AP Calculus AB & BC
Limit, türev, integral ve seriler: konu anlatımı, formüller ve hata teşhisli sorular.
AP Calculus Hibrit Dijital Sınav Yönergesi
Resmi College Board kılavuzuna göre Calculus sınav kuralları:
Bölüm I: Çoktan Seçmeli (MCQ) • 42 Soru
• Part A (29 Soru, 62 dk): Hesap makinesi KULLANILMAZ. Bluebook üzerinde seçenek eleme ve soru işaretleme yapılabilir.
• Part B (13 Soru, 38 dk): Grafik hesap makinesi (TI-84 Plus CE / TI-Nspire CX) ZORUNLU. Sayısal türev ve integral hesaplatılır.
• May 2027 sınavı biçimi (College Board AP Central; eski biçim 45 MCQ / 1 sa 45 dk idi).
Bölüm II: Serbest Yanıtlı (FRQ) • 6 Soru
• Part A (2 Soru, 30 dk): Grafik hesap makinesi ZORUNLU. Sorular Bluebook ekranından okunur, yanıtlar basılı kitapçığa tükenmez kalemle el yazısıyla yazılır.
• Part B (4 Soru, 60 dk): Hesap makinesi KULLANILMAZ. Tamamen analitik ispat ve denklem çözümü.
Kritik Calculus Formülleri (Tıkla ve Kopyala)
Şart: f fonksiyonu [a, b] aralığında sürekli ve (a, b) aralığında türevlenebilir olmalıdır.
Türev ve integralin birbirinin tersi olduğunu kanıtlar. Zincir kuralı çarpanı unutulmamalıdır.
Dönme eksenine göre dış yarıçap R(x) ile iç yarıçap r(x) farkının integrali alınır.
College Board kuralı: Pay ve paydanın limitlerinin ayrı ayrı 0 veya sonsuz olduğu önce açıkça yazılmalıdır!
Graphing Calculator Command Card: The Four Uses That Need No Shown Work
On the calculator-active parts (Section I Part B and Section II Part A) a graphing calculator may be used without showing the mathematics for four tasks only: graphing a function in a chosen window, solving an equation numerically, finding the derivative of a function at a point, and calculating a definite integral. The official directions say you must still write the setup (the equation, function or integral you used); any other built-in feature or program needs the mathematical steps written out.
| Task | TI-84 Plus CE | TI-Nspire CX (non-CAS) | What to write on paper |
|---|---|---|---|
| 1. Graph a function | Y= → type $f(x)$ → WINDOW (Xmin, Xmax, Ymin, Ymax) → GRAPH; quick fit: ZOOM → 0: ZoomFit |
Graphs page → type in f1(x)= → Menu → 4: Window/Zoom → 1: Window Settings |
The picture earns nothing by itself. State what you read from it and why, e.g. "$f(x) > 0$ on $1 < x < 3$ because ..." |
| 2. Solve an equation (zero or intersection) | Enter both sides as Y1, Y2 → 2nd TRACE (CALC) → 5: intersect; for $Y_1 = 0$ use 2: zero |
Graphs page → Menu → 6: Analyze Graph → Zero / Intersection; or Calculator page → Menu → 3: Algebra → 1: Numerical Solve, e.g. nSolve(38/(25+t^2)=1.282008,t) |
Write the equation, then the result: $C'(t) = 1.282008 \Rightarrow t = 2.154$ |
| 3. Derivative at a point | MATH → 8: nDeriv(, e.g. nDeriv(2sin(X)^2,X,1.3) |
Menu → 4: Calculus → 1: Numerical Derivative at a Point (variable, value, order) |
$\left.\frac{dr}{d\theta}\right|_{\theta = 1.3} = 1.031$, never "nDeriv(...)" |
| 4. Definite integral | MATH → 9: fnInt(, e.g. fnInt(Y1,X,0,4) |
Menu → 4: Calculus → 2: Numerical Integral (fill in the template) |
$\frac{1}{4}\int_0^4 C(t)\,dt = 2.778$, never "fnInt(...)" |
• Radian mode. TI-84:
MODE → RADIAN. TI-Nspire: select the RAD/DEG indicator at the top of the screen, or Home → Settings → Document Settings.• Store, do not retype. Keep intermediate values in memory (TI-84:
STO→; TI-Nspire: ctrl var). Final answers need three correct decimal places, rounded or truncated; at most one point per question is lost for inappropriate rounding.• Standard notation only. $\int_1^5 x^2\,dx$ may not be written as
fnInt(X^2,X,1,5).• Bluebook option. The directions allow a handheld graphing calculator or the calculator built into the Bluebook application.
Kaynaklar: resmi kurallar 2025 FRQ yönergesinden (ap25-frq-calculus-bc.pdf, s. 2) ve puanlama kılavuzunun genel notlarından (ap25-sg-calculus-bc.pdf, s. 2). TI-Nspire menü yolları yerel "TI-Nspire Calculator Guide.pdf" (s. 7, 11, 12; yayıncı doğrulanamadı) ile karşılaştırıldı. Bu rehber paket dışında tutulduğu için bağlantı verilmedi. Rehberin 11. sayfasında Numerical Integral "1:" olarak geçiyor. Doğru sıra "2:"dir; rehberin 7. sayfasındaki Sum = 3, Product = 4 sırası da bunu gösteriyor. Onaylı hesap makinesi listesi: College Board calculator policy.
Reference Library: Formula Sheets, Unit Circle and Technique Charts
One page: limits, derivative and integral rules, key theorems and geometry formulas.
Kaynak: © MyMathsCloud
25 pages: theorem statements with hypotheses, derivative and integral tables, BC series tests.
Kaynak: Yayıncı doğrulanamadı
Power, exponential, logarithmic, trigonometric and inverse trigonometric derivatives; product, quotient and chain rule with worked examples (Units 2–3).
Kaynak: © MyMathsCloud
First- and second-derivative sign analysis for extrema, concavity and inflection points (Unit 5).
Kaynak: Yayıncı doğrulanamadı
When to use parts, choosing $u$ and $dv$, repeated and cyclic cases (Topic 6.11, BC).
Kaynak: © MyMathsCloud
Substitution, parts, partial fractions and separable ODE drills (no answer key). Section III, trigonometric substitution, is beyond the AP course.
Kaynak: Yazar doğrulanamadı
Exact values at multiples of $\frac{\pi}{6}$ and $\frac{\pi}{4}$, quadrant signs. Needed on every no-calculator part.
Kaynak: © mymathscloud
Algebra, logarithm, trigonometric identity and geometry formulas behind many calculus steps.
Kaynak: © MyMathsCloud
Cone, cylinder and sphere formulas used in related-rates problems (Topic 4.5).
Kaynak: Yayıncı doğrulanamadı
• Derivatives: $\frac{d}{dx}\tan x = \sec^2 x$, $\frac{d}{dx}\sec x = \sec x\tan x$, $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$, $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$, $\frac{d}{dx}a^x = a^x\ln a$.
• Integrals: $\int \frac{dx}{x} = \ln|x| + C$, $\int \sec^2 x\,dx = \tan x + C$, $\int \frac{dx}{1 + x^2} = \arctan x + C$.
• Applications: average value $\frac{1}{b - a}\int_a^b f(x)\,dx$; washer $\pi\int_a^b \left(R(x)^2 - r(x)^2\right)dx$.
• BC geometry: arc length $\int_a^b \sqrt{1 + (f'(x))^2}\,dx$; parametric speed $\sqrt{(x'(t))^2 + (y'(t))^2}$; polar area $\frac{1}{2}\int_\alpha^\beta r^2\,d\theta$ with $x = r\cos\theta$, $y = r\sin\theta$.
