The Swiss mathematician
Jacob Bernoulli
(1654–1705), painted by his brother Niklaus in 1687. While studying
compound interest
in 1683, Bernoulli was the first to notice the constant e ≈ 2.718 — the number a bank balance approaches when interest is compounded not yearly, not monthly, not daily, but at every instant. Image: Niklaus Bernoulli (1662–1716) / Wikimedia Commons —
File:Jakob_Bernoulli.jpg, public domain.
What is the difference between
“growing 5% per year”
and “growing continuously at 5%” —
and does the distinction
actually change anything?
Standards: HSA-SSE.A.1.b, HSF-IF.C.7, HSF-LE.B.5. Section C continuation — students met e as a number in L12; today they meet it as the base of a function, and learn what the r in \(P_0 \cdot e^{rt}\) means. The Big Question lives here; SWBAT moves to the lesson-synthesis slide (s9) — do not read objectives here. Two minutes max on this slide. (D)-paced lesson — Activity 3 (graphing practice) is omitted; if you have extra time, assign as homework or return to it next lesson.
Prep Checklist
☐ Scientific calculators with an e^x key and parentheses keys — every student.
☐ Graphing technology available (Desmos is fine) for the Activity 1 colony comparison.
☐ Calculator pitfall — the exponent must be wrapped in parens: e^(0.03*t), never e^0.03*t, or the calculator computes e0.03·t.
☐ No printed materials — everything is on these slides.
Warm-Up5 min
Do Now
Complete on your own. Then confirm your answers with a partner.
Press the e^x key (often above ln; use SHIFT or 2nd).
display shows e^(
2
Type the entire exponent inside the parentheses, close with ), then =.
e^(3.4) → 30.04166
Try Saying
My calculator gave ___ for problem 1. I had to ___ to get the parentheses right.
5 min · Combined launch + work + synthesis.
Launch (1 min): "The constant e is approximately 2.718. Today we use it as a base for exponential models. Right now, just practice entering it on your calculator."
Work (3 min): Students evaluate the three expressions independently, then confirm with a partner. Leave the mini How-To table visible — that IS the reference for this warm-up.
Synthesis (1 min): The synthesis happens through the ACB wrong-answer feedback. Common pitfalls to surface aloud if you see them:
No parentheses around the exponent → calculator interprets e^9 · .31 for problem 2, which is wildly off.
Using 2.718 manually instead of the e^x key → precision loss (e.g., 23.34 vs. 30.04 on problem 1).
Rounding mid-step → final value drifts.
Keep the warm-up tight — the meat of the lesson is Activity 1. The constant e gets a proper MASL slide next (s3).
Prep check: Every student should have a scientific calculator with an e^x key. If a calculator only has ln, the e^x is usually the SHIFT/2nd of ln.
Amplify · 3 inputs
Order per problem: Text component (problem text) → Math Response named answer1 / answer2 / answer3 → Note with CL.
All three are numeric — use answer.numericValue with bands centered on the exact value. Feedback strings point to parentheses + precision, never to the answer.
Math Response input hint: Example: 30.04
Problem 1: Evaluate e^(3.4).
content:
when answer1.submitted and answer1.numericValue > 30.04 and answer1.numericValue < 30.05
"Correct! Using the e^x key with the exponent in parentheses gives about 30.04166. You're set up for the rest of the lesson."
when answer1.submitted and answer1.numericValue > 27.0 and answer1.numericValue < 27.2
"Almost! That looks like e^3 · 0.4 instead of e^(3.4). The whole exponent — including the decimal — goes inside the parentheses. Try again with e^(3.4)."
when answer1.submitted and answer1.numericValue > 23.0 and answer1.numericValue < 23.7
"Hmm — that's close to (2.718)^(3.4), which means you typed 2.718 by hand instead of using the e^x key. The e^x key carries more digits of precision. Use it and retry."
when answer1.submitted and answer1.numericValue > 100
"Not quite — that's much too large. Double-check that the exponent is 3.4, not 34 or 3.4·something. Use e^(3.4) with parentheses."
when answer1.submitted and isBlank(answer1.latex)
"Press the e^x key (often above ln — use SHIFT or 2nd), type 3.4 inside the parentheses, close with ), then =. What number comes out?"
when answer1.submitted
"Take another look at the parentheses. The full expression on your screen should read e^(3.4). The answer should sit a little above 30."
