Outrun the Storm

CED topics

• 4.4 Introduction to Related Rates
• 4.5 Solving Related Rates Problems

The free-response question also touches:

• 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing

Big idea and practices

Big idea: linked quantities have linked rates. Differentiate an equation true at every moment and it links the rates. CHA-3.D and CHA-3.E: calculate and interpret related rates; 4.4: the chain rule, every variable differentiated with respect to time. FRQ (b) previews 5.3 (FUN-4.A): the sign of ${\frac{d^2A}{dt^2}}$ says whether ${\frac{dA}{dt}}$ is increasing.

Both hooks are models. Storm: ${r(t) = {200 - 2t}}$ m, one phase, centered, at a steady speed, a simplification of the game's storm, which moves and shrinks in phases. The runner's 27 ft/s is Baseball Savant's Major League average Sprint Speed on a competitive play, held constant; the 90-foot square is Rule 2.01 of the Official Baseball Rules.

Goals: set up (picture, changing quantities, an equation true at every moment); differentiate, then substitute; give each rate's sign and meaning, with units; explain why a steady rate can give a changing one.

Practices: 1.E, differentiate and solve for the rate (4.4's skill); 2.A, circle, triangle, road, ladder, two runners: one move; 3.E, why the circumference shrinks steadily and the area doesn't (FRQ (c)); 3.F, each rate's sign in context (4.5's skill); 4.B, units on every rate; 4.C, ${\frac{dA}{dt}}$ and ${\frac{dr}{dt}}$.

45-minute plan

1. 0–5 minLaunch: read both hooks aloud; pairs read "Does it shrink steadily?" together and pick in it (or circle on the printout). Vote: more, less, or the same each second?
2. 5–12 minStorm view, projected or on devices, and its four prompts. Settle the vote with the band labels.
3. 12–20 minWhole class: the unroll and its three prompts, then the chain rule and the six steps. Write ${\frac{dA}{dt} = 2\pi r \frac{dr}{dt}}$ under “edge length × edge speed.”
4. 20–27 minWorked example together; Challenge 1, the runner, in pairs with the diamond.
5. 27–35 minChallenges 2–4 in pairs. After Challenge 2, set the rate at 150 m beside the exact band: the unroll’s sliver.
6. 35–40 minChallenge 6, the road, with its figure. Ask what moves before any equation.
7. 40–45 minExit ticket, then the reflection.

Homework: Challenges 5 and 7, the slope explorable, the FRQ, fluency, and mixed review. Score the FRQ with the guideline below next class.

Launch questions

• A 1 m ring off a circle of radius 100 m, or off radius 10 m: which has more area? Why?
• Second to third at 27 ft/s: closing on home at 27 ft/s? Faster or slower?
• A negative rate for the storm's area means? For the runner's distance to home?
• After predictions: the edge moves steadily. What would make the area shrink steadily too?

Misconceptions and how to respond

• “A steady edge means a steady area.” The most common prediction. Bands all 20 m wide; areas 7,600π, 6,800π, and on down, 800π less each time. A shorter edge takes less ground per meter.
• Dropping the sign. Watch for 200 on Challenge 3 and 4 on Challenge 4; the page answers “The storm is shrinking, so the rate is negative.” Have students write ${\frac{dr}{dt} = -2}$, sign and all, first. Challenge 6 wants a positive closing speed, and −2.8 gets a pointer.
• The runner's speed is the rate toward home. Watch for −27 on Challenges 1 and 7. ${\frac{ds}{dt} = \frac{x}{s} \cdot \frac{dx}{dt}}$, and ${\frac{x}{s}}$ is less than 1. At 67.5 feet from third it's 0.6.
• Substituting too early. ${r = 150}$ in ${A = \pi r^2}$ before differentiating makes a constant, derivative 0. The worked example says so in its fourth step.
• A rate is an amount. Watch for 600 on Challenge 2: the rate at 150 m for one second counts the band as long as its outer edge. The exact band is 4π smaller, the sliver ${\pi\delta^2}$ with ${\delta = 2}$, which the unroll shows at ${\delta = 20}$.
• Closing on the center is closing on the edge. Watch for 4.8 and 6.8 on Challenge 6. The gap is ${{s - r}}$, its rate ${{\frac{ds}{dt} - \frac{dr}{dt}}}$; the edge moving in at 2 m/s works against the player.

