Must-know sheet

Calculus BC must-know sheet

The rules, theorems, setups and series you should know cold for AP Calculus AB and BC, from limits to Taylor series. The real exam gives you no formula sheet, so all of this has to be memorized; items marked BC-only appear only on the BC exam.

Items marked BC only aren't on the Calculus AB exam.

Limits and continuity

Unit 1

lim (x→c) f(x) = L exists only if both one-sided limits equal L: Check the limit from the left and from the right; if they're different numbers, the limit does not exist. The value f(c) itself doesn't matter, and it can be missing.
Limit laws: sums, differences, products, constant multiples and quotients split apart: If each piece has a limit, the limit of the combination is the same combination of the limits. For a quotient, the limit of the bottom can't be 0.
Try direct substitution first: For polynomials and any function that's continuous at c, the limit is just f(c). If you get a nonzero number over 0, the limit is ∞, −∞ or does not exist, so check the sign on each side of c.
0/0 means rewrite, not "the answer is 0": 0/0 is an indeterminate form. Factor and cancel, multiply by a conjugate, combine fractions, use a trig identity, or use L'Hospital's Rule, then try substitution again.
lim (x→0) sin x / x = 1 and lim (x→0) (1 − cos x)/x = 0: These only hold with x in radians. Adjust to match the form: lim (x→0) sin(5x)/x = 5 · lim sin(5x)/(5x) = 5.
Squeeze Theorem: if g(x) ≤ f(x) ≤ h(x) near c and lim g = lim h = L, then lim f = L: Use it when f is trapped between two simpler functions. Classic case: −x² ≤ x²·sin(1/x) ≤ x², so lim (x→0) x²·sin(1/x) = 0.
Continuous at x = c: f(c) is defined, lim (x→c) f(x) exists, and the two are equal: On a free-response question, check and write all three conditions. If any one fails, f is not continuous at c.
Three types of discontinuity: removable, jump and infinite: Removable is a hole (the limit exists but f(c) is missing or different). Jump means the one-sided limits are different numbers. Infinite means a vertical asymptote.
Where the usual functions are continuous: Polynomials, sin x, cos x and eˣ are continuous everywhere. Rational functions, tan x, ln x and roots are continuous wherever they're defined, so only check zeros of denominators, domain edges and where a piecewise rule switches.
Removing a discontinuity or connecting a piecewise function: To fill a hole, redefine f(c) as the limit at c. To make a piecewise function continuous at a switch point, set the two pieces' limits equal there and solve for the unknown constant.
Vertical asymptote x = c: f(x) → ∞ or −∞ as x → c from at least one side: For a rational function, look for zeros of the denominator that don't cancel with the numerator. A zero that cancels gives a hole, not an asymptote.
Horizontal asymptote y = L: lim (x→∞) f(x) = L or lim (x→−∞) f(x) = L: For rational functions, compare the highest powers: bottom bigger gives y = 0, equal powers give the ratio of leading coefficients, top bigger gives no horizontal asymptote. Check both directions when roots, absolute values or exponentials are involved, since √(x²) = |x|.
Growth order as x → ∞: ln x ≪ xᵖ (p > 0) ≪ bˣ (b > 1): Logs grow slower than any positive power of x, and powers grow slower than exponentials. So ln x / x → 0 and x³/eˣ → 0 as x → ∞.

What a derivative is and when it exists

Unit 2

Average rate of change on [a, b] = (f(b) − f(a))/(b − a): It's the slope of the secant line through the two endpoints. The instantaneous rate at x = a is f′(a), the slope of the tangent line.
f′(a) = lim (h→0) [f(a + h) − f(a)]/h = lim (x→a) [f(x) − f(a)]/(x − a): This is the definition of the derivative. Learn to spot it in disguise: lim (h→0) [√(9 + h) − 3]/h is f′(9) for f(x) = √x, which is 1/6.
Estimating f′(c) from a table: Use the slope between the two data points closest to c on either side: [f(b) − f(a)]/(b − a). Show the subtraction and division, and give units.
Tangent line at x = a: y = f(a) + f′(a)(x − a): You need a point, f(a), and a slope, f′(a). The normal line is perpendicular to it, with slope −1/f′(a) when f′(a) ≠ 0.
Differentiable at c means continuous at c, but not the other way around: If f isn't continuous at c, it can't be differentiable there. A continuous function can still fail to have a derivative at a corner (like |x| at 0), a cusp or a vertical tangent.
Piecewise function differentiable at a switch point: Two things must hold: the pieces meet (continuity), and the pieces' slopes match there (left and right derivatives are equal). Set up both equations to solve for two unknown constants.
Notation: f′(x), y′, dy/dx and d/dx [f(x)] all mean the derivative: d²y/dx² and f″(x) mean the second derivative, and f⁽ⁿ⁾(x) is the nth derivative.

