Motion Masterclass Slides Unit: Instantaneous Change
Motion Masterclass
Analyzing Average Rates of Change
The Radar Gun Dilemma
"You drove 60 miles in 60 minutes. Your average speed was 60 mph."
"But at 12:34 PM, my radar gun clocked you at 82 mph."
The Big Question
How can we measure speed at a single instant if motion requires time to pass?
Recap: Average Rate of Change
The average rate of change of a function \(f(x)\) over the interval \([a, b]\) is the slope of the secant line connecting the two points.
\[ m_{sec} = \frac{f(b) - f(a)}{b - a} \]
(a, f(a)) (b, f(b))
Secant Line = Average Slope
The Falling Phone
Case Study
A phone is dropped from a height. Its distance \(d\) (meters) from the drop point after \(t\) seconds is given by:
\[ d(t) = 4.9t^2 \]
Task 1:
Find the average velocity between \(t = 1\) and \(t = 3\).
Task 2:
Predict the "speedometer reading" exactly at \(t = 1\).
Can an "Instant" be
zero seconds long?
If we want speed at \(t = 1\), our interval is \([1, 1]\).
\[ m = \frac{f(1) - f(1)}{1 - 1} = \frac{0}{0} \]
Undefined!
Conclusion: We can't calculate instantaneous speed directly using the slope formula. We need a new strategy.
The Speed Trap Worksheet The Speed Trap
Analyzing Average vs. Instantaneous Velocity
Name:
Date:
Background
The average velocity of an object over a time interval \([t_1, t_2]\) is the change in position divided by the change in time. Geometrically, this is the slope of the secant line connecting two points on a position-time graph.
1 Scenario: The High-Altitude Drop
A heavy ball is dropped from a cliff. Its distance fallen \(d(t)\) (in meters) after \(t\) seconds is modeled by: \(d(t) = 5t^2\)
1. Calculate the average velocity of the ball over the following time intervals:
a) Interval [1, 3] seconds
b) Interval [1, 2] seconds
The Zoom-In Strategy:
Let's try to find the speed exactly at \(t = 1\) second by looking at even smaller intervals starting at 1.
Time Interval \([1, 1 + h]\) Change in Time (\(h\)) Calculation \(\frac{d(1+h) - d(1)}{h}\) Avg. Velocity (m/s) [1, 1.1] 0.1 [1, 1.01] 0.01 [1, 1.001] 0.001
2. Based on your table, what do you predict is the exact speed at \(t = 1\)? Why?
2 Geometric Reflection
The graph below shows \(d(t) = 5t^2\). Point \(A\) is at \(t = 1\) and Point \(B\) is at \(t = 3\).
A (1, 5) B (3, 45)
3. Describe the visual difference between the secant line from \(t=1\) to \(t=3\) and the "true" slope of the curve at point \(A\).
4. As Point B moves closer and closer to Point A, what happens to the slope of the secant line?
Reflection: The Police Radar Gun
"A radar gun doesn't calculate your speed over a mile. It measures it instantly."
Explain in your own words why calculating the slope over a tiny, tiny interval (like 0.0001 seconds) is a better "radar gun" than calculating it over 10 seconds.
Motion Facilitation Guide Teacher Resource
Motion Facilitation
Lesson 1: Analyzing Average Rates of Change
UNIT
Instantaneous Change
Pedagogical Approach
This lesson focuses on creating cognitive dissonance . Students are comfortable with average speed (\(d/t\)), but the concept of speed at an exact point is logically paradoxical because an "instant" has zero duration, and division by zero is undefined.
The Goal:
Lead students to discover that as the time interval \(h\) approaches 0, the average velocity converges to a specific value. This value is the instantaneous velocity .
Key Discussion Points:
The Limit Concept: Even if we can't evaluate the speed at \(t=1\), we can observe what the speed is trending toward .Secant vs. Tangent: Introduce the idea that a secant line crosses twice, but a tangent line "kisses" the curve at a single point (geometrically defining the instantaneous slope).
Misconceptions
1. Dividing by Zero
Students may think "speed at an instant" is \(0/0\). Emphasize that we never actually reach the instant; we look at the trend as we get closer.
2. Interval Direction
Remind students that we could approach \(t=1\) from the left \([0.9, 1]\) or the right \([1, 1.1]\). Both should yield the same trend.
3. Units
Ensure students recognize units are \(meters/second\) (ratio of position change to time change).