Support and extension

Support

• Start with numbers: radii of 200, 180, and 160 meters have areas of 40,000π, 32,400π, and 25,600π square meters, so the first two 10-second bands are 7,600π and 6,800π.
• Picture before equation: letters for what changes, numbers for what doesn't.
• Six steps on the board, checked off per problem. Circle step 5: substitute last.
• Pairs: one runs the storm view, one predicts the next band's label.

Extension

• Show the band from ${r}$ in to ${r - \delta}$ has area exactly ${2\pi{\left(r - \tfrac{\delta}{2}\right)}\delta}$: middle circumference times thickness. Check: ${2\pi(140)(20) = 5{,}600\pi}$.
• Sphere: ${\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}}$, surface area times the radius's rate. Why is that “edge × speed” one dimension up?
• Where does the distance to home shrink at half the runner's speed? (At ${30\sqrt{3} \approx 51.962}$ feet from third.)
• From FRQ (b), ${\frac{dA}{dt} = 8\pi t - 800\pi}$. When is the area shrinking at half its starting rate? (At ${t = 50}$ seconds, when ${r = 100}$ meters.)

Without devices

• Print the lesson as the handout; students circle their prediction. The storm view prints at 0 s with its labels; the unroll as a filmstrip of three frames: cut open, half unrolled, and flat.
• Storm view: the nine band areas, 7,600π down to 1,200π square meters, on the board; students find the differences: 800π every time. Same drop, different sizes.
• Unroll: from the filmstrip's labeled edges, the flat trapezoid is ${20 \cdot \frac{300\pi + 260\pi}{2} = 5{,}600\pi}$; compare the band, ${\pi{\left(150^2 - 130^2\right)}}$.
• Slope explorable: sketch ${A(t)}$ from a table at 0, 20, 40, 60, and 80 seconds: 40,000π, 25,600π, 14,400π, 6,400π, and 1,600π square meters. The tangent slopes there are −800π, −640π, −480π, −320π, and −160π: negative, climbing toward 0.
• The answer key below has the page's write-ups.
• Collect the exit ticket on paper.

Answer key

Every problem in the order the lesson gives them, with the answer the page accepts and my full write-up. Answer boxes check only the final number, so ask students for their work.

1. Worked example

   No calculator · CED 4.4, 4.5

   The safe circle's radius ${r}$ shrinks at a steady 2 meters per second. How fast is the safe area ${A}$ changing when ${r = 150}$ meters? Give the exact rate in terms of ${\pi}$, with units.

   Worked on the page: -600

   1. Radius ${r}$ (meters) and area ${A}$ (square meters) change with time ${t}$ (seconds). Known: ${\dfrac{dr}{dt} = -2}$ m/s (shrinking) and ${r = 150}$ m. Wanted: ${\dfrac{dA}{dt}}$.
   2. An equation true at every moment: ${A = \pi r^2}$.
   3. Differentiate with respect to ${t}$. Since ${r}$ depends on ${t}$, the chain rule gives ${\dfrac{dA}{dt} = \orange{2\pi r} \dfrac{dr}{dt}}$: the edge's length, ${\orange{2\pi r}}$, times the radius's rate.
   4. Only now substitute. Putting ${r = 150}$ into ${A = \pi r^2}$ first makes the area a constant, whose derivative is 0.
   5. With ${r = 150}$ and ${\dfrac{dr}{dt} = -2}$: $\displaystyle \frac{dA}{dt} = 2\pi(150)(-2)$$\frac{dA}{dt}$$=2\pi (150)(-2)$$\displaystyle = \boxed{-600\pi} \approx -1{,}884.956$$=\boxed{-600\pi }\approx -1,884.956$ The safe area changes at ${-600\pi}$ square meters per second; negative means shrinking. In the ${k\pi}$ form the problems below use, ${k = -600}$.

