Is the Ball Ever Stopped?

CED topics

• 1.1 Introducing Calculus: Can Change Occur at an Instant?
• 2.1 Defining Average and Instantaneous Rates of Change at a Point
• 2.2 Defining the Derivative of a Function and Using Derivative Notation
• 2.3 Estimating Derivatives of a Function at a Point

Individual problems also preview:

• 5.1 Using the Mean Value Theorem

Big idea and practices

Big idea: rate of change at an instant, the limit of average rates over shrinking intervals: the derivative, the tangent slope. CED: CHA-1, CHA-2; the calculator item previews the Mean Value Theorem (FUN-1, topic 5.1).

Practices: 1.D, 1.E (difference quotients, the limit definition, table estimates); 2.B (graph, table, formula); 3.E (the velocity's sign just before and just after 2.5 seconds); 4.B, 4.C (units; ${f'(a)}$, ${\frac{dh}{dt}}$, limit notation).

45-minute plan

1. 0–5 minRead both hooks; pairs read "Is it stopped?" together and pick in it (or circle on the printout). Vote: stopped at the top?
2. 5–10 minChallenge 1 in pairs. Units before the number.
3. 10–18 minSecant explorable and its three prompts, then Challenge 2.
4. 18–26 minWhole class: difference quotient, limit, ${f'(a)}$ as tangent slope; ${h}$ versus ${\Delta t}$; tangent explorable.
5. 26–34 minPairs read the worked example thread together, then Challenge 3.
6. 34–40 minChallenge 4 with calculators, then the reveal.
7. 40–45 minExit ticket and reflection.

Pairs read each thread together; at each question, both answer before tapping Next.

Homework: the fluency set. Review Fluency 6, the AP form, next class.

Launch questions

• A pop-up hangs at the top. Stopped? For how long?
• Speed is distance over time. At one instant, what distance, over what time?
• 30 miles in 45 minutes: your speed at minute 20?
• Up and back down: is the average velocity zero? Does where it lands matter?

Misconceptions and how to respond

• “Stopped means stopped for a while.” Velocity drops by 32 ft/s every second; ${v = 0}$ only for an instant. Drag the tangent explorable through 2.5 seconds.
• Averaging heights. Watch for 86 on Challenge 1. Ask for units.
• Last row as answer. Watch for 15.984 on Challenge 2, the average on ${[2, 2.001]}$; they go to 16, which no row equals.
• Plugging in 0 first gives ${0/0}$. Expand, subtract, divide, then take the limit; Fluency 6 gives feedback on this.
• Trip average. Watch for 14 on Challenge 3: 56 feet in 4 seconds, not the reading at 4 seconds.
• “It comes back down, so the average is zero.” Watch for 2.5 on Challenge 4, right only for a catch at bat height. Landing 6 feet lower puts the answer past the peak.
• Rounding early. Watch for 2.535 on Challenge 4: 5.07 ÷ 2. Keep four decimals in ${T}$.
• Averaging from 0. Watch for 30 on the exit ticket, the average over the first 5 seconds.

Support and extension

Support

• Table the height at 0 through 5 seconds, 6, 70, 102, 102, 70, and 6 feet, and the average velocities between, 64, 32, 0, −32, and −64 ft/s. Where does the sign flip?
• Units before every quotient.
• Board expand, subtract, divide for the first limit; erase it for the next.
• Pairs: one runs the explorable, one records readouts and predicts the next.

Extension

• For ${h(t) = at^2 + bt + c}$, show the average velocity on ${[p, q]}$ is ${h'\!\left(\frac{p + q}{2}\right)}$. Why is Challenge 4's answer exactly half the landing time?
• Why does averaging over ${[2 - \Delta t, 2 + \Delta t]}$ give exactly 16 for every ${\Delta t}$? What does that say about centered table estimates?
• Back at its starting height at 5 seconds, falling at 80 ft/s, its launch speed: why? Use symmetry and ${v(t) = 80 - 32t}$.
• Why is the landing speed more than 80 ft/s? (${8\sqrt{106} \approx 82.365}$ ft/s: it falls 6 feet farther than it rose.)

