Exam track · in progress

AP Calculus BC

Board · College Board Units · 1–10 (AB + 9–10) Exam · MCQ + FRQ, scored 1–5 Status · Preview

Overview & units

🗺 AP Calculus BC — Additional Topics Beyond AB

Unit	Title	Exam %	Key Topics
BC-1	Advanced Integration	17–20% (shared)	Integration by parts, partial fractions, improper integrals
BC-2	Logistic & Euler's Method	6–12% (shared)	Logistic growth model, Euler's method for DE approximation
BC-3	Parametric & Vectors	11–12% BC-only	Parametric derivatives, arc length, vector-valued functions
BC-4	Polar Coordinates	5–6% BC-only	Polar graphs, polar area, intersections
BC-5	Sequences & Series	17–18% BC-only	Convergence tests (integral, comparison, ratio, alternating)
BC-6	Power, Taylor & Maclaurin	17–18% BC-only	Taylor/Maclaurin series, radius & interval of convergence, error bound

How to use this track

BC contains everything in AP Calculus AB plus the additional topics below — advanced integration, logistic models and Euler's method, parametric, polar and vector functions, and the full theory of sequences and series. The six BC chapters, worked free-response questions, and a 50-question practice exam are all here.

AP Calculus BC is a superset of AB. It shares Units 1–8 and adds:

Units 1–8

Everything in AP Calculus AB

Unit 9

Parametric, polar & vector-valued functions

Unit 10

Infinite sequences & series

The exam format matches AB (multiple choice 50%, free response 50%), and a BC result also reports an AB subscore.

⏱️ AP Calculus BC Exam Breakdown & Timing Strategy

Section I: Multiple Choice (50% of score, 45 questions, 105 min):
• Part A (60 min, 30 questions, NO calculator): ~27 AB questions + ~3 BC questions. Pacing: 2 min/question.
• Part B (45 min, 15 questions, Graphing calculator required): ~9 AB questions + ~6 BC questions. Pacing: 3 min/question.
Section II: Free Response (50% of score, 6 questions, 90 min):
• Part A (30 min, 2 questions, Graphing calculator required): Typically Question 1 (Rate in/out or area/volume) and Question 2 (Parametric/vector motion).
• Part B (60 min, 4 questions, NO calculator): Questions 3–5 cover differential equations/Euler, FTC graph analysis, and polar area; Question 6 is always Taylor Series & Error Bounds.
BC Scoring Thresholds: A composite score of roughly 62–66% earns a 5 on AP Calculus BC, and ~50–55% earns a 4. High performance on core AB topics combined with harvesting method points on Taylor series guarantees top scores.

💡 The Two Giant BC-Exclusive Capstone Pillars

Unit 9: Parametric, Polar & Vector Calculus (11–12% of exam):
Extends single-variable calculus to curves in the plane: parametric slope $\frac{dy/dt}{dx/dt}$ , second derivative $\frac{(dy/dx)'}{dx/dt}$ , vector velocity $\vec{v}(t) = \langle x', y' \rangle$ , scalar speed $\|\vec{v}\| = \sqrt{(x')^2+(y')^2}$ , arc length, and polar area $\frac{1}{2}\int r^2\,d\theta$ .
Unit 10: Infinite Sequences & Series (17–18% of exam):
The most challenging unit on the exam: convergence tests (Geometric, p-Series, Ratio, Limit Comparison, Alternating Series), absolute vs. conditional convergence, power series intervals of convergence, Taylor and Maclaurin polynomial constructions, and error bounding (Alternating Series Error vs. Lagrange Error Bound).

AP Calculus BC exam topic distribution across primary units and AB/BC split. Core AB topics account for ~60% of the composite score (and the AB Subscore), while BC-exclusive topics (Integration techniques, Euler/Logistic, Parametric/Polar/Vector, and Infinite Series) comprise the remaining ~40%.

Section I of AP Calculus BC consists of 45 Multiple Choice questions (Part A: 30 non-calculator questions in 60 minutes; Part B: 15 calculator-active questions in 45 minutes). Calculate the average time allocated per question for Part A versus Part B in minutes.

For Part A:

\frac{60\text{ minutes}}{30\text{ questions}} = 2.0\text{ minutes per question}

. For Part B:

\frac{45\text{ minutes}}{15\text{ questions}} = 3.0\text{ minutes per question}

. Part B grants 50% more time per question because graphing calculator input and multi-step interpretation are required.

📝 Chapter Quiz

Questions Level

Beyond AB

▲ Parametric & polar

For a parametric curve, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ . Polar area is $\displaystyle A=\tfrac12\int_\alpha^\beta r^2\,d\theta.$

▲ Series convergence

The ratio test: $\sum a_n$ converges absolutely if $\displaystyle\lim_{n\to\infty}\Big|\dfrac{a_{n+1}}{a_n}\Big|<1$ . Taylor series and their intervals of convergence are the centrepiece of Unit 10.

▲ BC-only topics

AP Calculus BC extends AB with additional integration techniques, convergence tests, parametric/polar calculus, and infinite series — approximately 40% of the exam is BC-exclusive material.

BC-only: Series

Taylor/Maclaurin, ratio test, alternating series, error bounds

BC-only: Parametric

Slope

\frac{dy/dt}{dx/dt}

, arc length, speed

=\sqrt{(x')^2+(y')^2}

BC-only: Polar

Area

\frac{1}{2}\int r^2\,d\theta

, arc length, slope in polar

BC-only: Techniques

Integration by parts, partial fractions, improper integrals, Euler's method

🎯 The AB Subscore & Strategic Scoring in BC

Two Scores from One Exam: Every student taking AP Calculus BC receives both an overall BC Exam Score (1–5) and an AB Subscore (1–5) based solely on the ~60% of the exam covering AB topics (Units 1–8).
The College Credit Safety Net: Most colleges award Calculus I credit for an AB Subscore of 4 or 5, even if your composite BC score is lower. Mastering core AB calculus provides an unbeatable foundation for both scores.
Predictable BC FRQ Real Estate: The 6 Free Response questions always dedicate exactly 2 questions to BC-only material: typically FRQ 2 or 5 on Parametric/Vector Motion or Polar Area, and FRQ 6 on Taylor Series and Error Bounds.