2. Challenge 1

   No calculator · CED 4.4, 4.5

   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: ${x}$ is the distance to third and ${s}$ the distance to home, in feet. ${s^2 = x^2 + 90^2}$, ${\dfrac{dx}{dt} = -27}$ ft/s, and at ${x = 67.5}$, ${s = 112.5}$ ft, since ${\sqrt{67.5^2 + 90^2} = 112.5}$.

   Answer: -16.2

   1. (Done for you.) ${s^2 = x^2 + 90^2}$ and ${\dfrac{dx}{dt} = -27}$ ft/s. At ${x = 67.5}$ ft, ${s = 112.5}$ ft: a 3-4-5 triangle times 22.5.
   2. Differentiate with respect to ${t}$. The constant ${90^2}$ gives 0, and the 2s cancel: $\displaystyle \cancel{2}s\frac{ds}{dt} = \cancel{2}x\frac{dx}{dt}$$\cancel{2}s\frac{ds}{dt}$$=\cancel{2}x\frac{dx}{dt}$$\displaystyle \frac{ds}{dt} = \frac{x}{s} \cdot \frac{dx}{dt}$$\frac{ds}{dt}$$=\frac{x}{s}\cdot \frac{dx}{dt}$divide by ${s}$
   3. Substitute: $\displaystyle \frac{ds}{dt} = \frac{67.5}{112.5}(-27)$$\frac{ds}{dt}$$=\frac{67.5}{112.5}(-27)$$\displaystyle = 0.6(-27)$$=0.6(-27)$$\displaystyle = \boxed{-16.2}$$=\boxed{-16.2}$ The distance to home shrinks at 16.2 ft/s, slower than 27 ft/s: the runner isn't heading straight for home.

   Common wrong answers

   • -27: That's the runner's speed along the base path, not along the line to home.
   • 16.2: The distance to home is shrinking, so the rate is negative.
   • -20.25: Divide by s = 112.5 ft, not the 90 ft leg.
   • -45: ds/dt = (x/s) · dx/dt, and x/s is less than 1.

3. Challenge 2

   No calculator · CED 4.5

   At ${r = 150}$ meters the safe area shrinks at ${600\pi}$ 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 ${k\pi}$ square meters and enter ${k}$.

   Answer: 596

   1. The band is the circle of radius 150 m minus the circle of radius 148 m: ${{\pi \cdot 150^2 - \pi \cdot 148^2}}$ square meters.
   2. Difference of squares, ${a^2 - b^2 = (a - b)(a + b)}$, with ${150 - 148 = 2}$ and ${150 + 148 = 298}$: $\displaystyle 150^2 - 148^2 = 2 \times 298$$150^2-148^2$$=2\times 298$$\displaystyle = 596$$=596$
   3. The rate, ${2\pi r \cdot 2 = 600\pi}$, treats the whole band as long as its outer edge. The inner edge is shorter, so the band is ${4\pi}$ smaller: the sliver ${\pi\delta^2}$, with ${\delta = 2}$ m the band's thickness.
   4. The storm takes exactly ${\boxed{596\pi} \approx 1{,}872.389}$ square meters, so ${k = 596}$.

   Common wrong answers

   • 600: That's the rate right now, 2πr · 2. The band is a little smaller, because the inner edge is shorter.
   • -596: An amount of area is positive.
   • 4: π(150 − 148)² is a circle of radius 2 m. Subtract the circles: π · 150² − π · 148².

4. Challenge 3

   No calculator · CED 4.4, 4.5

   Later the safe circle is down to ${r = 50}$ meters, still shrinking at 2 meters per second. Find the safe area's rate ${\dfrac{dA}{dt}}$. Write it as ${k\pi}$ square meters per second and enter ${k}$.

