Shadow Geometry Presentation Slides Shadow Geometry
Lesson 1: Similar Triangles & Proportions
The Projection Effect
Imagine holding a ruler in front of a flashlight. A small movement of the ruler translates into a massive sweep across the wall.
The Big Question:
How do we mathematically link the speed of the object to the speed of its giant projection?
Similarity →
The Anatomy of Similarity
Proportional Relationships
When two triangles share the same angles, their side lengths are proportional.
\[ \frac{\text{Height}_1}{\text{Base}_1} = \frac{\text{Height}_2}{\text{Base}_2} \]
Identify the shared vertex (usually the light source or observer).
Define your variables ($x$, $y$, $s$) clearly on your sketch.
Pro Tip: Avoid the Quotient Rule
Instead of differentiating \( \frac{y}{x} = \frac{H}{L} \), rearrange to product form:
Modeling Related Rates
1
Static Equation: Create the proportion.
2
Substitution: Replace constants (fixed heights).
3
Implicit Differentiation: Differentiate w.r.t time (\(t\)).
4
Evaluation: Plug in given rates and solve.
Quick Fire Check
"A 6ft man is walking toward a 15ft streetlight. If \(x\) is his distance from the pole and \(s\) is the length of his shadow, what is the proportional relationship?"
\[ \frac{6}{s} = \frac{15}{x+s} \]
\[ \frac{6}{x} = \frac{15}{s} \]
Think: Does the big triangle include the man's distance?
Proportion Power Worksheet Proportion Power
Related Rates: Shadow Geometry
NAME:
DATE:
Goal: Master the setup of similar triangle proportions and differentiate them accurately using implicit differentiation. Remember to define your variables with a sketch!
1
The Basics: Wall Projection
A 10cm tall candle is placed 30cm away from a vertical wall. A small figurine (height 4cm) is placed between the candle and the wall, exactly 12cm from the candle. If the figurine begins moving toward the wall at a rate of 2cm/s, how fast is the height of its shadow on the wall changing?
Sketch the triangles here
Variables & Givens
Proportion Setup
SHOW YOUR WORK (Differentiate and Solve):
2
The Spotlight Approach
A spotlight on the ground is shining on a wall 20 meters away. A woman 1.6 meters tall walks from the spotlight toward the wall at a speed of 1.2 m/s. How fast is the length of her shadow on the wall decreasing when she is 5 meters from the wall?
STEP 1: SKETCH & LABEL
STEP 2: DIFFERENTIATE & CALCULATE
Final Answer:
3
The Inverse Shadow (Challenge)
A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate of \(0.5 \, m^3/min\), find the rate at which the water level is rising when the water is 3m deep.
Relating \(r\) and \(h\) via Similarity:
Full Solution Path:
Refined Reasoning
Why do similar triangles show up in cone problems? Look at the cross-section of the cone. The relationship between height and radius is constant regardless of how much water is inside. This is the ultimate "proportion power" move!
Shadow Geometry Teacher Guide Shadow Geometry
Facilitator Guide | Lesson 1
Duration
55-60 min
Instructional Narrative
The primary hurdle for students in shadow-related rates is not the calculus, but the **geometric setup**. Students often struggle to visualize the overlapping triangles. This lesson focuses on "encoding" the visual scenario into a static proportion before applying the tools of differentiation.
Key Objectives
Identify similar triangles in related rates scenarios.
Setup proportions relating heights and distances.
Differentiate implicitly with respect to time.
Prerequisites
Implicit Differentiation.
Geometric Similarity (AA Postulate).
Chain Rule mastery.
Misconceptions
Pole vs Person: Students often put the person's distance \(x\) in the denominator of the total distance rather than \(x+s\). Encourage "Nested Triangle" sketches.
Constants vs Variables: Remind students: "If it changes over time, it's a variable. If it's fixed (the pole height), it's a constant."
Pacing & Execution
0-10 min
The Hook: Whiteboard Projection
Use a high-lumen flashlight or phone light. Place a ruler near the light and project it on the board. Move the ruler 1 inch toward the board; watch the shadow jump. Ask: "Why did the shadow move more than the ruler?"