⚠️ The Top 3 BC-Exclusive Scoring Traps

1. Parametric Second Derivative: Differentiate $\frac{dy}{dx}$ with respect to $t$ , then divide by $\frac{dx}{dt}$ . Never divide second derivatives directly: $\frac{d^2y}{dx^2} \neq \frac{y''(t)}{x''(t)}$ .
2. Lagrange vs. Alternating Error Bound: If an FRQ explicitly requires the Lagrange Error Bound, quoting the alternating series test bound (first omitted term) receives zero credit for the error bound point.
3. Evaluating Limits for Improper Integrals: On Section II, you must write $\lim_{b\to\infty}\int_a^b$ rather than plugging in $\infty$ as an endpoint.

Taylor polynomial approximations to

f(x) = e^x

centered at

x = 0

. As the polynomial degree increases from

P_1(x) = 1 + x

P_2(x) = 1 + x + \frac{1}{2}x^2

, the polynomial matches higher derivatives and hugs the exponential curve with rapidly diminishing error near the center.

A particle moves in the

xy

-plane with position given by

x(t) = t^3 - 3t

and

y(t) = 2t^2 + 1

for

t > 0

. Find the slope of the tangent line

\frac{dy}{dx}

t = 2

Compute individual derivatives:

\frac{dx}{dt} = 3t^2 - 3

and

\frac{dy}{dt} = 4t

. At

t = 2

\frac{dx}{dt} = 3(4) - 3 = 9

and

\frac{dy}{dt} = 4(2) = 8

. The parametric slope is

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{8}{9}

📝 Chapter Quiz

Questions Level

Advanced Integration Techniques

📌 AP Calc BC BC-1 — Advanced Integration 17–20% of exam · Integration by parts · Partial fractions · Improper integrals

BC extends the integral toolkit with integration by parts, partial fractions, and improper integrals over infinite or unbounded intervals.

▲ Reference — Three new tools

By parts: $\displaystyle\int u\,dv=uv-\int v\,du$ . Partial fractions split a rational integrand into simpler pieces. Improper integrals are limits: $\displaystyle\int_a^{\infty} f=\lim_{b\to\infty}\int_a^{b} f$ .

The improper integral

\int_1^\infty \frac{1}{x^2}\,dx = \lim_{b\to\infty} \left[1 - \frac{1}{b}\right] = 1

. Even though the domain extends infinitely along the horizontal axis, the accumulated area converges to exactly

1

By parts

uv-\displaystyle\int v\,du

Partial fractions

\dfrac{1}{(x-a)(x-b)}

Improper

\displaystyle\lim_{b\to\infty}\int_a^b f

Arctangent

\displaystyle\int\dfrac{dx}{x^{2}+a^{2}}=\dfrac1a\arctan\dfrac{x}{a}

⚠️ AP Exam Traps: Advanced Integration Techniques

Mandatory Limit Notation on FRQs: You cannot plug $\infty$ directly into an antiderivative bracket on the AP Exam. Writing $\left[-\frac{1}{x}\right]_1^\infty$ loses the setup point. You MUST replace $\infty$ with a dummy variable and take the limit: $\lim_{b\to\infty} \int_1^b \frac{1}{x^2}\,dx = \lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b$ .
Integration by Parts Sign Distribution: In $\int u\,dv = uv - \int v\,du$ , when performing repeated IBP or when $v$ contains a negative sign (e.g. $\int \sin(x)dx = -\cos(x)$ ), parenthesize carefully: $-(-uv)$ turns into a $+$ , and a negative outside distributed over an inner subtraction is the single most frequent calculation error.
Improper Integrals with Interior Discontinuities: Watch out for integrals like $\int_{-1}^1 \frac{1}{x^2}\,dx$ . The integrand is undefined at $x=0$ , so you cannot apply the Fundamental Theorem of Calculus directly. It must be split into $\int_{-1}^0 \frac{1}{x^2}\,dx + \int_0^1 \frac{1}{x^2}\,dx$ , both of which diverge to $\infty$ .
LIATE Priority Rule for Parts: Pick $u$ following Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. Differentiate $u$ and integrate $dv$ .

Worked example · By parts

Evaluate $\displaystyle\int x e^{x}\,dx$ with $u=x,\ dv=e^{x}dx$ . $x e^{x}-\int e^{x}\,dx=(x-1)e^{x}+C.$

Worked example — Integration by parts

Evaluate $\displaystyle\int x^2 e^x\,dx$ .

Apply IBP twice. Let $u=x^2,\,dv=e^x dx$ : $x^2e^x-2\int xe^x dx$ .

For $\int xe^x dx$ : $u=x,\,dv=e^x dx\Rightarrow xe^x-e^x$ .

\int x^2e^x\,dx=x^2e^x-2xe^x+2e^x+c=e^x(x^2-2x+2)+c.

Determine whether

\displaystyle\int_1^\infty\frac{1}{x^2}\,dx

converges, and if so find its value.

\int_1^b\frac{1}{x^2}dx=\left[-\frac{1}{x}\right]_1^b=1-\frac{1}{b}\to1

b\to\infty

. Converges to

1

Evaluate the definite integral

\displaystyle\int_0^\pi x \sin x\,dx

using integration by parts.

Let

u = x

and

dv = \sin x\,dx

. Then

du = dx

and

v = -\cos x

. Applying IBP:

\int x\sin x\,dx = -x\cos x - \int(-\cos x)\,dx = -x\cos x + \sin x

. Evaluating from

0

\pi

[-\pi\cos(\pi) + \sin(\pi)] - [-0\cos(0) + \sin(0)] = [-\pi(-1) + 0] - 0 = \pi

📝 Chapter Quiz

Questions Level

Logistic Models & Euler's Method

📌 AP Calc BC BC-2 — Logistic & Euler's Method 6–12% of exam · Logistic growth model · Euler's method · Long-run behaviour

BC adds a richer population model and a numerical method for approximating solutions to differential equations.