otherwise ""
Problem 2: Evaluate e^(9.31).
content:
when answer2.submitted and answer2.numericValue > 11041.0 and answer2.numericValue < 11042.5
"Correct! e^(9.31) ≈ 11,041.74. That's a big jump from problem 1 — small change in the exponent, huge change in the output. That's exponential behavior."
when answer2.submitted and answer2.numericValue > 2510 and answer2.numericValue < 2520
"Almost — that looks like e^9 · 0.31, which is what happens when the .31 falls outside the parentheses. The exponent 9.31 needs to live entirely inside: e^(9.31)."
when answer2.submitted and answer2.numericValue > 8000 and answer2.numericValue < 8200
"Not quite — that's close to e^9 alone. The .31 belongs inside the parentheses with the 9. Try e^(9.31) as one chunk."
when answer2.submitted and answer2.numericValue < 100
"Hmm — e^(9.31) should be in the thousands. Check that you pressed the e^x key (not just typed 'e' as a letter), and that 9.31 is inside the parentheses."
when answer2.submitted and isBlank(answer2.latex)
"Press the e^x key, type 9.31 inside the parentheses, close with ), then =. What does your calculator show?"
when answer2.submitted
"Take another look at the parentheses around 9.31. The display should read e^(9.31). The answer should be in the low eleven-thousands."
otherwise ""
Problem 3: Evaluate e^(2.137).
content:
when answer3.submitted and answer3.numericValue > 8.478 and answer3.numericValue < 8.480
"Correct! e^(2.137) ≈ 8.47891. Notice you needed all three decimal places of the exponent inside the parentheses to land on this value."
when answer3.submitted and answer3.numericValue > 7.38 and answer3.numericValue < 7.40
"Almost — that's e^2 alone. The .137 belongs inside the parentheses with the 2. Try e^(2.137) as one chunk."
when answer3.submitted and answer3.numericValue > 8.0 and answer3.numericValue < 8.4
"You're close — check that you typed all three decimal places (2.137, not 2.13 or 2.14). The full exponent goes inside the parentheses."
when answer3.submitted and answer3.numericValue > 9.0
"Hmm — take another look. The answer should sit between 8 and 9. Make sure the exponent on your screen is exactly 2.137 inside the parentheses."
when answer3.submitted and isBlank(answer3.latex)
"Press the e^x key, type 2.137 inside the parentheses, close with ), then =. Give it a try."
when answer3.submitted
"Take another look at the exponent. The display should read e^(2.137) with all three decimals inside the parens. The answer should sit between 8 and 9."
otherwise ""
Activity 1 · MASL2 min
Reading e aloud
Math · We Say · Meaning
★ Math (given)
\(e\)
▲ We Say
“e”
(single syllable, like the letter)
● Meaning
The mathematical constant approximately 2.718. It is the base of the natural exponential function and appears whenever a quantity grows or decays continuously — at every moment, not in discrete steps.
Try Saying
I placed ___ here because ___.
Try Saying
Does ___ belong with ___?
Try Saying
I don't think ___ belongs here because ___.
Standalone MASL for the constant before the formula. Per the MASL general-before-components ordering rule, students meet the symbol e on its own here, before they encounter the continuous growth factor \(e^{rt}\) on s5 and the full model \(P_0 \cdot e^{rt}\) on s7. Read aloud as the single letter “e” — never “Euler's number” with students at this stage. The “why approximately 2.718” story (the continuous-compounding limit) is reserved for future courses (Calculus). For now, treat e as a named constant like π: an irrational number with a fixed value that shows up whenever a quantity changes at every moment. No ACB — MASL slides are discussion only.
Activity 1 · Work Time10 min
Same Situation, Different Equations
Two scientists model the population of the same insect colony — a starting size of 9 thousand. They write two different functions for the population t months in.
Scientist 1 · once per month
\(f(t) = 9 \cdot (1.03)^{t}\)
9*(1.03)^t
Scientist 2 · continuously
\(g(t) = 9 \cdot e^{0.03\,t}\)
9*e^(0.03*t)
Calculator tip: For g(t), always type e^(0.03*t) with parens — without them the calculator computes e^0.03 · t, which is wrong.
t (months)
\(f(t) = 9\cdot(1.03)^{t}\)
\(g(t) = 9\cdot e^{0.03t}\)
difference
6
12
24
48
100
What do you notice about the populations in the two models?
click to reveal sample row ▶
t
\(f(t) = 9\cdot(1.03)^{t}\)
\(g(t) = 9\cdot e^{0.03t}\)
diff
6
10.74
10.78
~0.04
12
12.82
12.90
~0.08
24
18.26
18.49
~0.23
48
37.04
37.99
~0.95
100
172.40
180.79
~8.4
10 min · Activity 1 Work Time. Leave on screen during work.
