Source: AP Calculus BC, Section I (Parts A and B)
Difficulty: Advanced | Prerequisites: familiarity with derivatives and integration, comfort with geometric series and basic convergence ideas. Parametric and polar sections assume you know how to differentiate and integrate in Cartesian coordinates.
Series, differential equations, and parametric/polar curves are the topics that distinguish BC from AB. They make up a significant portion of the exam, and they tend to be the areas where students lose marks because the material feels less intuitive than straightforward differentiation and integration.
Series questions test whether you can recognise known series, determine convergence, find the radius of convergence, and construct Taylor polynomials. Differential equations test modelling (especially logistic growth) and numerical approximation via Euler's method. Parametric and polar questions test your ability to work in coordinate systems other than Cartesian. All three areas reward pattern recognition and formula fluency.
Memmorise the Maclaurin series for sin(x), cos(x), e^x, and 1/(1 - x). Know how to apply the ratio test for convergence and for finding the radius of convergence. Understand the logistic equation form dy/dt = ky(1 - y). Practise Euler's method by hand. For parametric curves, a particle is at rest when both dx/dt and dy/dt are zero simultaneously. For polar curves, know the tangent slope formula and the area formula.
Maclaurin series
A Taylor series centred at x = 0. The Maclaurin series of f(x) is the sum from n = 0 to infinity of [f^(n)(0) / n!] x^n. In simple terms, it is the polynomial that best approximates f near zero, extended to infinitely many terms.
Taylor polynomial (degree n)
A finite partial sum of the Taylor series, giving an approximation of f near the centre point. The third-degree Taylor polynomial about x = a is f(a) + f'(a)(x - a) + [f''(a)/2!](x - a)^2 + [f'''(a)/3!](x - a)^3.
Radius of convergence
The distance R from the centre of a power series within which the series converges. If the series is centred at c, it converges for |x - c| < R and diverges for |x - c| > R. Think of it as: how far from the centre the series still works.
Ratio test
Compute L = lim |a_(n+1) / a_n| as n goes to infinity. If L < 1, the series converges absolutely. If L > 1, it diverges. If L = 1, the test is inconclusive.
p-series
The series sum of 1/n^p. Converges when p > 1, diverges when p <= 1. In simple terms, the higher the power on n in the denominator, the faster the terms shrink, and p > 1 is fast enough.
Geometric series
The series sum of ar^n. Converges when |r| < 1, with sum a / (1 - r). Diverges when |r| >= 1.
Logistic differential equation
The equation dy/dt = ky(1 - y/L), or in the normalised form dy/dt = ky(1 - y). It models growth that slows as the population approaches a carrying capacity. The key features: the rate depends on y itself and on how far y is from the limit.
Euler's method
A numerical method for approximating solutions to differential equations. Starting from an initial point, take a step: y_new = y_old + (step size)(dy/dx evaluated at the old point). Repeat. Think of it as: walking along the tangent line in small steps.
Parametric equations
A pair of equations x(t) and y(t) that define a curve by tracing out points as t varies. A particle is at rest when both dx/dt = 0 and dy/dt = 0 at the same value of t.
Polar curve
A curve defined by r = f(theta), where r is the distance from the origin and theta is the angle. Area and tangent slopes use their own formulas, not the Cartesian ones.
The Maclaurin series for 1/(1 - x) is sum of x^n for n = 0 to infinity, valid for |x| < 1.
Replacing x with -x/4: sum of (-x/4)^n is a geometric series with ratio -x/4. This is 1 / (1 + x/4) = 4 / (4 + x). Evaluating at x = 3: sum = 4/7.
The series 1 - x^2/3! + x^4/5! - x^6/7! + ... with general term (-1)^n x^(2n) / (2n + 1)! is sin(x) / x. Recognise it by comparing to the Maclaurin series of sin(x) = x - x^3/3! + x^5/5! - ... and dividing every term by x.
Ratio test: for sum of 8^n / n!, the ratio a_(n+1)/a_n = 8/(n+1), which goes to 0. Converges.
Factorial vs polynomial growth: sum of n! / n^100 diverges because factorial growth dominates any polynomial.
Partial fraction / telescoping: sum of (n+1) / [n(n+2)(n+3)] can be decomposed and may telescope. Check whether partial sums have a finite limit.
p-series: sum of 1/n^(2p) converges when 2p > 1, i.e. p > 1/2.
Geometric series: sum of (p/2)^n converges when |p/2| < 1, i.e. |p| < 2.
To have both converge simultaneously: 1/2 < p < 2.
For the series sum of (x - 4)^(2n) / 3^n, treat w = (x - 4)^2 to get a geometric series in w/3. Converges when |w| < 3, i.e. |(x - 4)^2| < 3, i.e. |x - 4| < sqrt(3). The radius of convergence is sqrt(3).