▲ Reference — Growth & approximation

Logistic: $\dfrac{dP}{dt}=kP\!\left(1-\dfrac{P}{M}\right)$ , carrying capacity $M$ , fastest growth at $P=\tfrac{M}{2}$ . Euler's method steps forward: $y_{n+1}=y_n+h\,f(x_n,y_n)$ .

Logistic growth curve

P(t) = \frac{10}{1 + 9e^{-0.9t}}

with carrying capacity

M=10

. Growth is fastest at the inflection point where

P = M/2 = 5

and

\frac{d^2P}{dt^2} = 0

. The curve is concave up on

(0, 2.44)

and concave down on

(2.44, \infty)

Logistic ODE

kP\!\left(1-\tfrac{P}{M}\right)

Carrying capacity

\lim_{t\to\infty}P=M

Fastest growth

P=\tfrac{M}{2}

Euler step

y_n+h f(x_n,y_n)

⚠️ AP Exam Traps: Logistic Models & Euler's Method

Fastest Growth is at Population $P = M/2$ , NOT Time $t = M/2$ : When the exam asks 'for what population is the growth rate greatest?', the answer is always half the carrying capacity $M/2$ . If it asks 'at what time is the growth rate greatest?', substitute $P = M/2$ into the particular solution and solve for $t$ .
Factoring Non-Standard Logistic Forms: If given $\frac{dP}{dt} = 0.2P - 0.0005P^2$ , factor out $0.2P$ : $0.2P\left(1 - \frac{0.0005}{0.2}P\right) = 0.2P\left(1 - \frac{P}{400}\right)$ . The carrying capacity is $M = 400$ , not $0.0005$ or $0.2$ !
Euler Concavity Error Analysis: Free-response questions frequently ask whether Euler's method overestimates or underestimates the true solution $y(b)$ :
• If $y''(x) > 0$ (concave up), tangent lines lie below the curve $\Rightarrow$ Euler's method underestimates.
• If $y''(x) < 0$ (concave down), tangent lines lie above the curve $\Rightarrow$ Euler's method overestimates.
Never Combine Steps in Euler's Method: Always calculate the slope $f(x_n, y_n)$ and step explicitly: $y_{n+1} = y_n + \Delta x \cdot f(x_n, y_n)$ . AP readers require evidence of step-by-step arithmetic.

Worked example · Euler step

For $\dfrac{dy}{dx}=y,\ y(0)=1$ , take one Euler step with $h=0.5$ . $y(0.5)\approx 1+0.5\cdot 1=1.5.$

Worked example — Euler's method

Use Euler's method with step $h=0.1$ to approximate $y(0.2)$ for $\dfrac{dy}{dx}=y-x$ , $y(0)=1$ .

$y_1=y_0+hf(x_0,y_0)=1+0.1(1-0)=1.1$ .

$y_2=y_1+hf(x_1,y_1)=1.1+0.1(1.1-0.1)=1.1+0.1=1.2$ .

The logistic model

\dfrac{dP}{dt}=0.4P\!\left(1-\dfrac{P}{500}\right)

has

P(0)=50

. What is

\lim_{t\to\infty}P(t)

? For what

P

is the growth rate greatest?

Carrying capacity

K=500

\lim P=500

. Maximum growth rate at

P=K/2=250

Let

\dfrac{dy}{dx} = 2x - y

with initial condition

y(0) = 2

. Use Euler's method with two steps of equal size

h = 0.5

to approximate

y(1)

Step 1: At

(x_0, y_0) = (0, 2)

, slope

f(0, 2) = 2(0) - 2 = -2

. Then

y_1 = y_0 + h\cdot f(0, 2) = 2 + 0.5(-2) = 1

x_1 = 0.5

.
Step 2: At

(x_1, y_1) = (0.5, 1)

, slope

f(0.5, 1) = 2(0.5) - 1 = 1 - 1 = 0

. Then

y_2 = y_1 + h\cdot f(0.5, 1) = 1 + 0.5(0) = 1

x_2 = 1.0

.
Therefore,

y(1) \approx 1

📝 Chapter Quiz

Questions Level

Parametric & Vector-Valued Functions

📌 AP Calc BC BC-3 — Parametric & Vectors 11–12% of exam · Parametric derivatives & arc length · Vector-valued functions & motion

When a curve is traced by a parameter $t$ , motion in the plane is captured by component functions and their derivatives.

▲ Reference — Parametric calculus

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ . Speed is $\sqrt{(dx/dt)^{2}+(dy/dt)^{2}}$ , and arc length is $\displaystyle\int\sqrt{(dx/dt)^{2}+(dy/dt)^{2}}\,dt$ . A vector position $\mathbf r(t)=\langle x(t),y(t)\rangle$ has velocity $\mathbf r'(t)$ .

Parametric trajectory

x(t) = t^2, y(t) = t^3

(Cartesian form

y = x^{3/2}

). At

t=1

, the particle is at

(1,1)

with velocity vector

\vec{v}(1) = \langle 2, 3 \rangle

, yielding tangent line slope

\frac{dy}{dx} = \frac{3}{2} = 1.5

Slope

\dfrac{dy/dt}{dx/dt}

Speed

\sqrt{x'^{2}+y'^{2}}

Arc length

\displaystyle\int\sqrt{x'^{2}+y'^{2}}\,dt

Velocity

\langle x'(t),y'(t)\rangle

⚠️ AP Exam Traps: Parametric & Vector Calculus

The Second Derivative Formula: The second derivative with respect to $x$ is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{\frac{dx}{dt}}$ . It is NEVER $\frac{y''(t)}{x''(t)}$ ! You must differentiate $\frac{dy}{dx}$ with respect to $t$ and then divide the entire result by $\frac{dx}{dt}$ .
Speed vs. Velocity Vector: Velocity is a vector with direction: $\vec{v}(t) = \langle x'(t), y'(t) \rangle$ . Speed is a scalar magnitude: $\text{Speed} = \|\vec{v}(t)\| = \sqrt{(x'(t))^2 + (y'(t))^2}$ . Always check whether a question asks for speed or velocity.
Total Distance vs. Displacement:
• Total distance traveled is the integral of speed: $\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ .
• Displacement is the vector net change: $\langle x(b) - x(a), y(b) - y(a) \rangle$ .
Particle at Rest: A particle is at rest if and only if both components are simultaneously zero: $x'(t) = 0$ and $y'(t) = 0$ . If only $x'(t) = 0$ , the curve has a vertical tangent; if only $y'(t) = 0$ , the curve has a horizontal tangent.