• BC equations: Euler step $y_{\text{new}} = y_{\text{old}} + \Delta x\cdot\frac{dy}{dx}$; logistic $\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$, fastest growth at $P = \frac{L}{2}$.
• BC series: $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$, $\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$, $\cos x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}$, $\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n$ for $|x| < 1$; Lagrange bound $\left|R_n(x)\right| \le \frac{\max\left|f^{(n+1)}\right|}{(n+1)!}\,|x - a|^{n+1}$.
Calculus AB & BC Ünite Dağılımı ve Sınav Ağırlıkları
Çoktan seçmeli bölüm ünite ağırlıkları — kaynak: College Board AP Central (AB · BC), 10 Ekim 2026 kontrolü:
Bu Ayın 5 Kritik FRQ Çözüm Laboratuvarı
College Board puanlama rubriğinde (Scoring Guidelines) tam puan getiren kritik FRQ soru kalıpları:
• Taktik: "Parçacık ne zaman hızlanır?" sorusunda v(t) ve a(t) işaretlerinin aynı olduğu aralıkları işaret tablosuyla açıkla.
• Toplam katedilen yol için mutlaka mutlak değer integralini yaz: ∫ |v(t)| dt.
• Taktik: Teoremi uygulamadan önce şu cümleyi harfiyen yaz: "f(x) kapalı [a, b] aralığında sürekli ve açık (a, b) aralığında türevlenebilir olduğundan...". Bu cümle tek başına 1 tam puan değerindedir!
• Taktik: dy/dx = g(x)h(y) denkleminde değişkenleri ilk adımda ayırmazsan (Separation of Variables), kalan 5 puanın tamamı anında yanar! Entegrasyon sabiti + C'yi hemen ilk integral adımında koy.
• Taktik: g(x) = ∫[0 to x] f(t) dt verildiğinde, g'(x) = f(x) ve g''(x) = f'(x) olduğunu yazarak f grafiğinin altındaki alanları geometrik olarak hesapla.
AP Calculus Theoretical Foundations & Concept Mastery
Formal Definition of Continuity: A function $f(x)$ is continuous at $x = c$ if and only if three conditions are satisfied:
1. f(c) is defined (c is in the domain of f)
2. \(\lim_{x \to c} f(x) \text{ exists (i.e., }\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L\text{)}\)
3. \(\lim_{x \to c} f(x) = f(c)\)
The Derivative as an Instantaneous Rate of Change: The derivative is defined rigorously as the limit of the difference quotient:
\(f'(x) = \lim_{h \to 0} [f(x + h) - f(x)] / h = \lim_{x \to a} [f(x) - f(a)] / (x - a)\)
Theorem: Differentiability implies continuity ($f \text{ differentiable at } c \implies f \text{ continuous at } c$). The converse is FALSE (e.g., $f(x) = |x|$ at $x = 0$ has a sharp corner).
f'(c) = [f(b) - f(a)] / (b - a)
Geometric Interpretation: The instantaneous rate of change (tangent slope) equals the average rate of change (secant slope) at least once within the interval.
1. Find all critical points in $(a, b)$ where $f'(x) = 0$ or $f'(x)$ is undefined.
2. Evaluate $f(x)$ at all critical points AND at the endpoints $a$ and $b$.
3. The largest value is the absolute maximum; the smallest is the absolute minimum.
The Fundamental Theorem of Calculus bridges the two central branches of mathematical analysis:
FTC Part 1 (Derivative of an Accumulation Function): If $f$ is continuous on $[a, b]$, then $g(x) = \int_a^x f(t) dt$ is continuous on $[a, b]$, differentiable on $(a, b)$, and:
g'(x) = d/dx [ ∫_a^x f(t) dt ] = f(x)
Chain Rule Extension: \(d/dx [ \int _a^{u(x)} f(t) dt ] = f(u(x)) \cdot u'(x)\)
FTC Part 2 (Evaluation Theorem & Net Change):
∫_a^b f(x) dx = F(b) - F(a) where F'(x) = f(x)
Final Position = Initial Position + Displacement: s(b) = s(a) + ∫_a^b v(t) dt
Taylor Polynomial Centered at $x = c$:
\(P_n(x) = f(c) + f'(c)(x - c) + [f''(c)/2!](x - c)^2 + ... + [f^{(n)}(c)/n!](x - c)^n\)
For Maclaurin series ($c = 0$), memorize the 4 canonical power series:
• $e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... \quad (\forall x \in \mathbb{R})$
• $\sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... \quad (\forall x \in \mathbb{R})$
• $\cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... \quad (\forall x \in \mathbb{R})$
• $\frac{1}{1 - x} = \sum_{n=0}^\infty x^n = 1 + x + x^2 + x^3 + ... \quad (|x| < 1)$
\(|R_n(x)| = |f(x) - P_n(x)| \le [M / (n + 1)!] \cdot |x - c|^{n + 1}\)
Unit 1: Limits and Continuity
Intuitive Definition: We write \(\lim_{x \to c} f(x) = L\) if as $x$ approaches $c$ from both sides (left and right), $f(x)$ becomes arbitrarily close to the finite real number $L$. Notice that $x \ne c$; limits describe behavior near a point, completely independent of the value $f(c)$.
Let $f$ be a function defined on an open interval containing $c$, except possibly at $c$ itself. We say that \(\lim_{x \to c} f(x) = L\) if for every $\varepsilon > 0$, there exists a corresponding $\delta > 0$ such that:
0 < |x - c| < δ ⟹ |f(x) - L| < ε.
Proof Mechanism: Given a challenge error tolerance $\varepsilon$ around the target limit $L$, we must construct an input radius $\delta$ around $c$ such that all $x$ within the punctured interval $(c - \delta, c) \cup (c, c + \delta)$ map inside $(L - \varepsilon, L + \varepsilon)$.
Existence Criterion: The two-sided limit exists if and only if both one-sided limits exist and are equal:
\(\lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L\).
When evaluating $\lim_{x \to c} f(x)$, always begin with Direct Substitution:
• If $f(c) = L$ (a real number) and $f$ is continuous, $\lim_{x \to c} f(x) = f(c)$.
• If substitution yields $\frac{k}{0}$ ($k \ne 0$), the limit is infinite ($\pm\infty$) or does not exist (Vertical Asymptote).
• If substitution yields $\frac{0}{0}$, this is an Indeterminate Form! The limit exists and can be resolved through algebraic cancellation or trigonometric identities.
1. Factoring & Cancellation: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 6$.
2. Conjugate Rationalization: For radicals, multiply by $\frac{\sqrt{A} + B}{\sqrt{A} + B}$ to exploit the difference of squares: $(\sqrt{A} - B)(\sqrt{A} + B) = A - B^2$.
3. Complex Fractions (LCD Method): Multiply numerator and denominator by the common denominator to clear embedded fractions.
\(\lim_{x \to 0} [sin(x) / x] = 1\) and \(\lim_{x \to 0} [(1 - cos(x)) / x] = 0\).
Generalized: $\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b}$ and $\lim_{x \to 0} \frac{\tan(ax)}{\sin(bx)} = \frac{a}{b}$.
Theorem Statement: If $g(x) \le f(x) \le h(x)$ for all $x$ in an open interval containing $c$ (except possibly at $c$), and if:
\(\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L\),
then \(\lim_{x \to c} f(x) = L\).
1. Bounding inequality: $-1 \le \sin(1/x) \le 1$ for all $x \ne 0$.
2. Multiply across by $x^2 \ge 0$: $-x^2 \le x^2 \sin(1/x) \le x^2$.
3. Evaluate bounds: $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$.
4. Conclusion: By the Squeeze Theorem, $\lim_{x \to 0} x^2 \sin(1/x) = 0$.
The 3-Condition Continuity Test (College Board Rubric Requirement):
A function $f(x)$ is continuous at $x = c$ if and only if ALL THREE conditions hold:
1. f(c) is defined (c is in the domain of f).
