How do words about change become a differential equation?
You will be able to: Translate a rate statement into an equation with a sign, units and initial condition.
How do words about change become a differential equation?
A small display tank holds 16 liters. A programmed pump removes water at a rate equal to half the current volume per minute. When the tank holds 8 liters, the pump removes 4 liters per minute. Its setting responds to how much remains.
A useful starting point: Prerequisite: interpreting a derivative as a rate →
Words and symbols before equations
- V(t)
- Water volume in liters at time t minutes.
- dV/dt
- Instantaneous volume change, measured in liters per minute.
- Proportional
- A constant multiple of the current quantity.
- Initial condition
- A known amount at a specified time, here V(0)=16.
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
The positive outflow magnitude is 0.5V L/min. Because this removes water, the net rate is V′=−0.5V. The coefficient has units min⁻¹ so both sides have units L/min.
The equation describes a rate, not the amount itself. V(0)=16 supplies the initial amount. The curve V(t)=16e^(−0.5t) will be justified later by separation; differentiating it already checks the rate law.
The tank has a 2×2 dm base: area 4 dm². Since 1 L=1 dm³, depth h=V/4 dm. At t=0 the depth is 4 dm; later the same volume rule determines the visible water height.
This is an ideal feedback-controlled pump with no inflow. It is not a gravity-drain law. The exponential model stays positive at finite times and approaches zero as time grows; a real pump eventually stops or departs from the model.
A worked example, step by step
At an instant when V=10 L, find the net rate and the depth-change rate.
- Use V′=−0.5V because outflow decreases volume.
- Substitute V=10: V′=−5 L/min.
- The base area is fixed, so h′=V′/4=−1.25 dm/min.
- Interpret the signs: both volume and depth are decreasing at that instant, not by fixed amounts for every future minute.
V is an amount; V′ is a rate. Do not write V=−0.5V or assume the instantaneous rate stays constant.
If V halves, what happens to the outflow?
Compare with an explanation
It halves because the coefficient 0.5 min⁻¹ stays fixed.
Predict. Change one thing. Explain.
Compare t=0,2 and 4. Read volume, depth and outflow. Rotate the tank to inspect its fixed base. Explain why the same loss of volume always gives the same loss of depth, but the losses per minute slow down.
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 pump removes 20% of the current water per minute from a tank initially containing 15 L. Write the model and interpret its initial rate.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Define V in liters and t in minutes.
- 1 point: Write V′=−0.2V, with coefficient in min⁻¹.
- 1 point: State V(0)=15 L.
- 1 point: Compute V′(0)=−3 L/min and explain that it changes as V changes.
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 belongs on the left of a rate equation?
A derivative such as dV/dt.
RECALL 2Why is the coefficient negative here?
Water leaves, decreasing the stored amount.
RECALL 3What does the 3D depth encode?
Volume divided by the fixed base area, not the rate itself.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do words about change become a differential equation?
- V′=−kV, k=0.5 min⁻¹, V(0)=16 L.
- h=V/(4 dm²); h′=V′/(4 dm²).
Remember: V is an amount; V′ is a rate. Do not write V=−0.5V or assume the instantaneous rate stays constant.
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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