If a power series sum of a_n (x - 3)^n converges at x = 5 (distance 2 from centre), the radius is at least 2. It must converge at x = 2 (distance 1 from centre, which is less than 2).
If sum of a_n converges with a_n > 0, then sum of a_n / n also converges (since a_n / n <= a_n for n >= 1, by comparison test).
It is not necessarily true that |a_(n+1)/a_n| goes to 0, or that |a_n| < 1 for all n, or that the sum equals 0, or that sum of n a_n diverges.
The third-degree Taylor polynomial for f about x = 3, given f(3) = 2, f'(3) = -1, f''(3) = 6, f'''(3) = 12:
P_3(x) = 2 + (-1)(x - 3) + (6/2)(x - 3)^2 + (12/6)(x - 3)^3 = 2 - (x - 3) + 3(x - 3)^2 + 2(x - 3)^3.
Remember to divide each derivative by the corresponding factorial.
The standard form is dy/dt = ky(1 - y) or dy/dt = ky(1 - y/L).
The rate depends on y and on how far y is from the carrying capacity.
dy/dt = kt or dy/dt = ky are not logistic (they are linear or exponential growth).
dy/dt = kt(1 - t) has no y on the right side, so it is not a DE about y.
dy/dt = ky(1 - t) depends on t, not on how far y is from a limit.
Population growing at a constant rate dP/dt = 200 is linear growth, not exponential or logistic.
Given dy/dx = x - y, f(1) = 3, step size 0.5, two steps:
At x = 1: dy/dx = 1 - 3 = -2. New y = 3 + 0.5(-2) = 2. New x = 1.5.
At x = 1.5: dy/dx = 1.5 - 2 = -0.5. New y = 2 + 0.5(-0.5) = 1.75. New x = 2.
Approximation: f(2) is approximately 1.75. The answer 7/4 = 1.75.
A particle at position (x(t), y(t)) is at rest when both x'(t) = 0 and y'(t) = 0.
For x(t) = t^3 - 3t^2 and y(t) = 12t - 3t^2: x'(t) = 3t^2 - 6t = 3t(t - 2), y'(t) = 12 - 6t = 6(2 - t).
x'(t) = 0 at t = 0 or t = 2. y'(t) = 0 at t = 2. Both zero at t = 2.
At t = 2: x = 8 - 12 = -4, y = 24 - 12 = 12. The particle is at rest at (-4, 12).
For the polar curve r = 1 + 2 sin(theta), the slope dy/dx at theta = 0:
x = r cos(theta), y = r sin(theta). Use dy/dx = [dr/dtheta sin(theta) + r cos(theta)] / [dr/dtheta cos(theta) - r sin(theta)].
dr/dtheta = 2 cos(theta). At theta = 0: dr/dtheta = 2, r = 1, sin(0) = 0, cos(0) = 1.
dy/dx = [2(0) + 1(1)] / [2(1) - 1(0)] = 1/2.
Area enclosed by a polar curve: A = (1/2) integral of r^2 dtheta.
For the region between r = 2 cos(3 theta) and r = 2: set up the integral carefully, identifying where the curves intersect and which is outer/inner over each interval.
When f' is given as a graph consisting of line segments and a semicircle, find f(-5) given f(2) = 1:
f(-5) = f(2) - integral from -5 to 2 of f'(t) dt. Compute the integral as the net signed area of the graph of f' from -5 to 2.
The line segments and semicircle contribute geometric areas (triangle, semicircle). If the net area from -5 to 2 is (3 - 2 pi), then f(-5) = 1 - (3 - 2 pi) = 2 pi - 2.
If h(x) = h(2 - x) for all x, the function is symmetric about x = 1.
h'(1) = 0 (the derivative at the axis of symmetry is zero).
Differentiating h(x) = h(2 - x) gives h'(x) = -h'(2 - x), so h'(0) = -h'(2), meaning h'(0) and h'(2) are equal in magnitude but opposite in sign, not necessarily equal.
The integral from 0 to 2 of h(x) dx need not be positive; it depends on whether h takes negative values.
If a(t) = (t + 3) / sqrt(t^3 + 1) and v(0) = 5, then v(3) = 5 + integral from 0 to 3 of a(t) dt. Use a calculator to evaluate the integral; the result is approximately 11.710.
T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k\text{Ratio test: } L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\text{Geometric series: } \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}, \quad |r| < 1\text{Logistic: } \frac{dy}{dt} = ky(1 - y)\text{Euler step: } y_{n+1} = y_n + h \cdot f(x_n, y_n)\frac{dy}{dx}\bigg|_{\text{polar}} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}\text{Polar area: } A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\, d\thetaTaylor series are how calculators and computers evaluate trig and exponential functions: they compute a finite number of terms of the series to the required precision. Logistic differential equations model population growth in ecology, the spread of diseases in epidemiology, and the adoption of new technologies in marketing. Euler's method is a basic numerical solver used in engineering simulations when no closed-form solution exists.
