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Units 1 and 2

Is the ball ever stopped?

A pop-up fly hangs at the top of its arc. Average velocity, instantaneous velocity, and the limit that connects them.

About 45 min · 11 problemsStart →

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

Two questions about speed

The pop-up. A pop-up leaves the bat 6 feet off the ground, climbs, seems to hang, and lands on the grass. Is it stopped at the top?

The speedometer. 30 miles in 45 minutes averages 40 miles per hour. Speed is distance over time, but no time passes in an instant. What does the speedometer divide?

Both ask for a rate at one instant: the derivative, the number averages approach as the interval shrinks.

Predict first

Is it stopped?

Friend, Teammate, Coach, You

  1. Friend
    at the top it just hangs there for a sec. so it's stopped

Start with an average

Velocity is speed with a sign: positive up, negative down. Average velocity on [a,b]{[a, b]} is h(b)−h(a)b−a{\frac{h(b) - h(a)}{b - a}}, so it can be 0 or negative while the ball moves.

Challenge 1

No calculator
A pop-up's height after t{t} seconds is h(t)=6+80t−16t2{h(t) = {6 + 80t - 16t^2}} feet. Find its average velocity from t=1{t = 1} to t=2{t = 2} seconds.

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

Shrink the interval

Shrink the interval that starts at t=2{t = 2}. Each average is the slope of a secant line.

Average velocity on a shrinking intervalPop-up model: h(t)=6+80t−16t2{h(t) = 6 + 80t - 16t^2}. A synthetic model.
Interval width

Secant from 2 s to 3 s (width 1 s): slope 0 ft/s, the average rate of change of height over that interval.

Endpoints: h(2) = 102 ft and h(3) = 102 ft.

Try these:

  1. At width 1 s the slope is 0 ft/s: the ball is 102 feet up at both 2 and 3 seconds. Stopped at t=2{t = 2}?
  2. Step the width down once. The slope is 8 ft/s. Once more, and it's 12.8 ft/s. Heading where?
  3. At 0.001 s, what happens to the secant and the dashed tangent?

About the model

The curve is h(t)=6+80t−16t2{h(t) = {6 + 80t - 16t^2}} feet. The −16t2{-16t^2} is gravity: velocity drops 32 feet per second every second. The launch speed of 80 ft/s and the 6-foot start are made up; no air resistance.

Challenge 2

No calculator
Same ball: h(t)=6+80t−16t2{h(t) = {6 + 80t - 16t^2}} feet after t{t} seconds. Its average velocity on [2,2+Δt]{[2, 2 + \Delta t]} as Δt{\Delta t} shrinks: Δt (s)average velocity (ft/s)100.114.40.0115.840.00115.984\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){v(2)}, the velocity at t=2{t = 2} seconds: limΔt→0⁡h(2+Δt)−h(2)Δt{\displaystyle\lim_{\Delta t \to 0} \frac{h(2 + \Delta t) - h(2)}{\Delta t}}

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

Velocity at an instant

The average rate of change of f{f} on [a,b]{[a, b]}:

f(b)−f(a)b−a\orange{\frac{f(b) - f(a)}{b - a}}

Undefined at a=b{a = b}, so set b=a+Δt{b = a + \Delta t} and let Δt→0{\Delta t \to 0} from either side. If the limit exists, it's the derivative, the instantaneous rate of change:

f′(a)=limΔt→0⁡f(a+Δt)−f(a)Δt\displaystyle \teal{f'(a)} = \lim_{\Delta t \to 0} \orange{\frac{f(a + \Delta t) - f(a)}{\Delta t}}
  • Graph: the quotient is a secant slope; f′(a){f'(a)} is the tangent slope, the secants' limit.
  • Ball: v(t)=h′(t){v(t) = h'(t)}, in ft/s: output unit per input unit.
  • With a variable t{t}, you get a function: h′(t)=dhdt{h'(t) = \frac{dh}{dt}}. For y=f(x){y = f(x)}: f′(x)=dydx=y′{f'(x) = \frac{dy}{dx} = y'}.
  • Table: average rate of change over the nearest values around the point; centered beats one-sided.

A note on letters

Books and the AP exam call the step h{h}: f′(a)=limh→0⁡f(a+h)−f(a)h{f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}}. In this lesson h{h} is the ball's height, so the ball, car, and puck limits use Δt{\Delta t}. Fluency uses h{h}.

Drag the point; the tangent slope is the velocity.

Velocity at any instantPop-up model: h(t)=6+80t−16t2{h(t) = 6 + 80t - 16t^2}. A synthetic model.

