No calculator 1 — limits, derivatives, integrals

Fourteen questions, every one of them solvable by hand — which is the whole design. The two no-calculator parts of this exam are 35% and 33.3% of the score, more than two thirds between them, and that fluency is the part that decays fastest when practice happens with a device on the desk.

What is covered. Limits that need factoring before they mean anything; continuity at a point, asked as "what value makes this function continuous"; the product, quotient and chain rules; implicit differentiation; derivatives of trigonometric, exponential and logarithmic functions; u-substitution; both forms of the Fundamental Theorem; the definite integral as accumulation; average value; and a related-rates problem.

The theorems are asked about by their hypotheses, not their statements. Which function fails the Mean Value Theorem on a given interval — the answer turns on differentiability on the open interval, and the absolute value function is the classic failure at zero. What the Intermediate Value Theorem actually requires — continuity on the closed interval, and nothing more.

This matters more than it looks. On the free-response section, a justification earns its point by naming the theorem and checking the condition it needs. A student who knows the statement but not the hypotheses writes an answer that reads correctly and scores nothing.

Two questions require no calculation at all. What does $B^{\mathrm{\prime }}(t)$ represent when $B$ is a bacterial population; what does ${\int }_0^{10}R(t)dt$ represent when $R$ is a flow rate in gallons per minute. The integral of a rate is a total, and its units are the rate's units times time. These are free marks on Section I and they are routinely lost by students who start computing.

And one question separates the average value of a function from the average rate of change — the confusion this course generates most reliably, because both are averages and neither is the other.

How the explanations work. Each names the specific error behind each wrong option: the missing inner derivative, the sign lost in implicit differentiation, the $r^2$ dropped from a related-rates chain, a rate reported where a total was asked for. That last one is worth internalising: in the balloon problem here, one of the options is simply the given rate of change of volume, offered as though it were the rate of change of the radius.

What this does not do. It does not set a multi-part free-response problem, and no multiple-choice bank can. On that half of the exam the answer is the smaller part of the mark — most points are for the work and the justification, written in sentences, with units and with the theorem named. College Board publishes every past free-response question with the scoring guidelines that show exactly where each point is awarded, and reading two of those carefully is worth more than a week of extra practice problems.

Thirteen flashcards carry the derivative and integral rules, both forms of the Fundamental Theorem, and the hypotheses of the named theorems.

• Evaluate limits by hand, including those that require factoring first
• State the condition for continuity at a point and find the value that satisfies it
• Differentiate with the product, quotient and chain rules without dropping the inner derivative
• Use implicit differentiation and keep the sign
• Set up and solve a related-rates problem, differentiating a formula with respect to time
• Apply u-substitution and both forms of the Fundamental Theorem of Calculus
• Interpret a derivative and the integral of a rate in context, with units, without computing
• Distinguish the average value of a function from the average rate of change
• State the hypotheses of the Mean Value and Intermediate Value Theorems, and recognise when they fail

Air is pumped into a spherical balloon so that its volume increases at a constant rate of 12π cubic centimetres per second. At the instant when the radius is 2 centimetres, at what rate is the radius increasing? — Differentiating V = (4/3)πr³ with respect to time gives dV/dt = 4πr²·dr/dt, and the tell that you did it properly is that 4πr² is the surface area. One of the wrong options is simply the given rate of change of volume, offered as though it were the rate of change of the radius.