Without devices

• Print the lesson page as the handout. Students circle their predictions.
• Secant: board the averages on ${[2, 2 + \text{width}]}$ for widths of 1, 0.5, 0.2, 0.1, 0.05, 0.02, 0.01, 0.005, 0.002, and 0.001 s, in order: 0, 8, 12.8, 14.4, 15.2, 15.68, 15.84, 15.92, 15.968, and 15.984 ft/s. What do they approach? Prove it with Challenge 2.
• Tangent: sketch the parabola from the support table; lay a ruler along it at 1, 2.5, and 4 seconds.
• Grade with the answer key below.
• 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. Challenge 1

   No calculator · CED 1.1, 2.1

   A pop-up's height after ${t}$ seconds is ${h(t) = {6 + 80t - 16t^2}}$ feet. Find its average velocity from ${t = 1}$ to ${t = 2}$ seconds.

   Answer: 32

   1. Average velocity: ${\dfrac{h(2) - h(1)}{2 - 1}}$.
   2. Heights, in feet: $\displaystyle h(1) = 6 + 80 - 16$$h(1)$$=6+80-16$$\displaystyle = 70$$=70$$\displaystyle h(2) = 6 + 160 - 64$$h(2)$$=6+160-64$$\displaystyle = 102$$=102$
   3. ${\dfrac{102 - 70}{2 - 1} = \boxed{32}}$: the ball rises 32 feet in that second, an average velocity of 32 ft/s.

   Common wrong answers

   • 86: That's the average height, not change in height over change in time.
   • 48: That's the average from t = 0 to t = 2. Start at t = 1.
   • -32: Same order on top and bottom: h(2) − h(1) over 2 − 1.
   • 102: That's the height at t = 2, not a velocity.

2. Challenge 2

   No calculator · CED 1.1, 2.1, 2.2

   Same ball: ${h(t) = {6 + 80t - 16t^2}}$ feet after ${t}$ seconds. Its average velocity on ${[2, 2 + \Delta t]}$ as ${\Delta t}$ shrinks: $$\begin{array}{c|c} \Delta t \text{ (s)} & \text{average velocity (ft/s)} \\ \hline 1 & 0 \\ 0.1 & 14.4 \\ 0.01 & 15.84 \\ 0.001 & 15.984 \end{array}$$ Use the limit definition to find ${v(2)}$, the velocity at ${t = 2}$ seconds: ${\displaystyle\lim_{\Delta t \to 0} \frac{h(2 + \Delta t) - h(2)}{\Delta t}}$

   Answer: 16

   1. ${h(2) = 6 + 160 - 64 = 102}$.
   2. Expand ${h(2 + \Delta t)}$: $\displaystyle h(2 + \Delta t) = 6 + 80(2 + \Delta t) - 16(2 + \Delta t)^2$$h(2+\Delta t)$$=6+80(2+\Delta t)-16(2+\Delta t{)}^2$put in ${2 + \Delta t}$ for ${t}$$\displaystyle = 6 + 160 + 80\Delta t - 64 - 64\Delta t - 16(\Delta t)^2$$=6+160+80\Delta t-64-64\Delta t-16(\Delta t{)}^2$multiply out$\displaystyle = 102 + 16\Delta t - 16(\Delta t)^2$$=102+16\Delta t-16(\Delta t{)}^2$combine like terms
   3. Subtract ${h(2) = 102}$, divide by ${\Delta t}$, and factor it out of the top; since ${\Delta t \ne 0}$, it cancels: $\displaystyle h(2 + \Delta t) - h(2) = 16\Delta t - 16(\Delta t)^2$$h(2+\Delta t)-h(2)$$=16\Delta t-16(\Delta t{)}^2$$\displaystyle \frac{\cancel{\Delta t}(16 - 16\Delta t)}{\cancel{\Delta t}} = 16 - 16\Delta t$$\frac{\cancel{\Delta t}(16-16\Delta t)}{\cancel{\Delta t}}$$=16-16\Delta t$ Check: ${\Delta t = 0.001}$ gives ${15.984}$, the table's last row.
   4. As ${\Delta t \to 0}$, ${16 - 16\Delta t \to \boxed{16}}$: at ${t = 2}$ the ball rises at 16 ft/s.