(D)-COMPRESSED: IM's original activity uses four colonies (rates 1%, 3%, 7%, 15%) with separate graphs. We use ONE colony (the 3% case) here, then on s6 show all four pairs of graphs together as a single side-by-side reveal. Cuts ~10 min of redundant graph plotting while preserving the “the bigger the rate, the bigger the gap” insight.
Calculator pitfall: Without parens, e^0.03*t evaluates e^0.03 first, then multiplies by t — a linear function, not exponential. Show the parenthesized form on the warm-up How To if a pair gets stuck.
Targets (autograded column = g(t) only):t=6 → 10.78; t=12 → 12.90; t=24 → 18.49; t=48 → 37.99; t=100 → 180.79. The f(t) column students verify against their partner. Difference column is informal — observation, not autograded.
Collect and Display: listen for “the values are close,” “continuous is a tiny bit bigger,” “the gap grows.” Save the gap insight for s6 where we compare all four rates.
Amplify · 5 inputs (g(t) column)
Order per row: Text component (problem label) → Math Response named answer1…answer5 → Note with CL.
All five inputs autograde the g(t) column only. Use generous bands (±0.05) to absorb rounding from 2dp display vs. full precision.
content:
when answer1.submitted and answer1.numericValue > 10.73 and answer1.numericValue < 10.83
"Correct! g(6) = 9 · e^(0.18) ≈ 10.78. The discrete model gives 10.74 — just 0.04 less. At a 3% rate over 6 months, the two models barely diverge."
when answer1.submitted and answer1.numericValue > 10.69 and answer1.numericValue < 10.79
"Almost — you may have typed f(t) instead of g(t). 10.74 is the (1.03)^6 value. For g(6), use e^(0.18), calc entry 9*e^(0.03*6)."
when answer1.submitted and answer1.numericValue > 1.19 and answer1.numericValue < 1.22
"Hmm — take another look. You may have computed e^(0.03)*6 (parens around 0.03 only). The exponent is 0.03 times 6 = 0.18. Calc entry: 9*e^(0.03*6) with parens around 0.03*6."
when answer1.submitted and isBlank(answer1.latex)
"Start with the exponent: 0.03 · 6 = 0.18. Then enter 9*e^(0.18) — or in one step, 9*e^(0.03*6)."
when answer1.submitted
"Good start — check the calculator entry. You want 9*e^(0.03*6) with parens around the whole exponent. What do you get?"
otherwise ""
content:
when answer2.submitted and answer2.numericValue > 12.85 and answer2.numericValue < 12.95
"Correct! g(12) = 9 · e^(0.36) ≈ 12.90. The discrete model gives 12.82 — a gap of about 0.08. Twice the time, twice the gap (roughly)."
when answer2.submitted and answer2.numericValue > 12.77 and answer2.numericValue < 12.87
"You're close — that value is f(12), not g(12). 12.82 comes from (1.03)^12. For g(12), use e^(0.36)."
when answer2.submitted and answer2.numericValue > 0.10 and answer2.numericValue < 0.15
"Not quite — you may have left off the 9. The model has a leading factor of 9 (the starting population). Calc entry: 9*e^(0.03*12)."
when answer2.submitted and isBlank(answer2.latex)
"Compute the exponent first: 0.03 · 12 = 0.36. Then 9*e^(0.36). What do you get?"
when answer2.submitted
"Take another look — check whether you included the leading 9 and the parens around 0.03*12. Calc entry: 9*e^(0.03*12)."
otherwise ""
content:
when answer3.submitted and answer3.numericValue > 18.44 and answer3.numericValue < 18.54
"Correct! g(24) = 9 · e^(0.72) ≈ 18.49. The discrete model gives 18.26 — the gap is now about 0.23. Notice how it grows."
when answer3.submitted and answer3.numericValue > 18.21 and answer3.numericValue < 18.31
"Close — 18.26 is f(24), the discrete model. For g(24), the exponent is 0.72 (not the base 1.03 raised). Calc entry: 9*e^(0.03*24)."