Derivative rules

Units 2, 3

d/dx [c] = 0, d/dx [c·f] = c·f′, d/dx [f ± g] = f′ ± g′: Constants disappear, constant multiples stay in front, and you can differentiate a sum one term at a time.
Power rule: d/dx [xⁿ] = nxⁿ⁻¹: Works for any real power n. Rewrite roots and fractions as powers first, for example √x = x^(1/2) and 1/x³ = x⁻³.
Product rule: (fg)′ = f′g + fg′: Derivative of the first times the second, plus the first times the derivative of the second.
Quotient rule: (f/g)′ = (f′g − fg′)/g²: The order in the numerator matters because of the minus sign. Valid where g(x) ≠ 0.
Chain rule: d/dx [f(g(x))] = f′(g(x)) · g′(x): Differentiate the outside function, keep the inside the same, then multiply by the derivative of the inside. For example, d/dx [sin(x²)] = cos(x²) · 2x.
d/dx sin x = cos x, d/dx cos x = −sin x, d/dx tan x = sec²x: x must be in radians. With the chain rule: d/dx sin(u) = cos(u) · u′.
d/dx cot x = −csc²x, d/dx sec x = sec x tan x, d/dx csc x = −csc x cot x: The "co" functions (cos, cot, csc) all have a minus sign in their derivatives.
d/dx eˣ = eˣ and d/dx ln x = 1/x: With the chain rule: d/dx eᵘ = eᵘ · u′ and d/dx ln u = u′/u. Also d/dx ln|x| = 1/x for x ≠ 0.
d/dx aˣ = aˣ · ln a and d/dx logₐ x = 1/(x ln a): For any base a > 0, a ≠ 1. For example, d/dx 2ˣ = 2ˣ ln 2.
d/dx arcsin x = 1/√(1 − x²), d/dx arccos x = −1/√(1 − x²), d/dx arctan x = 1/(1 + x²): arcsin x is the same as sin⁻¹ x. Use the chain rule for an inside function: d/dx arctan(3x) = 3/(1 + 9x²).
d/dx arccot x = −1/(1 + x²), d/dx arcsec x = 1/(|x|√(x² − 1)), d/dx arccsc x = −1/(|x|√(x² − 1)): The less common three, and rarely tested (some books define arcsec differently and drop the absolute value), so focus on arcsin, arccos and arctan. As with the trig derivatives, the "co" versions get the minus sign.
Inverse function: if f(a) = b, then (f⁻¹)′(b) = 1/f′(a): Find the a that f sends to b (often from a table), then take the reciprocal of f′ there. Graphically, inverse functions have reciprocal slopes at reflected points.
Implicit differentiation: Differentiate both sides with respect to x, multiply every derivative of a y-term by dy/dx (chain rule), then solve for dy/dx. For d²y/dx², differentiate dy/dx again and substitute the dy/dx you already found.
Higher-order derivatives: f″ is the derivative of f′, f‴ is the derivative of f″, and so on. In motion, the second derivative of position is acceleration.

Using derivatives in context

Unit 4

Explaining f′(a) in context: Say it is the rate of change of the quantity, at the given input, in units of output per unit of input. If V(t) is gallons and t is minutes, V′(5) = −3 means that at t = 5 minutes the volume is decreasing at 3 gallons per minute.
Related rates: equation first, differentiate with respect to t, then substitute: Name the variables, write one equation that links them (geometry, similar triangles or the Pythagorean theorem), and differentiate both sides with respect to t. Only plug in the numbers for that instant after you differentiate.
Shape formulas you must bring yourself: Circle: A = πr², C = 2πr. Sphere: V = (4/3)πr³, surface area 4πr². Cylinder: V = πr²h. Cone: V = (1/3)πr²h. Right triangle: a² + b² = c².
Linearization: f(x) ≈ L(x) = f(a) + f′(a)(x − a) for x near a: Use the tangent line to estimate f just beyond a known point. If f is concave up near a (f″ > 0) the tangent line lies below the curve, so the estimate is an underestimate; concave down gives an overestimate.
L'Hospital's Rule: if lim f(x) and lim g(x) are both 0, or both ±∞, then lim f(x)/g(x) = lim f′(x)/g′(x): f and g must be differentiable near the point, and the new limit must exist (or be ±∞). Differentiate the top and bottom separately, not with the quotient rule, and on a free-response question show that the top and bottom each go to 0 (or ∞) first. Only the 0/0 and ∞/∞ forms are tested.