Answer Key: The Speed Trap
Part 1: The Falling Object (\(d(t) = 5t^2\))
1a. Interval [1, 3]
\(\frac{d(3) - d(1)}{3 - 1} = \frac{45 - 5}{2} = 20 \text{ m/s}\)
1b. Interval [1, 2]
\(\frac{d(2) - d(1)}{2 - 1} = \frac{20 - 5}{1} = 15 \text{ m/s}\)
Numerical Convergence Table:
Interval h Calculation Result [1, 1.1] 0.1 \((5(1.1)^2 - 5) / 0.1 = (6.05 - 5) / 0.1\) 10.5 m/s [1, 1.01] 0.01 \((5(1.01)^2 - 5) / 0.01 = (5.1005 - 5) / 0.01\) 10.05 m/s [1, 1.001] 0.001 \((5(1.001)^2 - 5) / 0.001 = (5.010005 - 5) / 0.001\) 10.005 m/s
Prediction: The exact speed at \(t=1\) is 10 m/s.
Part 2: Geometric Reflection
The Infinite Zoom Slides The Infinite Zoom
From Secant Lines to Tangent Lines
Lesson 02
The Zoom-In Challenge
Pick any smooth curve on a graph. Now, zoom in on a single point. Keep zooming.
What do you eventually see?
Normal
10x Zoom
100x Zoom
Local Linearity
The curve becomes a line.
The Tangent Line
1
A Tangent Line is the unique line that "touches" the curve at a point and has the same slope as the curve at that point.
2
It is the limit of the secant lines as the interval width (\(h\)) goes to zero.
Tangent vs Secant
Formalizing the Idea
The slope of the tangent line is the limit of the secant slope:
\[ m_{tan} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
As \(h\) (the gap) gets infinitely small, the secant line "becomes" the tangent line.
Microscopic Slopes Activity Microscopic Slopes
Secant to Tangent Transition
NAME: ___________________________
DATE: ___________________________
PHASE 1
The Magic of Local Linearity
Calculus works because smooth curves look like straight lines if you zoom in enough. This is called local linearity . In this section, we will visually estimate the slope of the curve \(f(x) = x^2\) at \(x = 1\).
NORMAL VIEW
A curved parabola: hard to find the slope.
ZOOMED VIEW (at x=1)
At 1000x zoom, the curve is a line!
1. Look at the ZOOMED VIEW. Use the background grid to estimate the slope of the blue line.
(Remember: Slope = Rise / Run)
2. This estimated slope is the "Instantaneous Rate of Change" at \(x = 1\). Based on your visual estimate, what is the value?
PHASE 2
The Shrinking Secant
A secant line connects two points, \((x, f(x))\) and \((x+h, f(x+h))\). As the distance \(h\) between the points shrinks, the secant line rotates.
\(h = 2\)
Wide interval
\(h = 1\)
Shrinking gap
\(h \to 0\)
Tangent Line
3. Why can we never let \(h\) actually equal zero in the slope formula \(\frac{f(x+h) - f(x)}{h}\)?
4. Draw a tangent line by hand at the three points indicated on the curve below. Use a ruler if possible.
Slope at Point 1 is:
Slope at Point 2 is:
Slope at Point 3 is:
Big Idea Connection
A tangent line is the "limit" of secant lines. As the points on a curve get closer together, the secant line's average slope approaches the curve's instantaneous slope.
Precision Power Slides Unit: Instantaneous Change
Precision Power
Estimating Derivatives Numerically
Closing the Gap
We've seen it geometrically. Now let's see it numerically.
"To find the slope at \(x = c\), we calculate average slopes over smaller and smaller intervals: \([c, c+h]\) as \(h\) shrinks."
The Shrinking \(h\)
Large h
0.1
Medium h
0.01
Tiny h
0.001
We are looking for convergence.
Example: \(f(x) = x^2\) at \(x = 3\)
\(h\) Interval Slope Calculation Avg Slope 0.1 [3, 3.1] \((3.1^2 - 3^2) / 0.1\) 6.1 0.01 [3, 3.01] \((3.01^2 - 3^2) / 0.01\) 6.01 0.001 [3, 3.001] \((3.001^2 - 3^2) / 0.001\) 6.001
The Convergence
As \(h \to 0\), the slopes approach a single, clean whole number.
6.000...
Pro-Tip: Using Tables
Don't type the formula manually 5 times! Use your graphing calculator's TABLE function.
Set \(Y_1 = f(x)\)
Set \(Y_2 = (Y_1(X+h) - Y_1(X)) / h\)
Change \(h\) to see the slope converge.
TABLE SETUP
X=3
Y1=9
Y2=6.0001
Looking good!
Convergence Found
Closing the Gap Worksheet Closing the Gap
Numerical Estimation of Instantaneous Rates
Name:
Date:
Objective: Use tables and small increments (\(h\)) to approximate the slope of the tangent line (the derivative) at a specific point and identify the value toward which the slopes converge.
