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Outrun the Storm · Unit 4

Does the safe zone shrink steadily?

A storm's edge closes in at a steady 2 meters per second. Related rates, with the shrinking storm and a runner racing from second base to third.

About 45 min · 17 problemsStart →

Checks grade only your final answer. The write-up unlocks when you're right or after two tries; hints are always open.

Two things closing in

The storm. In Fortnite Battle Royale, the Storm closes in on the Island in phases, shrinking the circle where players are safe. Our storm is a simplified model: one centered circle, r(t)=200−2t{r(t) = 200 - 2t} meters.

The runner. Second to third at a steady 27 ft/s on a 90 ft square. How fast is the distance to home shrinking?

Both: two quantities tied by an equation, one rate known, the other wanted. Related rates; the tool is the chain rule.

Predict first

Does it shrink steadily?

Friend, Teammate, Coach, You

  1. Friend
    the edge moves in 2 m every second, so the same area goes every second

Watch the storm close

Rings mark the edge every 10 s, so each band is 20 m wide. Labels are band areas in m²; the hatched band is the next 10 s. Slide the time:

The storm from aboveStorm model: r(t) = 200 − 2t m. A synthetic model, drawn in 10-second steps (20 m) so the bands show; the readout gives rates per second.
  • Solid: the edge now
  • Hatched: what the storm takes in the next 10 s
  • Thin rings: the edge every 10 s, each band labeled with its area in m²
Time

t = 0 s: r = 200 m, dr/dt = −2 m/s. Right now the area shrinks at 2πr · 2 = 800π ≈ 2,513 m²/s. The next 10 s take 7,600π ≈ 23,876 m².

  1. At 0 s, read the labels from the outside in: 7,600π, then 6,800π, and so on. Same width; same area?
  2. How much smaller is each band than the one outside it? Every time?
  3. At 25 s the area shrinks at 600π ≈ 1,885 m²/s, but the next 10 s take 5,600π, not 10 × 600π = 6,000π. Why less?
  4. At 50 s the radius is half its start. Compare the rate with 0 s.
Your prediction

Answer: less than the second before. Each band holds less area than the one outside it: the edge gets shorter as the circle shrinks.

About the model

A(t)=π(200−2t)2{A(t) = \pi{(200 - 2t)}^2} square meters for the first 90 seconds. The numbers are made up for this lesson.

Why the area slows down

In a short time Δt{\Delta t}, the edge moves in 2 Δt{2\,\Delta t} meters, taking a thin band. Unrolled, the band is nearly a rectangle: edge length 2πr{\orange{2\pi r}} times thickness. Below, the band the storm takes in the 10 seconds after t=25{t = 25}.

Unroll the bandThe storm model's next 10 s from t = 25 s: the band from r = 150 m in to r − δ = 130 m, cut along a radius and drawn to scale.
  • Shaded: the band, 5,600π m² however far it unrolls
  • Dashed, once flat: the rectangle 2πr · δ = 6,000π m²
  • Gray, crosshatched, once flat: the slivers the rectangle adds, πδ² = 400π m² together
Unroll

Cut open. Both edges keep their lengths: outer 2πr = 300π ≈ 942 m, inner 2π(r − δ) = 260π ≈ 817 m. Flat, the band is a trapezoid 20 m tall with area 20 · (300π + 260π)/2 = 5,600π ≈ 17,593 m², the band's own area π(150² − 130²). The rectangle 2πr · δ = 6,000π m² overshoots it by the sliver πδ² = 400π ≈ 1,257 m².

  1. Drag to Flat. Outer edge 300π m, inner 260π. Why is the inner shorter?
  2. Flat, it's a trapezoid with exactly the band's area, 5,600π m². The rectangle 2πr⋅δ{2\pi r \cdot \delta} is 6,000π. Where does the extra 400π come from?
  3. The extra is πδ2{\pi\delta^2}, with δ{\delta} the thickness. Halve δ{\delta}: about how much do the band and the extra shrink?

So the band is the rectangle less the sliver:

band=2πr δ⏟edge × thickness−πδ2⏟sliver\displaystyle \text{band} = \underbrace{\orange{2\pi r}\,\delta}_{\text{edge }\times\text{ thickness}} - \underbrace{\pi\delta^2}_{\text{sliver}}

With δ=2 Δt{\delta = 2\,\Delta t}, divide by Δt{\Delta t} and let Δt→0{\Delta t \to 0}: the sliver term vanishes, leaving edge length times edge speed. Without the picture: A=πr2{A = \pi r^2} at every moment and r{r} depends on t{t}, so by the chain rule

dAdt=dAdr⋅drdt\displaystyle \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}=2πr drdt\displaystyle = \orange{2\pi r}\,\frac{dr}{dt}differentiate πr2{\pi r^2}

2πr{\orange{2\pi r}} is the edge's length. drdt=−2{\frac{dr}{dt} = -2} m/s never changes, but r{r} does: a shorter edge takes less ground per meter it moves in.