   Common wrong answers

   • 15.984: That's the average on [2, 2.001]. Find the number the averages approach.
   • 102: That's the height at t = 2, not a velocity.
   • 0: That's the average on [2, 3]. Let Δt shrink toward 0.
   • 32: That's the average on [1, 2]. Find the velocity at the instant t = 2.

3. Worked example

   No calculator · CED 2.2

   The pop-up's height is ${h(t) = {6 + 80t - 16t^2}}$ feet after ${t}$ seconds, from bat to landing. Use the limit definition to find its velocity ${v(t)}$. Then: is the ball ever stopped, and if so, at what time?

   Worked on the page: 2.5

   1. Start from the limit definition: the velocity ${\teal{v(t)}}$ is the limit of the average velocity ${\orange{\dfrac{h(t + \Delta t) - h(t)}{\Delta t}}}$ as ${\Delta t \to 0}$.
   2. Expand ${h(t + \Delta t)}$: $\displaystyle h(t + \Delta t) = 6 + 80(t + \Delta t) - 16(t + \Delta t)^2$$h(t+\Delta t)$$=6+80(t+\Delta t)-16(t+\Delta t{)}^2$put in ${t + \Delta t}$ for ${t}$$\displaystyle = 6 + 80t + 80\Delta t - 16t^2 - 32t\Delta t - 16(\Delta t)^2$$=6+80t+80\Delta t-16t^2-32t\Delta t-16(\Delta t{)}^2$multiply out
   3. Subtract ${h(t) = 6 + 80t - 16t^2}$. Every term without a ${\Delta t}$ cancels: $\displaystyle h(t + \Delta t) - h(t) = \cancel{6} + \cancel{80t} + 80\Delta t - \cancel{16t^2} - 32t\Delta t - 16(\Delta t)^2 - \cancel{6} - \cancel{80t} + \cancel{16t^2}$$h(t+\Delta t)-h(t)$$=\cancel{6}+\cancel{80t}+80\Delta t-\cancel{16t^2}-32t\Delta t-16(\Delta t{)}^2-\cancel{6}-\cancel{80t}+\cancel{16t^2}$subtract each term$\displaystyle = 80\Delta t - 32t\Delta t - 16(\Delta t)^2$$=80\Delta t-32t\Delta t-16(\Delta t{)}^2$
   4. Divide by ${\Delta t}$ for the average velocity, then let ${\Delta t \to 0}$ for the velocity: $\displaystyle \orange{\frac{h(t + \Delta t) - h(t)}{\Delta t}} = 80 - 32t - 16\Delta t$$\frac{h(t+\Delta t)-h(t)}{\Delta t}$$=80-32t-16\Delta t$${\Delta t \ne 0}$$\displaystyle \teal{v(t)} = \teal{80 - 32t}$$v(t)$$=80-32t$let ${\Delta t \to 0}$ So ${v(t) = 80 - 32t}$ ft/s. Check: ${v(2) = 80 - 64 = 16}$, as in Challenge 2.
   5. Stopped means ${v(t) = 0}$: $\displaystyle 80 - 32t = 0$$80-32t$$=0$$\displaystyle t = \boxed{2.5}$$t$$=\boxed{2.5}$ Yes, at the top, 2.5 s, but only for an instant: ${v > 0}$ just before and ${v < 0}$ just after.