The big theorems, with every condition

Units 1, 5, 6

Intermediate Value Theorem (IVT): If f is continuous on [a, b] and k is any number between f(a) and f(b), then f(c) = k for at least one c in [a, b]. Example: f continuous, f(1) = 2 and f(3) = −1, so f(c) = 0 for some c between 1 and 3.
Extreme Value Theorem (EVT): If f is continuous on a closed interval [a, b], then f has both an absolute maximum and an absolute minimum on [a, b]. Without continuity or without a closed interval, there's no guarantee.
Mean Value Theorem (MVT): If f is continuous on [a, b] and differentiable on (a, b), then f′(c) = [f(b) − f(a)]/(b − a) for at least one c in (a, b). In words: at some moment the instantaneous rate equals the average rate.
Checking the MVT conditions from a stated fact: "f is differentiable" on an interval already gives you continuity there too. A table of values alone never shows continuity, so the problem must tell you.
Fundamental Theorem of Calculus, accumulation form: d/dx ∫ₐˣ f(t) dt = f(x): Holds when f is continuous on an interval containing a and x. If the upper limit is a function u(x), multiply by its derivative: d/dx ∫ from a to u(x) of f(t) dt = f(u(x)) · u′(x). If x is in the lower limit, flip the limits and add a minus sign first.
Fundamental Theorem of Calculus, evaluation form: ∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f: Holds when f is continuous on [a, b]. Read the other way, it says the integral of a rate of change is the total (net) change: ∫ₐᵇ F′(x) dx = F(b) − F(a).
Using a theorem on a free-response question: Name the theorem, show each condition holds for this function on this interval, then state the conclusion with numbers. The conditions are usually where the points are.

Analyzing graphs with derivatives

Unit 5

Critical point: x = c in the domain of f where f′(c) = 0 or f′(c) doesn't exist: Relative maximums and minimums can only happen at critical points. But a critical point isn't always one: x³ has f′(0) = 0 and keeps rising.
f′ > 0 on an interval means f is increasing; f′ < 0 means f is decreasing: Find where f′ is zero or undefined, then test the sign of f′ on each interval between those points.
First Derivative Test: At a critical point c where f is continuous: f′ changes from positive to negative means a relative max, negative to positive means a relative min, and no sign change means neither.
Second Derivative Test: If f′(c) = 0 and f″(c) < 0, f has a relative max at c; if f′(c) = 0 and f″(c) > 0, a relative min. If f″(c) = 0 the test tells you nothing, so use the First Derivative Test.
Concavity: f″ > 0 means concave up (f′ increasing); f″ < 0 means concave down (f′ decreasing): Concave up curves bend like a cup and concave down like a cap. You can also read concavity from whether the graph of f′ is rising or falling.
Point of inflection: where f is continuous and f″ changes sign: f″(c) = 0 by itself isn't enough: x⁴ has f″(0) = 0 but no inflection point. Equivalently, it's where f′ switches between increasing and decreasing.
Candidates Test for absolute extrema on [a, b]: For f continuous on a closed interval, evaluate f at every critical point inside the interval and at both endpoints. The largest value is the absolute max and the smallest is the absolute min; show the table of values as your justification.
One critical point on an interval: If a continuous function has only one critical point on an interval and it's a relative max (or min), it's also the absolute max (or min) on that interval. This is a common way to justify optimization answers.
Reading f from the graph of f′: f increases where f′ is above the axis and decreases where it's below. f has extrema where f′ crosses the axis, inflection points where f′ has a peak or valley, and is concave up where f′ is rising.
Optimization steps: Write the quantity to maximize or minimize, use the constraint to get it in one variable, and find its domain. Find the critical points, justify the max or min with the Candidates Test or a derivative test, and answer the exact question asked, with units.
Horizontal and vertical tangents on implicit curves: With dy/dx written as a fraction, a horizontal tangent is where the numerator is 0 and the denominator isn't; a vertical tangent is where the denominator is 0 and the numerator isn't. The point must also lie on the curve.