2. \(\lim_{x \to c} f(x) \text{ exists (i.e., }\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)\text{)}\).
3. \(\lim_{x \to c} f(x) = f(c)\).
• Removable Discontinuity (Hole): $\lim_{x \to c} f(x)$ exists, but $f(c)$ is undefined or $\lim \ne f(c)$. Can be repaired by redefining $f(c) = L$.
• Jump Discontinuity: $\lim_{x \to c^-} f(x) \ne \lim_{x \to c^+} f(x)$ (both one-sided limits are finite but unequal). Common in piecewise and $|x|/x$ functions.
• Infinite Discontinuity (Essential): At least one one-sided limit equals $\pm\infty$ (Vertical Asymptote). Occurs when denominator is zero and cannot be canceled.
Theorem Statement: If $f$ is continuous on the closed interval $[a, b]$, and $N$ is any value strictly between $f(a)$ and $f(b)$ (where $f(a) \ne f(b)$), then there exists at least one number $c \in (a, b)$ such that:
f(c) = N.
To earn all rubric points on an IVT FRQ, write these 3 exact sentences:
1. "Since $f(x)$ is continuous on the closed interval $[a, b]$..." (1 pt)
2. "Since $f(a) = \text{[val]} < 0 < \text{[val]} = f(b)$..." (1 pt)
3. "By the Intermediate Value Theorem, there exists at least one $c \in (a, b)$ such that $f(c) = 0$." (1 pt)
Horizontal Asymptotes (HA): The line $y = L$ is a horizontal asymptote of $y = f(x)$ if:
\(\lim_{x \to \infty } f(x) = L \quad\text{or}\quad \lim_{x \to -\infty } f(x) = L\).
A function can have at most TWO horizontal asymptotes (one as $x \to +\infty$ and another as $x \to -\infty$).
For $x < 0$, $\sqrt{x^2} = |x| = -x$.
Therefore, $\lim_{x \to -\infty} \frac{\sqrt{4x^2 + 1}}{3x - 2} = \lim_{x \to -\infty} \frac{|x|\sqrt{4 + 1/x^2}}{x(3 - 2/x)} = \lim_{x \to -\infty} \frac{-x\sqrt{4 + 1/x^2}}{x(3 - 2/x)} = -\frac{2}{3}$! (Not $+2/3$!)
Unit 1 Limits & Continuity: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 2: Differentiation — Definition and Fundamental Derivative Rules
The Derivative as a Function: The derivative of a function $f(x)$ with respect to $x$ is defined as the limit of the difference quotient as the interval width $h \to 0$:
\(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \frac{dy}{dx}\)
Alternative Definition at a Fixed Point $x = c$:
\(f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}\)
Many AP multiple-choice questions do not ask you to compute a limit by algebraic manipulation; instead, they give you a limit that represents $f'(c)$ in disguise!
Example: $\lim_{h \to 0} \frac{\cos(\frac{\pi}{3} + h) - \frac{1}{2}}{h}$ represents $\frac{d}{dx}[\cos x]$ evaluated at $x = \frac{\pi}{3}$. Since $f'(x) = -\sin x$, the limit equals $-\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$!
Tangent and Normal Line Equations:
• Tangent Line at $(c, f(c))$: $y - f(c) = f'(c)(x - c)$
• Normal Line (Perpendicular) at $(c, f(c))$: $y - f(c) = -\frac{1}{f'(c)}(x - c) \quad (f'(c) \ne 0)$.
The Golden Theorem: If a function $f$ is differentiable at $x = c$, then $f$ is continuous at $x = c$.
We must prove that $\lim_{x \to c} f(x) = f(c)$, which is equivalent to proving $\lim_{x \to c} [f(x) - f(c)] = 0$.
1. For all $x \ne c$, write the identity: $f(x) - f(c) = \frac{f(x) - f(c)}{x - c} \cdot (x - c)$.
2. Take limits as $x \to c$ on both sides using the Product Law of Limits:
$$\lim_{x \to c} [f(x) - f(c)] = \lim_{x \to c} \left[\frac{f(x) - f(c)}{x - c}\right] \cdot \lim_{x \to c} (x - c) = f'(c) \cdot 0 = 0$$.
3. Since $\lim_{x \to c} [f(x) - f(c)] = 0$, it follows that $\lim_{x \to c} f(x) = f(c)$. Hence $f$ is continuous at $c$. $\blacksquare$
A function can be perfectly continuous at a point but fail to be differentiable. The 4 causes of non-differentiability are:
1. Sharp Corner (Node): Left-hand derivative $\ne$ Right-hand derivative (e.g., $f(x) = |x|$ at $x=0$, where $f'_-(0) = -1$ and $f'_+(0) = +1$).
2. Cusp: One-sided derivative approaches $+\infty$ while the other approaches $-\infty$ (e.g., $f(x) = x^{2/3}$ at $x = 0$).
3. Vertical Tangent: Both one-sided derivatives approach $+\infty$ or both approach $-\infty$ (e.g., $f(x) = x^{1/3}$ at $x = 0$).
4. Discontinuity: If $f$ is not continuous at $c$, it cannot be differentiable at $c$.
The General Power Rule: For any real exponent $n \in \mathbb{R}$:
\(\frac{d}{dx}[x^n] = n \cdot x^{n - 1}\)
Radicals & Negative Exponents: Always rewrite roots and denominators into rational powers before taking derivatives:
• $\frac{d}{dx}\left[\frac{1}{x^3}\right] = \frac{d}{dx}[x^{-3}] = -3x^{-4} = -\frac{3}{x^4}$
• $\frac{d}{dx}[\sqrt[3]{x^2}] = \frac{d}{dx}[x^{2/3}] = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$
• Natural Exponential: $\frac{d}{dx}[e^x] = e^x$
• General Base Exponential ($a > 0$): $\frac{d}{dx}[a^x] = a^x \cdot \ln(a)$
• Natural Logarithm: $\frac{d}{dx}[\ln(x)] = \frac{1}{x} \quad (x > 0)$
• General Base Logarithm: $\frac{d}{dx}[\log_a(x)] = \frac{1}{x \ln(a)}$
Do NOT confuse $x^n$ (variable base, constant power $\implies$ Power Rule) with $a^x$ (constant base, variable power $\implies$ Exponential Rule):
• $\frac{d}{dx}[x^e] = e x^{e-1}$ (Power Rule)
• $\frac{d}{dx}[e^x] = e^x$ (Exponential Rule)
• $\frac{d}{dx}[e^\pi] = 0$ (Constant Rule! $e^\pi$ has no variable $x$!)
The derivatives of the six fundamental trigonometric functions must be memorized instantaneously:
The Product Rule: The derivative of a product is NOT the product of derivatives!
\(\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)\)
Let $P(x) = f(x)g(x)$. Using the limit definition:
$$P'(x) = \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}$$
Add and subtract the cross-term $f(x+h)g(x)$ in the numerator:
$$= \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h}$$
$$= \lim_{h \to 0} \left[ f(x+h) \cdot \frac{g(x+h) - g(x)}{h} + g(x) \cdot \frac{f(x+h) - f(x)}{h} \right]$$
Since $f$ is differentiable, it is continuous, so $\lim_{h \to 0} f(x+h) = f(x)$. Thus:
$$P'(x) = f(x)g'(x) + g(x)f'(x). \quad \blacksquare$$
The Quotient Rule: For $g(x) \ne 0$:
\(\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} = \frac{\text{Low } d(\text{High}) - \text{High } d(\text{Low})}{\text{Low}^2}\)
Because subtraction is not commutative, the numerator of the Quotient Rule MUST be $\mathbf{f'(x)g(x) - f(x)g'(x)}$ (Low $d$-High minus High $d$-Low). Swapping the terms to $f(x)g'(x) - f'(x)g(x)$ multiplies your answer by $-1$, which is always included as a deliberate distractor by College Board!
