A changing radius creates a changing area
How does a length rate become an area rate?
Entering midway? Circle area is πr². The chain rule connects a changing intermediate quantity with the output that depends on it.
Start with a concrete question
A circle radius changes at a constant signed rate. Compare instantaneous area rate with exact finite-step area change.
Try the initial inputs, then write a prediction before changing one setting. All calculations run locally on your CPU.
Go directly to the experiment · Review prerequisite mathematics and return here
What do you expect?
Will the same radius increase add the same area to a small circle and a large one?
Change one thing. Inspect what happens.
- At defaults compute dA/dr, then multiply by dr/dt.
- Compare exact and linear changes as the time interval shrinks.
- Start at zero radius with positive growth and distinguish zero instantaneous rate from positive later area.
- Try shrinking a small radius past zero and explain the rejected geometric step.
Your settingsMechanism
Calculation belowEvidence
Compare outcomes
Area as a function of radius, with the current point
Horizontal: Radius (m). Vertical: Area (m²).
- 1. A=πr² (solid line)
- 2. Current radius (points)
- 3. Future radius (points)
Axes use the displayed ranges.
Read plotted values as a table
| Series | Radius (m) | Area (m²) |
|---|---|---|
| A=πr² | 0 | 0 |
| A=πr² | 0.0625 | 0.01227 |
| A=πr² | 0.125 | 0.04909 |
| A=πr² | 0.1875 | 0.1104 |
| A=πr² | 0.25 | 0.1963 |
| A=πr² | 0.3125 | 0.3068 |
| A=πr² | 0.375 | 0.4418 |
| A=πr² | 0.4375 | 0.6013 |
| A=πr² | 0.5 | 0.7854 |
| A=πr² | 0.5625 | 0.994 |
| A=πr² | 0.625 | 1.227 |
| A=πr² | 0.6875 | 1.485 |
| A=πr² | 0.75 | 1.767 |
| A=πr² | 0.8125 | 2.074 |
| A=πr² | 0.875 | 2.405 |
| A=πr² | 0.9375 | 2.761 |
| A=πr² | 1 | 3.142 |
| A=πr² | 1.063 | 3.547 |
| A=πr² | 1.125 | 3.976 |
| A=πr² | 1.188 | 4.43 |
| A=πr² | 1.25 | 4.909 |
| A=πr² | 1.313 | 5.412 |
| A=πr² | 1.375 | 5.94 |
| A=πr² | 1.438 | 6.492 |
| A=πr² | 1.5 | 7.069 |
| A=πr² | 1.563 | 7.67 |
| A=πr² | 1.625 | 8.296 |
| A=πr² | 1.688 | 8.946 |
| A=πr² | 1.75 | 9.621 |
| A=πr² | 1.813 | 10.32 |
| A=πr² | 1.875 | 11.04 |
| A=πr² | 1.938 | 11.79 |
| A=πr² | 2 | 12.57 |
| A=πr² | 2.063 | 13.36 |
| A=πr² | 2.125 | 14.19 |
| A=πr² | 2.188 | 15.03 |
| A=πr² | 2.25 | 15.9 |
| A=πr² | 2.313 | 16.8 |
| A=πr² | 2.375 | 17.72 |
| A=πr² | 2.438 | 18.67 |
| A=πr² | 2.5 | 19.63 |
| A=πr² | 2.563 | 20.63 |
| A=πr² | 2.625 | 21.65 |
| A=πr² | 2.688 | 22.69 |
| A=πr² | 2.75 | 23.76 |
| A=πr² | 2.813 | 24.85 |
| A=πr² | 2.875 | 25.97 |
| A=πr² | 2.938 | 27.11 |
| A=πr² | 3 | 28.27 |
| A=πr² | 3.063 | 29.46 |
| A=πr² | 3.125 | 30.68 |
| A=πr² | 3.188 | 31.92 |
| A=πr² | 3.25 | 33.18 |
| A=πr² | 3.313 | 34.47 |
| A=πr² | 3.375 | 35.78 |
| A=πr² | 3.438 | 37.12 |
| A=πr² | 3.5 | 38.48 |
| A=πr² | 3.563 | 39.87 |
| A=πr² | 3.625 | 41.28 |
| A=πr² | 3.688 | 42.72 |
| A=πr² | 3.75 | 44.18 |
| A=πr² | 3.813 | 45.66 |
| A=πr² | 3.875 | 47.17 |
| A=πr² | 3.938 | 48.71 |
| A=πr² | 4 | 50.27 |
| A=πr² | 4.063 | 51.85 |
| A=πr² | 4.125 | 53.46 |
| A=πr² | 4.188 | 55.09 |
| A=πr² | 4.25 | 56.75 |
| A=πr² | 4.313 | 58.43 |
| A=πr² | 4.375 | 60.13 |
| A=πr² | 4.438 | 61.86 |
| A=πr² | 4.5 | 63.62 |
| A=πr² | 4.563 | 65.4 |
| A=πr² | 4.625 | 67.2 |
| A=πr² | 4.688 | 69.03 |
| A=πr² | 4.75 | 70.88 |
| A=πr² | 4.813 | 72.76 |
| A=πr² | 4.875 | 74.66 |
| A=πr² | 4.938 | 76.59 |
| A=πr² | 5 | 78.54 |
| Current radius | 2 | 12.57 |
| Future radius | 2.25 | 15.9 |
Follow the calculation
- A=πr²; dr/dt=0.5 m/s, so dA/dt=2πr×dr/dt=6.28319 m²/s.
- Future radius=2+0.5×0.5=2.25; exact area change=π(2.25)²−π(2)²=3.33794.
- Linear estimate = current rate×Δt=3.14159; remainder=π(rate×Δt)²=0.19635.
The chain rule converts a radius rate into an area rate. An instantaneous linear estimate omits a quadratic finite-step term.
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Explain the mechanism
Start with the meaning
Circle area is πr². The chain rule connects a changing intermediate quantity with the output that depends on it.
A circle radius changes at a constant signed rate. Compare instantaneous area rate with exact finite-step area change.
A worked explanation
The derivative dA/dr=2πr measures square metres of area per metre of radius. Multiplying by dr/dt gives square metres per second.
At r=2 and dr/dt=0.5, the area rate is 2π m²/s. Over 0.5 s the local estimate is π m², while exact change is π(2.25²−2²)=1.0625π m².
The finite-step remainder π(rate×time)² explains the difference. Negative radius is outside this geometric model, so a time step crossing zero is not accepted as a physical area prediction.
Symbols and relationship
dA/dt=(dA/dr)(dr/dt)=2πr(dr/dt). Exact ΔA=π(r+rate×Δt)²−πr² for nonnegative final radius.
Use the graph and trace together
The graph and calculations use your current inputs. Read axis labels and units before comparing positions or slopes. Change one input, predict the effect, and use the numerical trace to verify the plotted behavior. A displayed curve is an illustration; the formula and its assumptions determine which conclusions are justified.
Scope and edge cases
- Constant radius rate over the proposed step; no physical growth mechanism.
- An exact geometric formula does not establish that a real process maintains the assumed rate.
Try a fresh case
A circle has radius 3 m and radius rate 2 m/s. What is its instantaneous area rate divided by π, in m²/s?
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Take the investigation further
Choose any extension. These are project prompts for you to carry out; they are not automatically assessed.
School investigation
Compare equal radius changes on two circles.
College investigation
Derive related rates for volume and propagate uncertainty through the chain rule.
Workplace investigation
Check dimensions and domain constraints in a rate-conversion model.
Keep the evidence, choose your next step
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