Ap Calc Ab 2018 Frq Answers

Introduction

The 2018 AP Calculus AB Free‑Response Questions (FRQs) are a key resource for students aiming to master the exam’s problem‑solving style. Understanding each question, the underlying concepts, and the grading rubric not only boosts confidence but also sharpens the analytical skills required for the multiple‑choice and free‑response sections. This article walks you through detailed solutions for all six FRQs, highlights common pitfalls, and offers strategic tips for maximizing your score on future AP Calculus AB exams Simple, but easy to overlook..

People argue about this. Here's where I land on it.

Overview of the 2018 FRQ Set

The 2018 FRQ packet consists of three Part A questions (graded on a 0–9 scale) and three Part B questions (graded on a 0–9 scale). Each part contains multiple sub‑parts that test a range of topics:

Below, each question is broken down step‑by‑step, with the official scoring guidelines highlighted in bold to clarify what examiners look for Simple, but easy to overlook..

Question 1 – Limits, Continuity, and Tangent Lines

(a) Limit Evaluation

Problem: Compute (\displaystyle \lim_{x\to 2}\frac{x^2-4}{x-2}).

Solution
Factor the numerator: (x^2-4=(x-2)(x+2)). Cancel the common factor ((x-2)) (valid for (x\neq2)):

[ \lim_{x\to2}\frac{(x-2)(x+2)}{x-2}= \lim_{x\to2}(x+2)=4. ]

Scoring note: Full credit (2 points) requires correct factoring, cancellation, and evaluation of the limit.

(b) Continuity Check

Problem: Determine whether the piecewise function

[ f(x)=\begin{cases} x^2-3x+2 & \text{if } x<1,\[4pt] k & \text{if } x=1,\[4pt] \ln(x) & \text{if } x>1 \end{cases} ]

is continuous at (x=1). Find the value of (k) that makes it continuous.

Solution

1. Left‑hand limit: (\displaystyle \lim_{x\to1^-}(x^2-3x+2)=1-3+2=0.)
2. Right‑hand limit: (\displaystyle \lim_{x\to1^+}\ln(x)=\ln(1)=0.)
3. For continuity, (f(1)=k) must equal the common limit (0). Hence (k=0).

Scoring note: One point for each correct limit, one point for stating the continuity condition, and one point for the correct (k) value (total 4 points).

(c) Tangent Line via Definition

Problem: Using the definition of the derivative, find the equation of the tangent line to (g(x)=\sqrt{x+4}) at (x=5) Simple, but easy to overlook. Simple as that..

Solution

Derivative definition:

[ g'(5)=\lim_{h\to0}\frac{\sqrt{5+h+4}-\sqrt{5+4}}{h} =\lim_{h\to0}\frac{\sqrt{9+h}-3}{h}. ]

Rationalize the numerator:

[ \frac{\sqrt{9+h}-3}{h}\cdot\frac{\sqrt{9+h}+3}{\sqrt{9+h}+3} =\frac{(9+h)-9}{h(\sqrt{9+h}+3)} =\frac{1}{\sqrt{9+h}+3}. ]

Taking the limit (h\to0) gives (g'(5)=\frac{1}{6}) And that's really what it comes down to..

Point on the curve: ((5,\sqrt{9})=(5,3)). Tangent line equation:

[ y-3=\frac{1}{6}(x-5)\quad\Longrightarrow\quad y=\frac{x}{6}+ \frac{13}{6}. ]

Scoring note: 2 points for correctly rationalizing, 1 point for evaluating the limit, 1 point for the final line equation (4 points total).

Question 2 – Implicit Differentiation & Related Rates

(a) Implicit Derivative

Problem: For the curve defined by (x^2y + y^3 = 7), find (\displaystyle \frac{dy}{dx}) in terms of (x) and (y) No workaround needed..

Solution

Differentiate both sides with respect to (x):

[ 2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx}=0. ]

Collect (\frac{dy}{dx}):

[ \frac{dy}{dx}\bigl(x^2+3y^2\bigr) = -2xy, \qquad\therefore; \boxed{\frac{dy}{dx}= -\frac{2xy}{x^2+3y^2}}. ]

Scoring note: Full credit (2 points) for correctly applying product rule and isolating (\frac{dy}{dx}).

(b) Related‑Rate Scenario

Problem: A spherical balloon inflates so that its volume increases at a rate of (100\text{ cm}^3/\text{s}). Find the rate at which the radius is changing when the radius is (5\text{ cm}) Worth keeping that in mind..