Worked example · Parametric slope

For $x=t^{2},\ y=t^{3}$ , find $\dfrac{dy}{dx}$ . $\frac{dy}{dx}=\frac{3t^{2}}{2t}=\frac{3t}{2}.$

Worked example — Parametric arc length

Find the arc length of $x=t^2,\;y=\tfrac{2}{3}t^3$ from $t=0$ to $t=1$ . $\frac{dx}{dt}=2t,\quad\frac{dy}{dt}=2t^2.$ $L=\int_0^1\sqrt{4t^2+4t^4}\,dt=\int_0^1 2t\sqrt{1+t^2}\,dt=\left[\tfrac{2}{3}(1+t^2)^{3/2}\right]_0^1=\tfrac{2}{3}(2^{3/2}-1)=\tfrac{2(2\sqrt2-1)}{3}.$

A particle moves with

x(t)=\cos t,\;y(t)=\sin t

. Find the speed at

t=\pi/4

and the total distance travelled from

t=0

t=\pi

Speed

=\sqrt{\dot x^2+\dot y^2}=\sqrt{\sin^2t+\cos^2t}=1

(constant). Distance

=\int_0^\pi1\,dt=\pi

A curve is given parametrically by

x(t) = t^3 - 3t

and

y(t) = t^2

. Find

\dfrac{d^2y}{dx^2}

t = 2

First compute derivatives:

x'(t) = 3t^2 - 3

and

y'(t) = 2t

. The first derivative is

\frac{dy}{dx} = \frac{2t}{3t^2 - 3}

. Next differentiate with respect to

t

using the quotient rule:

\frac{d}{dt}\left[\frac{dy}{dx}\right] = \frac{(3t^2 - 3)(2) - (2t)(6t)}{(3t^2 - 3)^2} = \frac{6t^2 - 6 - 12t^2}{(3t^2 - 3)^2} = \frac{-6t^2 - 6}{(3t^2 - 3)^2}

. At

t = 2

3(2)^2 - 3 = 9

, so

\frac{d}{dt}\left[\frac{dy}{dx}\right] = \frac{-6(4)-6}{9^2} = \frac{-30}{81} = -\frac{10}{27}

. Finally,

\frac{d^2y}{dx^2} = \frac{-10/27}{x'(2)} = \frac{-10/27}{9} = -\frac{10}{243}

📝 Chapter Quiz

Questions Level

Polar Coordinates

📌 AP Calc BC BC-4 — Polar Coordinates 5–6% of exam · Converting polar/rectangular · Polar area formula · Polar curve derivatives

Polar coordinates locate points by radius and angle, and turn area into a sweep of the radius about the origin.

The polar curve $r=\sin(k\theta)$ traces a rose. Change $k$ and count the petals — odd $k$ gives $k$ petals, even $k$ gives $2k$ .

⊙ Explorer · Polar roseinteractive

Petal factor k = 3

petals—

Polar area is $\tfrac12\displaystyle\int r^{2}\,d\theta$ — one of the headline BC integrals.

▲ Reference — Polar calculus

Convert with $x=r\cos\theta,\ y=r\sin\theta$ . The area swept is $A=\tfrac12\displaystyle\int_\alpha^\beta r^{2}\,d\theta$ . Slope follows from treating $x$ and $y$ as parametric in $\theta$ .

Polar area is generated as the radial arm

r(\theta)

sweeps through angle

d\theta

, creating an infinitesimal circular sector with area

dA = \frac{1}{2} r^2\,d\theta

. Integrating across

[\alpha, \beta]

yields the total enclosed area

A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta

Conversion

x=r\cos\theta,\ y=r\sin\theta

Polar area

\tfrac12\displaystyle\int r^{2}\,d\theta

Circle

r=a

Cardioid

r=a(1\pm\cos\theta)

⚠️ AP Exam Traps: Polar Coordinates & Calculus

The Missing $\frac{1}{2}$ Factor: The single most frequent student error in polar integration is omitting the $\frac{1}{2}$ factor. Always write $A = \frac{1}{2} \int_\alpha^\beta r^2\,d\theta$ .
Area Between Two Polar Curves: If region $R$ lies between outer curve $r_1(\theta)$ and inner curve $r_2(\theta)$ , the area is $\frac{1}{2} \int [r_1^2 - r_2^2]\,d\theta$ . It is NEVER $\frac{1}{2}\int [r_1 - r_2]^2\,d\theta$ !
Polar Slope $\frac{dy}{dx}$ vs. $\frac{dr}{d\theta}$ : The derivative $\frac{dr}{d\theta}$ measures how fast the radius grows with respect to angle — it is NOT the slope of the tangent line. To find slope, apply the product rule to $y = r\sin\theta$ and $x = r\cos\theta$ : $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}.$
Petal Bounds: To find integration limits for a single petal, find consecutive angles where $r(\theta) = 0$ . For $r = \sin(2\theta)$ , $2\theta = 0 \Rightarrow \theta = 0$ and $2\theta = \pi \Rightarrow \theta = \pi/2$ , so the petal spans $[0, \pi/2]$ .

Worked example · Polar area

Find the area enclosed by $r=2$ . $\tfrac12\int_0^{2\pi}2^{2}\,d\theta=\tfrac12\cdot4\cdot2\pi=4\pi.$

Worked example — Polar area

Find the area enclosed by one petal of $r=\sin(2\theta)$ .

One petal: $0\leq\theta\leq\pi/2$ .

A=\tfrac{1}{2}\int_0^{\pi/2}\sin^2(2\theta)\,d\theta=\tfrac{1}{2}\int_0^{\pi/2}\frac{1-\cos(4\theta)}{2}\,d\theta=\frac{\pi}{8}.

