AP Calculus AB · Unit 4

Unit 4: Contextual Applications of Differentiation

How to read a derivative in context, with units, from a formula or a table

How velocity, speed and acceleration describe a particle moving on a line

How to set up and solve a related-rates problem

How to estimate a value with a tangent line and tell whether the estimate is too high or too low

How to use L’Hôpital’s rule on limits of the form 0/0 or infinity over infinity

On the exam

About an eighth of the multiple-choice section by our estimate, roughly 5 of the 42 questions, in both the no-calculator and the calculator parts

The free-response section has 6 questions, 2 with a graphing calculator and 4 without; any question that sets a derivative in a real context, such as a rate read from a table or a moving particle, uses the skills in this unit

Key terms

19 terms. Say what each one means, then open the card to check.

  • Instantaneous rate of change

    The derivative f′(a)f'(a): how fast ff is changing at the single input aa. It is the slope of the tangent line to the graph at that point.

    Example. If B(t)B(t) is the number of bees in a hive after tt days and B′(12)=40B'(12) = 40, the hive is gaining about 40 bees per day on day 12.

  • Units of a derivative

    The units of the output divided by the units of the input. A second derivative divides by the input units twice.

    Example. If D(h)D(h) is distance in miles after hh hours, D′(h)D'(h) is in miles per hour and D′′(h)D''(h) is in miles per hour per hour.

  • Average rate of change

    f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}: the change in output per unit of input over a whole interval, which is the slope of the secant line.

    Example. A kite at 20 meters when t = 1 second and at 35 meters when t = 6 seconds rose an average of 3 meters per second.

  • Estimating a derivative from a table

    Divide the change in output by the change in input across the two listed inputs closest to the point, one on each side of it.

    Example. If C(8)=31C(8) = 31 and C(12)=39C(12) = 39, then C′(10)≈39−3112−8=2C'(10) \approx \frac{39 - 31}{12 - 8} = 2.

  • Interpreting a derivative in context

    A full reading names the quantity, the moment, the rate with its units, and whether the quantity is rising or falling.

    Example. If S(t)S(t) is a store's stock in boxes on day tt, S′(3)=−12S'(3) = -12 means that on day 3 the stock is falling by 12 boxes per day.

  • Position function

    A function, such as x(t)x(t) or s(t)s(t), that gives an object's location on a line at time tt, measured from a fixed origin.

    Example. s(t)=4t−t2s(t) = 4t - t^2 meters puts an object 3 meters to the right of the origin at t=1t = 1.

  • Velocity

    The derivative of position, v(t)=x′(t)v(t) = x'(t). Its sign gives the direction: positive for right or up, negative for left or down.

    Example. For x(t)=t2−4tx(t) = t^2 - 4t, v(1)=−2v(1) = -2, so at t=1t = 1 the object moves left at 2 units per second.

  • Speed

    The absolute value of velocity, ∣v(t)∣|v(t)|. Speed has no direction and is never negative.

    Example. Velocities of 5 and -5 meters per second both mean a speed of 5 meters per second.

  • Acceleration

    The derivative of velocity, a(t)=v′(t)=x′′(t)a(t) = v'(t) = x''(t): how fast the velocity is changing.

    Example. If v(t)=3t2v(t) = 3t^2 feet per second, then a(2)=12a(2) = 12 feet per second per second.

  • Speeding up and slowing down

    Speed increases when velocity and acceleration have the same sign, and decreases when their signs are opposite.

    Example. An object with velocity -3 and acceleration -1 is speeding up, even though its acceleration is negative.

  • Change of direction

    A particle changes direction where its velocity changes sign, not at every point where its velocity is 0.

    Example. v(t)=(t−2)2v(t) = (t - 2)^2 is 0 at t=2t = 2 but never negative, so the particle does not turn around there.

  • At rest

    A particle is at rest at an instant when its velocity is 0.

    Example. For v(t)=t2−9v(t) = t^2 - 9 with t≥0t \geq 0, the particle is at rest only at t=3t = 3.

  • Differentiating with respect to time

    Every changing quantity is a function of tt, so the chain rule gives each one its own rate factor, such as dxdt\frac{dx}{dt}.

    Example. Differentiating x2+y2=100x^2 + y^2 = 100 with respect to tt gives 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.

  • Values at the instant

    Numbers that are true only at the moment asked about. Substitute them after differentiating, or a changing quantity is treated as a constant.

    Example. If a side is 4 cm at one moment, keep it as ss, differentiate, and only then put in s=4s = 4.

  • Local linear approximation

    Estimating f(x)f(x) near x=ax = a with the tangent line L(x)=f(a)+f′(a)(x−a)L(x) = f(a) + f'(a)(x - a).

    Example. With g(1)=5g(1) = 5 and g′(1)=2g'(1) = 2, g(1.1)≈5+2(0.1)=5.2g(1.1) \approx 5 + 2(0.1) = 5.2.

  • Overestimate or underestimate

    A tangent line lies below a concave-up graph, so it underestimates, and above a concave-down graph, so it overestimates.

    Example. x\sqrt{x} is concave down, so its tangent line at x=9x = 9 gives a value slightly above 9.5\sqrt{9.5}.

  • Indeterminate form

    A limit that substitutes to 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. The form alone does not decide the value; more work is needed.

    Example. lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2} substitutes to 00\frac{0}{0}, yet the limit is 4.

  • L’Hôpital’s rule

    If f(x)g(x)\frac{f(x)}{g(x)} has the form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}, with ff and gg differentiable and g′(x)≠0g'(x) \neq 0 near the point, its limit equals lim⁡f′(x)g′(x)\lim \frac{f'(x)}{g'(x)} when that limit exists or is infinite.

    Example. lim⁡x→0ex−1x\lim_{x \to 0} \frac{e^x - 1}{x} has the form 00\frac{0}{0}, and ex1→1\frac{e^x}{1} \to 1, so the limit is 1.

In this unit

  1. Rates in context and motion
  2. Related rates
  3. Tangent-line estimates and L’Hôpital’s rule
  4. Rates in context from a table
  5. Summary
  6. Mastery quiz