Complete Study Guide

AP Calculus BC
2014 Free-Response

Full solutions with concept foundations, key pitfalls, model answers, and TI-Nspire CX II step-by-step calculator guidance.

6 Questions Section II A + B 🔢 Calculator: Q1–2 No Calc: Q3–6
Section II — Part A
Calculator Required · 30 min
1
Grass Clippings — Exponential Decay Model
Calculator Avg Rate of Change Derivatives MVT / IVT Linear Approximation
\( A(t) = 6.687(0.931)^t \) pounds, \(0 \le t \le 30\) days
(a) Average Rate of Change
Core Concept: Average rate of change = \(\dfrac{A(30) - A(0)}{30 - 0}\). This is simply the slope of the secant line. Always include units.
⚠ Common Trap
Don't forget units! This is a rate — answer is in pounds per day. AP graders specifically check units here.
TI-Nspire CX II — Evaluate A(0) and A(30)
1. Open Calculator app (Home → New Doc → Add Calculator)
2. Type: 6.687×0.931^0 → enter
3. Type: 6.687×0.931^30 → enter
4. Compute: (ans - 6.687) / 30
A(0) = 6.687, A(30) ≈ 0.737 → (0.737 − 6.687)/30 ≈ −0.1983
✦ Model Answer
\(\dfrac{A(30)-A(0)}{30-0} = \dfrac{6.687(0.931)^{30} - 6.687}{30} \approx \dfrac{0.737 - 6.687}{30}\)
≈ −0.1983 pounds per day
(b) Find A′(15) and interpret
Core Concept: \(A'(t) = 6.687(0.931)^t \cdot \ln(0.931)\). The derivative of an exponential \(b^t\) is \(b^t \ln b\). This represents the instantaneous rate of change at \(t=15\).
Interpretation Key: Must say "rate", "at t = 15 days", and give sign meaning (decreasing). Don't just state the number.
TI-Nspire CX II — Compute Derivative Numerically
Method 1: Define function, then use d/dx template
Press menu → 4: Calculus → 1: Derivative
Type: d/dx(6.687×0.931^x)|x=15 → enter
Method 2: menu → 4 → 5: Numerical Derivative at a Point, enter function and x=15
A′(15) ≈ −0.1277 pounds per day
✦ Model Answer
\(A'(t) = 6.687(0.931)^t \ln(0.931)\)
\(A'(15) = 6.687(0.931)^{15}\ln(0.931) \approx -0.1277\) pounds per day

Interpretation: At \(t = 15\) days, the amount of grass clippings in the bin is decreasing at a rate of approximately 0.1277 pounds per day.

