How is a constant rate different from proportional change?
You will be able to: Distinguish additive change from a rate proportional to an amount or a difference.
How is a constant rate different from proportional change?
Two tanks each start with 16 L. One loses exactly 2 L each minute. The other uses the pump that removes half its current volume per minute. Both lose water, but their rate rules are different.
A useful starting point: How do words about change become a differential equation? →
Words and symbols before equations
- Constant rate
- The derivative is a fixed number.
- Relative rate
- The rate divided by the current amount, when that amount is nonzero.
- Net rate
- Inflow minus outflow.
- Equilibrium
- A constant amount for which the net rate equals zero.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. V′=−0.5V with V(0)=16 L. Feedback-controlled pump, no inflow; not gravity drainage. Tank base 2×2 dm, height 6 dm; depth=V/4. Volume stays positive at every finite time.
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.
- At 2 min: volume 5.88607 L, depth 1.47152 dm; net rate -2.94304 L/min. Positive outflow magnitude 2.94304 L/min.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
A constant loss gives V′=−2 and V=16−2t, valid physically until t=8 min. Equal time intervals remove equal amounts.
A proportional loss gives V′=−0.5V and V=16e^(−0.5t). Its relative rate V′/V is constant, but its absolute rate shrinks in magnitude as V falls.
With 3 L/min inflow and outflow 0.5V, the net law is V′=3−0.5V. Setting the rate to zero gives V=6 L. Below 6 the amount increases; above 6 it decreases.
Words such as “proportional to the difference” require parentheses. Cooling toward a fixed room temperature R can be modeled by T′=−k(T−R) with k>0, so warm objects cool and cooler objects warm.
| Feature | Constant loss | Proportional loss |
|---|---|---|
| Equation | V′=−a | V′=−kV |
| Equal intervals | Equal amounts lost | Equal fractional multipliers |
| Amount curve | Straight line until empty | Exponential while model applies |
A worked example, step by step
A tank gains 4 L/min and loses 0.25V L/min. Find its equilibrium and its rate at V=12 L.
- Write V′=4−0.25V, preserving inflow minus outflow.
- At equilibrium set 4−0.25V=0.
- Solve V=16 L; inflow and outflow both equal 4 L/min.
- At V=12, V′=4−3=1 L/min: volume increases toward equilibrium under this model.
A constant percentage rate is not a constant number of liters per minute. An equilibrium is an amount, not a time.
Does equilibrium mean no water moves?
Compare with an explanation
No. Inflow and outflow can be equal, producing zero net change.
Predict. Change one thing. Explain.
Read the outflow at t=0 and t=2. Compare with a hypothetical fixed 2 L/min outflow. Explain which curve would be straight and why the displayed tank follows the proportional law only.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At 2 min: volume 5.88607 L, depth 1.47152 dm; net rate -2.94304 L/min. Positive outflow magnitude 2.94304 L/min.
Original model; rounded readouts and finite sampled graphs. V′=−0.5V with V(0)=16 L. Feedback-controlled pump, no inflow; not gravity drainage. Tank base 2×2 dm, height 6 dm; depth=V/4. Volume stays positive at every finite time.
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 room remains at 20°C. An object cools with T′=−0.1(T−20), t in minutes. Interpret the rate at T=50°C and identify equilibrium.
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Compare with the answer and four-point rubric
- 1 point: Identify T−20 as temperature above the room.
- 1 point: Calculate T′=−0.1(30)=−3°C/min.
- 1 point: At T=20 the rate is zero, so 20°C is equilibrium.
- 1 point: For T<20 the derivative is positive; the object warms under this model.
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 produces a straight amount-versus-time graph?
A constant derivative.
RECALL 2How do you find an equilibrium?
Set the rate to zero and check the constant function.
RECALL 3Why use parentheses in a difference model?
The constant multiplies the whole difference from the reference amount.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How is a constant rate different from proportional change?
- Constant loss: V′=−a.
- Inflow/outflow: V′=a−kV; equilibrium V=a/k.
Remember: A constant percentage rate is not a constant number of liters per minute. An equilibrium is an amount, not a time.
Conditions: Original model; rounded readouts and finite sampled graphs. V′=−0.5V with V(0)=16 L. Feedback-controlled pump, no inflow; not gravity drainage. Tank base 2×2 dm, height 6 dm; depth=V/4. Volume stays positive at every finite time.
Refresh Kid · AP Calculus BC Unit 7 · Objectives FUN-7.A · Review edition
Framework, scope and review status
Mapped to College Board CED Topic 7.1, FUN-7.A. 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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