   Answer: -200

   1. ${\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}}$, now with ${r = 50}$ m and ${\dfrac{dr}{dt} = -2}$ m/s.
   2. Substitute: $\displaystyle \frac{dA}{dt} = 2\pi(50)(-2)$$\frac{dA}{dt}$$=2\pi (50)(-2)$$\displaystyle = \boxed{-200\pi} \approx -628.319$$=\boxed{-200\pi }\approx -628.319$ The area changes at ${-200\pi}$ square meters per second, so ${k = -200}$. A third the radius, a third the rate: the edge is a third as long and moves just as fast.

   Common wrong answers

   • 200: The storm is shrinking, so the rate is negative.
   • -600: That's the rate at r = 150 m. Use r = 50 m now.
   • 2500: That's the area, π · 50², not its rate.

5. Challenge 4

   No calculator · CED 4.4, 4.5

   The safe circle's circumference ${C}$ shrinks too. When ${r = 150}$ meters and ${\dfrac{dr}{dt} = -2}$ m/s, find ${\dfrac{dC}{dt}}$. Write it as ${k\pi}$ meters per second and enter ${k}$.

   Answer: -4

   1. ${C = 2\pi r}$, so ${\dfrac{dC}{dt} = 2\pi \dfrac{dr}{dt}}$.
   2. Substitute ${\dfrac{dr}{dt} = -2}$: $\displaystyle \frac{dC}{dt} = 2\pi(-2)$$\frac{dC}{dt}$$=2\pi (-2)$$\displaystyle = -4\pi \approx -12.566$$=-4\pi \approx -12.566$
   3. No ${r}$ is left, so the rate is the same at every radius: ${\dfrac{dC}{dt} = \boxed{-4\pi}}$ m/s. ${C}$ is linear in ${r}$ and ${A = \pi r^2}$ is not, so only the area's rate changes as the circle shrinks. So ${k = -4}$.

   Common wrong answers

   • 4: The storm is shrinking, so the rate is negative.
   • -2: That's dr/dt. C = 2πr, so the circumference changes 2π times as fast.
   • -600: That's the area's rate, in π units. Use C = 2πr.

6. Challenge 5

   No calculator · CED 4.4, 4.5

   The radius keeps shrinking at 2 meters per second. At what radius, in meters, is the safe area shrinking at ${300\pi}$ square meters per second?

   Answer: 75

   1. ${\dfrac{dA}{dt} = -300\pi}$ (shrinking) and ${\dfrac{dr}{dt} = -2}$ m/s.
   2. ${2\pi r{(-2)} = -300\pi}$, so ${-4\pi r = -300\pi}$.
   3. The radius is ${r = \boxed{75}}$ meters: half the worked example's radius, half its rate. In the model ${r(t) = {200 - 2t}}$, that is at ${t = 62.5}$ s, since ${{200 - 2(62.5)} = 75}$.

   Common wrong answers

   • 62.5: That's the time in the model. Find the radius.
   • 150: At 150 m it shrinks at 600π, twice as fast.
   • -75: A radius is positive. Shrinking at 300π means dA/dt = −300π, with dr/dt = −2.

7. Challenge 6

   No calculator · CED 4.4, 4.5

   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.

   Answer: 2.8

   1. Let ${x}$ be the distance along the road and ${s}$ the distance to the center, in meters: ${s^2 = x^2 + 60^2}$. At ${x = 80}$, ${s = \sqrt{6{,}400 + 3{,}600} = 100}$, and the gap is ${{s - r} = {100 - 90} = 10}$ m.
   2. Differentiate with respect to ${t}$; the 2s cancel: $\displaystyle \cancel{2}s\frac{ds}{dt} = \cancel{2}x\frac{dx}{dt}$$\cancel{2}s\frac{ds}{dt}$$=\cancel{2}x\frac{dx}{dt}$$\displaystyle \frac{ds}{dt} = \frac{x}{s} \cdot \frac{dx}{dt}$$\frac{ds}{dt}$$=\frac{x}{s}\cdot \frac{dx}{dt}$divide by ${s}$$\displaystyle = \frac{80}{100}(-6)$$=\frac{80}{100}(-6)$$\displaystyle = -4.8$$=-4.8$
   3. The gap is ${g = s - r}$, so $\displaystyle \frac{dg}{dt} = \frac{ds}{dt} - \frac{dr}{dt}$$\frac{dg}{dt}$$=\frac{ds}{dt}-\frac{dr}{dt}$$\displaystyle = -4.8 - (-2)$$=-4.8-(-2)$$\displaystyle = -2.8$$=-2.8$
   4. The gap closes at ${\boxed{2.8}}$ meters per second: the player gains 4.8 m/s on the center, but the edge moves away at 2 m/s.