Riemann sums and definite integrals

Unit 6

∫ₐᵇ f(x) dx is the net signed area between f and the x-axis: Area above the axis counts as positive and area below counts as negative. Its units are the units of f times the units of x, such as (gallons per hour) × hours = gallons.
Left, right and midpoint Riemann sums: add up height × width for each rectangle: Each rectangle's height is f at the left endpoint, the right endpoint or the midpoint of its subinterval. If the subintervals have different widths, multiply each height by its own width.
Trapezoidal sum: each piece is (1/2)(f(xᵢ₋₁) + f(xᵢ)) · Δxᵢ: With equal widths Δx, T = (Δx/2)[f(x₀) + 2f(x₁) + … + 2f(xₙ₋₁) + f(xₙ)]. It equals the average of the left and right sums.
Over- or underestimate?: f increasing: left sum under, right sum over (reversed when f is decreasing). f concave up: trapezoidal sum over, midpoint sum under (reversed when concave down).
Limit of a Riemann sum: ∫ₐᵇ f(x) dx = lim (n→∞) Σ (k = 1 to n) f(a + kΔx) · Δx, with Δx = (b − a)/n: To turn a sum into an integral, find Δx, the starting value a and the function. For example, lim (n→∞) Σ (k = 1 to n) (1 + 2k/n)² · (2/n) = ∫ from 1 to 3 of x² dx.
Integral properties: ∫ₐᵃ f(x) dx = 0. Swapping the limits flips the sign. ∫ₐᵇ f + ∫ from b to c of f = ∫ from a to c of f. Constants factor out, and the integral of a sum is the sum of the integrals.
Use geometry when the graph is made of simple shapes: Many definite integrals are areas of triangles, rectangles, trapezoids or semicircles. For example, ∫ from −r to r of √(r² − x²) dx is half a circle's area, πr²/2.
Accumulation function g(x) = ∫ₐˣ f(t) dt: g(a) = 0, g′(x) = f(x) and g″(x) = f′(x). So g increases where f > 0, has a relative max where f changes from positive to negative, and is concave up where f is increasing. Find values of g by adding and subtracting areas starting from a.

Antiderivatives and integration techniques

Unit 6

∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, for n ≠ −1: Add 1 to the power and divide by the new power. Always add + C to an indefinite integral.
∫ (1/x) dx = ln|x| + C: This is the n = −1 case the power rule can't handle. The absolute value matters when x can be negative.
∫ eˣ dx = eˣ + C and ∫ aˣ dx = aˣ/ln a + C: For a linear inside, divide by its slope: ∫ e³ˣ dx = (1/3)e³ˣ + C.
∫ sin x dx = −cos x + C and ∫ cos x dx = sin x + C: Check signs by differentiating your answer: the derivative of −cos x is sin x.
∫ sec²x dx = tan x + C, ∫ csc²x dx = −cot x + C: These come from reversing the derivatives of tan x and cot x.
∫ sec x tan x dx = sec x + C, ∫ csc x cot x dx = −csc x + C: These come from reversing the derivatives of sec x and csc x.
∫ tan x dx = −ln|cos x| + C = ln|sec x| + C, and ∫ cot x dx = ln|sin x| + C: Rewrite tan x as sin x / cos x and use u = cos x.
∫ 1/(1 + x²) dx = arctan x + C and ∫ 1/√(1 − x²) dx = arcsin x + C: With a constant a > 0: ∫ 1/(a² + x²) dx = (1/a) arctan(x/a) + C and ∫ 1/√(a² − x²) dx = arcsin(x/a) + C.
u-substitution reverses the chain rule: Pick an inside expression u whose derivative also appears (up to a constant), write du = u′(x) dx, and rewrite the whole integral in u. For a definite integral, change the limits to u(a) and u(b), or switch back to x before plugging in.
Long division first when the top's degree ≥ the bottom's: For example, (x² + 1)/(x + 1) = x − 1 + 2/(x + 1), which integrates to x²/2 − x + 2 ln|x + 1| + C.
Complete the square to reach an arctan: ∫ 1/(x² + 2x + 5) dx = ∫ 1/((x + 1)² + 4) dx = (1/2) arctan((x + 1)/2) + C.
Integration by parts: ∫ u dv = uv − ∫ v duBC only: Pick u to be the part that gets simpler when differentiated (logs and powers of x are good choices), and dv to be something you can integrate. Examples: ∫ x eˣ dx = x eˣ − eˣ + C and ∫ ln x dx = x ln x − x + C.
Partial fractions with distinct linear factorsBC only: Split the fraction into A/(x − a) + B/(x − b) and solve for A and B, then each piece integrates to a log. For example, 1/((x − 1)(x + 2)) = (1/3)/(x − 1) − (1/3)/(x + 2).
Improper integral: ∫ from a to ∞ of f(x) dx = lim (b→∞) ∫ₐᵇ f(x) dxBC only: Also use a limit when f blows up at a point in [a, b]: replace that point with a variable and approach it. If the limit is a finite number the integral converges; otherwise it diverges. Write the limit notation, since it's often worth a point.
∫ from 1 to ∞ of 1/xᵖ dx converges only when p > 1; ∫ from 0 to 1 of 1/xᵖ dx converges only when p < 1BC only: When p > 1, the first one equals 1/(p − 1). With p = 1, both diverge, since ln x is unbounded at each end.

