Time Derivative Slides Everything is Moving
Introduction to Differentiation with Respect to Time
Calculus II: Related Rates Workshop
The Independent Variable: Time (\(t\))
In Related Rates, variables like \(x\), \(y\), \(r\), and \(V\) are all functions of time (\(t\)).
Implicit Assumption:
\(x = x(t)\)
\(V = V(t)\)
The Notation Change
Instead of \(dy/dx\), we focus on \(dx/dt\) and \(dy/dt\).
Rates of change are always with respect to t.
The "Time" Chain Rule
When you differentiate any variable \(u\) with respect to \(t\):
\(\frac{d}{dt}[f(u)] = f'(u) \cdot \frac{du}{dt}\)
Example 1
\(\frac{d}{dt}[x^2] = \)
\(2x \frac{dx}{dt}\)
Example 2
\(\frac{d}{dt}[\pi r^2] = \)
\(2\pi r \frac{dr}{dt}\)
The 5-Step Protocol
01
SKETCH
Draw the scenario and label moving parts.
02
IDENTIFY
List known rates and target rate.
03
RELATE
Write an equation relating the variables.
04
DERIVE
Differentiate implicitly wrt time \(t\).
05
SOLVE
Substitute "snapshot" values and calculate.
The Snapshot Rule
NEVER substitute numbers that are changing until AFTER you have differentiated.
If you plug in \(x = 5\) before differentiating:
\(x^2 \rightarrow 5^2 \rightarrow 25\)
\(\frac{d}{dt}[25] = 0\) ← WRONG! The rate is lost!
Rate Translation Lab Worksheet Rate Translation Lab
Lesson 1: Differentiation with Respect to Time
NAME:
DATE:
Part 1: The Language of Rates
Translate each verbal description into standard Calculus notation using time \(t\) as the independent variable. Assume all lengths are in cm and time is in seconds.
1. The radius of a circle is expanding at a rate of 3 cm/s.
2. The volume of a sphere is decreasing at 10 cm³/s.
3. A person's distance from a wall is 15 cm and increasing at 5 cm/s.
4. The water level in a tank is dropping at 0.5 cm/s.
Part 2: Differentiating wrt Time (\(t\))
Differentiate both sides of each equation with respect to \(t\). Show every step of the chain rule.
5. \(A = \pi r^2\)
6. \(x^2 + y^2 = 25^2\)
7. \(V = \frac{4}{3}\pi r^3\)
Part 3: Constant vs. Instantaneous
Scenario: A 10-foot ladder is leaning against a wall. The bottom of the ladder is being pulled away at 2 ft/s. We want to know how fast the top is sliding down when the base is 6 feet from the wall.
Identify which of these values are "Constants" (fixed for the whole problem) and which are "Instantaneous" (only true at a specific moment).
Quantity Value Constant? Instantaneous? Length of ladder (\(L\)) 10 ft
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| Distance from base to wall (\(x\)) | 6 ft |
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| Rate base is pulled away (\(dx/dt\)) | 2 ft/s |
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Critical Thinking:
If the ladder was 10 feet long, and we differentiated \(x^2 + y^2 = 10^2\), what happens to the \(10^2\) on the right side? Why?
Related Rates Facilitation Guide Related Rates
Teacher Facilitation Guide
Unit Duration: 5 Sessions
Complexity: High
Subject: Calculus I/II
Instructional Blueprint
This sequence follows a "Hierarchy of Modeling" approach. Instead of teaching Related Rates as a single monolithic topic, we break it down by the geometric constraint that governs the problem. By categorizing problems this way, students learn to recognize the notation-to-geometry patterns before the calculus even begins.
L1
Notation Mastery
Moving from verbal "rates" to \(d(\text{var})/dt\).
L2
Pythagorean Relationships
Ladders, vehicles, and distance signs.
L3
Expansion Dynamics
Circles, Spheres, and Cubes.
L4
Variable Constraints
Cones and Similar Triangles (Variable Reduction).
L5
Angular Velocities
Trigonometric derivatives and tracking.
The "Snapshot" Rule
The single most common error is students substituting a variable (like \(x=5\)) before differentiating. Emphasize the "Snapshot" analogy: You can't see motion in a single still photo. You need the video (the derivative) before you can look at the frame (the value).