when answer3.submitted and answer3.numericValue > 27.0 and answer3.numericValue < 35.0
"Hmm — take another look. You may have raised 9 to a power instead of multiplying. The form is 9 TIMES e^(0.72), not 9^something."
when answer3.submitted and isBlank(answer3.latex)
"Exponent: 0.03 · 24 = 0.72. Calc entry: 9*e^(0.72). What comes out?"
when answer3.submitted
"Good start — verify the calculator entry: 9*e^(0.03*24). The answer should be a little under 19."
otherwise ""
content:
when answer4.submitted and answer4.numericValue > 37.94 and answer4.numericValue < 38.04
"Correct! g(48) = 9 · e^(1.44) ≈ 37.99. The discrete model gives 37.04 — the gap is up to about 0.95. The longer the time, the wider the spread."
when answer4.submitted and answer4.numericValue > 36.99 and answer4.numericValue < 37.09
"Almost — that value is f(48), not g(48). For the continuous model, use e^(1.44). Calc entry: 9*e^(0.03*48)."
when answer4.submitted and answer4.numericValue > 4.0 and answer4.numericValue < 5.0
"Not quite — you may have typed e^0.03 first (which is about 1.0305), then multiplied by 48. The exponent is 0.03 TIMES 48 = 1.44, all inside the e^( ) parens."
when answer4.submitted and isBlank(answer4.latex)
"Exponent: 0.03 · 48 = 1.44. Calc entry: 9*e^(1.44). What do you get?"
when answer4.submitted
"Take another look — check the parens. The whole exponent 0.03*48 must sit inside e^( ). Try 9*e^(0.03*48) again."
otherwise ""
content:
when answer5.submitted and answer5.numericValue > 180.74 and answer5.numericValue < 180.84
"Correct! g(100) = 9 · e^(3) ≈ 180.79. The discrete model gives 172.40 — the gap is now about 8.4. Over 100 months, the continuous model has pulled noticeably ahead."
when answer5.submitted and answer5.numericValue > 172.35 and answer5.numericValue < 172.45
"Close — 172.40 is f(100), the discrete model. For g(100), the exponent is 3 (which is 0.03 · 100). Calc entry: 9*e^(0.03*100)."
when answer5.submitted and answer5.numericValue > 7.0e+8
"Hmm — way too big. You may have typed 9^(0.03*100) by mistake (which would be 9^3 = 729, or worse). The form is 9 TIMES e^(3), not 9^3."
when answer5.submitted and answer5.numericValue > 9.0 and answer5.numericValue < 28.0
"Take another look — you may have dropped the parens. Without them, the calculator reads e^0.03 * 100 = about 103. The full exponent is 0.03 * 100 = 3. Calc entry: 9*e^(0.03*100)."
when answer5.submitted and isBlank(answer5.latex)
"Exponent: 0.03 · 100 = 3. Calc entry: 9*e^(3). What comes out?"
when answer5.submitted
"Good start — verify the calculator entry: 9*e^(0.03*100). The answer should be about 180."
otherwise ""
Math As A Second Language2 min
Reading the continuous growth factor
Math · We Say · Meaning
★ Math (given)
\(e^{rt}\)
▲ We Say
“e to the rt”
● Meaning
The growth factor when a rate r is applied continuously over time t. Example: an 8% continuous growth rate over time t is written \(e^{0.08\,\cdot\,t}\). The exponent multiplies the rate by the elapsed time.
Try Saying
I placed ___ here because ___.
Try Saying
Does ___ belong with ___?
Try Saying
I don't think ___ belongs here because ___.
MASL for the new form — pair with How-To on s7. This is the second MASL of the lesson: s3 introduced the constant e; this slide names the continuous growth factor \(e^{rt}\) before students see it inside the full \(P_{0}\cdot e^{rt}\) model on s7. Emphasize:r is the continuous rate per unit time — it is not the per-period growth factor itself. The factor for one unit of time is \(e^{r}\) (so for a 0.15 continuous rate per month, the one-month factor is \(e^{0.15}\approx 1.162\), not 1.15). We will formalize this distinction on s7 (How To). Per MASL standard: no ACB on MASL slides. The Math card is pre-placed as the given; the We Say and Meaning cards lock in here and become the reference students point back to during Activity 2 work time (s8).