Motion along a line

Units 4, 8

v(t) = x′(t) and a(t) = v′(t) = x″(t): Position x(t), velocity v(t) and acceleration a(t). Going the other way, velocity is the integral of acceleration, and position is the integral of velocity.
Speed = |v(t)|: Velocity has a sign that shows direction; speed doesn't. Positive velocity means moving right (or up), negative means left (or down), and v = 0 means at rest.
Changes direction where v(t) changes sign: v(t) = 0 isn't enough by itself; check that the sign of v really flips there.
Speeding up when v and a have the same sign; slowing down when they have opposite signs: A negative acceleration doesn't automatically mean slowing down. If v < 0 and a < 0, the particle is speeding up.
Displacement = ∫ₐᵇ v(t) dt; total distance traveled = ∫ₐᵇ |v(t)| dt: Displacement is the net change in position and can be negative. Distance counts every move as positive, so split at sign changes or integrate |v| on a calculator.
Position later = position now + integral of velocity: x(b) = x(a) + ∫ₐᵇ v(t) dt: The same pattern works for velocity: v(b) = v(a) + ∫ₐᵇ a(t) dt.
Average velocity = [x(b) − x(a)]/(b − a) = (1/(b − a)) ∫ₐᵇ v(t) dt: Average acceleration on [a, b] is [v(b) − v(a)]/(b − a).
Farthest left or right: Use the Candidates Test on position: compare x(t) at the endpoints and at every time where v changes sign.

Applications of integrals: accumulation, area and volume

Unit 8

Average value of f on [a, b] = (1/(b − a)) ∫ₐᵇ f(x) dx: This is the height of the rectangle with the same area. Don't confuse it with the average rate of change, [f(b) − f(a)]/(b − a).
Amount at time b = amount at time a + ∫ₐᵇ (rate in − rate out) dt: The integral of a rate gives the net change in the amount. Give units: a rate in people per hour integrated over hours gives people.
When is the amount biggest or smallest?: The amount increases while rate in > rate out and decreases while rate in < rate out. Use the Candidates Test with the endpoints and the times where the two rates are equal and the difference changes sign.
Area between curves in x: ∫ₐᵇ (top − bottom) dx: The limits are the x-values where the region starts and ends, often intersection points.
Area between curves in y: ∫ from c to d of (right − left) dy: Use this when the region is easier to describe left-to-right, with x written as a function of y. The limits are y-values.
Curves that cross: split at each intersection, or integrate |f(x) − g(x)|: Otherwise areas on opposite sides of the crossing cancel. On a calculator, ∫ₐᵇ |f(x) − g(x)| dx handles it in one step.
Volume with known cross sections: V = ∫ₐᵇ A(x) dx: s is the side length lying in the base, usually top − bottom (or right − left for slices perpendicular to the y-axis, integrating in y). A(x) is the area of one slice.
Cross-section areas with side s in the base: Square: s². Rectangle of height h: s·h. Equilateral triangle: (√3/4)s². Isosceles right triangle with a leg in the base: (1/2)s². Isosceles right triangle with the hypotenuse in the base: (1/4)s². Semicircle with diameter s: (π/8)s².
Disc method: V = π ∫ₐᵇ [R(x)]² dx: R is the radius from the axis of rotation to the edge of the region. Revolve around a horizontal line and integrate in x; revolve around a vertical line and integrate in y, with R written as a function of y.
Washer method: V = π ∫ₐᵇ ([R(x)]² − [r(x)]²) dx: R is the outer radius and r the inner radius, both measured from the axis. Square each radius before subtracting: R² − r², not (R − r)².
Revolving around a line other than an axis: Every radius is a distance from that line. Around y = −2 for a region above the line, R = f(x) + 2. Around y = 5 for a region below it, R = 5 − g(x).
Arc length of y = f(x) on [a, b] = ∫ₐᵇ √(1 + [f′(x)]²) dxBC only: f′ must be continuous on [a, b]. For a curve x = g(y), use ∫ √(1 + [g′(y)]²) dy with y-limits.