A′(15) ≈ −0.128 pounds per day (decreasing)
(c) Find t where A(t) = average value of A on [0, 30]
Core Concept: Average value of a function: \(\bar{A} = \dfrac{1}{30}\displaystyle\int_0^{30} A(t)\,dt\). Then solve \(A(t) = \bar{A}\).
⚠ Don't Confuse!
Average rate of change (part a) ≠ average value (part c). Average value requires an integral, not just two endpoint values.
TI-Nspire CX II — Numerical Integral + Solver
1. Compute average value:
menu → 4: Calculus → 3: Integral
Type: (1/30)×∫(6.687×0.931^x, x, 0, 30) → enter
≈ 2.483 pounds
2. Solve A(t) = 2.483:
menu → 3: Algebra → 1: Solve
Type: solve(6.687×0.931^t = 2.483, t) → enter
t ≈ 12.415 days
✦ Model Answer
\(\bar{A} = \dfrac{1}{30}\int_0^{30}6.687(0.931)^t\,dt \approx 2.483\)
Solve: \(6.687(0.931)^t = 2.483 \Rightarrow t \approx 12.415\) days
t ≈ 12.415 days
(d) Linear approximation for t > 30; find when L(t) = 0.5
Core Concept: Linear approximation at \(t = 30\):
\(L(t) = A(30) + A'(30)\cdot(t - 30)\)
Then solve \(L(t) = 0.5\).
Key: You must show the work — compute A(30), A′(30), write L(t), then solve. This is explicitly required.
TI-Nspire CX II — Compute A(30) and A′(30)
6.687×0.931^30 → A(30) ≈ 0.7369
d/dx(6.687×0.931^x)|x=30 → A′(30) ≈ −0.05025
L(t) = 0.7369 + (−0.05025)(t − 30)
Solve L(t) = 0.5:
0.5 = 0.7369 − 0.05025(t−30) → t ≈ 34.714 days
✦ Model Answer
\(A(30) \approx 0.737,\quad A'(30) \approx -0.050\)
\(L(t) = 0.737 - 0.050(t-30)\)
\(0.5 = 0.737 - 0.050(t-30) \Rightarrow t - 30 \approx 4.71 \Rightarrow t \approx 34.711\)
t ≈ 34.711 days
2
Polar Curves — Area, dx/dθ, Distance, Related Rates
Calculator Polar Area Polar Derivatives Related Rates
\(r = 3\) and \(r = 3 - 2\sin(2\theta)\), for \(0 \le \theta \le \pi\)
(a) Area of region R (inside r=3 AND inside r=3−2sin2θ)
Polar Area Formula: \(A = \dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta\)
Strategy: Find intersections first. The curves meet where \(3 = 3 - 2\sin(2\theta)\), i.e., \(\sin(2\theta)=0\), giving \(\theta = 0, \pi/2, \pi\). The shaded region (inside both curves for \(0 \le \theta \le \pi\)) uses the smaller r-value between the two curves at each angle.
⚠ Which curve is "inside"?
For \(0 < \theta < \pi/2\): \(\sin(2\theta) > 0\), so \(3-2\sin(2\theta) < 3\). The inner boundary is \(r = 3-2\sin(2\theta)\). For \(\pi/2 < \theta < \pi\): \(\sin(2\theta) < 0\), so \(3-2\sin(2\theta) > 3\). The inner boundary is \(r = 3\).
TI-Nspire CX II — Polar Area Integration
Region splits at θ = π/2. Use two integrals:
(1/2)×∫((3−2sin(2θ))^2, θ, 0, π/2) + (1/2)×∫(3^2, θ, π/2, π)
≈ 10.370 + 7.069 ≈ 17.28 ... verify with CAS
Or use menu → 4 → 3: Integral for each piece separately
✦ Model Answer
\(A = \frac{1}{2}\int_0^{\pi/2}(3-2\sin 2\theta)^2\,d\theta + \frac{1}{2}\int_{\pi/2}^{\pi}(3)^2\,d\theta\)
\(\approx \frac{1}{2}(20.740) + \frac{1}{2}(9\pi/2) \approx \boxed{17.279}\)
Area ≈ 17.279
(b) Find dx/dθ at θ = π/6 for r = 3−2sin(2θ)
Core Concept: In polar, \(x = r\cos\theta\). So:
\(\dfrac{dx}{d\theta} = \dfrac{dr}{d\theta}\cos\theta - r\sin\theta\)
Key Steps: First find \(r\) and \(dr/d\theta\) at \(\theta = \pi/6\), then substitute.
TI-Nspire CX II
Define: r(θ) := 3−2sin(2θ)
Then evaluate: d/dθ(r(θ)×cos(θ))|θ=π/6
dx/dθ at θ=π/6 ≈ −3.864
✦ Model Answer
\(r = 3-2\sin(2\theta),\quad \frac{dr}{d\theta} = -4\cos(2\theta)\)
At \(\theta = \pi/6\): \(r = 3-2\sin(\pi/3) = 3-\sqrt{3},\quad \frac{dr}{d\theta} = -4\cos(\pi/3) = -2\)
\(\frac{dx}{d\theta} = (-2)\cos(\pi/6) - (3-\sqrt{3})\sin(\pi/6)\)
\(= -\sqrt{3} - \dfrac{3-\sqrt{3}}{2} \approx -3.866\)
dx/dθ ≈ −3.866
(c) Rate of change of distance between curves at θ = π/3
Core Concept: The radial distance between the curves at angle \(\theta\) is \(D(\theta) = 3 - (3-2\sin 2\theta) = 2\sin 2\theta\). Find \(D'(\pi/3)\).
⚠ Setup Check
Only valid for \(0 < \theta < \pi/2\) where \(r=3\) is the outer curve. Confirm this at \(\theta = \pi/3\) ✓
✦ Model Answer
\(D(\theta) = 2\sin(2\theta)\)
\(D'(\theta) = 4\cos(2\theta)\)
\(D'(\pi/3) = 4\cos(2\pi/3) = 4\cdot(-1/2) = -2\)
dD/dθ = −2 at θ = π/3
(d) Find dr/dt at θ = π/6, given dθ/dt = 3
Core Concept: Chain rule: \(\dfrac{dr}{dt} = \dfrac{dr}{d\theta}\cdot\dfrac{d\theta}{dt}\)
✦ Model Answer
\(\frac{dr}{d\theta} = -4\cos(2\theta)\)
At \(\theta = \pi/6\): \(\frac{dr}{d\theta} = -4\cos(\pi/3) = -4\cdot\frac{1}{2} = -2\)
\(\frac{dr}{dt} = (-2)(3) = -6\)
dr/dt = −6
Section II — Part B
No Calculator · 60 min
3
Accumulation Function — Graph of f
No Calculator FTC Increasing/Concavity Quotient Rule Chain Rule
f defined on [−5, 4] as three line segments. Key points: (−5, 2), peak at (0, 2), zero-crossings, (4, −4).
\(g(x) = \displaystyle\int_{-3}^{x} f(t)\,dt\)
📊 Read the Graph First — Critical!