Activity 1 · Synthesis5 min
Discrete vs. continuous — when do they diverge?
For each colony, the two models give predictions that drift apart as the rate gets larger.
Colony 1 · 1%
\((1.01)^t\) \(e^{0.01t}\)
Colony 2 · 3%
\((1.03)^t\) \(e^{0.03t}\)
Colony 3 · 7%
\((1.07)^t\) \(e^{0.07t}\)
Colony 4 · 15%
\((1.15)^t\) \(e^{0.15t}\)
discrete \((1+r)^t\)continuous \(e^{rt}\)
“Which colony's two graphs look most alike?”Colony 1 (1% rate). The two models give nearly identical values even at \(t = 100\). With a small rate, applying it “every moment” vs. “once per period” barely matters.
“Which colony's two graphs look least alike?”Colony 4 (15% rate). By 100 months the continuous model is significantly larger. Bigger rate = more compounding moments = wider gap.
“If a third student modeled the same colony with \(9 \cdot (1.04)^t\), what would a continuous version look like?”\(9 \cdot e^{0.04t}\). Calc-input: 9*e^(0.04*t)
click to advance discussion ▶
Try Saying
The discrete and continuous models drift apart most when ___ because ___.
Quick discussion (~5 min). Takeaway: continuous compounding always grows slightly faster than per-period compounding at the same nominal rate, because the rate keeps acting on the new total at every moment. Smaller rate → smaller gap; larger rate → wider gap. Don't ask students to plot all four pairs — the strip above lets them picture the four comparisons at a glance. Tie back to Activity 1's table: with \(r = 0.03\) the gap was small (~8 thousand at \(t = 100\)); with \(r = 0.15\) the gap would dwarf the discrete value entirely. SWBAT stays on s9 — do not read it here. Standards: HSF-LE.B.5, HSF-IF.C.7.
Reference · How To2 min
How to read a continuous-growth model
General form: \(P(t) = P_0 \cdot e^{rt}\). Worked example: \(P(t) = 9 \cdot e^{0.15t}\) (insect colony, thousands, \(t\) in months).
Leave this slide on screen during s8 Activity 2 work time. Pure reference — no paired problem. Click each row to advance: action reveals, then example, then next row. Press Z or the Back-up button to undo a reveal.
CHECK 10 compliance: paired with the type-in work-time on s8. MASL → How-To pairing: s5 (the \(e^{rt}\) MASL trio) sits immediately before this slide, per the ordering standard.
Step 1 anchors students in P0 as the t = 0 value — not just “the number out front.” Step 2 is the trap: r is the continuous decimal rate, not the per-period factor \(e^r\). When students stumble on s8, redirect them here: “Look at the exponent, not the whole expression.” Step 3 names the time-unit dependency (per month vs. per year vs. per hour). Step 4 demonstrates calculator-input format with parentheses around the entire exponent expression — same rule as the warm-up. The colony example threads through Activity 1 → this How-To → the lesson synthesis on s9 for a consistent running context.
Activity 2 · Work Time10 min
Reading the parameters of a continuous-growth model
Each of the situations below is modeled with P0·e(rt). Fill in any missing parts of the function and state the continuous growth rate as a percent.
Bacteria. A colony starts with 50 bacteria at t = 0 hours. It is modeled by \(f(t) = 50 \cdot e^{0.25t}\), where t is measured in hours.
Identify the continuous growth rate per hour as a percent.
50*e^(0.25*t)
U.S. population (1964). The population was about 192 million at t = 0 years, growing at a continuous rate of 1.4% per year.
Complete the function \(P(t) = \underline{\phantom{000}} \cdot e^{\underline{\phantom{000}}\, t}\).
192*e^(0.014*t)
World population (1955). About 2.5 billion people at t = 0, modeled by \(W(t) = 2.5 \cdot e^{0.0168t}\), where t is measured in years.
Identify the continuous growth rate per year as a percent.
2.5*e^(0.0168*t)
click to reveal sample responses ▶
1. \(f(t) = 50 \cdot e^{0.25t}\). The continuous growth rate is 25% per hour (r = 0.25).
2. \(P(t) = 192 \cdot e^{0.014t}\). P0 = 192 million; r = 0.014, i.e. 1.4% per year.
3. \(W(t) = 2.5 \cdot e^{0.0168t}\). The continuous growth rate is 1.68% per year (r = 0.0168).