   Common wrong answers

   • 4.8: That's how fast you close on the center. The edge is moving too.
   • 6.8: The edge moves in, away from you, so it slows you down.
   • -2.8: The question asks how fast the gap is closing: enter a positive number.
   • 4: The player isn't running straight at the center, so the distance to the center shrinks slower than 6 m/s.
   • 1.6: The ratio is x/s = 80/100.

8. Challenge 7

   Calculator · CED 4.4, 4.5

   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?

   Answer: -12.075 (or -12.074)

   1. Find ${s}$ at ${x = 45}$ ft: $\displaystyle s = \sqrt{45^2 + 90^2}$$s$$=\sqrt{45^2+90^2}$$\displaystyle = \sqrt{10{,}125}$$=\sqrt{10,125}$$\displaystyle = 45\sqrt{5}$$=45\sqrt{5}$ Keep the exact value, not ${s \approx 100.623}$ ft.
   2. With ${\dfrac{dx}{dt} = -27}$: $\displaystyle \frac{ds}{dt} = \frac{x}{s} \cdot \frac{dx}{dt}$$\frac{ds}{dt}$$=\frac{x}{s}\cdot \frac{dx}{dt}$$\displaystyle = \frac{45}{45\sqrt{5}}(-27)$$=\frac{45}{45\sqrt{5}}(-27)$$\displaystyle = -\frac{27}{\sqrt{5}}$$=-\frac{27}{\sqrt{5}}$
   3. ${\dfrac{ds}{dt} \approx \boxed{-12.075}}$ ft/s (or ${-12.074}$, truncated). Nearer third, less of the runner's speed points toward home: 12.075 ft/s here, 16.2 ft/s at 67.5 ft from third.

   Common wrong answers

   • -27: That's the runner's speed along the base path, not along the line to home.
   • 12.075: The runner is getting closer to home, so the distance is shrinking: the rate is negative.
   • -16.2: That's the rate at 67.5 ft from third. Use x = 45 ft now.
   • -13.5: Divide by s = 45√5 ≈ 100.623 ft, not the 90 ft leg.

9. Fluency 1

   No calculator · CED 4.4

   A sphere's radius grows at 3 cm/s. How fast is its volume, ${V = \frac{4}{3}\pi r^3}$, growing when the radius is 10 cm? Write it as ${k\pi}$ cubic centimeters per second and enter ${k}$.

   Answer: 1200

   1. ${V = \frac{4}{3}\pi r^3}$, so ${\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}}$.
   2. With ${r = 10}$ and ${\dfrac{dr}{dt} = 3}$: $\displaystyle \frac{dV}{dt} = 4\pi(10)^2(3)$$\frac{dV}{dt}$$=4\pi (10{)}^2(3)$$\displaystyle = \boxed{1{,}200\pi} \approx 3{,}769.911$$=\boxed{1,200\pi }\approx 3,769.911$ The volume grows at ${1{,}200\pi}$ cubic centimeters per second, so ${k = 1{,}200}$.

   Common wrong answers

   • 400: That's 4πr² in π units. Multiply by dr/dt = 3 cm/s.
   • 4000: Differentiate r³ first: dV/dt = 4πr² · dr/dt.

10. Fluency 2

    No calculator · CED 4.4

    A square's side grows at 4 cm/s. How fast is its area growing when the side is 5 cm?

    Answer: 40

    1. ${A = x^2}$, so ${\dfrac{dA}{dt} = 2x\dfrac{dx}{dt}}$.
    2. ${\dfrac{dA}{dt} = 2(5)(4) = \boxed{40}}$ square centimeters per second.