Differential equations

Unit 7

Writing a differential equation from words: "The rate of change of y is proportional to y" becomes dy/dt = ky. "Proportional to the difference between 70 and y" becomes dy/dt = k(70 − y).
Verifying a solution: A solution is a function, not a number. Find the derivatives it needs, substitute into both sides, and check that they match.
Slope fields: At each grid point (x, y), draw a short segment with slope dy/dx from the equation. If dy/dx depends only on x, slopes are the same in each column; if only on y, the same in each row; where dy/dx = 0 the segments are flat.
Second derivative from a differential equation: Differentiate dy/dx with respect to x (implicitly, since y depends on x), then substitute the original dy/dx for every dy/dx. The sign of d²y/dx² tells you the concavity of the solution curve.
Tangent line estimate: y(a + Δx) ≈ y(a) + (dy/dx at the point) · Δx: Plug the known point into the differential equation to get the slope, then step along the tangent line.
Separation of variables: Move every y and dy to one side and every x and dx to the other, integrate both sides, and add + C. Use the initial condition to find C right away, then solve for y. Watch the domain of the answer.
dy/dt = ky has the solution y = y₀eᵏᵗ: y₀ is the amount at t = 0. k > 0 means exponential growth and k < 0 means decay. The doubling time (or half-life) is ln 2 / |k|.
dy/dt = k(y − A) has the solution y = A + (y₀ − A)eᵏᵗ: Models such as an object cooling toward room temperature A. Solve it by separation of variables; with k < 0, y approaches A over time.
Euler's method: yₙ₊₁ = yₙ + (dy/dx at (xₙ, yₙ)) · Δx, and xₙ₊₁ = xₙ + ΔxBC only: Take repeated tangent-line steps from the starting point, recomputing the slope at each new point. If solution curves are concave up the estimate is too low; concave down, too high.
Logistic model: dP/dt = kP(1 − P/L)BC only: L is the carrying capacity. With k > 0 and any starting value P(0) > 0, P(t) → L as t → ∞: P rises toward L if it starts below L and falls toward it if it starts above. You read these facts from the equation and the starting value, without solving it.
Logistic growth is fastest when P = L/2BC only: For 0 < P < L, dP/dt is largest at P = L/2, so a solution curve that passes through L/2 has its inflection point there. Factor other forms to find L: dP/dt = 0.2P − 0.001P² = 0.2P(1 − P/200), so L = 200, and dP/dt = kP(L − P) also has carrying capacity L.