Try Saying
In this model, P0 is ___ and r is ___ , so the population grows by ___% every ___ .
10 min · Activity 2 Work Time. Core formal-interpretation slide.
Students see the form P0·e(rt) for the first time as a parameter-naming exercise. The How To from slide 7 (\(P(t)=9 \cdot e^{0.15t}\)) should stay on screen — students can refer back to identify P0 and r.
The big idea:r is the decimal continuous rate. Multiply by 100 to get a percent. The per-period factorer is a different (slightly larger) number — that's the most common confusion.
Common error to watch for: students reading the entire expression e0.25 ≈ 1.284 as “the rate.” Redirect them to the exponent, not the whole expression.
Decay aside (optional, if time): If r is negative, you have continuous decay — e.g., e(−0.05t) is 5% continuous decay. We won't compute decay today, just naming.
Building on thinking (IM probe): “How could you use the example from slide 7 to help you identify the parts of each expression?”
Amplify · 4 inputs
Order per problem: Text component (problem text) → Math Response → Note with CL.
All four inputs are Math Responses. Use answer.numericValue with bands that accept BOTH the decimal form (e.g., 0.25) and the percent form (e.g., 25). Feedback should consistently emphasize that r is the decimal — multiply by 100 for a percent.
Math Response input hint: Example: 0.25 (or 25 for percent)
Problem 1: Bacteria. f(t) = 50 · e^(0.25t), t in hours. State the continuous growth rate per hour (as a decimal or percent).
content:
when answer1.submitted and answer1.numericValue > 0.245 and answer1.numericValue < 0.255
"Correct! r = 0.25, which is 25% continuous growth per hour. Read the exponent: 0.25t tells you r is 0.25. Multiply by 100 to get the percent."
when answer1.submitted and answer1.numericValue > 24.5 and answer1.numericValue < 25.5
"Correct! 25% is the continuous growth rate per hour. As a decimal, r = 0.25 — that's the number sitting next to t in the exponent."
when answer1.submitted and answer1.numericValue > 1.28 and answer1.numericValue < 1.29
"Hmm — take another look. 1.284 is e^(0.25), the per-hour growth factor for the whole expression. The continuous rate r is the number in the exponent itself: 0.25 (or 25%)."
when answer1.submitted and answer1.numericValue > 0.99 and answer1.numericValue < 1.01
"Not quite — e^0 = 1 is the factor at t = 0, not the rate. Look at the exponent: 0.25t. What is r?"
when answer1.submitted and isBlank(answer1.latex)
"Compare 50 · e^(0.25t) to the general form P₀ · e^(rt). What number is sitting in the r slot? Multiply by 100 to express as a percent."
when answer1.submitted
"Take another look at the *exponent*, not the whole expression. In e^(0.25t), the r is 0.25 — that's 25% as a continuous rate."
otherwise ""
Problem 2a: U.S. population (1964). Complete P(t) = ___ · e^(___ t). Enter the rate r in the exponent (as a decimal).
content:
when answer2.submitted and answer2.numericValue > 0.0135 and answer2.numericValue < 0.0145
"Correct! r = 0.014 because the problem says 1.4% per year, and 1.4% = 0.014 as a decimal. The exponent is 0.014t."
when answer2.submitted and answer2.numericValue > 1.35 and answer2.numericValue < 1.45
"Almost — you wrote 1.4 (the percent). In the exponent, r must be the *decimal*: 1.4% = 0.014. So the exponent is 0.014t."
when answer2.submitted and answer2.numericValue > 0.135 and answer2.numericValue < 0.145
"Hmm — take another look. 0.14 would be 14%, but the problem says 1.4%. Slide the decimal one more place: 1.4% = 0.014."
when answer2.submitted and answer2.numericValue > 1.0139 and answer2.numericValue < 1.0143
"You're close — 1.014 is the per-period factor e^(0.014) ≈ 1.0141. The continuous rate r itself is just 0.014 (the exponent multiplier)."
when answer2.submitted and isBlank(answer2.latex)
"The problem says 1.4% per year. Convert 1.4% to a decimal: divide by 100. What goes in the exponent of e^(__ t)?"
when answer2.submitted
"Re-read the problem: '1.4% per year' means r = 1.4 ÷ 100 = 0.014. The exponent is 0.014t."
otherwise ""
Problem 2b: U.S. population (1964). Enter the coefficient P₀ (in millions).