Convert the polar equation

r=4\cos\theta

to rectangular form and identify the curve.

Multiply both sides by

r

r^2=4r\cos\theta\Rightarrow x^2+y^2=4x\Rightarrow(x-2)^2+y^2=4

. Circle, centre

(2,0)

, radius

2

Find the slope of the line tangent to the cardioid

r = 1 + \cos\theta

\theta = \frac{\pi}{2}

Write rectangular coordinates in terms of

\theta

x = (1+\cos\theta)\cos\theta = \cos\theta + \cos^2\theta

, and

y = (1+\cos\theta)\sin\theta = \sin\theta + \sin\theta\cos\theta

. Differentiate:

\frac{dx}{d\theta} = -\sin\theta - 2\cos\theta\sin\theta

; at

\theta = \pi/2

\frac{dx}{d\theta} = -1 - 0 = -1

. Next,

\frac{dy}{d\theta} = \cos\theta + \cos^2\theta - \sin^2\theta

; at

\theta = \pi/2

\frac{dy}{d\theta} = 0 + 0 - 1 = -1

. Thus, the tangent slope is

\frac{dy}{dx} = \frac{-1}{-1} = 1

📝 Chapter Quiz

Questions Level

Infinite Sequences & Series — Convergence

📌 AP Calc BC BC-5 — Sequences & Series 17–18% of exam · Convergence tests · Alternating series error bound · Absolute vs. conditional

The heart of BC: deciding whether an infinite sum settles to a finite value, using a battery of convergence tests.

Watch the geometric series $\sum_{n\ge1}\left(\tfrac12\right)^{n}$ converge: each partial sum $S_N$ climbs toward the limit $1$ (the amber line).

⊙ Explorer · Partial sumsinteractive

Terms N = 4

partial sum S_N—

$S_N=1-\left(\tfrac12\right)^{N}\to1$ ; a geometric series converges because $|r|<1$ .

▲ Reference — Convergence tests

nth-term test: if $a_n\not\to0$ , the series diverges. A geometric series converges when $|r|<1$ ; a p-series when $p>1$ . The comparison, ratio, and alternating series tests handle the rest.

Partial sums

S_N = \sum_{n=1}^N (1/2)^n = 1 - (1/2)^N

climb monotonically toward the horizontal asymptote

S = 1

. Because

|r| = 1/2 < 1

, the infinite series converges to

\frac{a}{1-r} = \frac{1/2}{1 - 1/2} = 1

Geometric

\displaystyle\sum ar^{n}=\dfrac{a}{1-r}

p-series

converges if

p>1

Ratio test

\lim\left|\dfrac{a_{n+1}}{a_n}\right|<1

Alternating

terms

\downarrow 0

⚠️ AP Exam Traps: Infinite Series Convergence Tests

The $n$ -th Term Divergence Test Never Proves Convergence: If $\lim a_n \neq 0$ , the series diverges. However, if $\lim a_n = 0$ , the test is INCONCLUSIVE! The harmonic series $\sum \frac{1}{n}$ has terms approaching $0$ , but it diverges. Never write 'converges by the $n$ -th term test' on an AP exam.
Ratio Test Inconclusiveness: If $\lim |a_{n+1}/a_n| = 1$ , the Ratio Test fails completely. The Ratio Test cannot determine the convergence of $p$ -series or algebraic fractions (it always gives $1$ ). Use the Limit Comparison Test with a $p$ -series instead.
Alternating Series Test (AST) Hypotheses: To earn full credit for AST on an FRQ, you must explicitly state that (1) terms alternate in sign, (2) terms decrease in magnitude ( $|a_{n+1}| \le |a_n|$ ), and (3) $\lim_{n\to\infty} |a_n| = 0$ .
Alternating Error Bound: If an alternating series converges by AST, the truncation error when using the $N$ -th partial sum $S_N$ is bounded by the magnitude of the first omitted term: $|S - S_N| \le |a_{N+1}|$ .

Worked example · Geometric sum

Evaluate $\displaystyle\sum_{n=1}^{\infty}\left(\tfrac12\right)^{n}$ . $\frac{\tfrac12}{1-\tfrac12}=1.$

Worked example — Ratio test

Determine whether $\displaystyle\sum_{n=1}^\infty\dfrac{n!}{n^n}$ converges. $L=\lim_{n\to\infty}\frac{(n+1)!/(n+1)^{n+1}}{n!/n^n}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n=\frac{1}{e}<1.$

Ratio test: $L<1$ , so the series converges.

Find the interval of convergence of

\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^n x^n}{n}

Ratio test:

|x|

. Converges absolutely for

|x|<1

. At

x=1

: alternating harmonic series, converges. At

x=-1

: harmonic series, diverges. Interval of convergence:

-1<x\leq1

Determine whether the series

\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}

converges absolutely, converges conditionally, or diverges.

First test absolute convergence:

\sum\left|\frac{(-1)^n}{\sqrt{n}}\right| = \sum\frac{1}{n^{1/2}}

. This is a

p

-series with

p = 1/2 \le 1

, which diverges. Next test the original alternating series using AST: the terms alternate in sign,

b_n = \frac{1}{\sqrt{n}}

is strictly decreasing (

b_{n+1} < b_n

), and

\lim_{n\to\infty}\frac{1}{\sqrt{n}} = 0

. By the Alternating Series Test, the series converges. Since the series converges but does not converge absolutely, it converges conditionally.

📝 Chapter Quiz

Questions Level

Power, Taylor & Maclaurin Series

📌 AP Calc BC BC-6 — Power, Taylor & Maclaurin 17–18% of exam · Taylor/Maclaurin series · Radius & interval of convergence · Lagrange error

Functions can be rebuilt as infinite polynomials, exact within an interval of convergence — the capstone of the course.

The amber Maclaurin polynomial for $\sin x$ hugs the true curve (indigo) ever more widely as you add terms.

⊙ Explorer · Taylor polynomialinteractive

Terms = 3

polynomial degree—

$\sin x = x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\cdots$ — convergent for all $x$ .

