AP Calculus AB · Free-response skill

Rates in context from a table

The move. From a table of an invented quantity, estimate a derivative with units, say what it means in context, set up and solve a related-rates equation, and make a tangent-line estimate.

How to do it

  1. For a derivative from a table, use the inputs the question names, or else the two listed inputs closest to the point, one on each side. Divide the change in output by the change in input, and give units: output units per input unit.
  2. To say what a derivative means, name the quantity, the moment, the rate with its units, and whether the quantity is rising or falling.
  3. For a related-rates part, write the equation that links the quantities first, differentiate both sides with respect to time, and only then substitute the values for that moment. Give the answer with units.
  4. For a tangent-line estimate, write L(t)=f(a)+f′(a)(t−a)L(t) = f(a) + f'(a)(t - a) with the given values, then evaluate it. Show the expression, not just the number.
  5. Every step can be done by hand, so be ready for this kind of question whether or not a calculator is allowed.

Practice prompt Try it

An algae patch on a pond grows in a roughly circular shape. Its radius r(t)r(t), in meters, is a differentiable, increasing function of time tt, in days. Selected values are in the table.

Radius of the algae patch, r(t)r(t)

tt (days)025710
r(t)r(t) (meters)4.05.26.47.07.6

Use the table and the information given in each part.

  1. Use the values at t=5t = 5 and t=7t = 7 in the table to estimate r′(6)r'(6). Show the numbers you used, and give units.
  2. Tell what your answer to part (a) means for the algae patch.
  3. The area of the patch is A=πr2A = \pi r^2. Suppose that at t=6t = 6, r=6.7r = 6.7 meters and r′(6)=0.3r'(6) = 0.3 meter per day. Write an equation that links dAdt\frac{dA}{dt} and drdt\frac{dr}{dt}, then find dAdt\frac{dA}{dt} at t=6t = 6. Give units.
  4. Suppose r′(10)=0.15r'(10) = 0.15 meter per day. Use the tangent line to the graph of rr at t=10t = 10 to estimate r(10.5)r(10.5).

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 6 points

Model answer, annotated

(a) The closest times on either side of 6 are 5 and 7, so r′(6)≈r(7)−r(5)7−5=7.0−6.42=0.3r'(6) \approx \frac{r(7) - r(5)}{7 - 5} = \frac{7.0 - 6.4}{2} = 0.3 meter per day.

  • Point 1 Uses the bracketing times 5 and 7 and shows the difference quotient before the value.
  • Point 2 States the units as output units per input unit: meters per day.

(b) At t = 6 days, the radius of the algae patch is increasing at about 0.3 meter per day.

  • Point 3 Names the quantity, the moment, the rate with units and the direction.

(c) Differentiating A=πr2A = \pi r^2 with respect to tt gives dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r\frac{dr}{dt}. At t=6t = 6: dAdt=2π(6.7)(0.3)=4.02π≈12.63\frac{dA}{dt} = 2\pi(6.7)(0.3) = 4.02\pi \approx 12.63 square meters per day.

  • Point 4 Writes the related-rates equation before substituting the values for t = 6.
  • Point 5 Substitutes r = 6.7 and dr/dt = 0.3 and gives the rate with square units per day.

(d) The tangent line at t=10t = 10 is L(t)=7.6+0.15(t−10)L(t) = 7.6 + 0.15(t - 10), so r(10.5)≈7.6+0.15(0.5)=7.675r(10.5) \approx 7.6 + 0.15(0.5) = 7.675 meters.

  • Point 6 Builds the tangent line from r(10) = 7.6 and the slope 0.15, then evaluates at 10.5.