Related rates: quantities tied by an equation have rates tied by its derivative. Every time:

  1. Draw a picture. Name each changing quantity, with units.
  2. List the rates known and wanted. Shrinking means negative.
  3. Write an equation true at every moment.
  4. Differentiate both sides with respect to t{t}. Chain rule: each changing quantity brings its rate.
  5. Only now substitute the moment's values, and solve.
  6. Answer in context, with units and what the sign means.

Worked example, then your turn

Worked example

No calculator
The safe circle's radius r{r} shrinks at a steady 2 meters per second. How fast is the safe area A{A} changing when r=150{r = 150} meters? Give the exact rate in terms of π{\pi}, with units.
  1. Radius r{r} (meters) and area A{A} (square meters) change with time t{t} (seconds). Known: drdt=−2{\dfrac{dr}{dt} = -2} m/s (shrinking) and r=150{r = 150} m. Wanted: dAdt{\dfrac{dA}{dt}}.
  2. An equation true at every moment: A=πr2{A = \pi r^2}.
  3. Differentiate with respect to t{t}. Since r{r} depends on t{t}, the chain rule gives dAdt=2πrdrdt{\dfrac{dA}{dt} = \orange{2\pi r} \dfrac{dr}{dt}}: the edge's length, 2πr{\orange{2\pi r}}, times the radius's rate.
  4. Only now substitute. Putting r=150{r = 150} into A=πr2{A = \pi r^2} first makes the area a constant, whose derivative is 0.
  5. With r=150{r = 150} and drdt=−2{\dfrac{dr}{dt} = -2}: dAdt=2π⁢(150)(−2)\displaystyle \frac{dA}{dt} = 2\pi(150)(-2)=−600π≈−1,884.956\displaystyle = \boxed{-600\pi} \approx -1{,}884.956 The safe area changes at −600π{-600\pi} square meters per second; negative means shrinking. In the kπ{k\pi} form the problems below use, k=−600{k = -600}.

Runner, third, and home: a right triangle, right angle at third. Setup done; you finish.

HomeFirstSecondThirdx = 67.5 ft90 fts = 112.5 ft27 ft/s
The runner's distance to home plate

Challenge 1

No calculator
A runner goes from second base to third at a steady 27 ft/s on a square infield, 90 feet on a side, so the base path meets the third-to-home line at a right angle. When the runner is 67.5 feet from third, how fast is the distance to home plate changing, in feet per second? The first steps are done for you: x{x} is the distance to third and s{s} the distance to home, in feet. s2=x2+902{s^2 = x^2 + 90^2}, dxdt=−27{\dfrac{dx}{dt} = -27} ft/s, and at x=67.5{x = 67.5}, s=112.5{s = 112.5} ft, since 67.52+902=112.5{\sqrt{67.5^2 + 90^2} = 112.5}.

A number or fraction, like 12, 0.25, or 3/4.

Back to the storm

A rate is one instant; a second isn't.

Challenge 2

No calculator
At r=150{r = 150} meters the safe area shrinks at 600π{600\pi} square meters per second. In the next second the radius drops to 148 meters. Exactly how much area does the storm take in that second? Write it as kπ{k\pi} square meters and enter k{k}.

A number or fraction, like 12, 0.25, or 3/4.

Challenge 3

No calculator
Later the safe circle is down to r=50{r = 50} meters, still shrinking at 2 meters per second. Find the safe area's rate dAdt{\dfrac{dA}{dt}}. Write it as kπ{k\pi} square meters per second and enter k{k}.

A number or fraction, like 12, 0.25, or 3/4.

Does the circumference's rate change as the circle shrinks?

Challenge 4

No calculator
The safe circle's circumference C{C} shrinks too. When r=150{r = 150} meters and drdt=−2{\dfrac{dr}{dt} = -2} m/s, find dCdt{\dfrac{dC}{dt}}. Write it as kπ{k\pi} meters per second and enter k{k}.

A number or fraction, like 12, 0.25, or 3/4.

Backward: rate known, radius wanted.

Challenge 5

No calculator
The radius keeps shrinking at 2 meters per second. At what radius, in meters, is the safe area shrinking at 300π{300\pi} square meters per second?

A number or fraction, like 12, 0.25, or 3/4.

Run for the circle

Two things move at once: you and the edge.

r = 90 m60 m80 ms = 100 m6 m/s
The player's distance to the storm's center

Challenge 6

No calculator
A straight road passes 60 meters from the storm's center. A player 80 meters along the road from the point nearest the center runs toward that point at 6 meters per second. The safe circle's radius is 90 meters, still shrinking at 2 meters per second, and the player is outside it. How fast is the gap between the player and the edge closing? Enter a positive number of meters per second.

A number or fraction, like 12, 0.25, or 3/4.

The runner, halfway

Calculator. Then compare it with Challenge 1: closing on home faster or slower now, and why?

Challenge 7

Calculator
The runner again: second base to third at a steady 27 ft/s, 90-foot square. How fast is the distance to home plate changing when the runner is halfway, 45 feet from third?