Unit 2 Differentiation: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 3: Composite, Implicit, and Inverse Functions
Beyond elementary rules, calculus models interdependent systems via the Chain Rule, implicit relations, and inverse functions. Here are the core theorems, derivations, and exam-level techniques required for College Board 5/5 mastery:
The Chain Rule: If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite function $F(x) = f(g(x))$ is differentiable at $x$:
\(F'(x) = f'(g(x)) \cdot g'(x) \quad \text{or in Leibniz notation:} \quad \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
When differentiating multi-layered composite functions, work systematically from the outside in ("peeling the onion"):
For $y = f(g(h(x)))$, $\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$.
Example: $\frac{d}{dx}[\sin^4(3x^2)] = 4\sin^3(3x^2) \cdot \frac{d}{dx}[\sin(3x^2)] = 4\sin^3(3x^2) \cdot \cos(3x^2) \cdot (6x) = 24x\sin^3(3x^2)\cos(3x^2)$.
When an equation relates $x$ and $y$ implicitly (cannot be easily solved for $y=f(x)$), differentiate both sides with respect to $x$, treating $y$ as a differentiable function of $x$ (applying the Chain Rule $\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}$):
1. Differentiate both sides of the equation with respect to $x$, applying the product, quotient, and chain rules wherever $y$ appears.
2. Collect all terms containing $\frac{dy}{dx}$ (or $y'$) on one side and all other terms on the opposite side.
3. Factor out $\frac{dy}{dx}$ and solve for it algebraically: $\frac{dy}{dx} = \frac{\text{terms without } y'}{\text{coefficients of } y'}$.
4. For Second Derivative $\frac{d^2y}{dx^2}$: Differentiate $\frac{dy}{dx}$ with respect to $x$ (often using the Quotient Rule), then substitute the expression for $\frac{dy}{dx}$ back into the result to obtain $\frac{d^2y}{dx^2}$ purely in terms of $x$ and $y$.
Derivative of an Inverse Function: If $f$ is strictly monotonic and differentiable with inverse $g = f^{-1}$, and $f(b) = a$ (so $g(a) = b$), then:
\(g'(a) = (f^{-1})'(a) = \frac{1}{f'(g(a))} = \frac{1}{f'(b)} \quad \text{provided } f'(b) \ne 0\)
Geometric Insight: Tangent lines of inverse functions across $y = x$ have reciprocal slopes.
•
\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1•
\frac{d}{dx}[\arccos(x)] = -\frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1• \(\frac{d}{dx}[\arctan(x)] = \frac{1}{1 + x^2}, \quad x \in \mathbb{R}\)
• \(\frac{d}{dx}[\text{arcsec}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}}, \quad |x| > 1\)
Unit 3 Composite, Implicit & Inverse Functions: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 4: Contextual Applications of Differentiation — Dynamic Modeling & Rates
The Fundamental Interpretation Template for AP FRQs:
When asked to "explain the meaning of $f'(c) = k$ in the context of the problem including units", your answer must strictly contain three components:
1. Specific Moment in Time ($t = c$): Always anchor to the exact input value with units (e.g., "At time $t = 5$ minutes...").
2. Contextual Quantity & Direction of Change: If $k > 0$, state the quantity is increasing; if $k < 0$, state the quantity is decreasing (never say "decreasing at a rate of $-k$").
3. Rate Value with Ratio Units: $\text{Units of } f'(t) = \frac{\text{units of } f}{\text{units of } t}$ (e.g., "$1.8^\circ\text{C}$ per minute").
For a particle moving along a straight line with position $x(t)$ (or $s(t)$):
• Velocity: $v(t) = x'(t) = \frac{dx}{dt}$. ($v > 0 \implies$ moving right/up; $v < 0 \implies$ moving left/down).
• Acceleration: $a(t) = v'(t) = x''(t) = \frac{d^2x}{dt^2}$.
• Particle Changes Direction: Requires $v(t) = 0$ AND $v(t)$ must change sign across that critical moment!
• Speed: The absolute magnitude of velocity: $\text{Speed} = |v(t)|$.
Never look at acceleration alone! Speed is the absolute magnitude of velocity:
• A particle is speeding up if and only if velocity and acceleration have the SAME SIGN ($v(t) \cdot a(t) > 0$).
• A particle is slowing down if and only if velocity and acceleration have OPPOSITE SIGNS ($v(t) \cdot a(t) < 0$).
Example: If $v(t) = -6\text{ m/s}$ and $a(t) = -3\text{ m/s}^2$, the particle is moving in the negative direction and SPEEDING UP ($|-6| \to |-9| = 9\text{ m/s}$)!
In related rates problems, all geometric and physical quantities are implicit functions of time $t$. Differentiating with respect to time requires systematic application of the Chain Rule ($\frac{d}{dt}$):
1. Diagram & Variable Tagging: Draw a sketch. Label quantities that change with variables ($x, y, r, h, \theta$); label quantities that remain strictly constant throughout the motion with numbers (e.g., fixed ladder length $= 10\text{ ft}$).
2. Identify Given & Target Rates: Explicitly write down $\frac{d(\cdot)}{dt}$ given and $\frac{d(\cdot)}{dt}$ sought, noting signs ($\text{shrinking/decreasing} \implies \text{negative}$).
3. Geometric Equation: Write an equation relating the variables (Pythagorean $x^2 + y^2 = L^2$, Conical Volume $V = \frac{1}{3}\pi r^2 h$, Trig $\tan\theta = y/x$). If a variable is unneeded and its rate is unknown, use geometry (e.g., similar triangles $\frac{r}{h} = \frac{R}{H} \implies r = \frac{R}{H}h$) to eliminate it BEFORE differentiating!
4. Implicit Time Derivative: Differentiate both sides with respect to $t$ using the Chain Rule ($\frac{d}{dt}$).
5. Instantaneous Substitution: Substitute numerical values valid only at that specific instant and solve algebraically for the target rate.
Local Linearity: Differentiable curves look locally like straight lines when zoomed in near a point of tangency $(a, f(a))$. The tangent line approximation (linearization) is: $$L(x) = f(a) + f'(a)(x - a)$$
• Concave Down ($f''(x) < 0$ on the interval): The graph bends below its tangent line. The tangent line lies strictly ABOVE the curve $\implies L(x)$ is an OVERESTIMATE ($L(x) > f(x)$).
• Concave Up ($f''(x) > 0$ on the interval): The graph bends above its tangent line. The tangent line lies strictly BELOW the curve $\implies L(x)$ is an UNDERESTIMATE ($L(x) < f(x)$).
L'Hôpital's Rule Theorem: Suppose $f$ and $g$ are differentiable on an open interval containing $c$ (except possibly at $c$) and $g'(x) \ne 0$ near $c$. If $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$ (or both $\pm\infty$), then: $$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$
To earn full points on free-response questions, you must:
1. Separately verify and write: $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$.
2. DO NOT write $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}$! "$\frac{0}{0}$" is not a number; equating a limit to $\frac{0}{0}$ with an equals sign constitutes a mathematical error and results in the forfeiture of the point.
3. Explicitly declare that L'Hôpital's Rule applies, and differentiate the numerator and denominator independently (do NOT use the Quotient Rule!).
Unit 4 Contextual Applications: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 5: Analytical Applications of Differentiation — Optimization & Graph Analysis
Both theorems are existence theorems: they guarantee that a certain value exists, but do not provide an explicit formula to find it.