▲ Reference — Series for functions

The Taylor series about $a$ is $\displaystyle\sum \frac{f^{(n)}(a)}{n!}(x-a)^{n}$ ; centred at $0$ it is a Maclaurin series. Memorise the series for $e^{x}$ , $\sin x$ , $\cos x$ and $\tfrac{1}{1-x}$ , each with its interval of convergence.

Maclaurin polynomial approximations for

f(x) = \sin(x)

centered at

x=0

. The tangent line

P_1(x) = x

is accurate only near the origin, whereas the cubic polynomial

P_3(x) = x - \frac{x^3}{3!}

hugs the curve closely across

[-\pi/2, \pi/2]

Taylor

\displaystyle\sum\dfrac{f^{(n)}(a)}{n!}(x-a)^{n}

Exponential

e^{x}=\displaystyle\sum\dfrac{x^{n}}{n!}

Geometric

\dfrac{1}{1-x}=\displaystyle\sum x^{n}

Convergence

radius / interval

⚠️ AP Exam Traps: Taylor Series & Error Bounds

Don't Forget the Factorial Denominator: In the Taylor coefficient formula $c_n = \frac{f^{(n)}(a)}{n!}$ , the denominator is $n!$ , not $n$ . If given $f'''(2) = 18$ , the coefficient of $(x-2)^3$ is $\frac{18}{3!} = \frac{18}{6} = 3$ , not $18$ !
Alternating Series Error vs. Lagrange Error Bound:
• If a series alternates with terms decreasing in magnitude to zero, the truncation error is bounded simply by the first omitted term: $|R_n(x)| \le |a_{n+1}|$ .
• If the series is not alternating, or if the question explicitly asks for the Lagrange Error Bound, use: $|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$ , where $M = \max_{t \in [a, x]} |f^{(n+1)}(t)|$ .
Endpoints in the Interval of Convergence: Finding the radius of convergence $R$ via the Ratio Test does NOT determine the interval endpoints. You must independently test both endpoints $x = a - R$ and $x = a + R$ using other convergence tests (p-series, AST, or comparison).
Substitute into Known Series: When finding series like $\cos(x^2)$ or $x e^{-x}$ , never take multiple derivatives. Instead, substitute directly into the standard series for $\cos u$ or $e^u$ and simplify.

Worked example · Maclaurin of

e^{x}

Write the Maclaurin series for $e^{x}$ . $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}.$

Worked example — Lagrange error bound

The Maclaurin series for $\cos x$ is $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots$ . Use the Lagrange error bound to bound the error when approximating $\cos(0.1)$ with the first two non-zero terms.

Using $P_2(x)=1-\dfrac{x^2}{2}$ . Error $\leq\dfrac{|x|^4}{4!}\cdot\max|f^{(4)}|$ . Since $|f^{(4)}|=|\cos c|\leq1$ :

|E|\leq\frac{(0.1)^4}{24}=\frac{0.0001}{24}\approx4.2\times10^{-6}.

Find the first three terms of the Taylor series for

f(x)=\ln(1+x)

centred at

x=0

and state its radius of convergence.

f(0)=0

f'(0)=1

f''(0)=-1

\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots

Radius of convergence

R=1

A function

f

has derivatives of all orders with

f(2) = 3

f'(2) = -4

f''(2) = 6

, and

f'''(2) = -12

. Write the third-degree Taylor polynomial

P_3(x)

for

f

centered at

x = 2

Using the Taylor polynomial formula

P_3(x) = f(2) + f'(2)(x-2) + \frac{f''(2)}{2!}(x-2)^2 + \frac{f'''(2)}{3!}(x-2)^3

:
Substitute the given derivatives:

P_3(x) = 3 - 4(x-2) + \frac{6}{2!}(x-2)^2 + \frac{-12}{3!}(x-2)^3 = 3 - 4(x-2) + 3(x-2)^2 - 2(x-2)^3.

📝 Chapter Quiz

Questions Level

✶

Worked FRQs

Taylor series template

f(x)=\sum_{n=0}^\infty\dfrac{f^{(n)}(a)}{n!}(x-a)^n

Lagrange error bound

|R_n|\leq\dfrac{M|x-a|^{n+1}}{(n+1)!}

Polar area

A=\dfrac{1}{2}\int_\alpha^\beta r^2\,d\theta

Series convergence

Apply ratio test; check endpoints; state interval of convergence

Parametric trajectory for FRQ 1:

x(t) = t^2, y(t) = t^3 - 3t

. At parameter value

t=2

, the position is

(4, 2)

and the slope of the tangent line is

\frac{dy}{dx} = \frac{3t^2 - 3}{2t}\Big|_{t=2} = \frac{9}{4} = 2.25

FRQ 1 · Parametric motion

A curve is given by $x(t)=t^2$ , $y(t)=t^3-3t$ . Find the slope $\dfrac{dy}{dx}$ at $t=2$ .

Differentiate each: $\dfrac{dx}{dt}=2t$ , $\dfrac{dy}{dt}=3t^2-3$ . Then

\frac{dy}{dx}=\frac{3t^2-3}{2t}\Big|_{t=2}=\frac{12-3}{4}=\frac{9}{4}.

FRQ 2 · Interval of convergence

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}$ .

Ratio test: $\left|\dfrac{x^{n+1}/(n+1)}{x^n/n}\right|=|x|\cdot\dfrac{n}{n+1}\to|x|$ , so it converges for $|x|<1$ . Check endpoints: $x=1$ gives the divergent harmonic series; $x=-1$ gives the convergent alternating harmonic series.

\text{Interval of convergence: } [-1,\,1).

FRQ 3 · Taylor Polynomial & Lagrange Error Bound

A function $f$ has derivatives of all orders with $f(0) = 2$ , $f'(0) = -3$ , $f''(0) = 5$ , and $f'''(0) = -12$ . The fourth derivative satisfies $|f^{(4)}(x)| \le 48$ for all $x \in [0, 0.5]$ .