10-25 min
Modeling Proportion Power
Walk through Slide 3 & 4. Explicitly show the "No Quotient Rule" trick. Rearranging \( \frac{h}{x} = \frac{H}{L} \) to \( hL = Hx \) saves students from massive algebraic errors during differentiation.
25-45 min
Proportion Power Worksheet
Students work on Problems 1 & 2. Circulate and check specifically for the "Big Triangle" setup in Problem 2. Ensure they are using \(20\) as the total base, not just the distance from the wall.
45-55 min
Synthesis: The Cone Connection
Discuss Problem 3. Show how the cone's radius and height are just a 2D similar triangle in disguise. This bridges "shadows" to "volume" problems.
Worksheet Answer Key (Quick Ref)
PROBLEM 1
Shadow height decreases at \( \approx 0.44 \, cm/s \)
PROBLEM 2
Shadow decreases at \( 0.384 \, m/s \)
PROBLEM 3
\( dh/dt = \frac{2}{9\pi} \, m/min \)
Chasing Shadows Presentation Slides Chasing Shadows
Lesson 2: The Streetlight Problem
The Midnight Sprint
Walk past a streetlight at night. You're walking at a steady pace, but your shadow seems to race ahead of you.
Why the difference?
There are two different rates at play: the growth of the shadow itself, and the speed of the shadow's tip along the ground.
Distinguishing the Rates
Shadow Length (\(s\))
The actual distance from the person to the end of the shadow.
\( \frac{ds}{dt} \)
Rate of shadow growth
Shadow Tip (\(x+s\))
The total distance from the light source to the end of the shadow.
\( \frac{dx}{dt} + \frac{ds}{dt} \)
Velocity of the tip
The Standard Model
LIGHT (H) PERSON (h) x (Distance from Pole) s (Shadow Length)
Proportion: \( \frac{h}{s} = \frac{H}{x+s} \)
The Velocity Vector
The speed of the shadow's tip is always the sum of the person's speed AND the shadow's growth rate.
Tip Speed = \( \frac{dx}{dt} + \frac{ds}{dt} \)
If you walk at 5 ft/s and your shadow grows at 2 ft/s, the tip is moving at 7 ft/s!
Tip vs Length Practice Worksheet Tip vs. Length
The Streetlight Mastery Practice
SUBJECT: CALCULUS BC // UNIT 4
NAME:
Critical Distinction
Rate of Shadow Length Growth: \( \frac{ds}{dt} \)Rate of Shadow Tip Velocity: \( \frac{d}{dt}(x + s) = \frac{dx}{dt} + \frac{ds}{dt} \)
PROPORTION
\( \frac{h}{s} = \frac{H}{x+s} \)
1
The Runner's Shadow
A runner 6 ft tall is running away from a streetlight that is 18 ft high. The runner's speed is 10 ft/s.
Part A: Shadow Growth
Find the rate at which the length of the shadow is increasing.
Part B: Tip Speed
Find the speed at which the tip of the shadow is moving along the ground.
2
The Approaching Pedestrian
A person 1.8 meters tall is walking toward a lamp post at a rate of 1.5 m/s. The lamp is 5 meters above the ground. How fast is the tip of the shadow moving when the person is 3 meters from the post?
Calculation Space
3
Conceptual Synthesis
Two identical streetlights are 50 ft apart. A person walks from the first light toward the second at a constant speed \(v\). Describe the behavior of the shadow length as the person passes the midpoint between the two lights.
OBSERVATION:
MATHEMATICAL JUSTIFICATION:
"Calculus is the language of motion."