01 Investigating \(f(x) = x^2 + 2x\) at \(x = 1\)
Complete the table to find the average slope over shrinking intervals \([1, 1+h]\). Show your setup for the first calculation.
Slope Setup (\(h=0.1\))
Increment (\(h\)) Point \(x + h\) \(f(x+h)\) Avg. Slope (\(m\)) 0.1 1.1 0.01 1.01 0.001 1.001 0.0001 1.0001
Based on the table above, what value does the slope seem to be approaching?
02 Investigating \(g(x) = \sin(x)\) at \(x = 0\)
(Make sure your calculator is in RADIAN mode!)
Increment (\(h\)) \(g(0+h)\) Avg. Slope \(\frac{\sin(h) - \sin(0)}{h}\) 0.1 0.01 0.001
3. What number is the slope of \(\sin(x)\) approaching at \(x = 0\)?
4. Draw a rough sketch of \(\sin(x)\) near the origin. Does your numerical answer make visual sense? Explain.
Critical Thinking
5. If we found that the slopes for a function approach 4.9, 4.99, 4.999... what would you guess is the exact derivative?
6. Why do we need to check very small values of \(h\)? Why isn't \(h=0.5\) good enough for an estimate?
Closing the Gap Answer Key Answer Key
Numerical Estimation
Lesson 3: Closing the Gap
Problem 1: \(f(x) = x^2 + 2x\) at \(x = 1\)
Calculation Setup (h=0.1):
Slope = \([f(1.1) - f(1)] / 0.1 = [(1.1^2 + 2(1.1)) - (1^2 + 2(1))] / 0.1\)
Slope = \([3.41 - 3] / 0.1 = 0.41 / 0.1 = 4.1\)
h x + h f(x+h) Avg. Slope 0.1 1.1 3.41 4.1 0.01 1.01 3.0401 4.01 0.001 1.001 3.004001 4.001 0.0001 1.0001 3.0004... 4.0001
Convergence: The slope approaches 4.
Problem 2: \(g(x) = \sin(x)\) at \(x = 0\)
h (radians) \(\sin(h)\) Calculation: \(\sin(h)/h\) 0.1 0.09983 0.9983 0.01 0.00999 0.99998... 0.001 0.00099... 0.99999...
3. Result: Approaching 1.
4. Visual Sense: Yes, near \(x=0\), the graph of \(\sin(x)\) looks like the line \(y = x\), which has a slope of 1.
Synthesis Solutions
5. Guessing the derivative for 4.9, 4.99, 4.999...
The exact derivative is 5.
6. Why small \(h\) is necessary:
Large intervals like \(h=0.5\) pick up too much curvature and don't reflect the "local" behavior. As \(h \to 0\), the secant line aligns with the tangent line, providing a precise "snapshot" of the slope at that point.
Derivative Portrait Slides Visual Analysis
Derivative Portraits
Sketching the Function's Shadow
Lesson 04
The Big Mapping Rule
Original Function (\(f\))
Look at the SLOPE.
Derivative Function (\(f'\))
Look at the y-VALUE.
Why this works:
The derivative is just a collection of all the slopes of the original function. We plot those slopes as our new \(y\)-coordinates.
"The derivative is the story of the original's growth."
Mapping the Landmarks
1
Hill or Valley
Slope is 0.
\(f'(x)\) crosses the x-axis.
2
Going UP
Slope is Positive.
\(f'(x)\) is above the x-axis.
3
Going DOWN
Slope is Negative.
\(f'(x)\) is below the x-axis.
The Velocity Profile
Imagine driving over a hill. Your elevation is \(f(x)\). Your speedometer reading is \(f'(x)\).
Steepest part = highest peak on derivative.
Flat top = zero point on derivative.
Straight sections = constant lines on derivative.
f(x): Rollercoaster Track
f'(x): Velocity Graph
Vertical Alignment Matters!
Slope Sketcher Activity Slope Sketcher
Graphing the Derivative
NAME: ___________________________
DATE: ___________________________
LEVEL 1
The Snapshot Rule
For each graph of \(f(x)\) below, sketch the graph of its derivative \(f'(x)\) on the grid provided directly beneath it. Tip: Align your zeros first! If \(f(x)\) has a flat peak, \(f'(x)\) must be zero at that exact \(x\)-coordinate.
f(x): PARABOLA
SKETCH f'(x) HERE
Self-Check Checklist:
Where is the slope zero? (Mark this on the x-axis of \(f'\))
Where is the slope positive? (Sketch \(f'\) above the x-axis)
Where is the slope negative? (Sketch \(f'\) below the x-axis)
f(x): CUBIC
SKETCH f'(x) HERE
Reflect:
Observe the cubic curve. It has two "turning points" (a peak and a valley). What does this tell you about the number of times the derivative will cross the x-axis?