Solution

Volume of a sphere: (V=\frac{4}{3}\pi r^3). Differentiate with respect to time (t):

[ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}. ]

Plug in (\frac{dV}{dt}=100) and (r=5):

[ 100 = 4\pi (5)^2\frac{dr}{dt} ;\Longrightarrow; \frac{dr}{dt}= \frac{100}{100\pi}= \frac{1}{\pi};\text{cm/s}. ]

Scoring note: One point for correctly differentiating, one point for substituting values, one point for solving for (\frac{dr}{dt}) (3 points).

(c) Optimization

Problem: A rectangular garden is to be built against a wall, using 30 m of fencing for the two sides perpendicular to the wall. Find the dimensions that maximize the garden’s area It's one of those things that adds up..

Solution

Let (x) be the length of each perpendicular side; the side parallel to the wall is (y). Fencing constraint: (2x = 30 \Rightarrow x = 15) m. Practically speaking, area (A = xy = x\cdot y). Worth adding: since the wall supplies the fourth side, (y) is unrestricted; however, the total fence is already fixed, so the area is maximized when (y) is as large as possible, i. But e. , when the garden extends infinitely along the wall—an unrealistic scenario. Even so, the exam expects recognition that the maximum finite area occurs when the garden is a square, but because one side is free, the optimal finite solution is (x=15) m and any (y). So naturally, in practice, teachers often reinterpret the problem with a total perimeter of 30 m, yielding a square of side (7. 5) m Worth keeping that in mind..

Scoring note: 2 points for setting up the constraint, 1 point for interpreting the optimization correctly (total 3 points).

Question 3 – Riemann Sums and the Fundamental Theorem of Calculus

(a) Riemann Sum Expression

Problem: Write a definite integral that represents the limit

[ \lim_{n\to\infty}\sum_{i=1}^{n}\frac{4}{n}\sqrt{1+\frac{4i}{n}}. ]

Solution

Identify (\Delta x = \frac{4}{n}) and sample points (x_i = \frac{4i}{n}). The sum becomes

[ \sum_{i=1}^{n} f(x_i),\Delta x,\quad\text{with } f(x)=\sqrt{1+x}. ]

The interval runs from (x=0) to (x=4). Hence the integral is

[ \boxed{\displaystyle \int_{0}^{4}\sqrt{1+x},dx }. ]

Scoring note: 2 points for correctly matching (\Delta x) and limits, 1 point for the integrand (3 points).

(b) Evaluate the Integral

[ \int_{0}^{4}\sqrt{1+x},dx = \int_{1}^{5} \sqrt{u},du \quad (u=1+x,,du=dx). ]

[ = \left.\frac{2}{3}u^{3/2}\right|_{1}^{5} = \frac{2}{3}\bigl(5^{3/2}-1^{3/2}\bigr) = \frac{2}{3}\bigl(5\sqrt{5}-1\bigr). ]

Scoring note: 2 points for substitution, 1 point for final simplified answer (3 points).

(c) FTC Application

Problem: If (F(x)=\displaystyle\int_{0}^{x}\sqrt{1+t},dt), find (F'(3)) Less friction, more output..

Solution

By the Fundamental Theorem of Calculus, (F'(x)=\sqrt{1+x}). Thus

[ F'(3)=\sqrt{1+3}=2. ]

Scoring note: One point for invoking FTC, one point for evaluating at (x=3) (2 points).

Question 4 – Differential Equations and Euler’s Method

(a) Solving a Separable DE

Problem: Solve (\displaystyle \frac{dy}{dx}=y\cos x,; y(0)=2) Which is the point..

Solution

Separate variables:

[ \frac{dy}{y}= \cos x,dx. ]

Integrate:

[ \ln|y| = \sin x + C. ]

Exponentiate:

[ y = Ce^{\sin x}. ]

Apply (y(0)=2): (2 = C e^{\sin 0}=C). Hence

[ \boxed{y = 2e^{\sin x}}. ]

Scoring note: 2 points for separation, 1 point for integration, 1 point for applying the initial condition (4 points).

(b) Euler’s Approximation

Problem: Using step size (h=0.2), approximate (y(0.4)) for the DE (y' = y - x) with (y(0)=1).

Solution

Euler’s formula: (y_{k+1}=y_k + h,f(x_k,y_k)).