Parametric, vector and polar

Unit 9

Parametric slope: dy/dx = (dy/dt)/(dx/dt), where dx/dt ≠ 0BC only: A horizontal tangent is where dy/dt = 0 and dx/dt ≠ 0; a vertical tangent is where dx/dt = 0 and dy/dt ≠ 0.
Parametric second derivative: d²y/dx² = [d/dt (dy/dx)] / (dx/dt)BC only: Differentiate dy/dx with respect to t, then divide by dx/dt again. Forgetting that division is the classic mistake.
Parametric arc length = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dtBC only: This is the length of the path for a ≤ t ≤ b. If the curve retraces itself, it counts the retraced part again, which is what you want for distance traveled.
Vector-valued functions: work one component at a timeBC only: For r(t) = ⟨x(t), y(t)⟩, differentiate or integrate each component separately. An integral of a vector function picks up a constant in each component.
Velocity ⟨x′(t), y′(t)⟩, acceleration ⟨x″(t), y″(t)⟩, speed √((x′(t))² + (y′(t))²)BC only: Speed is the length (magnitude) of the velocity vector. The particle moves right when x′(t) > 0 and up when y′(t) > 0.
Position from velocity: x(b) = x(a) + ∫ₐᵇ x′(t) dt, and the same for yBC only: Do each coordinate separately and give the position as a point or vector. Total distance traveled is ∫ₐᵇ (speed) dt.
Polar and rectangular: x = r cos θ, y = r sin θ, r² = x² + y², tan θ = y/xBC only: Use these to change between the two systems and to set up polar derivatives.
Polar slope: dy/dx = (dy/dθ)/(dx/dθ) = (r′ sin θ + r cos θ)/(r′ cos θ − r sin θ), where r′ = dr/dθBC only: Write y = r(θ) sin θ and x = r(θ) cos θ and use the product rule on each.
Reading dr/dθBC only: If r and dr/dθ have the same sign, the point is moving away from the origin as θ increases; opposite signs mean moving toward the origin. Likewise dy/dθ > 0 means y is increasing.
Area inside one polar curve = (1/2) ∫ from α to β of r² dθBC only: Choose α and β so the curve traces the region exactly once, often between angles where r = 0.
Area between two polar curves = (1/2) ∫ from α to β of (R² − r²) dθBC only: R is the outer curve and r the inner one. Find the limits by solving where the curves meet, and square each radius before subtracting.

Series convergence tests

Unit 10

A series converges when its partial sums Sₙ approach a finite numberBC only: Sₙ = a₁ + a₂ + … + aₙ. If lim Sₙ = S, the series converges to S; otherwise it diverges.
Geometric series: Σ (n = 0 to ∞) arⁿ = a/(1 − r) when |r| < 1BC only: a is the first term and r is the common ratio. With a ≠ 0, it diverges when |r| ≥ 1. Example: Σ (n = 0 to ∞) 3(1/2)ⁿ = 3/(1 − 1/2) = 6.
nth term test: if lim aₙ ≠ 0 (or doesn't exist), Σ aₙ divergesBC only: It can only prove divergence. If lim aₙ = 0, the test says nothing; the harmonic series has terms going to 0 and still diverges.
p-series: Σ 1/nᵖ converges if p > 1 and diverges if p ≤ 1BC only: The harmonic series Σ 1/n (p = 1) diverges. Σ 1/n² converges, and so does Σ 1/(n√n).
Integral testBC only: If aₙ = f(n) where f is positive, continuous and decreasing for x ≥ 1, then Σ aₙ and ∫ from 1 to ∞ of f(x) dx both converge or both diverge. The integral's value is not the sum of the series.
Direct comparison test (positive terms)BC only: If 0 ≤ aₙ ≤ bₙ and Σ bₙ converges, then Σ aₙ converges. If 0 ≤ aₙ ≤ bₙ and Σ aₙ diverges, then Σ bₙ diverges. Smaller than a convergent series, or bigger than a divergent one, is what you need.
Limit comparison test (positive terms)BC only: If aₙ > 0, bₙ > 0 and lim aₙ/bₙ = c with 0 < c < ∞, then Σ aₙ and Σ bₙ both converge or both diverge. Compare with a p-series or geometric series that has the same dominant terms.
Alternating series testBC only: Σ (−1)ⁿbₙ with bₙ > 0 converges if bₙ is decreasing and lim bₙ = 0. It only proves convergence; if it fails, use the nth term test or another test.
Ratio test: L = lim (n→∞) |aₙ₊₁/aₙ|BC only: L < 1: converges absolutely. L > 1 (or ∞): diverges. L = 1: no conclusion, so try another test. Best for factorials and powers like 2ⁿ.
Absolute vs. conditional convergenceBC only: If Σ |aₙ| converges, Σ aₙ converges absolutely (and so converges). If Σ aₙ converges but Σ |aₙ| diverges, it converges conditionally; the alternating harmonic series 1 − 1/2 + 1/3 − … is the standard example.
Choosing a testBC only: Check the nth term first. Then: a common ratio means geometric; 1/nᵖ means p-series; factorials or nth powers suggest the ratio test; alternating signs suggest the alternating series test; a rational or root expression suggests comparing with a p-series.