4. Challenge 3

   No calculator · CED 2.1, 2.2

   A car leaving a parking lot has gone ${s(t) = 3t^2 + 2t}$ feet after ${t}$ seconds, for ${0 \le t \le 10}$. What does its speedometer read at ${t = 4}$ seconds, in feet per second? It only moves forward, so the reading is its velocity, ${\displaystyle\lim_{\Delta t \to 0} \frac{s(4 + \Delta t) - s(4)}{\Delta t}}$. Step one, done for you: $\displaystyle s(4 + \Delta t) = 3(4 + \Delta t)^2 + 2(4 + \Delta t)$$s(4+\Delta t)$$=3(4+\Delta t{)}^2+2(4+\Delta t)$$\displaystyle = 56 + 26\Delta t + 3(\Delta t)^2$$=56+26\Delta t+3(\Delta t{)}^2$

   Answer: 26

   1. Done for you: $\displaystyle s(4 + \Delta t) = 3(4 + \Delta t)^2 + 2(4 + \Delta t)$$s(4+\Delta t)$$=3(4+\Delta t{)}^2+2(4+\Delta t)$$\displaystyle = 48 + 24\Delta t + 3(\Delta t)^2 + 8 + 2\Delta t$$=48+24\Delta t+3(\Delta t{)}^2+8+2\Delta t$multiply out$\displaystyle = 56 + 26\Delta t + 3(\Delta t)^2$$=56+26\Delta t+3(\Delta t{)}^2$combine like terms
   2. Subtract ${s(4) = 48 + 8 = 56}$: $\displaystyle s(4 + \Delta t) - s(4) = 26\Delta t + 3(\Delta t)^2$$s(4+\Delta t)-s(4)$$=26\Delta t+3(\Delta t{)}^2$
   3. Divide by ${\Delta t}$ and factor it out of the top; since ${\Delta t \ne 0}$, it cancels: $\displaystyle \frac{\cancel{\Delta t}(26 + 3\Delta t)}{\cancel{\Delta t}} = 26 + 3\Delta t$$\frac{\cancel{\Delta t}(26+3\Delta t)}{\cancel{\Delta t}}$$=26+3\Delta t$
   4. As ${\Delta t \to 0}$, ${26 + 3\Delta t \to \boxed{26}}$: the speedometer reads 26 ft/s, about 18 mph. The trip average is only ${56 \div 4 = 14}$ ft/s.

   Common wrong answers

   • 14: That's the trip average, 56 feet in 4 seconds. Find the speed at the instant t = 4.
   • 29: That's the average on [4, 5], Δt = 1. Let Δt shrink toward 0.
   • 56: That's the distance by t = 4, not a speed.

5. Challenge 4

   Calculator · CED 2.1, 5.1

   The pop-up's height is ${h(t) = {6 + 80t - 16t^2}}$ feet after ${t}$ seconds, and its velocity is ${v(t) = 80 - 32t}$ ft/s. It leaves the bat 6 feet above the ground and lands on the ground (${h = 0}$). At what time ${t}$, in seconds, does its velocity equal its average velocity over the whole flight?

   Answer: 2.537 (or 2.536)

   1. It lands when ${h(t) = 0}$: $\displaystyle 6 + 80t - 16t^2 = 0$$6+80t-16t^2$$=0$$\displaystyle 16t^2 - 80t - 6 = 0$$16t^2-80t-6$$=0$times ${-1}$ The positive root is the landing time ${T}$, in seconds: $\displaystyle T = \frac{80 + \sqrt{6{,}784}}{32}$$T$$=\frac{80+\sqrt{6,784}}{32}$quadratic formula$\displaystyle = \frac{10 + \sqrt{106}}{4} \approx 5.074$$=\frac{10+\sqrt{106}}{4}\approx 5.074$
   2. Average velocity over the flight, in ft/s: $\displaystyle \frac{h(T) - h(0)}{T - 0} = \frac{0 - 6}{T} \approx -1.183$$\frac{h(T)-h(0)}{T-0}$$=\frac{0-6}{T}\approx -1.183$ Negative: the ball ends 6 feet lower.
   3. Set ${v(t)}$ equal to that average and solve for ${t}$, in seconds: $\displaystyle 80 - 32t = -\frac{6}{T}$$80-32t$$=-\frac{6}{T}$$\displaystyle t = \frac{80 + 6/T}{32} \approx 2.537$$t$$=\frac{80+6/T}{32}\approx 2.537$
   4. Cross-check: for quadratic ${h}$, the average velocity over an interval equals the velocity at its midpoint, and ${\dfrac{T}{2}}$ is 2.53695… s. So ${t \approx \boxed{2.537}}$ s (or 2.536 s, truncated), just after the peak at 2.5 s. As a polynomial, ${h}$ is continuous on ${[0, T]}$ and differentiable on ${(0, T)}$, so the Mean Value Theorem guarantees this time exists.

