AP Calculus AB · Lesson 1

Rates in context and motion

Know Know it

The derivative f′(a)f'(a) is the instantaneous rate of change of ff at aa. Its units are the units of the output per one unit of the input. If H(t)H(t) is the height of the tide, in feet, tt hours after midnight, then H′(t)H'(t) is in feet per hour and H′′(t)H''(t) is in feet per hour per hour.

From a table, estimate a derivative with a difference quotient f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}, using the two listed inputs closest to the point, one on each side.

Motion on a line. If x(t)x(t) is a particle's position, its velocity is v(t)=x′(t)v(t) = x'(t) and its acceleration is a(t)=v′(t)=x′′(t)a(t) = v'(t) = x''(t). Its speed is ∣v(t)∣|v(t)|. Positive velocity means moving right (or up); negative means moving left (or down). The particle is at rest when v(t)=0v(t) = 0, and it changes direction only where v(t)v(t) changes sign.

Speeding up or slowing down. Speed increases when vv and aa have the same sign and decreases when they have opposite signs.

Motion on a line at a glance
QuantityHow to get itWhat its sign tells you
Positionx(t)x(t)Which side of the origin the particle is on
Velocityv(t)=x′(t)v(t) = x'(t)Direction: positive is right, negative is left
Accelerationa(t)=v′(t)a(t) = v'(t)Whether the velocity is increasing or decreasing
Speed∣v(t)∣|v(t)|Never negative; compare the signs of v and a to see if it grows

Calculator On the exam

The multiple-choice section has two parts: 29 of the 42 questions allow no calculator, then 13 require a graphing calculator, though not every one of them needs it. The free-response section is split the same way: a graphing calculator is required on the first 2 of the 6 questions and not allowed on the last 4. You may bring an approved handheld graphing calculator; the exam’s testing app also has a graphing calculator, available only in the calculator parts. Check the current calculator policy before test day.

So learn to read rates from formulas and tables by hand. When a calculator is allowed, use it to evaluate a derivative at a point, find where a velocity equals 0, or check an estimate.

Trap Negative acceleration does not mean slowing down

A particle with v=−5v = -5 and a=−2a = -2 is speeding up: it moves left, and its velocity grows more negative, so its speed grows. Never judge speeding up from the acceleration alone. Compare the signs of vv and aa.

Worked example Easy

Let T(m)T(m) be the temperature, in degrees Celsius, at the center of a loaf of bread mm minutes after it goes into an oven. Which is the best interpretation of T′(20)=1.5T'(20) = 1.5?

  1. At 20 minutes, the temperature at the center is 1.5 degrees Celsius
  2. At 20 minutes, the center is warming at 1.5 degrees Celsius per minuteAnswer
  3. Over the first 20 minutes, the center warmed an average of 1.5 degrees Celsius per minute
  4. The center warms by 1.5 degrees Celsius every 20 minutes

How to solve it

  1. The prime means a rate: T′(20)T'(20) is how fast TT changes at m=20m = 20.
  2. Units: output units over input units, so degrees Celsius per minute.
  3. The sign is positive, so the temperature is rising at that moment.

Why each choice is right or wrong

  • A. Incorrect. A temperature of 1.5 degrees would be T(20)=1.5T(20) = 1.5; the prime means a rate of change, not the temperature itself.
  • B. Correct. T′(20)T'(20) is the rate of change of the temperature at m=20m = 20, in degrees Celsius per minute, and it is positive, so the center is warming.
  • C. Incorrect. An average over the first 20 minutes is an average rate of change on an interval, not the derivative at the single moment m = 20.
  • D. Incorrect. The units of the derivative are degrees per minute, so 1.5 is a change per minute, not per 20 minutes.

Explain Explain it

Task verbs: interpret, determine

To interpret a derivative, say four things: the quantity, the moment, the rate with units, and the direction. For H′(3)=−0.8H'(3) = -0.8: at 3 a.m. the height of the tide is falling at 0.8 foot per hour.

To interpret a second derivative, talk about the rate, not the quantity. H′′(3)<0H''(3) < 0 says the rate H′H' is decreasing at 3 a.m.; with H′(3)<0H'(3) < 0, that means the tide is falling faster and faster.

To determine whether a particle speeds up, find v(t)v(t) and a(t)a(t) at that time and compare their signs.

Worked example Medium

A particle moves along the x-axis with position x(t)=t3−9t2+24tx(t) = t^3 - 9t^2 + 24t for t≥0t \geq 0. Which describes the particle at t=2.5t = 2.5?

  1. Moving right and speeding up
  2. Moving right and slowing down
  3. Moving left and slowing down
  4. Moving left and speeding upAnswer

How to solve it

  1. v(t)=x′(t)=3t2−18t+24v(t) = x'(t) = 3t^2 - 18t + 24, so v(2.5)=18.75−45+24=−2.25v(2.5) = 18.75 - 45 + 24 = -2.25.
  2. a(t)=v′(t)=6t−18a(t) = v'(t) = 6t - 18, so a(2.5)=15−18=−3a(2.5) = 15 - 18 = -3.
  3. v is negative, so the particle moves left. v and a are both negative, so it speeds up.

Why each choice is right or wrong

  • A. Incorrect. Moving right needs v(2.5)>0v(2.5) > 0, but v(2.5)=−2.25v(2.5) = -2.25, so the particle moves left.
  • B. Incorrect. v(2.5)=−2.25v(2.5) = -2.25 is negative, so the particle is not moving right, and with a(2.5)=−3a(2.5) = -3 the signs match, so it is not slowing down.
  • C. Incorrect. Moving left is right, but v(2.5)=−2.25v(2.5) = -2.25 and a(2.5)=−3a(2.5) = -3 have the same sign, so the particle is speeding up, not slowing down.
  • D. Correct. v(2.5)=−2.25<0v(2.5) = -2.25 < 0, so it moves left, and a(2.5)=−3a(2.5) = -3 has the same sign as vv, so its speed is increasing.

Interpret Your turn

R(t)R(t) is the volume of water in a reservoir, in millions of gallons, tt days after a dry spell begins. Say what R′(10)=−2.4R'(10) = -2.4 means in context. Then say what R′′(10)>0R''(10) > 0 tells you about how the volume is changing at that time.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says that 10 days after the dry spell begins, the volume of water is decreasing at 2.4 million gallons per day
  • Says that R′′(10)>0R''(10) > 0 means the rate of change of the volume is increasing at t=10t = 10 (the water is going down more slowly)

Model answer

On day 10 of the dry spell, the volume of water in the reservoir is decreasing at a rate of 2.4 million gallons per day. Since R′′(10)>0R''(10) > 0, the rate R′R' is increasing at t=10t = 10: at that moment it is negative and getting closer to 0, so the reservoir is losing water more slowly.

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