From the graph: f is piecewise linear. Key values to extract:

  • f(−5) = 2, f(−3) = 0, f(0) = 2, f(2) = 0, f(4) = −4
  • f > 0 on (−5, −3) ∪ (−3, 2), f < 0 on (2, 4)
  • f is increasing (slope > 0) on (−5, 0), decreasing on (0, 4)
⚠ Graph Reading Trap
Re-read the graph carefully — g′ = f and g″ = f′. "g is concave down" means f′ < 0, i.e., f is decreasing.
(a) Find g(3)
FTC: \(g(3) = \displaystyle\int_{-3}^{3} f(t)\,dt\). Compute as areas of triangles/trapezoids under the graph.
✦ Model Answer
\(g(3) = \int_{-3}^{3} f(t)\,dt = \int_{-3}^{0}f\,dt + \int_{0}^{2}f\,dt + \int_{2}^{3}f\,dt\)
• On [−3, 0]: triangle with base 3, height 2 → area = \(\frac{1}{2}(3)(2)=3\)
• On [0, 2]: triangle with base 2, height 2 → area = \(\frac{1}{2}(2)(2)=2\)
• On [2, 3]: triangle, f goes 0 to −2 (linear), area = \(\frac{1}{2}(1)(−2)=−1\)
\(g(3) = 3 + 2 + (-1) = 4\)
g(3) = 4
(b) Where is g increasing AND concave down?
Key Translations via FTC:
• g increasing ⟺ g′ > 0 ⟺ f > 0
• g concave down ⟺ g″ < 0 ⟺ f′ < 0 ⟺ f is decreasing
✦ Model Answer
f > 0: (−5, −3) ∪ (−3, 2) → g increasing on (−5, 2)
f decreasing: (0, 4) → g concave down on (0, 4)
Intersection: \((0, 2)\)
g is increasing and concave down on (0, 2)

Reason: g′ = f > 0 and g″ = f′ < 0 on (0, 2) because f is positive and decreasing there.