Assessment Strategy
Focus 50% of the grade on Step 1-3 (Setup).
Penalize missing \(d(\text{var})/dt\) notation during differentiation.
Require units in the final answer.
Lesson Facilitation Notes
L1: The Chain Rule Pivot
Why it sticks: Students are used to \(dy/dx\), not \(d/dt\).
Key Question: "Is \(x\) a constant or a function of time?" Ask this every time they see a variable. In Related Rates, nothing is a constant unless it physically cannot change (like the height of a lamp post).
L2: Interpreting Signs
Why it sticks: Negative rates vs negative positions.
Use the "Approach vs. Depart" rule. If distance is shrinking, \(dr/dt < 0\). This is critical for Car-Intersection problems where cars are moving toward the origin.
L4: The Conical Ratio
Why it sticks: The variable reduction step.
Students will try to use the Product Rule on \(V = 1/3 \pi r^2 h\). While mathematically correct, it leads to a dead end because they won't know \(dr/dt\) or \(dh/dt\). Force the substitution of the similar triangle ratio before differentiating.
L5: The Secant Trick
Why it sticks: Calculating \(\sec^2 \theta\).
When differentiating \(\tan \theta\), students get \(\sec^2 \theta \cdot d\theta/dt\). Don't let them find \(\theta\) in degrees. Teach them to look at the Triangle Snapshot to find \(\sec \theta = \text{hyp}/\text{adj}\), then square it. It's much cleaner!
Pythagorean Rates Slides Triangular Motion
Pythagorean Relationships in Related Rates
Lesson 2: Right-Triangle Modeling
The Master Equation
Most right-triangle scenarios involve distances changing over time.
\[x^2 + y^2 = z^2\]
x y z
The "Time" Derivative
Differentiating both sides with respect to \(t\):
\[2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}\]
\[x \frac{dx}{dt} + y \frac{dy}{dt} = z \frac{dz}{dt}\]
Pro-tip: The \(2\)'s always cancel out!
Which Direction?
Positive Rate (+)
Distance is INCREASING.
(e.g., Moving away, growing, sliding out)
Negative Rate (-)
Distance is DECREASING.
(e.g., Approaching, shrinking, sliding down)
Case Study: The Ladder
A 13-ft ladder leans against a wall. The bottom is pulled out at 5 ft/s. How fast is the top sliding down when the base is 12 ft away?
Key Insight
The ladder length is CONSTANT (\(z = 13\)), so its rate of change is ZERO (\(dz/dt = 0\)).
The Equation
\(x^2 + y^2 = 13^2\)
\(x \frac{dx}{dt} + y \frac{dy}{dt} = 0\)
Plug and solve for \(dy/dt\)...
Pythagorean Modeling Worksheet Triangle Tactics
Workshop 2: Pythagorean Modeling
NAME:
DATE:
1
The Gravity Experiment
A 17-foot ladder is leaning against a vertical wall. If the bottom of the ladder is pulled away from the wall at a constant rate of 3 ft/s, how fast is the top of the ladder sliding down the wall when the bottom is 8 feet from the wall?
STEP 1: SKETCH & LABEL
Draw right triangle here
STEP 2: GIVENS & TARGET
GIVEN
FIND
STEP 3-5: CALCULATE
2
The Intersection
Car A is traveling west at 50 mph and Car B is traveling north at 60 mph. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when Car A is 0.3 miles and Car B is 0.4 miles from the intersection?
Note on Signs: If they are headed toward the intersection, their distances from it are decreasing . Think about what that means for the signs of \(dx/dt\) and \(dy/dt\)!
Show your full modeling process:
Final Answer (include units):
3
The Docking Maneuver
A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 meter higher than the bow. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 meters from the dock?
Expansion Rates Slides Growth Patterns
Area and Volume Expansion Rates
Lesson 3: Multi-Dimensional Related Rates
The Toolkit
Circle
Area
\(A = \pi r^2\)
Sphere Volume
\(V = \frac{4}{3}\pi r^3\)
Cube Volume
\(V = s^3\)
2D Expansion
Consider an oil spill spreading in a perfect circle.
\(\frac{d}{dt}[A] = \frac{d}{dt}[\pi r^2]\)
\(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)
Interpretation:
The rate of area increase (\(dA/dt\)) depends on two things:
1. How fast the radius grows (\(dr/dt\))
2. How large the circle already is (\(r\))
The Balloon Paradox
If you blow air into a balloon at a constant rate (\(dV/dt\) is fixed)...