Taylor and power series, and error bounds

Unit 10

Taylor polynomial: Pₙ(x) = f(a) + f′(a)(x − a) + f″(a)/2! · (x − a)² + … + f⁽ⁿ⁾(a)/n! · (x − a)ⁿBC only: It matches f and its first n derivatives at x = a. Centered at a = 0, it's called a Maclaurin polynomial.
Coefficient of (x − a)ᵏ = f⁽ᵏ⁾(a)/k!BC only: So f⁽ᵏ⁾(a) = k! × (that coefficient). Use this to read derivative values straight from a series.
eˣ = 1 + x + x²/2! + x³/3! + … = Σ (n = 0 to ∞) xⁿ/n!BC only: Converges for every real x.
sin x = x − x³/3! + x⁵/5! − … = Σ (n = 0 to ∞) (−1)ⁿ x²ⁿ⁺¹/(2n + 1)!BC only: Converges for every real x. Only odd powers, since sin x is an odd function.
cos x = 1 − x²/2! + x⁴/4! − … = Σ (n = 0 to ∞) (−1)ⁿ x²ⁿ/(2n)!BC only: Converges for every real x. Only even powers, since cos x is an even function. It's the derivative of the sin x series.
1/(1 − x) = 1 + x + x² + x³ + … = Σ (n = 0 to ∞) xⁿ, for −1 < x < 1BC only: It's a geometric series with ratio x. Rewrite other fractions to match: 1/(2 − x) = (1/2) · 1/(1 − x/2) = Σ xⁿ/2ⁿ⁺¹ for |x| < 2.
Series built from those: ln(1 + x) and arctan xBC only: ln(1 + x) = x − x²/2 + x³/3 − … for −1 < x ≤ 1, from integrating 1/(1 + x). arctan x = x − x³/3 + x⁵/5 − … for −1 ≤ x ≤ 1, from integrating 1/(1 + x²) = 1 − x² + x⁴ − ….
Making new series from known onesBC only: Substitute (replace x with −x² in eˣ to get e^(−x²)), multiply by a power of x, or differentiate or integrate term by term. Differentiating and integrating keep the radius of convergence; a substitution changes the interval to match, so 1/(1 + x²) needs |x²| < 1.
Radius and interval of convergenceBC only: Apply the ratio test with x left in, and solve L < 1 to get |x − a| < R. Then plug in each endpoint and test those two series separately; differentiating or integrating a series can change what happens at the endpoints.
Alternating series error bound: |S − Sₙ| ≤ |aₙ₊₁|BC only: If the series passes the alternating series test, stopping after n terms leaves an error no bigger than the first term you left out.
Lagrange error bound: |f(x) − Pₙ(x)| ≤ M · |x − a|ⁿ⁺¹/(n + 1)!BC only: M is the largest value of |f⁽ⁿ⁺¹⁾(z)| for z between a and x. For sin x and cos x, M = 1 always works.
Which error bound?BC only: Use the alternating series bound when the series alternates with terms shrinking to 0. Use the Lagrange bound for any Taylor polynomial when you can find M for the next derivative.

Exam habits and justification language

Units 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

There's no formula sheet: Every rule on this page has to be memorized, including geometry formulas for related rates and volume.
Calculator: graph, solve, find a derivative at a point, find a definite integral: These are the four calculator skills the exam expects. Write the setup on paper (for example ∫ from 0 to 4 of v(t) dt) before giving the number, since a calculator answer with no setup can lose credit. Write it in math notation, not calculator syntax like fnInt(.
Decimal answers must be accurate to three decimal places: Round or truncate to at least three decimal places. Store long values in the calculator instead of retyping rounded ones, so early rounding doesn't throw off your final answer.
Radians, always: Every trig derivative and integral rule assumes radians, so keep your calculator in radian mode.
Justify with f′ or f″, not with "the graph": Say exactly what changes sign: "f has a relative max at x = 3 because f′ changes from positive to negative there." Name the function you're describing instead of writing "it".
Answer in context, with units: Say what the number means, at what time, and in what units, for example "the water is rising at 2.4 cm per minute at t = 5 minutes."
You don't need to simplify numbers: Unless the question says otherwise, an unsimplified numerical answer like 3² − 2(3) earns the credit. If you do simplify and make an arithmetic slip, you can lose the point, so it's often safer to stop.