(c) h(x) = g(x)/(5x), find h′(3)
Quotient Rule: \(h'(x) = \dfrac{g'(x)\cdot 5x - g(x)\cdot 5}{(5x)^2} = \dfrac{5x\cdot f(x) - 5g(x)}{25x^2}\)
Need at x = 3: g(3) = 4 (from part a), g′(3) = f(3). Read f(3) from the graph!
✦ Model Answer
From graph: f(3) = −2 (line from (2,0) to (4,−4), slope = −2, so f(3) = −2)
\(h'(3) = \dfrac{5(3)\cdot f(3) - 5\cdot g(3)}{(5\cdot 3)^2} = \dfrac{15(-2) - 5(4)}{225} = \dfrac{-30-20}{225} = \dfrac{-50}{225}\)
h′(3) = −50/225 = −2/9
(d) p(x) = f(x²−x), slope of tangent at x = −1
Chain Rule: \(p'(x) = f'(x^2-x)\cdot(2x-1)\)
At x = −1: argument = (−1)²−(−1) = 2. So need f′(2).
f′(2) = slope of f at t = 2 (read from graph).
✦ Model Answer
At x = −1: \(x^2 - x = 1+1 = 2\), so we need \(f'(2)\)
From graph: segment from (2, 0) to (4, −4) has slope \(\frac{-4-0}{4-2} = -2\). So \(f'(2) = -2\)
\(p'(-1) = f'(2)\cdot(2(-1)-1) = (-2)(-3) = 6\)
Slope = 6
4
Train A & B — Table Data, Trapezoidal Sum, Related Rates
No Calculator MVT IVT Trapezoidal Sum Related Rates
Train A: \(v_A(t)\) given by table. Train B: \(v_B(t) = t^2 - 5t + 60\). At t=2: A is 300m east, B is 400m north of Origin.
t (min)025812
\(v_A(t)\) (m/min)010040−120−150
(a) Average acceleration on [2, 8]
Average acceleration = \(\dfrac{v_A(8)-v_A(2)}{8-2}\). Units: m/min².
✦ Model Answer
\(\dfrac{v_A(8)-v_A(2)}{8-2} = \dfrac{-120-100}{6} = \dfrac{-220}{6} = -\dfrac{110}{3}\)
−110/3 meters per minute per minute
(b) Is v_A = −100 m/min somewhere on (5, 8)?
IVT: Since \(v_A\) is differentiable (hence continuous), and \(v_A(5) = 40 > -100 > -120 = v_A(8)\), by the IVT there exists some c ∈ (5, 8) with \(v_A(c) = -100\).
⚠ Must Cite IVT by Name
You must explicitly say "by the Intermediate Value Theorem" and verify continuity + that the value is between the endpoints.
✦ Model Answer

Yes. \(v_A\) is differentiable, so continuous on [5, 8]. Since \(v_A(5) = 40\) and \(v_A(8) = -120\), and \(-120 < -100 < 40\), by the Intermediate Value Theorem there exists \(t \in (5, 8)\) with \(v_A(t) = -100\) m/min.

