AP Calculus AB and BC

AP Calculus Sub Plans With Full Answer Keys

Four AP Calculus sub plans, one for each amount of notice you get, built around interactives a substitute can run without teaching any calculus: drag a point, read a slider, watch a solid take shape. Each one ends with an exit ticket and an answer key with every derivative and integral worked out.

What a calculus sub plan has to do

A sub plan for AP Calculus fails for a specific reason: the substitute almost certainly cannot check a derivative or an integral, so anything that needs grading comes back undone, and anything that needs explaining collapses the moment a student asks a question. A plan that survives has three properties. A student can start it without being taught, it produces written work you can check when you are back, and the adult in the room only has to keep the period moving.

The four plans below meet all three. Each one is built around a single interactive that explains itself on screen, paired with a written task that forces a hand computation, so the screen time has a paper trail. They are ordered by how much notice you have.

No notice at all: the Tangent Line Tracer

You found out at six in the morning. Project or share one link and stop there.

Hand out or project: /interactives/tangent-line-tracer, left on its default curve, f(x) = x squared. Nothing needs to be changed. A point rides the curve with a line touching it at one spot, and a second, empty panel below fills in as the point is dragged.

Drag the point on the top curve until the readout says x = -2. Write down that x-value and the slope number the tool reports for the line touching the curve there.
Do the same at x = -1, then x = 0, then x = 1, then x = 2. Five x-values, five slope readings, written down in a column.
Look at the five pairs of numbers. Compare each slope to two times its own x-value. Write one sentence stating the rule that connects x to the slope.
Without dragging again, use that sentence to predict the slope reading at x = 3. Write the prediction, then drag to x = 3 and check it against the tool.

Exit ticket

Write the rule you found in the form slope = ___, and use only that rule, no dragging, to state what the tool would read at x = -5.

Answer key. The curve is f(x) = x squared, and its derivative by the power rule is:

f(x) = x^2 \quad\Rightarrow\quad f'(x) = 2x

x	Slope reading f'(x)
-2	-4
-1	-2
0	0
1	2
2	4
3	6
-5 (predicted, not dragged)	-10

A few hours' notice: Secant to Tangent

Enough time to write a short handout, still nothing to build.

Hand out or project: /interactives/secant-to-tangent, left on its default curve f(x) = x squared, with a fixed point at x = 0 and a second, movable point starting one unit away. The straight line through both points is the secant line; the slider that pulls the second point toward the first is labeled h.

Before touching the slider, compute the slope of the secant line by hand: it connects the point at x = 0 to the point at x = 1, so use rise over run with f(0) and f(1). Write the number down, then check it against the tool's readout.
Drag h down to 0.5. Write the new secant slope. Repeat at h = 0.25, then h = 0.1, then the smallest value the slider allows.
Write one sentence: as h keeps shrinking, what single number are the secant slopes closing in on?
Pick one other curve from the tool's menu and one other fixed point, then repeat steps 1 through 3 on that new pair.

Exit ticket

State the number the secant slopes were closing in on as h shrank toward zero, and name what that number is called in calculus.

Answer key. With the fixed point at a = 0 and f(x) = x squared, the difference quotient simplifies before you even take a limit:

\frac{f(0+h) - f(0)}{h} = \frac{h^2 - 0}{h} = h

So the secant slope reading equals h exactly at every step: 1 at h = 1, 0.5 at h = 0.5, 0.25 at h = 0.25, and 0.1 at h = 0.1, closing in on 0 as h closes in on 0. That limit is the derivative at a = 0, and the general version confirms the same rule found in the first plan:

f'(a) = \lim_{h \to 0} \frac{(a+h)^2 - a^2}{h} = \lim_{h \to 0} (2a + h) = 2a

A full day's notice: the sliding ladder

A full day is enough to ask for a written product shaped like a free response answer.

Hand out or project: /interactives/related-rates-scene, left on its default scene, a 10-foot ladder leaning against a wall. The base slides away from the wall at a constant 2 feet per second, and dragging the base position moves the top of the ladder down the wall to match.