If a function $f$ satisfies two mandatory preconditions:
1. $f$ is continuous on the closed interval $[a, b]$, and
2. $f$ is differentiable on the open interval $(a, b)$,
then there exists at least one number $c \in (a, b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ Geometric Interpretation: The instantaneous rate of change (tangent line slope) at $c$ equals the average rate of change (secant line slope) across $[a, b]$.
• Rolle's Theorem: If additionally $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
On College Board FRQs, you receive ZERO points for applying MVT if you do not explicitly state: "Since $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, by the Mean Value Theorem...". If $f$ has a cusp, vertical tangent, or jump discontinuity anywhere on the interval, MVT CANNOT BE APPLIED!
If $f$ is continuous on a closed interval $[a, b]$, then $f$ MUST attain both an absolute (global) maximum and an absolute (global) minimum on $[a, b]$.
The 3-Step Candidates Test for Absolute Extrema:
1. Find all critical points $c \in (a, b)$ where $f'(c) = 0$ or $f'(c)$ is undefined.
2. Evaluate $f(x)$ at all critical points $c$ within the interval.
3. Evaluate $f(x)$ at the endpoints $x = a$ and $x = b$. The largest resulting value is the absolute maximum; the smallest is the absolute minimum!
Monotonicity Criteria:
• If $f'(x) > 0$ on an open interval, then $f$ is strictly increasing on that interval.
• If $f'(x) < 0$ on an open interval, then $f$ is strictly decreasing on that interval.
• If $f'(x) = 0$ on an interval, then $f$ is constant on that interval.
Let $c$ be a critical point of a continuous function $f$:
1. Local Maximum: If $f'(x)$ changes from positive to negative at $c$, then $f$ has a relative maximum at $c$.
2. Local Minimum: If $f'(x)$ changes from negative to positive at $c$, then $f$ has a relative minimum at $c$.
3. No Extremum: If $f'(x)$ does not change sign at $c$ (e.g., positive on both sides, like $f(x) = x^3$ at $x = 0$), then $c$ is neither a local maximum nor a local minimum.
Concavity and the Second Derivative:
• If $f''(x) > 0$ on an interval (or equivalently, $f'(x)$ is increasing), the graph of $f$ is concave up (holds water; tangent lines lie BELOW the curve).
• If $f''(x) < 0$ on an interval (or equivalently, $f'(x)$ is decreasing), the graph of $f$ is concave down (spills water; tangent lines lie ABOVE the curve).
A point $(c, f(c))$ is a point of inflection if and only if:
1. $f$ is continuous at $c$, AND
2. The graph of $f$ changes concavity at $c$ (i.e., $f''(x)$ changes sign from positive to negative OR negative to positive).
Critical Trap: Having $f''(c) = 0$ is a necessary candidate condition, but NOT sufficient! For example, for $f(x) = x^4$, $f''(0) = 0$, but $f''(x) = 12x^2 \ge 0$ for all $x$, so there is NO sign change and NO inflection point at $x = 0$!
Suppose $f''(x)$ is continuous near $c$ and $f'(c) = 0$:
• If $f''(c) < 0$, then $f$ is concave down at $c \implies f$ has a relative maximum at $c$.
• If $f''(c) > 0$, then $f$ is concave up at $c \implies f$ has a relative minimum at $c$.
• If $f''(c) = 0$, the test is INCONCLUSIVE! You must revert to the First Derivative Test.
The most heavily tested skill on AP Calculus exams is extracting behavior of $f$ when given the graph of $f'$:
• Where $f'$ is above the x-axis ($f' > 0$) $\iff f$ is increasing.
• Where $f'$ is below the x-axis ($f' < 0$) $\iff f$ is decreasing.
• Where $f'$ crosses the x-axis from above to below $\iff f$ has a local maximum.
• Where $f'$ crosses the x-axis from below to above $\iff f$ has a local minimum.
• Where $f'$ is increasing (slope of $f' > 0$) $\iff f'' > 0 \iff f$ is concave up.
• Where $f'$ is decreasing (slope of $f' < 0$) $\iff f'' < 0 \iff f$ is concave down.
• Where $f'$ has a horizontal tangent / relative peak or valley $\iff f'$ changes between increasing and decreasing $\iff f$ has a point of inflection.
Optimization problems require finding the absolute maximum or minimum value of a modeled quantity subject to physical geometric or economic constraints:
1. Variables & Primary Equation: Identify the quantity to be maximized/minimized (e.g., Volume $V$, Area $A$, Cost $C$).
2. Secondary Constraint Equation: Relate the variables using given restrictions (e.g., Surface area $= 100\text{ cm}^2$, Perimeter $= 40\text{ m}$) and substitute to express the primary equation in terms of a single variable ($f(x)$).
3. Feasible Domain: Determine physical boundaries on the variable (e.g., $0 < x < L/2$).
4. Differentiation & Critical Points: Compute $f'(x)$, set $f'(x) = 0$, and solve for critical values.
5. Absolute Extrema Justification: On a closed interval, use the Candidates Test. On an open interval with only ONE critical point, use the First or Second Derivative Test to prove it is an absolute extremum: "Since $f'(c) = 0$ and $f'(x) > 0$ for $x < c$ and $f'(x) < 0$ for $x > c$, $x = c$ is the unique absolute maximum."
Unit 5 Analytical Applications of Differentiation: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Calculus AB/BC — Unit 6: Integration and Accumulation of Change
Differential calculus measures how fast a quantity changes; integral calculus reconstructs how much has changed. This unit moves from Riemann-sum approximations to the definite integral, the two parts of the Fundamental Theorem of Calculus, accumulation functions, and the core antidifferentiation techniques. Unit 6 carries 15–20% of the exam score on both AP Calculus AB and BC (College Board, AP Students course pages). Sections marked BC ONLY are tested only on AP Calculus BC.
Core idea: If $r(t)$ is a rate of change, then the area between the graph of $r$ and the $t$-axis represents the accumulated change in the underlying quantity. Units always multiply: $\text{(liters/minute)} \times \text{(minutes)} = \text{liters}$. Area below the axis counts as negative accumulation.
• Left Riemann sum: $L = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x_i$
• Right Riemann sum: $R = \sum_{i=1}^{n} f(x_i)\,\Delta x_i$
• Midpoint Riemann sum: $M = \sum_{i=1}^{n} f\!\left(\frac{x_{i-1} + x_i}{2}\right)\Delta x_i$
• Trapezoidal sum: $T = \sum_{i=1}^{n} \frac{f(x_{i-1}) + f(x_i)}{2}\,\Delta x_i = \frac{L + R}{2}$
Worked table example (unequal subintervals):
| $t$ (min) | 0 | 3 | 4 | 8 |
| $v(t)$ (m/min) | 10 | 14 | 13 | 9 |
Right Riemann sum for $\int_0^8 v(t)\,dt$: $3(14) + 1(13) + 4(9) = 42 + 13 + 36 = 91$ meters. Trapezoidal sum: $3\cdot\frac{10 + 14}{2} + 1\cdot\frac{14 + 13}{2} + 4\cdot\frac{13 + 9}{2} = 36 + 13.5 + 44 = 93.5$ meters.
• $f$ increasing ⟹ $L$ underestimates, $R$ overestimates.
• $f$ decreasing ⟹ $L$ overestimates, $R$ underestimates.
• $f$ concave up ⟹ $T$ overestimates, $M$ underestimates.
• $f$ concave down ⟹ $T$ underestimates, $M$ overestimates.