    Common wrong answers

    • 25: That's the area, 5², not its rate.
    • 16: That squares the rate. Differentiate A = x² first: dA/dt = 2x · dx/dt.
    • 20: dA/dt = 2x · dx/dt: keep the 2 from differentiating x².

11. Fluency 3

    No calculator · CED 4.4

    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.

    Answer: -5/6

    1. ${x^2 + y^2 = 169}$. When ${x = 5}$, ${y = \sqrt{169 - 25} = 12}$ ft.
    2. Differentiate with respect to ${t}$. The constant 169 gives 0, and the 2s cancel: $\displaystyle \cancel{2}x\frac{dx}{dt} + \cancel{2}y\frac{dy}{dt} = 0$$\cancel{2}x\frac{dx}{dt}+\cancel{2}y\frac{dy}{dt}$$=0$$\displaystyle \frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt}$$\frac{dy}{dt}$$=-\frac{x}{y}\cdot \frac{dx}{dt}$subtract, then divide by ${y}$
    3. $\displaystyle \frac{dy}{dt} = -\frac{5}{12}(2)$$\frac{dy}{dt}$$=-\frac{5}{12}(2)$$\displaystyle = \boxed{-\dfrac{5}{6}}$$=\boxed{-\frac{5}{6}}$ The top slides down at ${\frac{5}{6}}$ ft/s, 10 inches per second.

    Common wrong answers

    • 5/6: The top slides down, so the rate is negative.
    • -4.8: Check the ratio: dy/dt = −(x/y) · dx/dt, with x = 5 and y = 12.
    • -0.833: Enter the exact value, as a fraction.

12. Fluency 4

    No calculator · CED 4.4

    A cube's edge grows at 2 cm/s. How fast is its volume growing when the edge is 3 cm?

    Answer: 54

    1. ${V = x^3}$, so ${\dfrac{dV}{dt} = 3x^2\dfrac{dx}{dt}}$.
    2. ${\dfrac{dV}{dt} = 3(3)^2(2) = \boxed{54}}$ cubic centimeters per second.

    Common wrong answers

    • 27: 27 is both the volume, 3³, and 3x² at x = 3. Multiply 3x² by dx/dt = 2.
    • 8: That cubes the rate. Differentiate V = x³ first: dV/dt = 3x² · dx/dt.

13. Fluency 5

    No calculator · CED 4.4

    A circle's area grows at ${10\pi}$ square centimeters per second when its radius is 5 cm. How fast is its circumference growing then? Write it as ${k\pi}$ centimeters per second and enter ${k}$.

    Answer: 2

    1. ${A = \pi r^2}$, so ${\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}}$. With ${r = 5}$ and the area's rate, ${10\pi}$: $\displaystyle 10\pi = 2\pi(5)\frac{dr}{dt}$$10\pi$$=2\pi (5)\frac{dr}{dt}$$\displaystyle \frac{dr}{dt} = 1$$\frac{dr}{dt}$$=1$divide by ${10\pi}$ The radius grows at 1 cm/s.
    2. ${C = 2\pi r}$, so $\displaystyle \frac{dC}{dt} = 2\pi\frac{dr}{dt}$$\frac{dC}{dt}$$=2\pi \frac{dr}{dt}$$\displaystyle = 2\pi(1)$$=2\pi (1)$$\displaystyle = \boxed{2\pi} \approx 6.283$$=\boxed{2\pi }\approx 6.283$ The circumference grows at ${2\pi}$ centimeters per second, so ${k = 2}$.

    Common wrong answers

    • 1: That's dr/dt. C = 2πr grows 2π times as fast.
    • 10: That's the area's rate, in π units. Find dr/dt first, then dC/dt = 2π · dr/dt.

14. Fluency 6

    No calculator · CED 4.4

    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?