Before dragging anything, write the relationship between the base distance x and the wall height y for a 10-foot ladder: it is the Pythagorean relationship, x squared plus y squared equals 100.
Differentiate that relationship with respect to time by hand. Both x and y are functions of time, so both need the chain rule. Write the resulting equation on paper.
Drag the base to x = 6 feet. Read the height y the tool displays and the rate it reports for the top of the ladder. Using your equation from step 2, the given rate of 2 feet per second, and the x and y values at this position, solve by hand for the unknown rate, then check it against the tool.
Repeat the drag-and-check at x = 8 feet.
Write a short paragraph, in the shape of a free response answer: state the relation, show the differentiated equation, substitute the known values, and report the final rate with its correct sign and units.

Exit ticket

Is the top of the ladder sliding down faster when the base is 6 feet from the wall, or when the base is 1 foot from the wall? Answer using the formula, not the picture on screen.

Answer key. The relation and its derivative with respect to time:

x^2 + y^2 = 100 \quad\Rightarrow\quad 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}

Base x (ft)	Height y (ft)	dy/dt (ft/s)
1	sqrt(99) is about 9.95	about -0.201
6	8	-1.5
8	6	-8/3, about -2.667

At x = 6 the magnitude of dy/dt is 1.5 feet per second; at x = 1 it is only about 0.201 feet per second. The top of the ladder is sliding down faster when the base is at 6 feet, because the ladder is closer to lying flat and the same horizontal speed forces a bigger vertical response.

Several days, planned absence: Solid of Revolution Builder

With real notice, ask for the fullest written product, due the day you return.

Hand out or project: /interactives/solid-of-revolution-builder, left on its default setup, the region under y equals the square root of x, from x = 0 to x = 4, rotated around the x-axis.

Before touching any control, sketch the flat region by hand on graph paper: the curve, the x-axis, from x = 0 to x = 4.
Write one sentence explaining why this region produces stacked disks rather than washers when it spins: there is a single bounding curve and no inner curve to subtract.
Write the radius of one disk at position x: it equals the height of the curve there.
Write the volume integral for the whole solid, then evaluate it by hand.
Re-select the setup to watch the solid build itself through a short automatic rotation animation, then check its volume readout against the hand answer.
If time allows, switch the tool to its second setup, a region bounded by two curves, and repeat steps 1 through 4, deciding by hand whether the region needs a disk, a washer, or a shell before checking which one the tool selected.

Exit ticket for the day

In your own words, when does a solid need a washer instead of a plain disk?

Written product due the day you return

One full volume computation, shown step by step, for each setup completed. Two setups minimum.

Answer key, first setup. The radius at position x is the square root of x, so the squared radius is simply x:

V = \pi \int_{0}^{4} \left(\sqrt{x}\right)^2\, dx = \pi \int_{0}^{4} x\, dx = \pi \left[\frac{x^2}{2}\right]_{0}^{4} = \pi\left(\frac{16}{2} - 0\right) = 8\pi

Answer key, second setup: the region between y equals x and y equals x squared, from x = 0 to x = 1, about the x-axis. The line sits above the parabola on that interval, so the line gives the outer radius and the parabola gives the inner radius, and this one needs a washer:

V = \pi \int_{0}^{1} \left(x^2 - \left(x^2\right)^2\right) dx = \pi \int_{0}^{1} \left(x^2 - x^4\right) dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_{0}^{1} = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}

A solid needs a washer instead of a plain disk whenever two curves bound the region, so revolving it leaves a hole down the middle: an inner radius, from the second curve, has to be subtracted from the outer radius, the way the second setup's integral does above. The first setup needed only a disk because its region touched the axis directly, with no inner curve to subtract.