Definition: If $f$ is continuous on $[a, b]$, $\Delta x = \frac{b - a}{n}$ and $x_k = a + k\,\Delta x$, then $$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k)\,\Delta x.$$
1. Identify $\Delta x$ (the factor outside $f$, e.g. $\frac{3}{n}$) ⟹ $b - a = 3$.
2. Identify $x_k = a + k\Delta x$ inside $f$ (e.g. $1 + \frac{3k}{n}$) ⟹ $a = 1$, $b = 4$.
3. Read off $f$: $\lim_{n\to\infty}\sum_{k=1}^{n}\frac{3}{n}\sqrt{1 + \frac{3k}{n}} = \int_1^4 \sqrt{x}\,dx$ (equivalently $\int_0^3 \sqrt{1 + x}\,dx$).
FTC Part 1: If $f$ is continuous on an interval containing $a$, then the accumulation function $g(x) = \int_a^x f(t)\,dt$ is differentiable and $g'(x) = f(x)$.
$$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f\big(u(x)\big)\,u'(x) \qquad \frac{d}{dx}\int_{v(x)}^{u(x)} f(t)\,dt = f\big(u(x)\big)\,u'(x) - f\big(v(x)\big)\,v'(x)$$ Example: $\frac{d}{dx}\int_1^{x^3}\cos(t^2)\,dt = 3x^2\cos(x^6)$.
• $g(x)$ = signed area under $f$ from $a$ to $x$ (use geometry: triangles, trapezoids, semicircles).
• $g'(x) = f(x)$: $g$ increases where $f > 0$; relative max of $g$ where $f$ changes from $+$ to $-$.
• $g''(x) = f'(x)$: $g$ is concave up where $f$ is increasing; inflection points of $g$ occur where $f$ changes from increasing to decreasing (relative extrema of $f$).
• $g(a) = 0$ always; for $x \lt a$, $g(x) = -\int_x^a f(t)\,dt$.
• $\int_a^a f(x)\,dx = 0$ and $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$
• Additivity: $\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx$
• Linearity: $\int_a^b \big[k f(x) \pm g(x)\big]dx = k\int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$
• Constants: $\int_a^b c\,dx = c(b - a)$
• Symmetry: $f$ odd ⟹ $\int_{-a}^{a} f(x)\,dx = 0$; $f$ even ⟹ $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$
Net Change form (the FRQ workhorse): $$F(b) = F(a) + \int_a^b F'(x)\,dx$$ Amount at time $b$ = amount at time $a$ + accumulated change. For tanks: $W(t) = W(0) + \int_0^t \big[\text{rate in} - \text{rate out}\big]\,ds$.
• $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \;(n \ne -1)$ • $\int \frac{1}{x}\,dx = \ln|x| + C$
• $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ • $\int a^x\,dx = \frac{a^x}{\ln a} + C$
• $\int \sin x\,dx = -\cos x + C$ • $\int \cos x\,dx = \sin x + C$
• $\int \sec^2 x\,dx = \tan x + C$ • $\int \csc^2 x\,dx = -\cot x + C$
• $\int \sec x\tan x\,dx = \sec x + C$ • $\int \csc x\cot x\,dx = -\csc x + C$
• $\int \frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C$ • $\int \frac{dx}{1 + x^2} = \arctan x + C$
1. Choose $u$ = the "inside" function whose derivative (up to a constant) also appears.
2. Compute $du = u'(x)\,dx$ and solve for the piece of the integrand you need (e.g. $x\,dx = \frac{1}{2}du$).
3. If any $x$ remains, back-substitute from the definition of $u$ (e.g. $u = x - 1 \Rightarrow x = u + 1$).
4. Definite integrals: change the limits to $u$-values ($x = a \Rightarrow u = u(a)$) and never return to $x$; or return to $x$ and use the original limits — never mix the two.
Example: $\int_1^e \frac{(\ln x)^2}{x}\,dx = \int_0^1 u^2\,du = \frac{1}{3}$.
Completing the Square (Topic 6.10): $\int \frac{dx}{x^2 + 4x + 13} = \int \frac{dx}{(x + 2)^2 + 3^2} = \frac{1}{3}\arctan\!\left(\frac{x + 2}{3}\right) + C$, using $\int \frac{du}{u^2 + a^2} = \frac{1}{a}\arctan\frac{u}{a} + C$.
Example: $\int xe^{3x}\,dx$: $u = x$, $dv = e^{3x}dx$, $v = \frac{1}{3}e^{3x}$ ⟹ $\frac{x}{3}e^{3x} - \frac{1}{9}e^{3x} + C$.
Tabular method: for $x^n e^{kx}$, $x^n \sin(kx)$, repeatedly differentiate $u$ and integrate $dv$, alternating signs $+, -, +, \dots$
$\frac{5x + 1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$ ⟹ $5x + 1 = A(x + 2) + B(x - 1)$.
$x = 1$: $A = 2$; $x = -2$: $B = 3$ ⟹ $\int = 2\ln|x - 1| + 3\ln|x + 2| + C$.
• Infinite interval: $\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty}\int_a^b f(x)\,dx$. Example: $\int_0^\infty xe^{-x^2}dx = \lim_{b\to\infty}\left[\frac{1}{2} - \frac{1}{2}e^{-b^2}\right] = \frac{1}{2}$.
• Infinite discontinuity: at an endpoint use a one-sided limit, e.g. $\int_0^4 \frac{dx}{\sqrt{4 - x}} = \lim_{b \to 4^-}\left[-2\sqrt{4 - x}\right]_0^b = 4$; at an interior point, split the integral and evaluate each piece separately.
• The integral converges if the limit is a finite number; otherwise it diverges.
• $p$-integrals: $\int_1^\infty \frac{dx}{x^p}$ converges iff $p > 1$; $\int_0^1 \frac{dx}{x^p}$ converges iff $p \lt 1$.
2. Units: The units of $\int_a^b r(t)\,dt$ are (units of $r$) × (units of $t$). Include units whenever the question says "indicate units of measure."
3. Interpretation template: "$\int_2^5 R(t)\,dt = 40$ means that a total of 40 gallons of oil leaked from the pipeline over the time interval $2 \le t \le 5$ hours." Mention the quantity, the interval, and the units.
4. Net change / amount: Write the complete expression first: "$W(6) = 120 + \int_0^6 \big[E(t) - L(t)\big]\,dt$", then the calculator value to three decimal places.
5. "Justify" with accumulation functions: "$g$ has a relative maximum at $x = 2$ because $g'(x) = f(x)$ changes from positive to negative at $x = 2$." For absolute extrema, show a Candidates Test with all endpoint and critical values.
6. Over/underestimate: "The left Riemann sum is an underestimate because $r$ is increasing on the interval." The reason must refer to the behavior of the function, not to the picture.
Unit 6 Integration and Accumulation of Change: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Calculus AB/BC — Unit 7: Differential Equations
A differential equation (DE) states how a quantity changes; solving it recovers the quantity itself. This unit builds the language of modeling with rates, checks candidate solutions, reads slope fields, and solves separable equations exactly — including the exponential growth/decay model behind population, cooling, and radioactive decay problems. Unit 7 carries 5–10% of the exam score on both AB and BC (College Board), and a separable DE appears as a free-response question almost every year. Sections marked BC ONLY (Topic 7.5 Euler's Method and Topic 7.9 Logistic Models) are tested only on AP Calculus BC.
Definition: A differential equation is an equation that contains a derivative of an unknown function. A solution is a function that makes the equation true on an interval. The general solution contains an arbitrary constant $C$ (a family of curves); a particular solution is the single member that satisfies an initial condition such as $y(0) = 4$.