    Answer: 5

    1. At 10 seconds, ${y = 30}$ m and ${x = 40}$ m, so ${D = \sqrt{40^2 + 30^2} = 50}$ m.
    2. Differentiate with respect to ${t}$; the 2s cancel: $\displaystyle \cancel{2}D\frac{dD}{dt} = \cancel{2}x\frac{dx}{dt} + \cancel{2}y\frac{dy}{dt}$$\cancel{2}D\frac{dD}{dt}$$=\cancel{2}x\frac{dx}{dt}+\cancel{2}y\frac{dy}{dt}$$\displaystyle 50\frac{dD}{dt} = 40(4) + 30(3)$$50\frac{dD}{dt}$$=40(4)+30(3)$put in the values$\displaystyle = 250$$=250$
    3. ${\dfrac{dD}{dt} = \dfrac{250}{50} = \boxed{5}}$ m/s, at every moment: the triangle stays 3-4-5, so ${D = 5t}$.

    Common wrong answers

    • 7: That adds the speeds. At right angles, use D² = x² + y².
    • 50: That's the distance at t = 10, not its rate.

15. Review 1

    From Is the Ball Ever Stopped?, Fluency 1 · No calculator · CED 2.1

    Find the average rate of change of ${f(x) = x^2 + 3x}$ on ${[1, 4]}$.

    Answer: 8

    1. ${f(4) = 16 + 12 = 28}$ and ${f(1) = 1 + 3 = 4}$.
    2. Difference quotient: $\displaystyle \frac{f(4) - f(1)}{4 - 1} = \frac{28 - 4}{3}$$\frac{f(4)-f(1)}{4-1}$$=\frac{28-4}{3}$$\displaystyle = \frac{24}{3}$$=\frac{24}{3}$
    3. The average rate of change is 8.

    Common wrong answers

    • 24: Divide by the width of the interval, 4 − 1 = 3.
    • -8: Same order on top and bottom: f(4) − f(1) over 4 − 1.

16. Review 2

    From Is the Ball Ever Stopped?, Fluency 4 · No calculator · CED 2.2

    Use the limit definition to find ${f'(3)}$ for ${f(x) = x^2 - 4x}$.

    Answer: 2

    1. ${f(3) = 9 - 12 = -3}$.
    2. $\displaystyle f(3 + h) = (3 + h)^2 - 4(3 + h)$$f(3+h)$$=(3+h{)}^2-4(3+h)$$\displaystyle = 9 + 6h + h^2 - 12 - 4h$$=9+6h+h^2-12-4h$multiply out$\displaystyle = -3 + 2h + h^2$$=-3+2h+h^2$
    3. Subtract ${f(3) = -3}$, then divide by ${h}$: $\displaystyle f(3 + h) - f(3) = 2h + h^2$$f(3+h)-f(3)$$=2h+h^2$$\displaystyle \frac{f(3 + h) - f(3)}{h} = 2 + h$$\frac{f(3+h)-f(3)}{h}$$=2+h$${h \ne 0}$
    4. As ${h \to 0}$, ${2 + h \to 2}$, so ${f'(3) = \boxed{2}}$.

    Common wrong answers

    • -3: That's f(3), a function value, not a rate.
    • 3: That's the average on [3, 4], h = 1. Let h go to 0.

17. Exit ticket

    No calculator · CED 4.4

    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 ${k\pi}$ square centimeters per second and enter ${k}$.

    Answer: 120

    1. ${A = \pi r^2}$, so ${\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}}$.
    2. Substitute: $\displaystyle \frac{dA}{dt} = 2\pi(20)(3)$$\frac{dA}{dt}$$=2\pi (20)(3)$$\displaystyle = \boxed{120\pi} \approx 376.991$$=\boxed{120\pi }\approx 376.991$ The area grows at ${120\pi}$ square centimeters per second (positive: the ripple spreads), so ${k = 120}$.

    Common wrong answers

    • 400: That's the area, π · 20², in π units, not its rate.
    • 60: dA/dt = 2πr · dr/dt: keep the 2 from differentiating r².
    • 6: That's the circumference's rate, 2π · dr/dt. The area is A = πr².

