How does a carrying capacity change proportional growth?
You will be able to: Interpret logistic rates, equilibria and long-term behavior without solving the equation.
How does a carrying capacity change proportional growth?
A small habitat can support about 100 organisms. When resources are abundant, growth is roughly proportional to the current population. Near 100, competition slows the increase. Above 100, the model predicts decline.
A useful starting point: How can observed data determine a growth or decay constant? →
Words and symbols before equations
- Carrying capacity K
- The positive equilibrium approached by positive solutions in this model.
- Intrinsic rate r
- A positive coefficient measured in inverse time.
- Logistic law
- P′=rP(1−P/K).
- Equilibrium
- A constant population at which P′=0.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. P′=0.5P(1−P/100), t in days. K=100 and r=0.5/day fixed; P is treated continuously. Teal dashed line is capacity. Exact curves include separate 0 and 100 equilibria. The rate graph distinguishes population from growth rate.
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.
- P₀=20; initial rate=8 organisms/day. At 6 days P=83.3925. Reaches 50 at t=2.77259 days; peak positive rate 12.5 organisms/day.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
The factors have different jobs: P supplies population-dependent growth, while 1−P/K represents the remaining fraction of capacity. The model assumes constant r and K and treats population as continuous.
For r>0 and K>0, equilibria are P=0 and P=K. If 0<P<K the derivative is positive; ifP>K it is negative. A positive solution tends toward K as time grows. The zero solution stays at 0 and does not tend toK.
Another common notation is P′=kP(K−P). It is the same law when r=kK. The coefficients have different units: r is inverse time, while k is inverse population per time.
For the model’s exact displayed curves, P(t)=K/[1+(K/P₀−1)e^(−rt)] when P₀>0. You can interpret the required behavior from rate signs without deriving this formula. At P₀=0 we display the separate equilibrium.
| Feature | Exponential | Logistic |
|---|---|---|
| Rate law | P′=rP | P′=rP(1−P/K) |
| Positive equilibrium | None for r>0 | K |
| Long-term positive solution | Unbounded ideal growth | Approaches K |
A worked example, step by step
For P′=0.02P(100−P), find the carrying capacity and the rates at P=20 andP=120.
- The positive equilibrium isK=100 organisms; the other is 0.
- At 20, P′=0.02(20)(80)=32 organisms per time unit.
- At 120, P′=0.02(120)(−20)=−48 organisms per time unit.
- The first population grows and the second declines toward 100 under the ideal law; here r=kK=2 per time unit.
Carrying capacity is an amount, not the coefficient k. A population starting at zero remains zero in this model.
If a population begins above capacity, must it first grow?
Compare with an explanation
No. Its logistic derivative is negative, so it declines toward capacity.
Predict. Change one thing. Explain.
Try initial populations 0,20,100 and 120 with K=100. Compare derivative signs and the capacity line. Use the formula’s exponential term to justify each long-term conclusion.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
P₀=20; initial rate=8 organisms/day. At 6 days P=83.3925. Reaches 50 at t=2.77259 days; peak positive rate 12.5 organisms/day.
Original model; rounded readouts and finite sampled graphs. P′=0.5P(1−P/100), t in days. K=100 and r=0.5/day fixed; P is treated continuously. Teal dashed line is capacity. Exact curves include separate 0 and 100 equilibria. The rate graph distinguishes population from growth rate.
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 questionFor P′=0.4P(1−P/80), identify equilibria, the sign of the rate at P=40 andP=100, and the limit for P₀=100.
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Compare with the answer and four-point rubric
- 1 point: Equilibria areP=0 andP=80.
- 1 point: At 40, P′=0.4(40)(1/2)=8, positive.
- 1 point: At 100, P′=0.4(100)(−1/4)=−10, negative.
- 1 point: A positive solution starting at 100 decreases toward 80; it does not approach zero.
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 is the carrying capacity?
The positive equilibrium K for this constant-parameter model.
RECALL 2How are r and k related?
r=kK when the model is written as kP(K−P).
RECALL 3What exception matters for the limiting value?
An initial population of zero remains at zero.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does a carrying capacity change proportional growth?
- P′=rP(1−P/K)=kP(K−P), r=kK.
- For positive P₀, P→K as t→∞; for P₀=0, P=0.
Remember: Carrying capacity is an amount, not the coefficient k. A population starting at zero remains zero in this model.
Conditions: Original model; rounded readouts and finite sampled graphs. P′=0.5P(1−P/100), t in days. K=100 and r=0.5/day fixed; P is treated continuously. Teal dashed line is capacity. Exact curves include separate 0 and 100 equilibria. The rate graph distinguishes population from growth rate.
Refresh Kid · AP Calculus BC Unit 7 · Objectives FUN-7.H · Review edition
Framework, scope and review status
Mapped to College Board CED Topic 7.9, FUN-7.H. 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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