• "The rate of change of $y$ is proportional to $y$" ⟹ $\frac{dy}{dt} = ky$
• "...proportional to the square root of $y$" ⟹ $\frac{dy}{dt} = k\sqrt{y}$
• "...inversely proportional to $y^2$" ⟹ $\frac{dy}{dt} = \frac{k}{y^2}$
• "...proportional to the difference between $T$ and the room temperature $20$" ⟹ $\frac{dT}{dt} = k(T - 20)$ (Newton's Law of Cooling)
• "...jointly proportional to $P$ and to $(M - P)$" ⟹ $\frac{dP}{dt} = kP(M - P)$ (logistic, BC)
The phrase "rate of change of $y$" always becomes the derivative $\frac{dy}{dt}$, never $y$ itself.
Example: Is $y = 3x^2$ a solution of $\frac{dy}{dx} = \frac{2y}{x}$ for $x > 0$? Left side: $\frac{dy}{dx} = 6x$. Right side: $\frac{2(3x^2)}{x} = 6x$. Both sides agree, so yes.
Example: For which $k$ does $y = e^{kx}$ satisfy $y'' - y' - 6y = 0$? Substituting gives $e^{kx}(k^2 - k - 6) = 0$, so $k = 3$ or $k = -2$.
Core idea: At each grid point $(x, y)$, the DE $\frac{dy}{dx} = f(x, y)$ gives the slope of the solution curve passing through that point. Drawing a short segment with that slope at many points produces a slope field. Every solution curve is tangent to the segments it passes through.
Worked slope field: $\frac{dy}{dx} = x - y$ on the integer grid $-3 \le x \le 3$, $-3 \le y \le 3$. Note the horizontal segments along the line $y = x$ (where $x - y = 0$) and the segments of slope $1$ along $y = x - 1$.
1. Where are the segments horizontal? Set $f(x, y) = 0$. For $x - y$, this is the line $y = x$.
2. Does the slope depend on $x$ only, $y$ only, or both? If every segment in a vertical column is identical, $f$ depends only on $x$; if every segment in a horizontal row is identical, $f$ depends only on $y$ (an autonomous DE).
3. Test one point with a nonzero slope (e.g. $(1, -1)$: $x - y = 2$, a steep positive segment) to separate answer choices.
4. Vertical segments mean the slope is undefined (a zero denominator), e.g. $y = 0$ for $\frac{dy}{dx} = -\frac{x}{y}$.
• Equilibrium solutions: constant functions $y = c$ with $f(x, c) = 0$ for all $x$. For $\frac{dy}{dx} = x(y^2 - 9)$, these are $y = 3$ and $y = -3$.
• Concavity: differentiate the DE implicitly. For $\frac{dy}{dx} = x - y$: $\frac{d^2y}{dx^2} = 1 - \frac{dy}{dx} = 1 - x + y$. Always substitute $\frac{dy}{dx}$ back in.
• Long-run behavior: for $\frac{dy}{dx} = 1 - y$, solutions above $y = 1$ decrease toward $1$ and solutions below increase toward $1$, so $\lim_{x \to \infty} y = 1$ for every solution.
Example: $\frac{dy}{dx} = x + y$, $y(0) = 1$, $h = 0.5$.
Step 1: slope at $(0, 1)$ is $1$ ⟹ $y(0.5) \approx 1 + 0.5(1) = 1.5$.
Step 2: slope at $(0.5, 1.5)$ is $2$ ⟹ $y(1) \approx 1.5 + 0.5(2) = 2.5$.
(The exact solution is $y = 2e^x - x - 1$, so $y(1) = 2e - 2 \approx 3.437$.)
1. Separate: $\frac{1}{h(y)}\,dy = g(x)\,dx$ (all $y$ with $dy$, all $x$ with $dx$).
2. Integrate both sides and write $+\,C$ immediately: $\int \frac{dy}{h(y)} = \int g(x)\,dx + C$.
3. Use the initial condition right away to find $C$ (it is usually easiest before solving for $y$).
4. Solve for $y$ and state the domain: the largest open interval containing the initial $x$-value on which the solution is defined and differentiable.
$\frac{dy}{y} = 2x\,dx$ ⟹ $\ln|y| = x^2 + C$.
$x = 0, y = 3$: $\ln 3 = 0 + C$ ⟹ $C = \ln 3$.
$|y| = e^{x^2 + \ln 3} = 3e^{x^2}$. Since $y(0) = 3 > 0$, take the positive branch: $y = 3e^{x^2}$ for all real $x$.
• $k \lt 0$: exponential decay; half-life $\frac{\ln 2}{|k|}$.
• $k$ is the relative growth rate: $\frac{dy/dt}{y} = k$. For $\frac{dP}{dt} = 0.03P$, the population grows at $3\%$ of its current size per year.
Example: $\frac{dT}{dt} = -0.1(T - 20)$, $T(0) = 90$ ⟹ $T(t) = 20 + 70e^{-0.1t}$, so $T(10) = 20 + 70e^{-1} \approx 45.75$.
• If $0 \lt P(0) \lt M$, $P$ increases toward $M$; if $P(0) > M$, $P$ decreases toward $M$.
• The population grows fastest when $P = \frac{M}{2}$ (the inflection point of the solution curve when $P(0) \lt \frac{M}{2}$).
• Equilibrium solutions: $P = 0$ (unstable) and $P = M$ (stable).
Wrong: $\ln|y| = \frac{x^2}{2}$ ⟹ $y = e^{x^2/2} + C$. This "late $C$" produces a function that does not satisfy the DE ($\frac{d}{dx}\big[e^{x^2/2} + 2\big] = xe^{x^2/2} \ne x\big(e^{x^2/2} + 2\big)$).
Right: $\ln|y| = \frac{x^2}{2} + C$ ⟹ $|y| = e^{C}e^{x^2/2}$ ⟹ $y = Ae^{x^2/2}$ (the constant is multiplicative after exponentiating).
2. Keep the absolute value: $\int \frac{dy}{y} = \ln|y| + C$. Resolve the sign with the initial condition: if $y(0) = -2$ for $\frac{dy}{dx} = y\cos x$, then $y = -2e^{\sin x}$. Writing $\ln y$ and declaring "no solution because $\ln(-2)$ is undefined" is a costly error.
3. Use the initial condition correctly. Substitute both coordinates into the equation that contains $C$; for $\ln|y| = x^2 + C$ with $y(0) = 3$, $C = \ln 3$, not $C = 3$. When the solution involves a square root, keep the branch that matches the sign of $y_0$.
4. Shifted variables: for $\frac{dy}{dx} = x^2(y - 1)$, integrate to $\ln|y - 1| = \frac{x^3}{3} + C$; the "$-1$" stays with $y$ throughout.
5. State the domain when asked: the open interval containing the initial $x$-value on which the solution is defined.
Unit 7 Differential Equations: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Calculus AB/BC — Unit 8: Applications of Integration
Unit 6 taught you how to compute a definite integral; Unit 8 teaches you what a definite integral measures. The same idea — slice, approximate each slice, add the slices with an integral — produces average values, displacement and total distance, accumulated amounts, areas between curves, and volumes of solids. Unit 8 carries 10–15% of the AP Calculus AB exam score and 5–10% of the BC score (College Board AP Central, 2026-27), and area/volume questions appear on the free-response section almost every year. Topics 8.1–8.12 are tested on both AB and BC; the section marked BC ONLY (Topic 8.13 Arc Length) is tested only on AP Calculus BC.
Definition: If $f$ is continuous on $[a, b]$, the average value of $f$ on $[a, b]$ is $$f_{avg} = \frac{1}{b - a}\int_a^b f(x)\,dx.$$ Geometrically, $f_{avg}$ is the height of the rectangle on $[a, b]$ whose area equals the (signed) area under $f$.
Example: $f(x) = x^2$ on $[1, 4]$: $\int_1^4 x^2\,dx = 21$, so $f_{avg} = \frac{21}{3} = 7$, attained at $c = \sqrt{7} \approx 2.646$.