Students often think a series converges just because its terms approach zero. The harmonic series (sum of 1/n) has terms going to zero but diverges. The terms going to zero is necessary but not sufficient.
When building a Taylor polynomial, students frequently forget to divide by the factorial. The coefficient of (x - a)^n is f^(n)(a) / n!, not f^(n)(a).
Students confuse the radius of convergence with the interval of convergence. The radius is a number R. The interval is (c - R, c + R), possibly including one or both endpoints, which must be checked separately.
In Euler's method, students sometimes use the new y to evaluate the slope at the same step. You evaluate dy/dx at the old point, then use that slope to compute the new y.
Recognising a known Maclaurin series (sin, cos, e^x, 1/(1-x)) from its written-out terms is tested regularly. Practise spotting these patterns.
The ratio test is the most commonly tested convergence tool on BC. Know it by reflex.
Taylor polynomial questions almost always involve computing P_3 or P_4 at a given centre point. The factorial division is the step where marks are lost.
Euler's method with two steps is a favourite format. Work these by hand until the process is automatic.
Parametric "particle at rest" is a classic: set both velocity components to zero and solve simultaneously.
Polar tangent slope at theta = 0 simplifies nicely because sin(0) = 0 and cos(0) = 1. The exam exploits this.
True or false: the Maclaurin series for e^x is sum of x^n / n! from n = 0 to infinity.
Fill in the blank: a geometric series with ratio r converges when |r| < ____.
True or false: in Euler's method, you evaluate dy/dx at the new point to find the slope.
Fill in the blank: a particle described by parametric equations is at rest when both ____ and ____ equal zero.
True or false: the p-series sum of 1/n^(1/2) converges.
Answers: 1. True. 2. 1. 3. False (you evaluate at the old point). 4. dx/dt and dy/dt. 5. False (p = 1/2 < 1, so it diverges).
Q: The Maclaurin series for f(x) is sum from n = 0 to infinity of (-x/4)^n. What is f(3)?
A: 4/7. This is a geometric series with ratio -3/4. Sum = 1 / (1 + 3/4) = 1 / (7/4) = 4/7. Answer: (C).
Q: Which of the following series converge? I. sum of 8^n / n!, II. sum of n! / n^100, III. sum of (n+1) / [n(n+2)(n+3)].
A: I and III only. Series I converges by the ratio test (ratio goes to 0). Series II diverges (factorial growth dominates). Series III converges by partial fractions and comparison. Answer: (D).
Q: What is the radius of convergence of sum of (x - 4)^(2n) / 3^n?
A: sqrt(3). Rewrite as a geometric series in (x - 4)^2 / 3. Converges when |(x - 4)^2| < 3, so |x - 4| < sqrt(3). Answer: (C).
Q: Which is a logistic differential equation? (A) dy/dt = kt, (B) dy/dt = ky, (C) dy/dt = kt(1-t), (D) dy/dt = ky(1-t), (E) dy/dt = ky(1-y).
A: (E). The logistic form has the rate proportional to y times (1 minus y or y/L). Answer: (E).
Q: Using Euler's method with step size 0.5, starting at f(1) = 3 with dy/dx = x - y, approximate f(2).
A: 7/4. Step 1: slope = 1 - 3 = -2, new y = 3 + 0.5(-2) = 2. Step 2: slope = 1.5 - 2 = -0.5, new y = 2 + 0.5(-0.5) = 1.75 = 7/4. Answer: (C).
Q: For x(t) = t^3 - 3t^2 and y(t) = 12t - 3t^2, at which point is the particle at rest?
A: (-4, 12). Both derivatives are zero at t = 2. Evaluating: x(2) = -4, y(2) = 12. Answer: (A).
Q: What is the slope of the tangent to r = 1 + 2 sin(theta) at theta = 0?
A: 1/2. Using the polar slope formula with theta = 0: dy/dx = (0 + 1) / (2 - 0) = 1/2. Answer: (B).
Q: For the power series sum of a_n(x - 3)^n converging at x = 5, which must be true?
A: The series converges at x = 2. Since x = 5 is distance 2 from the centre, the radius is at least 2. x = 2 is distance 1, which is within the radius. Answer: (D).
Q: The third-degree Taylor polynomial for f about x = 3, given f(3) = 2, f'(3) = -1, f''(3) = 6, f'''(3) = 12?
A: 2 - (x - 3) + 3(x - 3)^2 + 2(x - 3)^3. Coefficients: -1/1!, 6/2! = 3, 12/3! = 2. Answer: (E).
Series connect to integration through the integral test (an improper integral determines convergence of the corresponding series). Taylor polynomials are built from derivatives, tying this material back to the differentiation notes. Differential equations use both derivatives and integration (separation of variables, integrating factors). Parametric curves are integrated to find arc length and area, linking back to the integration techniques doc.
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