LEVEL 2
Reverse Engineering
Now, let's try it backwards. I'll give you the graph of the velocity (derivative). Can you sketch a possible position graph (original function)?
f'(x): THE DERIVATIVE
SKETCH A POSSIBLE f(x) HERE
Think About It:
If the derivative is a straight line with a positive slope (like above), it means the original function's slope is constantly increasing. What kind of shape gets steeper and steeper?
Describe your reasoning:
The Calculus Dialect Slides Final Lesson
The Calculus Dialect
Interpreting Notation in Context
Who's Who in Notation?
Lagrange Notation
"PRIME"
\[ f'(x) \]
Pros:
Short, quick to write, emphasizes that the derivative is a new function.
Leibniz Notation
"D-Y D-X"
\[ \frac{dy}{dx} \]
Pros:
"The rate of change of \(y\) with respect to \(x\)."
Clear about variables and units. Reminds us of \(\Delta y / \Delta x\).
The Logic of Units
The units of a derivative are always:
Output Units
Input Units
A
Scenario
Population \(P\) at time \(t\) (years).
Unit: People per Year
B
Scenario
Cost \(C\) (\(\$\)) to build \(x\) houses.
Unit: Dollars per House
Translation Challenge
Let \(T(h)\) be the temperature in °C at a height of \(h\) meters above sea level.
Interpret: \(T'(2000) = -0.01\)
Wrong Interpretation
"The temperature is -0.01 degrees."
No context, wrong units.
Perfect Interpretation
"At an altitude of 2000 meters, the temperature is decreasing at a rate of 0.01 °C per meter ."
Speaking Derivative Worksheet Speaking Derivative
Context and Notation Practice
NAME: ___________________________
DATE: ___________________________
Part 1
The Language of Change
"A derivative isn't just a number; it's a relationship between two units. The units of the derivative are always the units of the output divided by the units of the input."
Scenario Notation (\(dy/dx\)) Units of Derivative Fuel \(F\) (gal) in a tank after time \(t\) (hr). \(dF/dt\) Gallons per Hour Cost \(C\) (\(\$\)) to produce \(x\) tons of steel. ____________ Pressure \(P\) (psi) at depth \(d\) (ft) underwater. ____________ Weight \(W\) (kg) of a tree after \(t\) years. ____________
Part 2
The Interpreter's Task
Write a complete sentence to explain the meaning of each derivative statement. Include units and indicate whether the value is increasing or decreasing.
1. \(P(t)\) is the population of bees in a hive at time \(t\) (days). Interpret: \(P'(10) = -200\)
2. \(V(r)\) is the volume of a balloon (cm³) with radius \(r\) (cm). Interpret: \(V'(5) \approx 314\)
3. \(C(x)\) is the total cost (\(\$\)) to print \(x\) copies of a book. Interpret: \(C'(500) = 12\)
Part 3
The Calculus Consultant
A biologist is studying a bacterial colony. Let \(B(t)\) be the number of bacteria at time \(t\) (minutes).
4. What does it mean for the colony if \(B'(t) > 0\)?
5. If \(B'(t) = 0\), what is happening to the population at that exact moment?
6. The biologist says: "The rate of growth is slowing down." Does this mean the population is decreasing? Explain.
Speaking Derivative Answer Key Answer Key
Speaking Derivative
Lesson 5: Context & Notation
Part 1: Notation & Units
Scenario Notation (\(dy/dx\)) Units Cost \(C\) (\(\$\)) for \(x\) tons of steel. dC / dx Dollars per Ton Pressure \(P\) (psi) at depth \(d\) (ft). dP / dd PSI per Foot Weight \(W\) (kg) of tree after \(t\) years. dW / dt Kilograms per Year
Part 2: Sentence Building
1. \(P'(10) = -200\)
"On the 10th day, the population of bees is decreasing at a rate of 200 bees per day."
2. \(V'(5) \approx 314\)
"When the radius of the balloon is 5 cm, the volume is increasing at a rate of approximately 314 cm³ per cm of radius."
3. \(C'(500) = 12\)
"At a production level of 500 books, the cost to print the next book (marginal cost) is 12 dollars per book."
Part 3: The Calculus Consultant
4. \(B'(t) > 0\)
The population is growing/increasing.
5. \(B'(t) = 0\)
The population is at a standstill/constant. At this exact moment, it is neither growing nor shrinking (often at a peak or valley).
6. "Growth is slowing down" vs. "Decreasing"
No, it does not necessarily mean the population is decreasing. It means the rate of growth is getting smaller (e.g., growing by 10 per min, then 5 per min). The population is still growing, just not as fast. In calculus terms, the first derivative is positive, but the second derivative would be negative.