   Common wrong answers

   • 2.5: That's the peak, where v = 0. The average is 0 only for a landing at the starting height; this ball lands 6 feet lower.
   • 5.074: That's the landing time. Find when v(t) equals the flight's average velocity.
   • -1.183: That's the average velocity itself. Find the time when v(t) equals it.
   • 2.535: That's 5.07 ÷ 2: T was rounded too early. Keep four decimals in T; round only at the end.

6. 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.

7. Fluency 2

   No calculator · CED 2.1, 2.3

   A runner's distance from the start, ${D(t)}$, in meters, is measured every 10 seconds. Use the values at ${t = 10}$ and ${t = 30}$ to estimate ${D'(20)}$, the velocity at ${t = 20}$. $$\begin{array}{c|cccc} t \text{ (s)} & 0 & 10 & 20 & 30 \\ \hline D(t) \text{ (m)} & 0 & 62 & 128 & 188 \end{array}$$

   Answer: 6.3

   1. ${[10, 30]}$ is centered at ${t = 20}$, so its average rate of change estimates ${D'(20)}$.
   2. From the table: $\displaystyle \frac{D(30) - D(10)}{30 - 10} = \frac{188 - 62}{20}$$\frac{D(30)-D(10)}{30-10}$$=\frac{188-62}{20}$$\displaystyle = \frac{126}{20}$$=\frac{126}{20}$
   3. ${\dfrac{126}{20} = \boxed{6.3}}$, so ${D'(20)}$ is about 6.3 meters per second.

   Common wrong answers

   • 6.4: That's the average from t = 0 to t = 20. Use t = 10 and t = 30.
   • 6.6: That uses t = 10 and t = 20. Use t = 10 and t = 30.
   • 6: That uses t = 20 and t = 30. Use t = 10 and t = 30.
   • 126: Divide by the time between the readings, 30 − 10 = 20 seconds.

8. Fluency 3

   No calculator · CED 2.1

   For ${f(x) = 3x^2}$, simplify ${\dfrac{f(2 + h) - f(2)}{h}}$, then evaluate it at ${h = 0.1}$.

   Answer: 12.3

   1. ${f(2) = 12}$. Expand: $\displaystyle f(2 + h) = 3(2 + h)^2$$f(2+h)$$=3(2+h{)}^2$$\displaystyle = 3(4 + 4h + h^2)$$=3(4+4h+h^2)$$\displaystyle = 12 + 12h + 3h^2$$=12+12h+3h^2$
   2. Subtract ${f(2)}$, then divide by ${h}$: $\displaystyle f(2 + h) - f(2) = 12h + 3h^2$$f(2+h)-f(2)$$=12h+3h^2$$\displaystyle \frac{f(2 + h) - f(2)}{h} = 12 + 3h$$\frac{f(2+h)-f(2)}{h}$$=12+3h$${h \ne 0}$
   3. At ${h = 0.1}$: ${12 + 3(0.1) = \boxed{12.3}}$, the average rate of change of ${f}$ on ${[2, 2.1]}$.

   Common wrong answers

   • 12: 12 is the limit, f′(2). Evaluate at h = 0.1.
   • 1.23: That's f(2.1) − f(2). Divide by h = 0.1.

9. 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.

10. Fluency 5

    No calculator · CED 2.2

    Use the limit definition to find ${f'(x)}$ for ${f(x) = 5x^2 + 1}$. Then evaluate ${f'(-2)}$.

    Answer: -20

    1. $\displaystyle f(x + h) = 5(x + h)^2 + 1$$f(x+h)$$=5(x+h{)}^2+1$$\displaystyle = 5x^2 + 10xh + 5h^2 + 1$$=5x^2+10xh+5h^2+1$multiply out
    2. Subtract ${f(x) = 5x^2 + 1}$, then divide by ${h}$: $\displaystyle f(x + h) - f(x) = 10xh + 5h^2$$f(x+h)-f(x)$$=10xh+5h^2$$\displaystyle \frac{f(x + h) - f(x)}{h} = 10x + 5h$$\frac{f(x+h)-f(x)}{h}$$=10x+5h$${h \ne 0}$
    3. As ${h \to 0}$, this goes to ${f'(x) = 10x}$, so ${f'(-2) = \boxed{-20}}$.