Yes — justified by IVT (continuous, −100 is between 40 and −120)
(c) Position of train A at t=12; trapezoidal approximation
Position: \(x_A(12) = 300 + \displaystyle\int_2^{12} v_A(t)\,dt\)
Trapezoidal Rule: \(\int_a^b f\,dt \approx \sum \dfrac{(f(t_{i})+f(t_{i+1}))}{2}\cdot\Delta t\)
Use the THREE subintervals given in the table: [2,5], [5,8], [8,12]
✦ Model Answer
\(\int_2^{12} v_A\,dt \approx \frac{100+40}{2}(3) + \frac{40+(-120)}{2}(3) + \frac{-120+(-150)}{2}(4)\)
\(= 70(3) + (-40)(3) + (-135)(4) = 210 - 120 - 540 = -450\)
Position ≈ 300 + (−450) = −150 m from Origin Station
x_A(12) ≈ −150 meters (i.e., 150 m west of Origin)
(d) Rate of change of distance between A and B at t=2
Setup: Let \(x\) = east position of A, \(y\) = north position of B. Distance: \(D = \sqrt{x^2+y^2}\).
\(\dfrac{dD}{dt} = \dfrac{x\cdot\dot x + y\cdot\dot y}{D}\)
At t = 2: x = 300, y = 400, D = √(300²+400²) = 500.
ẋ = v_A(2) = 100, ẏ = v_B(2) = 4−10+60 = 54
⚠ Compute v_B(2) carefully
\(v_B(2) = (2)^2 - 5(2) + 60 = 4 - 10 + 60 = 54\) m/min. Easy arithmetic error here!
✦ Model Answer
\(D = \sqrt{300^2 + 400^2} = 500\)
\(\frac{dD}{dt}\bigg|_{t=2} = \frac{300(100)+400(54)}{500} = \frac{30000+21600}{500} = \frac{51600}{500} = 103.2\)
dD/dt = 103.2 meters per minute
5
Region R — Area, Volume, Perimeter
No Calculator Area between curves Volume (washer) Arc length
R is bounded by \(y = xe^{x^2}\), \(y = -2x\), and \(x = 1\)
🔍 Find the intersection first!
\(xe^{x^2} = -2x \Rightarrow x(e^{x^2}+2)=0 \Rightarrow x=0\) (since \(e^{x^2}+2 > 0\))

So the region spans from x = 0 to x = 1. At x = 0.5: y = 0.5e^{0.25} ≈ 0.64 (top), y = −1 (bottom). So top curve: \(xe^{x^2}\), bottom curve: \(-2x\).