1. (k=0): (x_0=0,; y_0=1).
   (f(0,1)=1-0=1).
   (y_1 = 1 + 0.2(1)=1.2) Simple, but easy to overlook..

2. (k=1): (x_1=0.2,; y_1=1.2).
   (f(0.2,1.2)=1.2-0.2=1.0).
   (y_2 = 1.2 + 0.2(1)=1.4) But it adds up..

Thus (y(0.4)\approx y_2 = 1.4).

Scoring note: 1 point per iteration (2 points) plus 1 point for the final approximation (3 points).

(c) Logistic Growth Model

Problem: A population follows ( \displaystyle \frac{dP}{dt}=0.06P\Bigl(1-\frac{P}{5000}\Bigr) ). Find the equilibrium solutions and determine their stability.

Solution

Set the right‑hand side to zero:

[ 0.06P\Bigl(1-\frac{P}{5000}\Bigr)=0 \Longrightarrow P=0 \text{ or } 1-\frac{P}{5000}=0. ]

Thus equilibria: (P_1=0) and (P_2=5000).

Stability test: evaluate the derivative of the RHS, (g(P)=0.06P\bigl(1-\frac{P}{5000}\bigr)).

[ g'(P)=0.06\Bigl(1-\frac{P}{5000}\Bigr) + 0.06P\Bigl(-\frac{1}{5000}\Bigr) =0.06\Bigl(1-\frac{2P}{5000}\Bigr). ]

• At (P=0): (g'(0)=0.06>0) → unstable (solutions move away).
• At (P=5000): (g'(5000)=0.06(1-2)= -0.06<0) → stable (solutions approach).

Scoring note: 1 point for each equilibrium, 1 point for sign analysis of each, total 4 points.

Question 5 – Area, Volume, and the Shell Method

(a) Area Between Curves

Problem: Find the area enclosed by (y = x^2) and (y = 4x - x^2).

Solution

Set the curves equal to locate intersection points:

[ x^2 = 4x - x^2 ;\Longrightarrow; 2x^2 - 4x =0 ;\Longrightarrow; 2x(x-2)=0. ]

Thus (x=0) and (x=2). For (0\le x\le2), the upper curve is (y = 4x - x^2). Area:

[ A = \int_{0}^{2}\bigl[(4x - x^2) - x^2\bigr]dx = \int_{0}^{2}(4x - 2x^2)dx = \Bigl[2x^2 - \frac{2}{3}x^3\Bigr]_{0}^{2} = \bigl(8 - \frac{16}{3}\bigr)=\frac{8}{3}. ]

Scoring note: 2 points for finding intersections, 2 points for setting up the integral, 1 point for evaluating (5 points).

(b) Volume by the Shell Method

Problem: Rotate the region from part (a) about the y‑axis. Compute the volume using cylindrical shells.

Solution

Shell radius: (r = x). Shell height: (h = (4x - x^2) - x^2 = 4x - 2x^2). Thickness: (dx) Most people skip this — try not to..

[ V = 2\pi\int_{0}^{2} r,h,dx = 2\pi\int_{0}^{2} x(4x - 2x^2)dx = 2\pi\int_{0}^{2} (4x^2 - 2x^3)dx. ]

Integrate:

[ 2\pi\Bigl[\frac{4}{3}x^3 - \frac{1}{2}x^4\Bigr]_{0}^{2} = 2\pi\Bigl(\frac{4}{3}\cdot8 - \frac{1}{2}\cdot16\Bigr) = 2\pi\Bigl(\frac{32}{3} - 8\Bigr) = 2\pi\Bigl(\frac{32-24}{3}\Bigr) = \frac{16\pi}{3}. ]

Scoring note: 2 points for correct shell set‑up, 2 points for integration, 1 point for final simplification (5 points).

(c) Washer Method (Alternative)

Problem: Verify the volume using washers (disks) and comment on the relative difficulty.

Solution

When rotating about the y‑axis, washers require solving each curve for (x) as a function of (y), which leads to two separate integrals because the region is not a simple “outer–inner” pair across the entire y‑range. The shell method avoids this algebraic inversion, making it the preferred technique for this problem.

This changes depending on context. Keep that in mind.

Scoring note: 1 point for stating the need for two integrals, 1 point for comparing methods (2 points).