Free response: scoring guideline

1. Part (a) · CED 4.4

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

   Answer: -1884.956 (or -1884.955)

   Scoring

   2 points. 1 point: ${r(25) = 200 - 2(25) = 150}$, and ${\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}}$ from ${A = \pi r^2}$. 1 point: the value, with units (square meters per second): $\displaystyle A'(25) = 2\pi(150)(-2)$$A^{\prime }(25)$$=2\pi (150)(-2)$$\displaystyle = -600\pi \approx -1{,}884.956$$=-600\pi \approx -1,884.956$

2. Part (b) · CED 5.3

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

   Scoring

   2 points. 1 point: ${\dfrac{d^2r}{dt^2} = 0}$ (${r}$ changes at a constant rate), so the product rule on ${\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}}$ gives $\displaystyle \frac{d^2A}{dt^2} = 2\pi\left(\frac{dr}{dt}\right)^2 + 2\pi r\frac{d^2r}{dt^2}$$\frac{d^{2}A}{d{t}^2}$$=2\pi {\left(\frac{dr}{dt}\right)}^2+2\pi r\frac{d^{2}r}{d{t}^2}$$\displaystyle = 2\pi(-2)^2$$=2\pi (-2{)}^2$$\displaystyle = 8\pi$$=8\pi$ 1 point: ${\dfrac{d^2A}{dt^2} = 8\pi}$ is positive, so ${\dfrac{dA}{dt}}$ is increasing at ${t = 25}$: negative and rising toward 0, so the area shrinks more and more slowly. Also accept ${\dfrac{dA}{dt} = 8\pi t - 800\pi}$, increasing since its slope, ${8\pi}$, is positive: $\displaystyle \frac{dA}{dt} = -4\pi r(t)$$\frac{dA}{dt}$$=-4\pi r(t)$$\displaystyle = -4\pi(200 - 2t)$$=-4\pi (200-2t)$$\displaystyle = 8\pi t - 800\pi$$=8\pi t-800\pi$

3. Part (c) · CED 4.4, 4.5

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

   Answer: -12.566

   Scoring

   2 points. 1 point: the value, with units (meters per second). ${C = 2\pi r}$, so $\displaystyle \frac{dC}{dt} = 2\pi\frac{dr}{dt}$$\frac{dC}{dt}$$=2\pi \frac{dr}{dt}$$\displaystyle = 2\pi(-2)$$=2\pi (-2)$$\displaystyle = -4\pi \approx -12.566$$=-4\pi \approx -12.566$ 1 point: ${C}$ is linear in ${r}$, so ${\dfrac{dC}{dt}}$ has no ${r}$ and is the same at every time. ${\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}}$ is proportional to ${r}$, which shrinks: a shorter edge takes less area per meter it moves in.

4. Part (d) · CED 4.4, 4.5

   At ${t = 40}$, 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.

   Answer: -4

   Scoring

   3 points. 1 point: with ${x}$ the distance along the road and ${s}$ to the center, ${s^2 = x^2 + 50^2}$, so ${s = 130}$ meters at ${x = 120}$; ${r(40) = {200 - 2(40)} = 120}$, so the player is outside, ${{s - r} = 10}$ meters from the edge. 1 point: the distance to the center changes at ${-6}$ meters per second: $\displaystyle 2s\frac{ds}{dt} = 2x\frac{dx}{dt}$$2s\frac{ds}{dt}$$=2x\frac{dx}{dt}$$\displaystyle \frac{ds}{dt} = \frac{x}{s} \cdot \frac{dx}{dt}$$\frac{ds}{dt}$$=\frac{x}{s}\cdot \frac{dx}{dt}$divide by ${2s}$$\displaystyle = \frac{120}{130}(-6.5)$$=\frac{120}{130}(-6.5)$$\displaystyle = -6$$=-6$ 1 point, with units: $\displaystyle \frac{d}{dt}(s - r) = \frac{ds}{dt} - \frac{dr}{dt}$$\frac{d}{dt}(s-r)$$=\frac{ds}{dt}-\frac{dr}{dt}$$\displaystyle = -6 - (-2)$$=-6-(-2)$$\displaystyle = -4$$=-4$ The distance to the edge changes at ${-4}$ meters per second, shrinking at 4 meters per second.