Chasing Shadows Answer Key Answer Key: Chasing Shadows
Teacher Reference // Lesson 2
PROBLEM 1
The Runner's Shadow
Part A: Shadow Growth (\(ds/dt\))
Proportion: \( \frac{6}{s} = \frac{18}{x+s} \)
Cross-multiply: \( 6(x+s) = 18s \Rightarrow 6x + 6s = 18s \Rightarrow 6x = 12s \)
Simplify: \( x = 2s \)
Differentiate: \( \frac{dx}{dt} = 2 \frac{ds}{dt} \)
Given \( \frac{dx}{dt} = 10 \), then \( 10 = 2 \frac{ds}{dt} \Rightarrow \frac{ds}{dt} = 5 \, \text{ft/s} \)
Part B: Tip Speed
Velocity of tip = \( \frac{d}{dt}(x+s) = \frac{dx}{dt} + \frac{ds}{dt} \)
Velocity of tip = \( 10 + 5 \)
Answer: 15 ft/s
PROBLEM 2
The Approaching Pedestrian
Setup: \( \frac{1.8}{s} = \frac{5}{x+s} \)
\( 1.8x + 1.8s = 5s \Rightarrow 1.8x = 3.2s \)
\( s = \frac{1.8}{3.2}x = \frac{9}{16}x \)
Differentiate: \( \frac{ds}{dt} = \frac{9}{16} \frac{dx}{dt} \)
Calculation: Given \( \frac{dx}{dt} = -1.5 \, \text{m/s} \) (approaching)
\( \frac{ds}{dt} = \frac{9}{16}(-1.5) \approx -0.84375 \, \text{m/s} \)
Tip Speed: \( -1.5 + (-0.84375) = -2.34375 \, \text{m/s} \)
Answer: Tip is moving at 2.34 m/s toward the post.
PROBLEM 3
Conceptual Synthesis
The Midpoint Behavior:
As the person walks from Light A to Light B, the shadow from Light A grows longer (positive rate), while the shadow from Light B (cast behind them) grows shorter (negative rate). At the midpoint, the two shadows will have identical lengths. As they pass the midpoint, the "primary" shadow they are walking toward will shrink to zero, while the "secondary" shadow cast by the light behind them will begin to dominate and grow rapidly.
Rising Angles Presentation Slides Rising Angles
Lesson 3: Trigonometric Rates of Change
The Rocket Tracker
A rocket launches vertically. You are in the viewing stands 3 miles away with a high-speed camera.
The Engineering Challenge:
As the rocket accelerates, the angle of your camera must change faster and faster. How do we program the motor to stay perfectly focused?
θ
The Differentiation Toolbox
Sine / Cosine
\[ \frac{d}{dt} \sin \theta = \cos \theta \frac{d\theta}{dt} \]
Used when hypotenuse is a known rate.
Tangent
\[ \frac{d}{dt} \tan \theta = \sec^2 \theta \frac{d\theta}{dt} \]
The "Go-To" for fixed distance observers.
Secant
\[ \frac{d}{dt} \sec \theta = \sec \theta \tan \theta \frac{d\theta}{dt} \]
Vital for rotating beam problems.
The Execution Flow
01
Choose the trig ratio that uses the constant side (usually the adjacent).
02
Differentiate both sides with respect to \(t\).
03
Find the value of \(\sec \theta\) or other ratios at the instant using the Pythagorean Theorem.
04
Solve for \(\frac{d\theta}{dt}\) in radians per second.
The Radian Rule
Calculus only works with Radians.
If your final answer for \(\frac{d\theta}{dt}\) is 2, it means 2 radians per second. If the question asks for degrees, you must convert!
Rocket Tracking Guided Notes Rocket Tracking
Guided Notes // Trigonometric Rates
NAME:
Angle of Elevation Focus
The Setup
"A rocket is launched vertically from a point on the ground. An observer stands 4000 meters away from the launch pad. At the moment the rocket's altitude is 3000 meters, it is rising at a rate of 200 m/s."
Sketch & Label (\(x, y, \theta\))
1 Choose the Relationship
Which trigonometric ratio connects the altitude (\(y\)) , the distance (\(x\)) , and the angle (\(\theta\)) ?
\( \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \)
2 Differentiate w.r.t Time (\(t\))
Apply implicit differentiation. Don't forget the Chain Rule for \(\theta\).