(a) Third-degree Taylor polynomial: $P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 = 2 - 3x + \frac{5}{2}x^2 - 2x^3.$

(b) Approximation at $x = 0.2$ : $P_3(0.2) = 2 - 3(0.2) + 2.5(0.04) - 2(0.008) = 2 - 0.6 + 0.1 - 0.016 = 1.484.$

(c) Lagrange Error Bound: By the Lagrange Error Bound formula: $|f(0.2) - P_3(0.2)| \le \frac{\max |f^{(4)}(t)|}{4!}|0.2 - 0|^4 \le \frac{48}{24}(0.0016) = 2(0.0016) = 0.0032.$

FRQ 4 · Polar Area & Rate of Change

A polar curve is defined by $r = 3 + 2\cos\theta$ for $0 \le \theta \le 2\pi$ .

(a) Area enclosed by the curve: $A = \frac{1}{2}\int_0^{2\pi} (3 + 2\cos\theta)^2\,d\theta = \frac{1}{2}\int_0^{2\pi} (9 + 12\cos\theta + 4\cos^2\theta)\,d\theta.$

Using $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ , the integral yields $\frac{1}{2}(18\pi + 4\pi) = 11\pi$ .

(b) Value of $\frac{dr}{d\theta}$ at $\theta = \frac{\pi}{3}$ : $\frac{dr}{d\theta} = -2\sin\theta \implies \frac{dr}{d\theta}\Big|_{\pi/3} = -2\left(\frac{\sqrt{3}}{2}\right) = -\sqrt{3}.$

Since $r(\pi/3) = 3 + 2(1/2) = 4 > 0$ and $\frac{dr}{d\theta} < 0$ , the distance between the curve and the origin is decreasing at $\theta = \frac{\pi}{3}$ .

⚠️ AP Reader Scoring Traps & BC FRQ Conventions

FRQ 6 Taylor Series Template: Always memorize the 4 foundational Maclaurin series: $e^x$ , $\sin x$ , $\cos x$ , and $\frac{1}{1-x}$ . When asked to find series for related functions, manipulate, differentiate, or integrate term-by-term. Never recalculate derivatives from scratch unless explicitly instructed.
Lagrange Bound Justification: On Lagrange error bound problems, clearly identify each piece of $\frac{M}{(n+1)!}|x-c|^{n+1}$ : state $n$ , $(n+1)!$ , the interval $[c, x]$ , and the bound $M$ . Omitting the factorial or the absolute value power forfeits the point.
Explaining Definite Integrals in Context: When asked to 'interpret the meaning of $\int_0^4 \sqrt{(x')^2 + (y')^2}\,dt$ ', you must state 3 distinct components: (1) what the value represents (total distance traveled), (2) the correct units (e.g. meters), and (3) the exact time interval (from $t = 0$ to $t = 4$ seconds).

An AP Calculus BC student scores 36 out of 45 on Section I (Multiple Choice) and 38 out of 54 on Section II (Free Response). Compute the student's composite score (scaled out of 108 total points) according to standard AP weighting (Section I = 1.20 × raw score, Section II = 1.00 × raw score).

\text{Section I Composite} = 1.20 \times 36 = 43.2

\text{Section II Composite} = 1.00 \times 38 = 38.0

\text{Total Composite Score} = 43.2 + 38.0 = 81.2\text{ out of 108}

(approximately 75.2%), which comfortably qualifies for an AP score of 5 (typically requiring ~66–72 composite points depending on the exam curve).

📝 Chapter Quiz

Questions Level

⊘

Exam practice

For $x=\cos t,\ y=\sin t$ , find $\dfrac{dy}{dx}$ at $t=\tfrac{\pi}{2}$ .

FRQ

$\dfrac{dy}{dx}=\dfrac{\cos t}{-\sin t}=-\cot t$ ; at $t=\tfrac{\pi}{2}$ , $\cot\tfrac{\pi}{2}=0$ .

\frac{dy}{dx}=0.

Does $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ converge or diverge?

MCQ

A $p$ -series with $p=2>1$ , so it converges (in fact to $\pi^2/6$ ).

Find the area enclosed by the polar curve $r=2$ .

FRQ

$A=\tfrac12\displaystyle\int_0^{2\pi} r^2\,d\theta=\tfrac12\int_0^{2\pi}4\,d\theta=\tfrac12\cdot 4\cdot 2\pi=4\pi$ — the circle of radius 2.

Differentiate $y=x^4-3x^2$ .

MCQ

y'=4x^3-6x.

Evaluate $\displaystyle\int_0^{2} 6x^2\,dx$ .

MCQ

\big[2x^3\big]_0^{2}=16.

Evaluate $\displaystyle\lim_{x\to 0}\frac{\tan x}{x}$ .

MCQ

1.

Differentiate $y=e^{x^2}$ .

MCQ

y'=2x\,e^{x^2}.

Evaluate $\displaystyle\int x\,e^{x^2}\,dx$ .

MCQ

Let $u=x^2$ , $x\,dx=\tfrac12\,du$ :

\tfrac12 e^{x^2}+C.

Differentiate $y=\arctan x$ .

MCQ

y'=\frac{1}{1+x^2}.

Evaluate $\displaystyle\int \frac{1}{1+x^2}\,dx$ .

MCQ

\arctan x+C.

Evaluate $\displaystyle\int x\,e^{x}\,dx$ .

FRQ

By parts:

e^{x}(x-1)+C.

Evaluate $\displaystyle\int \ln x\,dx$ .

FRQ

x\ln x-x+C.

Evaluate $\displaystyle\int x\sin x\,dx$ .

FRQ

By parts, $u=x,\ dv=\sin x\,dx$ :

-x\cos x+\sin x+C.

Evaluate $\displaystyle\int \frac{1}{x^2-1}\,dx$ .

FRQ

Partial fractions $\tfrac12\big(\tfrac{1}{x-1}-\tfrac{1}{x+1}\big)$ :

\tfrac12\ln\left|\frac{x-1}{x+1}\right|+C.

Evaluate $\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx$ .

FRQ

Improper: $\big[-\tfrac1x\big]_1^{\infty}=0-(-1)$ :

1.

Evaluate $\displaystyle\int_0^{\infty} e^{-x}\,dx$ .