content:
when answer2b.submitted and answer2b.numericValue > 191.5 and answer2b.numericValue < 192.5
"Correct! P₀ = 192 million. That's the value of P(t) at t = 0, because e^(0) = 1, so P(0) = 192 · 1 = 192."
when answer2b.submitted and answer2b.numericValue > 191000000 and answer2b.numericValue < 193000000
"Right idea — but enter it as 192 (the problem says 'in millions'). The coefficient itself is 192, and the units are millions."
when answer2b.submitted and answer2b.numericValue > 0.013 and answer2b.numericValue < 0.015
"Hmm — 0.014 is r, the rate in the exponent. P₀ is the *initial value*, the number out in front of e. Re-read the problem: how many million people in 1964?"
when answer2b.submitted and isBlank(answer2b.latex)
"The problem says '192 million at t = 0'. Compare to P₀ · e^(rt). What is P₀?"
when answer2b.submitted
"P₀ is the initial value — the number multiplied by e^(rt), not the rate. Re-read the first sentence of the problem."
otherwise ""
Problem 3: World population (1955). W(t) = 2.5 · e^(0.0168t), t in years. State the continuous growth rate per year (as a decimal or percent).
content:
when answer3.submitted and answer3.numericValue > 0.0163 and answer3.numericValue < 0.0173
"Correct! r = 0.0168, which is 1.68% continuous growth per year. Read the exponent: 0.0168t tells you r directly."
when answer3.submitted and answer3.numericValue > 1.63 and answer3.numericValue < 1.73
"Correct! 1.68% continuous growth per year. As a decimal, r = 0.0168 — the number multiplying t in the exponent."
when answer3.submitted and answer3.numericValue > 1.016 and answer3.numericValue < 1.018
"Hmm — take another look. 1.017 is e^(0.0168), the per-year growth factor for the whole expression. The continuous rate r is the exponent multiplier itself: 0.0168 (or 1.68%)."
when answer3.submitted and answer3.numericValue > 0.16 and answer3.numericValue < 0.18
"Almost — 0.168 would be 16.8%, but the exponent reads 0.0168 (one more zero). Slide the decimal one place: r = 0.0168 = 1.68%."
when answer3.submitted and isBlank(answer3.latex)
"Compare 2.5 · e^(0.0168t) to the general form P₀ · e^(rt). What number is in the r slot? Multiply by 100 to express as a percent."
when answer3.submitted
"Look at the *exponent*, not the whole expression. In e^(0.0168t), the r is 0.0168 — multiply by 100 to get 1.68%."
otherwise ""
Lesson Synthesis~3 min
Two forms, two meanings
Once per period
\[ f(t) = a \cdot b^{\,t} \]
Growth is applied once per period — the form you used in Lessons 7–12. Colony 4: \(9 \cdot (1.15)^{t}\).
At every moment
\[ f(t) = a \cdot e^{\,rt} \]
Growth is applied continuously, at every moment. Colony 4 (continuous version): \(9 \cdot e^{0.15\,t}\).
Both forms can describe the same situation — but only one captures growth that happens at every moment.
“Which form would you use to model a quantity that compounds once per year?”\(a \cdot b^{t}\) with \(b = 1 + r\). The growth is applied on a fixed schedule — once at the end of each year. The continuous form is overkill.
“Which form would scientists use for radioactive decay?”\(a \cdot e^{rt}\) with a negative \(r\). Decay happens at every moment, not on a daily timer — atoms don't wait until midnight to decay.
“Is \(e\) ever used to count the layers of a paper folded in half \(t\) times?”No. That's a discrete process — use \((1/2)^{t}\). \(e\) is reserved for continuous change.
click to advance discussion ▶
Try Saying
I would use the form with e when ___ because ___. I would use \(a \cdot b^{t}\) when ___.
I Can… (SWBAT)
I understand that \(e\) is used in exponential models when we assume the growth rate is applied at every moment.
I can identify the initial value \(P_0\) and the continuous rate \(r\) in a model of the form \(P_0 \cdot e^{rt}\).