(a) Find the area of R
\(A = \displaystyle\int_0^1 \left[xe^{x^2} - (-2x)\right]dx = \int_0^1 (xe^{x^2}+2x)\,dx\)
Key Antiderivative: \(\int xe^{x^2}dx = \frac{1}{2}e^{x^2} + C\) (u-sub: u = x²)
✦ Model Answer
\(A = \left[\frac{1}{2}e^{x^2} + x^2\right]_0^1 = \left(\frac{e}{2}+1\right) - \left(\frac{1}{2}+0\right)\)
\(= \frac{e}{2}+1-\frac{1}{2} = \frac{e}{2}+\frac{1}{2} = \frac{e+1}{2}\)
Area = (e+1)/2
(b) Volume when R is rotated about y = −2
Washer Method: \(V = \pi\displaystyle\int_0^1\left[R(x)^2 - r(x)^2\right]dx\)
• Outer radius: distance from axis y=−2 to top curve \(xe^{x^2}\): \(\quad R = xe^{x^2}+2\)
• Inner radius: distance from y=−2 to bottom curve \(-2x\): \(\quad r = -2x+2 = 2-2x\)
⚠ Radius = (curve) − (axis line)
Since the axis is y = −2, each radius = y_curve − (−2) = y_curve + 2. Don't forget the +2!
✦ Model Answer (write but do not evaluate)
\(V = \pi\int_0^1\left[(xe^{x^2}+2)^2 - (2-2x)^2\right]dx\)
V = π∫₀¹ [(xe^(x²)+2)² − (2−2x)²] dx
(c) Perimeter of R (write, do not evaluate)
Perimeter = arc length of top curve + arc length of bottom curve + vertical segment at x=1
Arc length: \(\displaystyle\int_a^b\sqrt{1+(y')^2}\,dx\)
Three pieces:
1. Top curve \(y = xe^{x^2}\) from x=0 to x=1: \(y' = e^{x^2}+2x^2e^{x^2}\)
2. Bottom curve \(y = -2x\) from x=0 to x=1: \(y' = -2\) → arc length = \(\sqrt{1+4} = \sqrt{5}\)
3. Vertical segment at x=1: length = \(xe^{x^2}\big|_{x=1} - (-2)\big|_{x=1} = e + 2\)
✦ Model Answer
\(P = \int_0^1\sqrt{1+(e^{x^2}+2x^2e^{x^2})^2}\,dx + \int_0^1\sqrt{1+4}\,dx + (e+2)\)
\(= \int_0^1\sqrt{1+e^{2x^2}(1+2x^2)^2}\,dx + \sqrt{5} + e + 2\)
P = ∫₀¹ √(1+e^(2x²)(1+2x²)²) dx + √5 + e + 2
6
Taylor Series — Radius, Derivative Series, Geometric Series
No Calculator Ratio Test Geometric Series Taylor/Maclaurin
\(\displaystyle f(x) = \sum_{n=1}^{\infty} \frac{(-1)^n \cdot 2(x-1)^n}{n}\), converges for \(|x-1| < R\)
(a) Find the radius of convergence R
Ratio Test: \(\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| < 1\) for convergence.
✦ Model Answer
\(\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{2(x-1)^{n+1}}{n+1}\cdot\frac{n}{2(x-1)^n}\right| = |x-1|\cdot\frac{n}{n+1} \to |x-1|\)
For convergence: \(|x-1| < 1\), so \(R = 1\)
R = 1
(b) First three nonzero terms and general term of Taylor series for f ′(x)
Differentiate term-by-term: \(f'(x) = \displaystyle\sum_{n=1}^{\infty}\frac{d}{dx}\left[\frac{(-1)^n\cdot 2(x-1)^n}{n}\right]\)
✦ Model Answer
\(f'(x) = \sum_{n=1}^{\infty}\frac{(-1)^n\cdot 2n(x-1)^{n-1}}{n} = \sum_{n=1}^{\infty}(-1)^n\cdot 2(x-1)^{n-1}\)
First three terms (n = 1, 2, 3):
\(= -2 + 2(x-1) - 2(x-1)^2 + \cdots\)
General term: \((-1)^n \cdot 2(x-1)^{n-1}\)
f′(x) = −2 + 2(x−1) − 2(x−1)² + ··· ; general term: (−1)ⁿ·2(x−1)^(n−1)
(c) Identify geometric series, find closed form for f ′, then integrate to get f
Geometric Series: \(\displaystyle\sum_{n=1}^{\infty}(-1)^n\cdot 2(x-1)^{n-1} = \sum_{n=0}^{\infty}2\cdot(-(x-1))^n\cdot\frac{1}{?}\)
Re-index: let \(m = n-1\), get \(\displaystyle\sum_{m=0}^{\infty}(-1)^{m+1}\cdot 2(x-1)^m = -2\sum_{m=0}^{\infty}(-(x-1))^m\)
Geometric sum formula: \(\displaystyle\sum_{m=0}^\infty r^m = \dfrac{1}{1-r}\) for \(|r| < 1\).
Here \(r = -(x-1) = 1-x\), so the sum = \(\dfrac{1}{1-(1-x)} = \dfrac{1}{x}\)
✦ Model Answer
\(f'(x) = -2\cdot\dfrac{1}{1-(1-x)} = \dfrac{-2}{x}\)
Integrate: \(f(x) = \int\dfrac{-2}{x}\,dx = -2\ln|x| + C\)
Find C: at x=1, f(1) = 0 (series = 0 when x=1 since every term is 0). So C = 0.
\(f(x) = -2\ln x\) for \(|x-1| < 1\), i.e., \(0 < x < 2\)
f′(x) = −2/x ; f(x) = −2 ln x for |x−1| < 1
Scoring Checklist

Each FRQ is worth 9 points. Key things graders always look for:

What Graders CheckWhy It Matters
✅ Units on every answer involving a rateExplicit point deduction if missing (Q1a, Q1b, Q4a, Q4d)
✅ "By the IVT / MVT" when citing theoremsMust name the theorem (Q4b)
✅ Show work for linear approximationProblem explicitly says "show the work" (Q1d)
✅ Write but do not evaluate (Q5b, Q5c)Full credit for correct setup; evaluating wastes time
✅ Interpret derivatives in context (Q1b)Must mention "decreasing", the rate, and the units
✅ Reason for Q3bMust state f > 0 AND f is decreasing, not just state the interval
✅ C = 0 justified in Q6cUse the fact that f(1) = 0 from the series