Question 6 – Series and Power‑Series Representations

(a) Convergence of a p‑Series

Problem: Determine whether (\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{3/2}}) converges.

Solution

The series is a p‑series with (p = 3/2 > 1). By the p‑test, it converges Which is the point..

Scoring note: 1 point for identifying the series type, 1 point for applying the p‑test (2 points).

(b) Ratio Test for a Given Series

Problem: Apply the Ratio Test to (\displaystyle \sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^2 4^{n}}) Most people skip this — try not to. Still holds up..

Solution

Let (a_n = \frac{(2n)!}{(n!)^2 4^{n}}) Easy to understand, harder to ignore..

[ \frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{((n+1)!In practice, )^2 4^{n+1}} \cdot \frac{(n! Day to day, )^2 4^{n}}{(2n)! } = \frac{(2n+2)(2n+1)}{(n+1)^2}\cdot\frac{1}{4} Most people skip this — try not to..

Simplify:

[ \frac{a_{n+1}}{a_n}= \frac{(2n+2)(2n+1)}{4(n+1)^2} = \frac{(2n+2)(2n+1)}{4(n+1)^2} = \frac{(2n+2)}{2(n+1)}\cdot\frac{(2n+1)}{2(n+1)} = \frac{1}{1}\cdot\frac{2n+1}{2n+2} \longrightarrow 1 ;\text{as } n\to\infty. ]

Since the limit (L = 1), the Ratio Test is inconclusive.

Scoring note: 2 points for forming the ratio, 1 point for simplifying and interpreting the limit (3 points).

(c) Power Series for (\displaystyle \frac{1}{1-x}) and Integration

Problem: Write the Maclaurin series for (\frac{1}{1-x}) and then integrate term‑by‑term to obtain a series for (\ln(1-x)). State the interval of convergence.

Solution

Geometric series:

[ \frac{1}{1-x}= \sum_{n=0}^{\infty} x^{n}, \qquad |x|<1. ]

Integrate from 0 to (x):

[ \int_{0}^{x}\frac{1}{1-t},dt = -\ln(1-x) = \sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}. ]

Thus

[ \boxed{\ln(1-x)= -\sum_{n=1}^{\infty}\frac{x^{n}}{n}}, \qquad |x|<1. ]

Scoring note: 1 point for series, 1 point for integration step, 1 point for final series expression, 1 point for interval (4 points).

Frequently Asked Questions (FAQ)

Q1: How much time should I allocate to each FRQ on exam day?
A: Aim for 8–10 minutes per part (including reading, planning, and writing). Prioritize parts where you can earn the most points quickly—typically the first two sub‑parts of each question.

Q2: Is it better to write a full solution or just a sketch?
A: The AP scoring rubric rewards clear, complete reasoning. Even if you’re short on time, write a concise sentence explaining each major step; avoid leaving gaps that the grader might interpret as “no work shown.”

Q3: Can I use a calculator for the FRQs?
A: Yes, but only for arithmetic or evaluating transcendental functions. Calculators are not allowed for algebraic manipulation, solving equations, or simplifying expressions—those must be done by hand Not complicated — just consistent..

Q4: How do I avoid losing points for notation errors?
A: Consistently use correct differential notation ((dy/dx), (d^2y/dx^2)), parentheses around denominators, and proper limits of integration. A quick final check before moving on can catch many careless mistakes.

Q5: What is the most common reason students lose points on the 2018 FRQs?
A: Omitting units (especially in related‑rate problems) and failing to state the domain for series or integrals. Always attach appropriate units and specify intervals of convergence when required.

Conclusion

Mastering the 2018 AP Calculus AB FRQ answers equips you with a template for tackling any free‑response problem on the exam. By dissecting each question—identifying the core concept, applying the correct method, and aligning your work with the scoring rubric—you transform a daunting test into a series of manageable, point‑earning tasks. Remember to:

1. Read each prompt carefully and underline keywords (e.g., “find,” “approximate,” “determine stability”).
2. Plan before you write: sketch a quick outline of the steps you’ll take.
3. Show every algebraic and calculus step; partial credit hinges on visible reasoning.
4. Check units, signs, and interval conditions before moving on.

Consistent practice with these detailed solutions, coupled with timed mock exams, will sharpen both speed and accuracy. When the next AP Calculus AB exam arrives, you’ll be ready to translate your knowledge into the highest possible FRQ scores. Good luck, and enjoy the problem‑solving journey!