How can observed data determine a growth or decay constant?
You will be able to: Determine k from two observations and calculate doubling time or half-life.
How can observed data determine a growth or decay constant?
A sample drops from 80 g to 40 g in 3 days. Instead of guessing a daily subtraction, use its constant proportional decay assumption to find the exponential rate.
A useful starting point: Why does a constant relative rate produce an exponential? →
Words and symbols before equations
- Half-life H
- Time needed to multiply a positive amount by 1/2.
- Doubling time D
- Time needed to multiply it by 2.
- Natural logarithm ln
- The inverse of the exponential e^x.
- Calibration
- Choosing model parameters from observations.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. y′=ky, y(0)=y₀>0, with time in hours. Exact y=y₀e^(kt); graph scale adapts. The doubling or half-life may lie outside the four-hour display.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- y=40e^(0.2t). At 4 h, amount=89.0216 units and rate=17.8043 units/h. Doubling time=3.46574h.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
From y(t)=y₀e^(kt), divide an observed amount by y₀, take ln and divide by elapsed time. For observations at times t₁ and t₂, k=ln(y₂/y₁)/(t₂−t₁), assuming positive amounts.
If decay halves in H time units, e^(kH)=1/2, so k=−ln 2/H. For growth, e^(kD)=2 gives D=ln 2/k with k>0.
For 80 g halving every 3 days, k=−ln 2/3 day⁻¹ and M(6)=80(1/2)²=20 g. Each half-life removes half of what remains, not another fixed 40 g.
Two observations determine k under the assumed exponential model; they do not prove that nature follows it for all times. Additional data and context are needed to assess that assumption.
A worked example, step by step
An ideal population rises from 100 to 150 in 2 hours. Find k and its doubling time.
- Use 150=100e^(2k), hence 1.5=e^(2k).
- Take logs: k=ln(1.5)/2≈0.202733 h⁻¹.
- Set 2=e^(kD), so D=ln 2/k≈3.419 h.
- Interpret this as the model’s time to double any positive starting amount while the same relative rate holds.
Half-life is constant only under the fixed exponential model. Do not subtract the same mass every half-life.
After two half-lives, what fraction remains?
Compare with an explanation
One quarter: (1/2)×(1/2).
Predict. Change one thing. Explain.
Set negative and positive k values. Use the displayed characteristic time to predict a halving or doubling, and relate it to e^(kt). The graph window may end before that event.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
y=40e^(0.2t). At 4 h, amount=89.0216 units and rate=17.8043 units/h. Doubling time=3.46574h.
Interpret the parameter
k=0.2 per hour; y′/y=k.
One-hour multiplier=e^k=1.2214.
The exponential is positive for every finite real t.
The four-hour plot is a window, not an unlimited prediction.
Original model; rounded readouts and finite sampled graphs. y′=ky, y(0)=y₀>0, with time in hours. Exact y=y₀e^(kt); graph scale adapts. The doubling or half-life may lie outside the four-hour display.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the rate units, derivative signs, initial condition, domain or solution check. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA positive amount triples in 4 hours under y′=ky. Find k, its doubling time, and the factor after 8 hours.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: From 3=e^(4k), k=ln 3/4 h⁻¹.
- 1 point: Doubling time is ln 2/k=4ln 2/ln 3.
- 1 point: This is approximately 2.524 h.
- 1 point: After 8 h the multiplier is e^(8k)=9, because there are two 4 hour tripling intervals.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1How do two observations determine k?
Take the logarithm of their positive ratio and divide by elapsed time.
RECALL 2What fraction remains after n half-lives?
(1/2)^n.
RECALL 3Do two data points prove an exponential law?
No; they calibrate it under an assumption.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can observed data determine a growth or decay constant?
- k=ln(y₂/y₁)/(t₂−t₁).
- H=−ln 2/k for k<0; D=ln 2/k for k>0.
Remember: Half-life is constant only under the fixed exponential model. Do not subtract the same mass every half-life.
Conditions: Original model; rounded readouts and finite sampled graphs. y′=ky, y(0)=y₀>0, with time in hours. Exact y=y₀e^(kt); graph scale adapts. The doubling or half-life may lie outside the four-hour display.
Refresh Kid · AP Calculus BC Unit 7 · Objectives FUN-7.F, FUN-7.G · Review edition
Framework, scope and review status
Mapped to College Board CED Topic 7.8, FUN-7.F, FUN-7.G. CED and Fall 2026 clarifications checked September 17, 2026. All nine BC topics are included, with Euler’s method and logistic interpretation. The correction to FUN-7.B.2 is reflected: an equation may have infinitely many solutions. Focused explanations, examples, visual models and practice are original Refresh Kid work.
Rate and amount units, signs, initial data, lost equilibria, solution intervals and model assumptions are explicit. Slope fields and Euler polygons are finite illustrations; exact algebra supports identities and limits. The logistic explicit formula supports the visualization; required interpretation can be done from the rate law without deriving that formula. The smooth examples have unique solutions; no universal uniqueness claim is made.
Organic Chemistry Tutor video titles and destinations were checked, not the full videos. Khan Academy’s destination was checked; its JavaScript lesson contents were not fully readable by the research tool. OpenStax sections 4.1–4.4 supplied conceptual cross-checks. No provider questions, artwork or scripts were copied. No affiliation or endorsement is implied.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. The original tank reuses self-hosted Three.js with its MIT license retained. The 2×2 dm base connects water depth to volume: 1 dm³=1 L. This tank uses a feedback-controlled pump, not a gravity-drain law. Its 3D height and labeled 2D graphs use the same exact exponential formula. Rotation changes only the view, and complete explanations remain available without WebGL.
Independent teacher review and student usability testing remain pending. Technical checks do not certify mathematical accuracy, full accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. This multi-unit archive is not assigned as a complete Unit 7 task.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
Want to work through this with a tutor?
Bring your question about How can observed data determine a growth or decay constant? Your explanation and answers remain free to access.