Round or truncate to 3 decimal places, as on the AP exam.

Watch the slope flatten

Safe area over the storm's first 80 seconds. The tangent's slope at the dot is dAdt{\teal{\frac{dA}{dt}}}, in m²/s. Drag the dot:

Safe area over timeStorm area model: A(t)=π(200−2t)2{A(t) = \pi{(200 - 2t)}^2}. A synthetic model.

Drag across the graph, or use the slider, to move the point.

Time

Tangent at 0 s: safe area 125,663.706 m², slope −2,513.274 square meters per second.

  1. At 0 s the slope is −800π ≈ −2,513 m²/s. Drag to 25 s and read the slope. Seen it before?
  2. Keep going to 80 s. The slope stays negative; what happens to it? Which way does the graph bend?
  3. At 0 s the tangent line reaches 0 at 50 s: at the starting rate, the area would be gone then. Drag to 50 s. How much is left?

You'll need this for part (b) below.

Free response

This one is written like an AP free-response question. The parts with a number check your answer here. Write every part out in full, with your work and units, the way you would on the exam: that's what gets scored.

The storm's safe zone is a circle whose radius r(t){r(t)} decreases at a constant 2 meters per second. At t=0{t = 0}, r=200{r = 200} meters. The area of the safe zone is A(t){A(t)} square meters and its circumference is C(t){C(t)} meters, where t{t} is measured in seconds.

Part (a)

Calculator
Find dAdt{\dfrac{dA}{dt}} at t=25{t = 25}. Show the computations that lead to your answer, and indicate units of measure.

Round or truncate to 3 decimal places, as on the AP exam.

Part (b)

Calculator
Is dAdt{\dfrac{dA}{dt}} increasing or decreasing at t=25{t = 25}? Give a reason for your answer.

Write this part out in sentences. It's scored on your reasoning, so there's no answer box.

Part (c)

Calculator
Find dCdt{\dfrac{dC}{dt}} at t=25{t = 25}, and explain why the circumference shrinks at a constant rate while the area does not. Indicate units of measure.

Round or truncate to 3 decimal places, as on the AP exam.

Part (d)

Calculator
At t=40{t = 40}, a player is on a straight road that passes 50 meters from the center of the safe zone. The player is 120 meters along the road from the point on it nearest the center and is running toward that point at 6.5 meters per second. At what rate is the distance between the player and the edge of the safe zone changing at that moment? Show the computations that lead to your answer, and indicate units of measure.

A number or fraction, like 12, 0.25, or 3/4.

Fluency

Each time: equation, differentiate with respect to t{t}, then substitute.

Fluency 1

No calculator
A sphere's radius grows at 3 cm/s. How fast is its volume, V=43πr3{V = \frac{4}{3}\pi r^3}, growing when the radius is 10 cm? Write it as kπ{k\pi} cubic centimeters per second and enter k{k}.

A number or fraction, like 12, 0.25, or 3/4.

Fluency 2

No calculator
A square's side grows at 4 cm/s. How fast is its area growing when the side is 5 cm?

A number or fraction, like 12, 0.25, or 3/4.

Fluency 3

No calculator
A 13-foot ladder leans against a wall; its base slides away at 2 ft/s. How fast is the top's height changing when the base is 5 feet from the wall? Enter the exact value: a fraction like 3/4 works.

A number or fraction, like 12, 0.25, or 3/4.

Fluency 4

No calculator
A cube's edge grows at 2 cm/s. How fast is its volume growing when the edge is 3 cm?

A number or fraction, like 12, 0.25, or 3/4.

Fluency 5

No calculator
A circle's area grows at 10π{10\pi} square centimeters per second when its radius is 5 cm. How fast is its circumference growing then? Write it as kπ{k\pi} centimeters per second and enter k{k}.

A number or fraction, like 12, 0.25, or 3/4.

Fluency 6

No calculator
Two runners leave the same corner of a field at once, one north at 3 m/s, one east at 4 m/s. How fast is the distance between them growing 10 seconds later?

A number or fraction, like 12, 0.25, or 3/4.

Mixed review

From earlier lessons: Is the Ball Ever Stopped?

Review 1

No calculator
Find the average rate of change of f(x)=x2+3x{f(x) = x^2 + 3x} on [1,4]{[1, 4]}.

A number or fraction, like 12, 0.25, or 3/4.

Review 2

No calculator
Use the limit definition to find f′(3){f'(3)} for f(x)=x2−4x{f(x) = x^2 - 4x}.

A number or fraction, like 12, 0.25, or 3/4.

Exit ticket

Exit ticket

No calculator
A circular ripple spreads on a pond; its radius grows at 3 cm/s. How fast is the area inside it growing when the radius is 20 cm? Write it as kπ{k\pi} square centimeters per second and enter k{k}.

A number or fraction, like 12, 0.25, or 3/4.

Reflect

  1. Why can the edge move steadily while the area doesn't?
  2. When was a rate negative, and what did the sign tell you?
  3. Was your prediction right? Why does the area shrink that way?

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