Why does a constant relative rate produce an exponential?
You will be able to: Solve and interpret y′=ky in context, including units of k.
Why does a constant relative rate produce an exponential?
An ideal population grows at a rate equal to 10% of its current size per hour. A larger population adds more organisms per hour, even though its relative rate stays the same.
A useful starting point: Why must a particular solution include its interval? →
Words and symbols before equations
- Relative rate k
- The ratio y′/y, with units inverse time.
- Exponential model
- y(t)=y₀e^(kt) for y′=ky.
- Growth
- Positive k for positive y₀.
- Decay
- Negative k for positive y₀.
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
Separate dy/y=k dt for y≠0 and integrate to ln∣y∣=kt+C. Applying y(0)=y₀ gives y=y₀e^(kt); y₀=0 supplies the equilibrium separately.
For positive initial amounts, k>0 causes growth and k<0 causes decay. The derivative is ky, so absolute rate changes with the amount while relative rate remains k.
If k=0.1 h⁻¹, the one-hour multiplier is e^0.1≈1.10517. A continuous relative rate of 10% per hour differs from multiplying by 1.10 exactly once per hour.
The draining tank uses the same mathematics with k=−0.5 min⁻¹. Its depth obeys the same exponential factor because its base area stays fixed. Unlimited growth and perpetual draining are idealizations valid only while assumptions hold.
A worked example, step by step
An amount starts at 80 and obeys y′=0.2y with time in hours. Find its amount and rate at t=3.
- Use y(t)=80e^(0.2t) from the initial value.
- At 3 h, y=80e^0.6≈145.77 units.
- The rate is y′=0.2y≈29.15 units/h.
- The amount is larger than 80 and the current rate is larger than the initial 16 units/h, consistent with positive proportional growth.
The exponent kt must be dimensionless. A continuous relative rate is not the same as a discrete percentage multiplier.
If k=0, what happens?
Compare with an explanation
The amount stays constant: y=y₀.
Predict. Change one thing. Explain.
Change k through negative, zero and positive values while keeping y₀ fixed. Compare the curve and endpoint rate. Explain what the sign of k changes and why the amount stays positive.
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 substance obeys M′=−0.3M, M(0)=50 g, with time in days. Find M(2), M′(2) and their meanings.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Write M(t)=50e^(−0.3t).
- 1 point: M(2)=50e^(−0.6)≈27.44 g.
- 1 point: M′(2)=−15e^(−0.6)≈−8.23 g/day.
- 1 point: The first is the amount remaining; the negative second value is its instantaneous rate of decrease.
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 1What stays constant in an exponential model?
The relative rate k, not usually the absolute rate.
RECALL 2Why does k have inverse-time units?
kt is dimensionless and ky has amount/time units.
RECALL 3Can a positive exponential decay reach zero at finite time?
No, within the ideal continuous model.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why does a constant relative rate produce an exponential?
- y(t)=y₀e^(kt).
- y′/y=k for y≠0.
Remember: The exponent kt must be dimensionless. A continuous relative rate is not the same as a discrete percentage multiplier.
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.
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