FRQ

\big[-e^{-x}\big]_0^{\infty}=0-(-1)=1.

Evaluate $\displaystyle\sum_{n=0}^{\infty}\Big(\tfrac12\Big)^{n}$ .

MCQ

\frac{1}{1-\tfrac12}=2.

Evaluate $\displaystyle\sum_{n=0}^{\infty}\Big(\tfrac13\Big)^{n}$ .

MCQ

\frac{1}{1-\tfrac13}=\frac32.

Evaluate $\displaystyle\sum_{n=1}^{\infty}\Big(\tfrac25\Big)^{n}$ .

MCQ

First term $\tfrac25$ , ratio $\tfrac25$ :

\frac{2/5}{1-2/5}=\frac23.

Does $\displaystyle\sum_{n=1}^{\infty}\frac1n$ converge?

MCQ

The harmonic series:

\text{diverges}.

Does $\displaystyle\sum_{n=0}^{\infty}(0.9)^{n}$ converge, and to what?

MCQ

Geometric with $|r|<1$ :

\frac{1}{1-0.9}=10.

Does $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ converge?

MCQ

$p$ -series with $p=\tfrac12\le 1$ :

\text{diverges}.

Does $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n+1}$ converge?

MCQ

Terms $\to 1\neq 0$ , so by the $n$ th-term test:

\text{diverges}.

Evaluate $\displaystyle\sum_{n=0}^{\infty} 3\Big(\tfrac14\Big)^{n}$ .

MCQ

\frac{3}{1-\tfrac14}=4.

For what $x$ does $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ converge?

FRQ

Ratio test: $\big|\tfrac{x}{n+1}\big|\to 0<1$ always:

\text{all real } x.

Write the Maclaurin series for $e^{x}$ .

MCQ

e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+x+\tfrac{x^2}{2}+\cdots.

Give the first three nonzero terms of the Maclaurin series for $\sin x$ .

FRQ

x-\frac{x^3}{6}+\frac{x^5}{120}.

Give the first three terms of the Maclaurin series for $\cos x$ .

FRQ

1-\frac{x^2}{2}+\frac{x^4}{24}.

Use a two-term Maclaurin polynomial for $e^{x}$ to approximate $e^{0.1}$ .

FRQ

$e^{x}\approx 1+x$ at $x=0.1$ :

1+0.1=1.1.

Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty} x^{n}$ .

FRQ

Geometric, converges for $|x|<1$ :

(-1,1).

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}$ .

FRQ

Ratio test gives $|x|<1$ . Endpoints: $x=1$ diverges (harmonic), $x=-1$ converges (alternating):

[-1,1).

Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ .

FRQ

Telescoping: $\tfrac1n-\tfrac{1}{n+1}$ :

\sum_{n=1}^{\infty}\Big(\tfrac1n-\tfrac{1}{n+1}\Big)=1.

Does $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ converge?

MCQ

The alternating harmonic series satisfies the alternating series test:

\text{converges}.

Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{2^n}$ .

FRQ

Geometric in $\tfrac{x}{2}$ ; converges for $\big|\tfrac{x}{2}\big|<1$ :

R=2.

For $x=t^2,\ y=t^3$ , find $\dfrac{dy}{dx}$ .

MCQ

\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}.

For $x=e^{t},\ y=e^{2t}$ , find $\dfrac{dy}{dx}$ .

MCQ

\frac{dy}{dx}=\frac{2e^{2t}}{e^{t}}=2e^{t}.

For $x=t^2,\ y=2t$ , find $\dfrac{dy}{dx}$ .

MCQ

\frac{dy}{dx}=\frac{2}{2t}=\frac1t.

For $x=3t,\ y=t^2-1$ , find $\dfrac{dy}{dx}$ .

MCQ

\frac{dy}{dx}=\frac{2t}{3}.

Find the arc length of $x=t,\ y=t$ for $t\in[0,1]$ .

FRQ

L=\int_0^{1}\sqrt{1^2+1^2}\,dt=\sqrt2.

For $x=t^3,\ y=t^2$ , find $\dfrac{dy}{dx}$ at $t=1$ .

MCQ

\frac{2t}{3t^2}\Big|_{t=1}=\frac23.

Find the speed of a particle with $x=t^2,\ y=t^2$ at $t=1$ .

FRQ

Speed $=\sqrt{(2t)^2+(2t)^2}=2\sqrt2\,t$ at $t=1$ :

2\sqrt2.

Convert the polar point $\big(2,\tfrac{\pi}{2}\big)$ to rectangular coordinates.

MCQ

$x=2\cos\tfrac{\pi}{2}=0,\ y=2\sin\tfrac{\pi}{2}=2$ :

(0,2).

Find the area enclosed by $r=4$ .

MCQ

\tfrac12\int_0^{2\pi}16\,d\theta=16\pi.

Convert the rectangular point $(0,-3)$ to polar coordinates.

MCQ

$r=3$ , and the point is on the negative $y$ -axis:

\big(3,\tfrac{3\pi}{2}\big).

Find the area enclosed by the cardioid $r=1+\cos\theta$ .

FRQ

\tfrac12\int_0^{2\pi}\big(1+\cos\theta\big)^2\,d\theta=\frac{3\pi}{2}.

Convert the rectangular point $(1,1)$ to polar coordinates.

MCQ

$r=\sqrt2$ , $\theta=\arctan 1=\tfrac{\pi}{4}$ :

\big(\sqrt2,\tfrac{\pi}{4}\big).

Find the area enclosed by $r=3$ .

MCQ

\tfrac12\int_0^{2\pi}9\,d\theta=9\pi.

The polar curve $r=2\cos\theta$ is a circle. Find its enclosed area.

FRQ

It is a circle of radius 1; equivalently $\tfrac12\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta$ :

\pi.

Solve $\dfrac{dy}{dx}=3y$ with $y(0)=1$ .

FRQ

y=e^{3x}.

Solve $\dfrac{dy}{dx}=-2y$ with $y(0)=10$ .

MCQ

Exponential decay, $k=-2$ :

y=10e^{-2x}.

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