Formula
\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
Result
\(\frac{dr}{dt} = \frac{dV/dt}{4\pi r^2}\)
As radius \(r\) increases, the rate of expansion \(dr/dt\) drops rapidly.
Sanity Check: Units
Radius Rate (\(dr/dt\))
Length / Time
cm/s, ft/min, m/hr
Volume Rate (\(dV/dt\))
Length³ / Time
cm³/s, ft³/min, gallons/hr
Test Your Logic
A snowball is melting. Its Surface Area (\(S = 4\pi r^2\)) is decreasing at 2 cm²/min.
Should \(dr/dt\) be positive or negative?
NEGATIVE (-)
Expansion Dynamics Worksheet Expansion Dynamics
WORKSHOP 3: AREA & VOLUME RATES
Observer:
Timestamp:
1
The Spill Spread
Oil is leaking from a tanker and spreading in a circular pattern. The area of the spill is increasing at a constant rate of 100 cm²/min. How fast is the radius of the spill increasing when the radius is 10 cm?
Givens & Targets
\(\frac{dA}{dt} = \)
Find \(\frac{dr}{dt}\) when \(r = \)
The Relation
\(A = \pi r^2\)
Calculations
dr/dt = _________
2
Inflation Control
Helium is being pumped into a spherical weather balloon at a rate of 4.5 cubic feet per minute. At what rate is the radius of the balloon increasing when the diameter is 6 feet?
Watch Out! The problem gives you the diameter, but your formula \(V = \frac{4}{3}\pi r^3\) uses the radius. Convert before substituting!
Show complete 5-step process:
3
Dimensional Coupling
A block of ice in the shape of a cube is melting. The side length \(s\) of the cube is decreasing at a rate of 0.2 cm/min. At what rate is the Volume of the cube changing when the Surface Area is 150 cm²?
1. Sub-Goal: Find 's'
Use Surface Area \(S = 6s^2\) to find the side length at this snapshot.
2. Identify Rates
\(ds/dt = \)
\(dV/dt = ?\)
Final Differentiation & Solution:
Final Volume Rate:
Constraint Modeling Slides The Constraint Ratio
Cones and Similar Triangles
Lesson 4: Reducing Variable Complexity
The 3-Variable Trap
The volume of a cone is:
\(V = \frac{1}{3}\pi r^2 h\)
Problem: We have three variables (\(V, r, h\)) all changing with time.
r h
The "Fixed Ratio"
Similar Triangles
The ratio of the liquid's radius (\(r\)) to its height (\(h\)) is always equal to the tank's radius (\(R\)) to height (\(H\)).
\(\frac{r}{h} = \frac{R}{H}\)
Goal: Get rid of \(r\)
\(r = \left(\frac{R}{H}\right)h\)
Substitute into Volume
\(V = \frac{1}{3}\pi \left[ \left(\frac{R}{H}\right)h \right]^2 h\)
Simplify FIRST
Always simplify your expression into a single variable term before you touch the derivative.
The Pro Way
\(V = (\text{constant}) \cdot h^3\)
\(\frac{dV}{dt} = 3(\text{constant}) \cdot h^2 \frac{dh}{dt}\)
No more product rule! No more \(r\)!
Application: Shadow Problems
As a man walks away from a lamp post, his shadow length (\(s\)) and distance from the post (\(x\)) are linked by nested triangles.
\(\frac{H_{\text{lamp}}}{x + s} = \frac{h_{\text{man}}}{s}\)
x (dist) s (shadow)
Conical Constraints Workshop Worksheet Conical Constraints
Workshop 4: Ratio-Based Modeling
STUDENT:
SESSION:
1
The Gravity Reservoir
Water is dripping out of a conical tank with a top diameter of 20 meters and a height of 30 meters. If the water level is dropping at a rate of 0.2 m/min when the water is 10 meters deep, at what rate is the water leaking out of the tank?
A. The Tank Ratio
Express \(r\) in terms of \(h\) based on the tank's dimensions.