3 The "Instant" Snapshot
When \(y = 3000\) and \(x = 4000\), what is \(\sec \theta\)?
4 Final Solution
Solve for \(\frac{d\theta}{dt}\) in radians per second.
\(\frac{d\theta}{dt} =\)
Concept Shift
If the rocket stays at the same vertical speed, does the camera need to rotate faster or slower as the rocket gets higher? Justify your answer using the \(\sec^2 \theta\) term in your derivative.
Rising Angles Teacher Reference Teacher Reference: Rising Angles
Lesson 3 Strategy & Key
UNIT 4: RELATED RATES
Strategic Focus
The biggest hurdle in trig-related rates is the substitution of the constant side . Students often try to use sine or cosine, which makes the differentiation much harder because the hypotenuse is also changing.
Pro-Tip:
"Always aim for the trig function that puts the constant value in the denominator of your ratio."
Common Errors to Watch For:
Chain Rule Neglect: Forgetting the \(d\theta/dt\) when differentiating \(\tan \theta\).Angle Confusion: Plugging in 3000 for \(\theta\) instead of using it to find \(\sec \theta\).Unit Blindness: Forgetting that all calc-derived rates are in radians.
Rocket Tracking Solution Key
1. The Equation
\[ \tan \theta = \frac{y}{4000} \]
2. Differentiation
\[ \sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{4000} \cdot \frac{dy}{dt} \]
3. Instantaneous Values
At \(y=3000\), \(x=4000\), hypotenuse \(z=5000\).
\[ \cos \theta = \frac{4000}{5000} = \frac{4}{5} \Rightarrow \sec \theta = \frac{5}{4} \]
\[ \sec^2 \theta = \left(\frac{5}{4}\right)^2 = \frac{25}{16} \]
4. Final Solve
\[ \frac{25}{16} \cdot \frac{d\theta}{dt} = \frac{1}{4000} \cdot (200) \]
\[ \frac{25}{16} \cdot \frac{d\theta}{dt} = \frac{1}{20} \]
\[ \frac{d\theta}{dt} = \frac{1}{20} \cdot \frac{16}{25} = \frac{16}{500} = \mathbf{0.032 \, rad/s} \]
Light Speed Presentation Slides Light Speed
Lesson 4: Rotating Beams
The Shoreline Illusion
A lighthouse beam rotates at a steady pace. But if you watch the spot of light as it travels down a long, straight beach...
The Paradox:
The further the beam gets from the lighthouse, the faster it seems to travel, eventually reaching nearly infinite speed.
Modeling the Beam
Static Model
If the lighthouse is \(d\) units from the shore and the light spot is \(x\) units down the shore:
\( x = d \cdot \tan \theta \)
Angular Velocity (\(d\theta/dt\))
This is usually constant (e.g., 2 revolutions per minute).
Differentiating for Velocity (\(dx/dt\))
1
\( \frac{dx}{dt} = d \cdot \sec^2 \theta \cdot \frac{d\theta}{dt} \)
Notice the \(\sec^2 \theta\) term. As \(\theta \to 90^\circ\), \(\sec \theta \to \infty\). This explains why the beam "accelerates" down the shore!
Case Study: The Police Beacon
"A police car with a rotating light is parked 50 feet from a long straight wall."
Rotation Rate: 30 rev/min
Distance to Wall (\(d\)): 50 ft
Step 1: Convert Rate to Radians
\( 30 \times 2\pi = 60\pi \text{ rad/min} \)
At \(\theta = 45^\circ\) (\(\pi/4\) rad), what is the speed of the beam?
The Infinity Check
If the light rotates at a fixed rate, is there a maximum speed for the light spot on the wall?
YES
NO
(Mathematically speaking, the beam can exceed the speed of light! Ask a physics teacher how that works.)
Lighthouse Sweep Activity Sheet The Shoreline Sweep
Activity // Rotating Beams & Related Rates
DUE DATE
The Lighthouse Scenario
A lighthouse is located on a small island 3 kilometers away from the nearest point \(P\) on a straight shoreline. The lighthouse light rotates at a constant rate of 8 revolutions per minute .