Lesson Synthesis (~3 min). Use Colony 4 (\(9 \cdot (1.15)^{t}\) vs. \(9 \cdot e^{0.15t}\)) as the running example — it's the colony from Activity 1 where the two graphs drifted apart the most, so students have a felt sense of why the continuous form matters. Cold-call through the three syn-queue questions. Q1 lands the per-period intuition; Q2 introduces decay (negative \(r\)) without computing it; Q3 polices the boundary — \(e\) is not a substitute for every exponential. Read the SWBAT aloud after the third reveal — per MASL standard, SWBAT lives on synthesis, not on the title slide.
Cool-Down5 min
Two Population Predictions
Independently — describe how each model predicts the population will grow.
The functions below each model the population of a city, in thousands, t years after 2010.
Model P
\(P(t) = 8.75 \cdot e^{0.01t}\)
8.75*e^(0.01*t)
Model Q
\(Q(t) = 8.75 \cdot (1.01)^{t}\)
8.75*(1.01)^t
Describe how each model predicts the population will grow.
Cool-Down · 5 min. Exit ticket — students must work independently. Leave this slide on screen during the exit ticket.
Expected response:“Both start at 8,750 in 2010. P applies a 1% rate continuously at every moment. Q applies a 1% rate once per year. With a rate this small, the two predictions stay very close — but P always grows slightly faster.”
Calibration: A full-credit answer touches three pieces: (1) shared starting value of 8,750 in 2010, (2) P applies the rate continuously / at every moment, and (3) Q applies the rate once per year / annually. Two of three is partial; one of three is a nudge.
Watch for students who describe only one model (most common) or who treat the two functions as having different starting populations because the forms look different. The CL feedback nudges toward the missing contrast without naming the answer.
Standard met: HSF-LE.B.5.
Tomorrow: Lesson 14 — last SDP lesson in this arc.
Amplify · 1 input
Order: Text component (problem text) → Text Response named answer → Note with CL in </>.
No auto-grade — written description. CL uses answer.content matches with keyword patterns to acknowledge the contrast students draw between the two models and nudge incomplete answers toward the missing piece.
Prompt: P(t) = 8.75·e^(0.01t) and Q(t) = 8.75·(1.01)^t each model the population of a city, in thousands, t years after 2010. Describe how each model predicts the population will grow.
content:
when answer.submitted and answer.content matches "8,?750" and answer.content matches "continuous|every moment|every instant" and answer.content matches "once|each year|annually|per year"
"Strong answer — you named the shared starting value, P's continuous rate, and Q's per-year rate. That's the full contrast."
when answer.submitted and answer.content matches "continuous|every moment|every instant"
"You're close — you described one model. When does Q apply the 1% rate? Add the contrast for the second model."
when answer.submitted and answer.content matches "once|each year|annually|per year"
"Good start — you described Q. Now describe when P applies the 1% rate — is it once per year, or something else?"
when answer.submitted and isBlank(answer.content)
"Both start at 8,750 in 2010. Describe *when* the 1% is applied in each model."
when answer.submitted
"Take another look — compare *when* the growth is applied in each model: every moment, or once per year?"
otherwise ""
Lesson Summary
Exponential models with base e
Many real-world situations involve growth or decay that happens continuously — at every moment, not on a fixed schedule. For those, we use models with base e.
The form
\[f(t) = P_{0} \cdot e^{rt}\]
P0 = initial value (when t = 0). r = continuous rate per unit time (as a decimal). t = elapsed time.
Algae example
24 sq ft of algae growing 8% continuously per day:
\[A(t) = 24 \cdot e^{0.08\,t}\]
24*e^(0.08*t)
After 10 days: ≈ 53.4 sq ft.
When NOT to use it
Folding a paper in half t times is
\[\left(\tfrac{1}{2}\right)^{t}, \text{ not } e^{(\ldots)}.\]
e is for continuous change only — not for processes that happen in discrete steps.
Small-rate shortcut
When r is small, \(e^{r} \approx 1 + r\). That is why \(e^{0.01\,t}\) and \((1.01)^{t}\) give such close values for small t.
Lesson Summary slide — read aloud or post as a student-facing reference. Reinforces today's boundary: discrete step-by-step processes use \(a \cdot b^{t}\); continuous moment-by-moment processes use \(P_{0} \cdot e^{rt}\). Foreshadow future courses (pre-calculus / calculus): students will derive why \(\left(1 + \tfrac{r}{n}\right)^{nt} \to e^{rt}\) as \(n \to \infty\) — the limit that defines continuous compounding. For now, the small-rate shortcut \(e^{r} \approx 1 + r\) is the intuition that ties today's two forms together.
