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AP Calculus BC FRQ Practice Guide

How to attack free-response questions and earn easy partial credit.

⏱️ ~3-4 hours of focused work∬ AP Calculus BC

The AP Exam gives you 6 free-response questions. Each tests predictable patterns the College Board has reused for 20+ years. Master these templates, and you'll know exactly what to write when you see an FRQ on exam day.

The 6 FRQ archetypes

Archetype 1: Rate In / Rate Out (AB-shared)

Pattern: Water flows into/out of a tank or container. You're given a rate function and asked to find accumulated water, average rate, or when the container overflows.

Template:

  1. Set up accumulation integral: amount=∫ab(rate in−rate out) dt\text{amount} = \int_a^b (\text{rate in} - \text{rate out}) \, dt.
  2. Evaluate the integral.
  3. Interpret units (gallons, liters, cubic feet).

Worked Example:

Water flows into a tank at a rate of (20+t)(20 + t) gallons/hour, and drains at 1515 gallons/hour, for 0≤t≤100 \leq t \leq 10. The tank initially holds 5050 gallons.

(a) Find the amount of water in the tank at t=10t = 10.

Solution: Amount=50+∫010[(20+t)−15] dt=50+∫010(5+t) dt\text{Amount} = 50 + \int_0^{10} [(20 + t) - 15] \, dt = 50 + \int_0^{10} (5 + t) \, dt =50+[5t+t22]010=50+(50+50)=150 gallons= 50 + \left[ 5t + \frac{t^2}{2} \right]_0^{10} = 50 + (50 + 50) = 150 \text{ gallons}

(b) Find the average rate of water flow (net) over [0,10][0, 10].

Solution: Average rate=110∫010(5+t) dt=10010=10 gallons/hour\text{Average rate} = \frac{1}{10} \int_0^{10} (5 + t) \, dt = \frac{100}{10} = 10 \text{ gallons/hour}


Archetype 2: Particle Motion (AB-shared + BC parametric/vector)

Pattern: A particle moves along a curve. Given position, velocity, or parametric equations, find speed, direction, arc length, or displacement.

Template:

  • Speed: (dx/dt)2+(dy/dt)2\sqrt{(dx/dt)^2 + (dy/dt)^2}.
  • Arc length: ∫ab(dx/dt)2+(dy/dt)2 dt\int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt.
  • Displacement: ∫abv(t) dt\int_a^b v(t) \, dt (can be negative).
  • Distance traveled: ∫ab∣v(t)∣ dt\int_a^b |v(t)| \, dt (always positive).

Worked Example (BC):

A particle moves along a curve with x(t)=t2x(t) = t^2, y(t)=t3y(t) = t^3, for 0≤t≤20 \leq t \leq 2.

(a) Find the speed of the particle at t=1t = 1.

Solution: dxdt=2t,dydt=3t2\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 Speed=(2⋅1)2+(3⋅12)2=4+9=13\text{Speed} = \sqrt{(2 \cdot 1)^2 + (3 \cdot 1^2)^2} = \sqrt{4 + 9} = \sqrt{13}

(b) Find the arc length of the curve from t=0t = 0 to t=2t = 2.

Solution: L=∫02(2t)2+(3t2)2 dt=∫024t2+9t4 dt=∫02t4+9t2 dtL = \int_0^2 \sqrt{(2t)^2 + (3t^2)^2} \, dt = \int_0^2 \sqrt{4t^2 + 9t^4} \, dt = \int_0^2 t\sqrt{4 + 9t^2} \, dt Using u=4+9t2u = 4 + 9t^2, du=18t dtdu = 18t \, dt: L=118∫440u du=118⋅23u3/2∣440=127[403/2−8]≈9.07L = \frac{1}{18} \int_4^{40} \sqrt{u} \, du = \frac{1}{18} \cdot \frac{2}{3} u^{3/2} \big|_4^{40} = \frac{1}{27} [40^{3/2} - 8] \approx 9.07


Archetype 3: Polar Curves (BC-only)

Pattern: A curve is defined in polar form r=f(θ)r = f(\theta). Find area enclosed, area between curves, or arc length.

Template:

  • Area: A=12∫αβr2 dθA = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta.
  • Area between: A=12∫αβ(router2−rinner2) dθA = \frac{1}{2} \int_{\alpha}^{\beta} (r_{\text{outer}}^2 - r_{\text{inner}}^2) \, d\theta.

Worked Example:

Find the area enclosed by r=2+2cos⁡θr = 2 + 2\cos \theta.

Solution: The curve is a cardioid. By symmetry, it completes one loop for 0≤θ≤2π0 \leq \theta \leq 2\pi. A=12∫02π(2+2cos⁡θ)2 dθ=12∫02π(4+8cos⁡θ+4cos⁡2θ) dθA = \frac{1}{2} \int_0^{2\pi} (2 + 2\cos \theta)^2 \, d\theta = \frac{1}{2} \int_0^{2\pi} (4 + 8\cos \theta + 4\cos^2 \theta) \, d\theta Using cos⁡2θ=1+cos⁡2θ2\cos^2 \theta = \frac{1 + \cos 2\theta}{2}: A=12[4θ+8sin⁡θ+2(θ+sin⁡2θ2)]02π=12⋅12π=6πA = \frac{1}{2} \left[ 4\theta + 8\sin \theta + 2(\theta + \frac{\sin 2\theta}{2}) \right]_0^{2\pi} = \frac{1}{2} \cdot 12\pi = 6\pi


Archetype 4: Area and Volume (AB-shared)

Pattern: Find area between curves or volume by rotating about an axis.