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

Time

Tangent at 1 s: height 70 ft, slope 48 ft/s.

Try these:

  1. 70 feet up twice: 48 ft/s at 1 second and −48 ft/s at 4 seconds. Same speed, opposite directions.
  2. Where is the tangent flat? The velocity there, just before, and just after?
  3. At 5 seconds it's back at its starting height. Its speed then versus at launch?

Worked example, then your turn

Worked example

No calculator
The pop-up's height is h(t)=6+80t−16t2{h(t) = {6 + 80t - 16t^2}} feet after t{t} seconds, from bat to landing. Use the limit definition to find its velocity v(t){v(t)}. Then: is the ball ever stopped, and if so, at what time?

Teammate, Coach, Friend

  1. Teammate
    where do we even start, for any time t{t}?

Back to the speedometer. You finish the limit.

Challenge 3

No calculator
A car leaving a parking lot has gone s(t)=3t2+2t{s(t) = 3t^2 + 2t} feet after t{t} seconds, for 0≤t≤10{0 \le t \le 10}. What does its speedometer read at t=4{t = 4} seconds, in feet per second? It only moves forward, so the reading is its velocity, limΔt→0⁡s(4+Δt)−s(4)Δt{\displaystyle\lim_{\Delta t \to 0} \frac{s(4 + \Delta t) - s(4)}{\Delta t}}. Step one, done for you: s(4+Δt)=3(4+Δt)2+2(4+Δt)\displaystyle s(4 + \Delta t) = 3(4 + \Delta t)^2 + 2(4 + \Delta t)=56+26Δt+3(Δt)2\displaystyle = 56 + 26\Delta t + 3(\Delta t)^2

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

The whole flight

Challenge 4

Calculator
The pop-up's height is h(t)=6+80t−16t2{h(t) = {6 + 80t - 16t^2}} feet after t{t} seconds, and its velocity is v(t)=80−32t{v(t) = 80 - 32t} ft/s. It leaves the bat 6 feet above the ground and lands on the ground (h=0{h = 0}). At what time t{t}, in seconds, does its velocity equal its average velocity over the whole flight?

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

Your prediction

Answer: negative. The ball lands 6 feet lower than it started. Zero needs a landing at the starting height.

The reveal

Velocity matches the average once, just after the top.

Mean Value Theorem (topic 5.1): if h{h} is continuous on [a,b]{[a, b]} and differentiable on (a,b){(a, b)}, then h′(c)=h(b)−h(a)b−a{h'(c) = \frac{h(b) - h(a)}{b - a}} for some c{c} in (a,b){(a, b)}. Compare your answer with the landing time.

Your prediction

Answer: for an instant. v=0{v = 0} at the top, but the ball never rests. Velocity goes from positive to negative, changing by 32 ft/s every second.

Fluency

Fluency 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.

Fluency 2

No calculator
A runner's distance from the start, D(t){D(t)}, in meters, is measured every 10 seconds. Use the values at t=10{t = 10} and t=30{t = 30} to estimate D′(20){D'(20)}, the velocity at t=20{t = 20}. t (s)0102030D(t) (m)062128188\begin{array}{c|cccc} t \text{ (s)} & 0 & 10 & 20 & 30 \\ \hline D(t) \text{ (m)} & 0 & 62 & 128 & 188 \end{array}

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

Fluency 3

No calculator
For f(x)=3x2{f(x) = 3x^2}, simplify f(2+h)−f(2)h{\dfrac{f(2 + h) - f(2)}{h}}, then evaluate it at h=0.1{h = 0.1}.

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

Fluency 4

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.

Fluency 5

No calculator
Use the limit definition to find f′(x){f'(x)} for f(x)=5x2+1{f(x) = 5x^2 + 1}. Then evaluate f′(−2){f'(-2)}.

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

Fluency 6

No calculator
Evaluate limh→0⁡(4+h)3−64h{\displaystyle\lim_{h \to 0} \frac{(4 + h)^3 - 64}{h}}, the derivative of some function f{f} at some number a{a}.

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

Exit ticket

Exit ticket

No calculator
A sliding hockey puck has gone s(t)=40t−2t2{s(t) = 40t - 2t^2} feet after t{t} seconds, for 0≤t≤10{0 \le t \le 10}. Use the limit definition to find how fast it moves at t=5{t = 5} seconds.

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

Reflect

A sentence each:

  1. Average velocity versus velocity at an instant: what's the difference?
  2. Why not use an interval of width 0? What instead?
  3. Which prediction was off, and why?

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