Flow Complexity: Cones
In a cylinder, the drain speed is constant for a desired level change.
But in a Cone...
Because \(dV/dt\) depends on \(h^2\), your required pump speed must change based on the current depth to maintain a steady drop!
The Net Conic Eq
\[ R_{\text{in}} - R_{\text{out}} = \frac{\pi h^2}{4} \frac{dh}{dt} \]
"As an engineer, you'd need a computer to constantly adjust the valve as the tank empties."
The Final Project
Managing the Chemical Mixing Plant
Identify
Determine the current net volume change.
Simulate
Solve for level rates at different depths.
Prevent
Set alarm thresholds for overflow risk.
Reactor Breach Case Study Reactor Breach Analysis
Industrial Systems Division // Net-Flow Protocol
Case File #
CALC-2026
Emergency Summary
A chemical mixing plant is experiencing a valve failure. Reactor Tank 7 (a cylinder with a radius of \( 5 \text{ ft} \)) is receiving toxic inflow at a constant rate of \( 40 \text{ ft}^3/\text{min} \). To prevent a pressure spike, the fluid level must not rise faster than \( 0.2 \text{ ft}/\text{min} \).
Core Requirement
\[ \frac{dh}{dt} \leq 0.2 \]
Task 1: Determining Minimum Extraction Rate
Calculate the minimum rate at which the emergency drain pump (\( \text{Rate}_{\text{out}} \)) must operate to satisfy the safety constraint.
SHOW ALL WORKINGS HERE
Task 2: The Conical Backup Tank
The excess fluid is being diverted into a conical containment pit with a height of \( 20 \text{ ft} \) and a top radius of \( 10 \text{ ft} \). Suppose the net flow into this pit is exactly \( 15 \text{ ft}^3/\text{min} \).
A. Sensitivity Check
At what depth is the level rising faster: when \( h = 5 \text{ ft} \) or \( h = 15 \text{ ft} \)? Explain your reasoning using the conic derivative formula.
B. Critical Calculation
Find the exact rate \( \frac{dh}{dt} \) in the pit at the instant the fluid is \( 10 \text{ ft} \) deep.
Engineer's Recommendation
Based on your calculations, should the plant manager be more concerned about the pressure in the cylinder or the conical pit as time progresses? Support your choice with a brief comparison of how their rates change over time.
Official Site Audit // Verified By Calculus Dept
Flow Expert Assessment Flow Expert
FINAL ASSESSMENT
Volume Flow Dynamics
Examinee:
Date:
01
Spherical Pressure
A weather balloon is losing helium at a constant rate of \( 8 \text{ ft}^3/\text{hr} \). How fast is the radius of the balloon decreasing at the instant when the surface area of the balloon is \( 36\pi \text{ ft}^2 \)?
02
Cylindrical Reservoir
A cylindrical reservoir with a base diameter of \( 20 \text{ ft} \) is being drained. If the water level is dropping at a rate of \( 0.1 \text{ ft/min} \), find the rate of volume change (\( dV/dt \)) of the water in the reservoir.
03
The Conical Sieve
Liquid is poured into a conical sieve at a rate of \( 5 \text{ cm}^3/\text{sec} \) while simultaneously leaking out of the bottom at \( 2 \text{ cm}^3/\text{sec} \). The sieve has a top radius of \( 6 \text{ cm} \) and a height of \( 12 \text{ cm} \).
Find the rate at which the liquid level is rising when the liquid is \( 4 \text{ cm} \) deep.
Final Solution Workspace
Calculus Level III // Assessment Complete