Template:

  • Area between: ∫ab(top−bottom) dx\int_a^b (\text{top} - \text{bottom}) \, dx or ∫cd(right−left) dy\int_c^d (\text{right} - \text{left}) \, dy.
  • Volume (disk): V=π∫ab[R(x)]2 dxV = \pi \int_a^b [R(x)]^2 \, dx.
  • Volume (washer): V=π∫ab([R(x)]2−[r(x)]2) dxV = \pi \int_a^b ([R(x)]^2 - [r(x)]^2) \, dx.

💡 Tip: Always identify the axis of rotation. If rotating about y=ky = k (not the xx-axis), adjust the radius: R=∣f(x)−k∣R = |f(x) - k|.


Archetype 5: Series and Convergence (BC-only)

Pattern: Determine if a series converges, find its sum, estimate error, or find the interval of convergence for a power series.

Template for convergence:

  1. Geometric: Does it have the form ∑arn\sum ar^n? If ∣r∣<1|r| < 1, it converges to a1−r\frac{a}{1-r}.
  2. pp-series: Does it have the form ∑1np\sum \frac{1}{n^p}? Converges iff p>1p > 1.
  3. Ratio test: Compute L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. If L<1L < 1, converges; L>1L > 1, diverges; L=1L = 1, test inconclusive.
  4. Alternating: If (−1)nan(-1)^n a_n where ana_n is decreasing and lim⁡an=0\lim a_n = 0, then converges (but may not converge absolutely).

Template for Lagrange error: ∣Rn(x)∣≤M∣x−a∣n+1(n+1)!|R_n(x)| \leq \frac{M |x - a|^{n+1}}{(n+1)!} where M=max⁡∣f(n+1)(t)∣M = \max |f^{(n+1)}(t)| on the interval.

Worked Example:

(a) Determine if ∑n=1∞n2n\sum_{n=1}^{\infty} \frac{n}{2^n} converges. If so, justify your test.

Solution: Use the ratio test. lim⁡n→∞∣an+1an∣=lim⁡n→∞(n+1)/2n+1n/2n=lim⁡n→∞n+12n=12<1\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)/2^{n+1}}{n/2^n} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} < 1 By the Ratio Test, the series converges.

(b) Find the interval of convergence for ∑n=0∞(x−2)nn+1\sum_{n=0}^{\infty} \frac{(x-2)^n}{n+1}.

Solution: Use the ratio test on (x−2)n(x - 2)^n. lim⁡n→∞∣(x−2)n+1n+2⋅n+1(x−2)n∣=∣x−2∣lim⁡n→∞n+1n+2=∣x−2∣\lim_{n \to \infty} \left| \frac{(x-2)^{n+1}}{n+2} \cdot \frac{n+1}{(x-2)^n} \right| = |x - 2| \lim_{n \to \infty} \frac{n+1}{n+2} = |x - 2| Converges when ∣x−2∣<1|x - 2| < 1, so 1<x<31 < x < 3 (open interval).

Test x=1x = 1: ∑(−1)nn+1\sum \frac{(-1)^n}{n+1} — alternating series, decreasing to 0, converges.

Test x=3x = 3: ∑1n+1=∑1n\sum \frac{1}{n+1} = \sum \frac{1}{n} (harmonic) — diverges.

IOC: [1,3)[1, 3).


Archetype 6: Taylor Polynomials and Power Series (BC-only)

Pattern: Write a Taylor polynomial about x=ax = a, find Lagrange error bound, or use a known series to approximate a value.

Template: Pn(x)=∑k=0nf(k)(a)k!(x−a)kP_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x - a)^k

Worked Example:

Write the 4th-degree Taylor polynomial for f(x)=exf(x) = e^x centered at x=0x = 0. Use it to approximate e0.1e^{0.1} and find a bound on the error.

Solution: All derivatives of exe^x are exe^x, so f(k)(0)=1f^{(k)}(0) = 1 for all kk. P4(x)=1+x+x22+x36+x424P_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} P4(0.1)=1+0.1+0.005+0.000167+0.0000042≈1.1052P_4(0.1) = 1 + 0.1 + 0.005 + 0.000167 + 0.0000042 \approx 1.1052

Error bound: ∣R4(0.1)∣≤M(0.1)55!|R_4(0.1)| \leq \frac{M(0.1)^5}{5!} where M=max⁡∣et∣M = \max |e^t| on [0,0.1][0, 0.1]. Since et≤e0.1<1.11e^t \leq e^{0.1} < 1.11 on [0,0.1][0, 0.1]: ∣R4(0.1)∣<1.11⋅(0.1)5120≈0.000000925|R_4(0.1)| < \frac{1.11 \cdot (0.1)^5}{120} \approx 0.000000925

⚠️ FRQ trap: Always compute the maximum of the derivative explicitly. Use calculus or endpoint analysis; don't just guess a bound.


Scoring rubric checklist

Every FRQ rubric demands:

  1. Setup: Clearly state the integral, differential equation, or formula you're using.
  2. Execution: Show all algebraic steps; don't jump to the answer.
  3. Justification: If asked to justify, cite the theorem by name (MVT, IVT, First Derivative Test, Ratio Test, etc.).
  4. Units: Always include them on applied problems.
  5. Answer: Box your final answer or clearly state it in a sentence.

Ready to drill? Pick one archetype, solve 2-3 problems, then move to the next. Browse more FRQ examples →. You've got the patterns — now execute. 🎯

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