A
Determine the angular velocity \(d\theta/dt\) in radians per minute.
B
Find the speed of the beam along the shore when the light is 1 km from \(P\).
C
Find the speed of the beam when the light is 5 km from \(P\).
[Diagram Construction Space]
Step 1: The Calculus Translation
Givens
Rates
Formula
Step 2: Differentiation
Differentiate \(x = 3 \tan \theta\) with respect to time \(t\)...
Step 3: Calculating at Specific Moments
WHEN \(x = 1\) km:
Finding \(\sec^2 \theta\):
WHEN \(x = 5\) km:
Finding \(\sec^2 \theta\):
The Physical Reality
As the light spot moves further down the shore, the distance it must cover for every 1 degree of rotation increases. Based on your calculations above, does the velocity of the light spot increase linearly or non-linearly ? How can you tell?
Light Speed Answer Key Answer Key: The Shoreline Sweep
Lesson 4 Solution Guide
TEACHER ONLY
Part A: Angular Velocity
\( \frac{d\theta}{dt} = 8 \, \text{rev/min} \times 2\pi \, \text{rad/rev} = \mathbf{16\pi \, \text{rad/min}} \)
General Differentiation
\( x = 3 \tan \theta \)
\[ \frac{dx}{dt} = 3 \sec^2 \theta \cdot \frac{d\theta}{dt} \]
Part B: At \(x = 1\) km
Hypotenuse \(c = \sqrt{3^2 + 1^2} = \sqrt{10}\)
\( \sec \theta = \frac{\sqrt{10}}{3} \)
\( \sec^2 \theta = \frac{10}{9} \)
\( \frac{dx}{dt} = 3 \left(\frac{10}{9}\right) (16\pi) \)
\( \frac{160\pi}{3} \approx 167.55 \, \text{km/min} \)
Part C: At \(x = 5\) km
Hypotenuse \(c = \sqrt{3^2 + 5^2} = \sqrt{34}\)
\( \sec \theta = \frac{\sqrt{34}}{3} \)
\( \sec^2 \theta = \frac{34}{9} \)
\( \frac{dx}{dt} = 3 \left(\frac{34}{9}\right) (16\pi) \)
\( \frac{544\pi}{3} \approx 569.67 \, \text{km/min} \)
Reflection Answer:
The velocity increases non-linearly . As the light spot moves from 1km to 5km (a 5x distance increase), the velocity increases from ~167 to ~570 km/min (a ~3.4x increase). The presence of the \(\sec^2 \theta\) term in the derivative ensures that the rate of change of \(x\) increases drastically as \(\theta\) approaches \(\pi/2\).
Calculus Gauntlet Challenge Cards Calculus Gauntlet
Level 5 Challenge Cards
Stage 1: Linear Shadows HARD
The Double Projection
A light source is 10m above the ground. A 2m tall man walks toward the light at 1m/s. Behind the light source, 5m away, is a giant mirror. Find the rate at which the man's shadow on the wall (behind the light) is changing when he is 4m from the light source.
SIMILAR TRIANGLES / PROPORTIONS
Stage 2: Angular Pursuit ELITE
The Balloon Tracker
A hot air balloon rises vertically from a point \(A\) at 5 m/s. An observer is at point \(B\), 100m from \(A\). Simultaneously, the observer begins walking toward \(A\) at 2 m/s. Find the rate of change of the angle of elevation \(\theta\) of the balloon after 10 seconds.
TRIG / QUOTIENT RULE / VARIABLE ADJACENT
Stage 3: Circular Beams BOSS
The Curved Wall
A searchlight 100 ft from a straight wall is rotating at 1 rad/s. However, the wall is actually a curve defined by \(y = x^2/100\). As the beam hits the point \((100, 100)\), how fast is the spot moving along the surface of the wall?