    Common wrong answers

    • 21: That's f(−2). Find f′(x), then substitute x = −2.
    • 20: Check the sign: f′(x) = 10x, and x is −2.

11. Fluency 6

    No calculator · CED 2.2

    Evaluate ${\displaystyle\lim_{h \to 0} \frac{(4 + h)^3 - 64}{h}}$, the derivative of some function ${f}$ at some number ${a}$.

    Answer: 48

    1. It is ${f'(4)}$ for ${f(x) = x^3}$, because ${f(4) = 4^3 = 64}$.
    2. Expand ${(4 + h)^3}$, subtract 64, then divide by ${h}$: $\displaystyle (4 + h)^3 = 64 + 48h + 12h^2 + h^3$$(4+h{)}^3$$=64+48h+12h^2+h^3$$\displaystyle (4 + h)^3 - 64 = 48h + 12h^2 + h^3$$(4+h{)}^3-64$$=48h+12h^2+h^3$$\displaystyle \frac{(4 + h)^3 - 64}{h} = 48 + 12h + h^2$$\frac{(4+h{)}^3-64}{h}$$=48+12h+h^2$${h \ne 0}$
    3. As ${h \to 0}$, the quotient goes to 48: ${f'(4) = \boxed{48}}$.

    Common wrong answers

    • 64: 64 is f(4), a function value. The limit is the rate f′(4).
    • 0: h = 0 gives 0/0, which says nothing yet. And (4 + h)³ isn't 64 + h³: expand fully to 64 + 48h + 12h² + h³, simplify, then let h go to 0.

12. Exit ticket

    No calculator · CED 2.1, 2.2

    A sliding hockey puck has gone ${s(t) = 40t - 2t^2}$ feet after ${t}$ seconds, for ${0 \le t \le 10}$. Use the limit definition to find how fast it moves at ${t = 5}$ seconds.

    Answer: 20

    1. ${s(5) = 200 - 50 = 150}$.
    2. Expand ${s(5 + \Delta t)}$: $\displaystyle s(5 + \Delta t) = 40(5 + \Delta t) - 2(5 + \Delta t)^2$$s(5+\Delta t)$$=40(5+\Delta t)-2(5+\Delta t{)}^2$put in ${5 + \Delta t}$ for ${t}$$\displaystyle = 200 + 40\Delta t - 50 - 20\Delta t - 2(\Delta t)^2$$=200+40\Delta t-50-20\Delta t-2(\Delta t{)}^2$multiply out$\displaystyle = 150 + 20\Delta t - 2(\Delta t)^2$$=150+20\Delta t-2(\Delta t{)}^2$combine like terms
    3. Subtract ${s(5) = 150}$, divide by ${\Delta t}$, and factor it out of the top; since ${\Delta t \ne 0}$, it cancels: $\displaystyle s(5 + \Delta t) - s(5) = 20\Delta t - 2(\Delta t)^2$$s(5+\Delta t)-s(5)$$=20\Delta t-2(\Delta t{)}^2$$\displaystyle \frac{\cancel{\Delta t}(20 - 2\Delta t)}{\cancel{\Delta t}} = 20 - 2\Delta t$$\frac{\cancel{\Delta t}(20-2\Delta t)}{\cancel{\Delta t}}$$=20-2\Delta t$
    4. As ${\Delta t \to 0}$, ${20 - 2\Delta t \to \boxed{20}}$. At ${t = 5}$ the puck is moving at 20 ft/s.

    Common wrong answers

    • 30: That's the average over the first 5 seconds. Find the speed at the instant t = 5.
    • 150: That's the distance by t = 5, not a speed.
    • 18: That's the average on [5, 6], Δt = 1. Let Δt shrink toward 0.