What does not work, and why

A worksheet of derivatives with no interactive attached. Nothing stops a student from copying a neighbor's answers line by line, and a substitute who cannot check a derivative cannot catch it.
A video with a note-taking sheet. Students transcribe the video into the sheet without processing any of it, and you get thirty completed sheets and no learning.
Anything that needs an account created during the period. A substitute cannot troubleshoot thirty logins. Every plan above opens with a bare link, on purpose.
A full practice test with no structure. Without a required written product attached to each problem, you get a stack of blank or guessed multiple choice answers and nothing you can use to see what was actually understood.

What to leave in the sub folder before you need it

Write this once, while you are not scrambling, and it works for any absence you did not see coming.

The four links above, printed or bookmarked: tangent-line-tracer, secant-to-tangent, related-rates-scene, and solid-of-revolution-builder, so nobody has to search for them.
One index card per plan stating only the exit ticket question and its one-line answer, so a substitute can confirm a class actually finished without reading any student work.
A single instruction on top of the folder: pick the plan matching how much notice this absence gave, hand out the matching card, and do not attempt to explain any of the math yourself.
Your actual class roster with any students who need a different device or format flagged in advance, since that decision cannot be made at six in the morning by someone who has never met the class.

The reason calculus sub plans are usually bad is not that a good one is hard to design. It is that it gets written in a rush by someone who feels terrible. Build the folder now, and every plan in it will hold a full period without anyone in the room needing to know what a derivative is.

Worked examples

Worked example

The difference quotient at a vertex

For f(x) = x squared with the fixed point at a = 0, find the secant slope for a general h, then find the exact derivative at a = 0 by letting h shrink to 0.

Write the difference quotient with a = 0: the secant slope between (0, f(0)) and (h, f(h)) is (f(0+h) - f(0)) / h.
Substitute f(x) = x squared: f(0+h) = h squared and f(0) = 0, so the quotient is (h squared - 0) / h.
Simplify by dividing out one factor of h: (h squared) / h = h, for any h not equal to 0.
Take the limit as h approaches 0: the secant slope, which equals h exactly, approaches 0. That limit is f'(0) by definition.

f'(0) = 0, matching the power rule result f'(x) = 2x evaluated at x = 0.

Worked example

The sliding ladder at two base positions

A 10-foot ladder leans against a wall, base distance x and wall height y related by x squared plus y squared equals 100. The base slides away from the wall at dx/dt = 2 feet per second. Find dy/dt when x = 6, and again when x = 8.

Differentiate x squared plus y squared equals 100 with respect to time: 2x (dx/dt) + 2y (dy/dt) = 0.
Solve for the unknown rate: dy/dt = -(x/y)(dx/dt).
At x = 6: find y from the original relation, y = square root of (100 minus 36) = square root of 64 = 8. Substitute: dy/dt = -(6/8)(2) = -1.5 feet per second.
At x = 8: find y again, y = square root of (100 minus 64) = square root of 36 = 6. Substitute: dy/dt = -(8/6)(2) = -8/3, about -2.667 feet per second.

dy/dt = -1.5 ft/s at x = 6, and dy/dt = -8/3 ft/s (about -2.667 ft/s) at x = 8. The negative sign means the top is moving down.

Frequently asked questions

What if the class period doesn't allow enough time for every step?

Drop the last list item in whichever plan you are running. Each one is written so the first three or four steps alone still produce a complete written product; the final step is an extension for a class that finishes early, not a requirement.

Do these sub plans work if there is no projector, only student devices?

Yes. Every interactive runs on a single device the same way it runs on a shared screen, and the written steps do not depend on the class watching one display together. A projector only helps with the paired or whole-class prediction step in the tangent line tracer plan, the one for no notice at all.

How much should a substitute grade before the teacher is back?

Nothing. Every exit ticket above has a one-line answer a substitute can check against an index card without understanding the underlying calculus, but full grading of the written product is meant to wait for the teacher, since it requires judging a shown derivation rather than matching a number.

Do these plans work for a BC-only topic?

The four here are all AB and BC content, chosen because every AP Calculus student has met them. A BC-only topic can use the same shape, one interactive plus a required hand computation plus a one-line exit ticket, even without a matching guide written for it yet.