TRIG / ARC LENGTH / CURVILINEAR MOTION
Stage 4: Mixed Realities ULTIMATE
The Solar Eclipse
A circular disk (radius 10cm) is moving toward a light source (5m away) at 20cm/s. The light projects the disk's shadow onto a screen 10m from the light. Find the rate of change of the AREA of the shadow when the disk is 2m from the light source.
GEOMETRY / AREA DERIVATIVES / SHADOWS
Gauntlet Workspace Sheet The Gauntlet Workspace
Calculus // Final Mastery Session
TEAM:
STAGE 1
SCORE: ___ / 10
Ans:
STAGE 2
SCORE: ___ / 10
Ans:
STAGE 3
SCORE: ___ / 10
Ans:
STAGE 4
SCORE: ___ / 10
Ans:
Strategy Reflection:
Which stage required the most complex geometric modeling? What was the "Aha!" moment for your team?
Gauntlet Solution Guide Teacher Resource Gauntlet Solutions
Teacher Guide // Lesson 5
Stage 1: The Double Projection
Concept: Similar triangles with shifting origins.
Relationship: \( \frac{2}{s} = \frac{10}{x+s} \Rightarrow 2x+2s = 10s \Rightarrow 2x = 8s \Rightarrow s = 0.25x \)
Differentiation: \( \frac{ds}{dt} = 0.25 \frac{dx}{dt} \)
Given \( \frac{dx}{dt} = -1 \) m/s, then \( \frac{ds}{dt} = -0.25 \) m/s.
Answer: Shadow length is decreasing at 0.25 m/s.
Stage 2: The Balloon Tracker
Concept: Tangent with two changing variables (\(x\) and \(y\)).
Relationship: \( \tan \theta = \frac{y}{x} \)
After 10s: \( y = 5(10) = 50 \)m, \( x = 100 - 2(10) = 80 \)m. \( z = \sqrt{50^2 + 80^2} = 10\sqrt{89} \)
Deriv: \( \sec^2 \theta \frac{d\theta}{dt} = \frac{x(dy/dt) - y(dx/dt)}{x^2} \)
Values: \( \sec^2 \theta = (10\sqrt{89}/80)^2 = 89/64 \)
\( (89/64) \frac{d\theta}{dt} = \frac{80(5) - 50(-2)}{80^2} = \frac{400+100}{6400} = \frac{5}{64} \)
Answer: \( \frac{d\theta}{dt} = \frac{5}{89} \approx 0.056 \) rad/s.
Stage 3: The Curved Wall
Concept: Parametric rates along a curve.
Spot moves along \(y = x^2/100\). Path speed \( v = \sqrt{(dx/dt)^2 + (dy/dt)^2} \).
At \((100, 100)\), \( dy/dx = 2x/100 = 2 \). So \( dy/dt = 2(dx/dt) \).
\( v = \sqrt{(dx/dt)^2 + (2dx/dt)^2} = \sqrt{5} \frac{dx}{dt} \).
From trig: \( x = 100 \tan \theta \Rightarrow \frac{dx}{dt} = 100 \sec^2 \theta \frac{d\theta}{dt} \).
At \((100,100)\), \(\theta = \pi/4\), \(\sec^2 \theta = 2\). \( \frac{dx}{dt} = 100(2)(1) = 200 \).
Answer: \( v = 200\sqrt{5} \approx 447.2 \) ft/s.
Stage 4: The Solar Eclipse
Concept: Area rates of change via radial proportions.
Disk radius \(r = 10\). Shadow radius \(R\). \( \frac{10}{x} = \frac{R}{10} \Rightarrow R = \frac{100}{x} \).
\( A = \pi R^2 = \pi (100/x)^2 = 10000\pi x^{-2} \).
\( \frac{dA}{dt} = -20000\pi x^{-3} \frac{dx}{dt} \).
At \(x = 200\) cm (\(2\)m), \( \frac{dx}{dt} = 20 \) cm/s.
\( \frac{dA}{dt} = -20000\pi (200)^{-3} (20) = -400000\pi / 8000000 = -\pi/20 \).
Answer: Area is decreasing at \(\pi/20 \approx 0.157\) cm²/s.