B. Variable Prep
\(h = \)
\(dh/dt = \)
FIND \(dV/dt\)
Calculations: Substitute Ratio & Differentiate
Leakage Rate: _________
2
Shadow Velocity
A man 6 feet tall walks away from a lamp post 15 feet high at a speed of 5 ft/s.
Sketch & Label Similar Triangles
Label: Post (15), Man (6), Distance (x), Shadow (s)
Establishing the Ratio
Tip: Solve for 's' in terms of 'x' before differentiating.
Part (a) Solve for \(ds/dt\):
Part (b) Solve for Tip Speed:
Hint: Tip Speed is \(\frac{d}{dt}(x + s)\)
Does the shadow grow faster or slower as the man speeds up?
Trigonometric Rates Slides Angular Velocity
Trigonometric Rates & Angle Elevation
Lesson 5: Tracking Dynamic Motion
Differentiating Angles
When \(\theta\) is a function of time \(t\), we must apply the chain rule to every trig function.
\(\frac{d}{dt}[\sin \theta] = \cos \theta \cdot \frac{d\theta}{dt}\)
\(\frac{d}{dt}[\tan \theta] = \sec^2 \theta \cdot \frac{d\theta}{dt}\)
The "Snapshot" Angle
Remember: \(\frac{d\theta}{dt}\) is the Angular Velocity (usually in radians/time).
Critical Warning:
Calculus requires angles to be in RADIANS. Convert degrees immediately!
Tracking a Rocket
An observer stands 2000 ft from a launch pad. A rocket rises vertically at 500 ft/s.
Step 3: Relate
\(\tan \theta = \frac{h}{2000}\)
Why tangent? We know Adjacent (2000) and Opposite (h).
θ 2000 ft h
The Derivation
\(\frac{d}{dt} [\tan \theta] = \frac{d}{dt} \left[ \frac{h}{2000} \right]\)
\(\sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{2000} \frac{dh}{dt}\)
Trick for \(\sec^2 \theta\)
Instead of finding \(\theta\), use the triangle to find \(\sec \theta = \frac{\text{hyp}}{\text{adj}}\).
Final Step
Isolate \(d\theta/dt\) by multiplying by \(\cos^2 \theta\).
Synthesis
The Takeaway
Related rates aren't just for geometry; they are the foundation for tracking, aviation, and orbital mechanics.
Trigonometry links linear & angular motion
Chain rule handles the time dimension
Radians are required for all Calculus
Final Mastery Question:
As an object flies directly over you, is the rate of angle change at its maximum or minimum?
MAXIMUM
Trigonometric Horizons Worksheet Trigonometric Horizons
Workshop 5: Angular Velocity & Elevation
IDENTIFIER:
1
The Cape Canaveral View
An observer is 5 kilometers from a launch pad. A rocket is launched vertically and maintains a constant speed of 0.8 km/s. How fast is the observer's angle of elevation changing when the rocket is 12 kilometers high?
Model Setup
Known: \(dy/dt = \)
Find: \(d\theta/dt\) when \(y = \)
Equation:
\(\tan \theta = \frac{y}{5}\)
Pro-Tip: Find \(\sec \theta\)
When \(y = 12\) and \(x = 5\), the hypotenuse is \(z = \sqrt{5^2 + 12^2} = 13\). Use this for \(\sec \theta\)!
Differentiation & Calculation
Final Rate (rad/s):
2
The Prison Wall
A searchlight 100 meters from a long straight wall is rotating at a constant rate of 0.5 radians per minute. How fast is the beam of light moving along the wall when the beam makes an angle of \(\pi/4\) with the perpendicular to the wall?
Geometric Hint: Let \(x\) be the distance from the point on the wall closest to the searchlight. The searchlight is 100m away (constant). Use \(\tan \theta = x/100\).
Show Modeling Steps:
3
Radar Tracking
A plane flies horizontally at an altitude of 3 miles and a speed of 400 mph. It passes directly over a radar station. At what rate is the angle of elevation changing 1 minute after it passes over the station?
Unit Alert!
The speed is in miles per hour, but the time given is 1 minute. You must find the horizontal distance \(x\) first!
Phase 1: Finding Distance x
x = speed × time
Phase 2: Snapshot Triangle
Find \(\theta\) or \(\sec^2 \theta\) for this moment.
Phase 3: Differentiate & Solve for \(d\theta/dt\)
Final Output
Include units in rad/hr or rad/min
________