• Velocity from acceleration: $v(t) = v(0) + \int_0^t a(s)\,ds$.
• Position from velocity: $x(t) = x(0) + \int_0^t v(s)\,ds$.
• Displacement on $[a, b]$: $\int_a^b v(t)\,dt$ (signed; can be negative or zero).
• Total distance traveled on $[a, b]$: $\int_a^b |v(t)|\,dt$ (never negative).
• Average velocity on $[a, b]$: $\frac{1}{b - a}\int_a^b v(t)\,dt$.
Displacement: $\int_0^4 v(t)\,dt = \frac{4}{3}$.
Distance: split at the sign changes $t = 1$ and $t = 3$: $\left|\frac{4}{3}\right| + \left|-\frac{4}{3}\right| + \left|\frac{4}{3}\right| = 4$.
On a calculator-active question, simply enter $\int_0^4 |v(t)|\,dt$ directly.
Net Change Theorem: if $r(t)$ is the rate of change of a quantity $Q$, then $\int_a^b r(t)\,dt = Q(b) - Q(a)$ is the net change in $Q$ from $t = a$ to $t = b$. Units: (units of $r$) × (units of $t$) — e.g. (gallons/minute) × (minutes) = gallons.
• "When is the amount a maximum on $[0, T]$?" ⟹ Candidates Test: evaluate $A$ at $t = 0$, $t = T$, and every $t$ where $R(t) = D(t)$.
Example: $E(t) = 12$, $L(t) = 3t$, $A(0) = 100$: $A'(t) = 0$ at $t = 4$; $A(0) = 100$, $A(4) = 124$, $A(6) = 118$, so the maximum is $124$.
Vertical slices (Topic 8.4, functions of $x$): a slice at $x$ has height $f(x) - g(x)$ (top minus bottom) and width $dx$: $$A = \int_a^b [f(x) - g(x)]\,dx, \quad f(x) \ge g(x) \text{ on } [a, b].$$ The limits $a$ and $b$ are usually intersection points: for $y = 2x$ and $y = x^2$, solve $2x = x^2$ to get $x = 0$ and $x = 2$, so $A = \int_0^2 (2x - x^2)\,dx = \frac{4}{3}$.
Example: $y = x^3$ and $y = x$ meet at $x = -1, 0, 1$. $\int_{-1}^{0} (x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$, while the single integral $\int_{-1}^{1} (x - x^3)\,dx$ incorrectly gives $0$.
Core idea: if every cross section perpendicular to the $x$-axis has area $A(x)$, then $V = \int_a^b A(x)\,dx$. The side length $s(x)$ of each cross section is the length of the slice in the base: $s(x) = (\text{top}) - (\text{bottom})$. For sections perpendicular to the $y$-axis, use $s(y) = x_{right} - x_{left}$ and $V = \int_c^d A(y)\,dy$.
• Square: $A = s^2$ (Topic 8.7)
• Rectangle with height $k$ times its base: $A = k s^2$ (Topic 8.7)
• Isosceles right triangle with a leg in the base: $A = \frac{1}{2}s^2$ (Topic 8.8)
• Isosceles right triangle with the hypotenuse in the base: $A = \frac{1}{4}s^2$ (Topic 8.8)
• Equilateral triangle: $A = \frac{\sqrt{3}}{4}s^2$ (Topic 8.8)
• Semicircle with diameter $s$ in the base: $A = \frac{1}{2}\pi\left(\frac{s}{2}\right)^2 = \frac{\pi}{8}s^2$ (Topic 8.8)
• About the $y$-axis: slice horizontally and write $x$ in terms of $y$. For $x = \sqrt{y}$, $0 \le y \le 4$: $V = \pi\int_0^4 y\,dy = 8\pi$.
• About another line (Topic 8.10): $R = |\text{axis} - \text{curve}|$. Region between $y = x^2$ and $y = 4$ about $y = 4$: $R = 4 - x^2$, $V = \pi\int_{-2}^{2} (4 - x^2)^2\,dx = \frac{512\pi}{15}$.
• About $y = 2$ (above the region): distances are measured downward from the axis, so $R = 2 - x^2$ (the lower curve is farther) and $r = 2 - x$: $V = \pi\int_0^1 \left[(2 - x^2)^2 - (2 - x)^2\right]dx = \frac{8\pi}{15}$.
• About $y = -1$ (below the region): $R = (\text{top}) + 1$, $r = (\text{bottom}) + 1$.
1. Write the full expression, e.g. $V = \pi\int_0^1 \left[(2 - x^2)^2 - (2 - x)^2\right]dx$, then the value.
2. Find intersection points to three decimal places when they come from a calculator, and store them; rounding limits early can cost the answer point.
3. For accumulation questions, show $A(t) = A_0 + \int_0^t [R(s) - D(s)]\,ds$ and justify a maximum with the Candidates Test.
4. For total distance, show $\int_a^b |v(t)|\,dt$ explicitly.
Unit 8 Applications of Integration: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Calculus BC — Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions BC ONLY
Units 2–8 described curves as $y = f(x)$. Unit 9 frees the curve from that restriction: a point can be driven by a parameter $t$ (parametric and vector-valued functions) or located by a distance and an angle (polar coordinates). Every tool you already own — the derivative as a rate, the definite integral as an accumulated sum — is reused, but each now has a new formula. The whole unit is tested only on AP Calculus BC and carries 10–15% of the BC exam score (College Board AP Central, AP Calculus BC course page, 2026-27). Topics 9.1–9.9 below follow the current Course and Exam Description; a particle-motion or polar-area free-response question appears on most BC exams.
Definition: a parametric curve is a pair of functions $x = x(t)$, $y = y(t)$; as the parameter $t$ increases, the point $(x(t), y(t))$ traces the curve in a definite direction (the orientation). If $x$ and $y$ are differentiable and $\frac{dx}{dt} \ne 0$, the chain rule $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$ gives $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$
• $\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 3t^2 - 3$, so $\frac{dy}{dx} = \frac{3t^2 - 3}{2t}$. At $t = 2$: $\frac{dy}{dx} = \frac{9}{4}$, and the point is $(4, 2)$, so the tangent line is $y - 2 = \frac{9}{4}(x - 4)$.
• Horizontal tangent: $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \ne 0$ ⟹ $t = \pm 1$, at the points $(1, -2)$ and $(1, 2)$.
• Vertical tangent: $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \ne 0$ ⟹ $t = 0$, at the point $(0, 0)$.
The second derivative is the derivative of the slope function with respect to $x$. Because $\frac{dy}{dx}$ is written in terms of $t$, apply the same quotient rule again: $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}.$$
• $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} + \frac{3}{2t^2}$.
• Divide by $\frac{dx}{dt} = 2t$: $\frac{d^2y}{dx^2} = \frac{3(t^2 + 1)}{4t^3}$. At $t = 2$: $\frac{15}{32} \gt 0$, so the curve is concave up there.
Over a short time $dt$ the point moves $dx = x'(t)\,dt$ horizontally and $dy = y'(t)\,dt$ vertically, so by Pythagoras the piece of curve has length $\sqrt{(x')^2 + (y')^2}\,dt$. If $x'$ and $y'$ are continuous and the curve is traced once for $a \le t \le b$, $$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt.$$
$(x')^2 + (y')^2 = (1 - \cos t)^2 + \sin^2 t = 2 - 2\cos t = 4\sin^2\left(\frac{t}{2}\right)$, and $\sin\left(\frac{t}{2}\right) \ge 0$ on $[0, 2\pi]$, so $L = \int_0^{2\pi} 2\sin\left(\frac{t}{2}\right)dt = \left[-4\cos\left(\frac{t}{2}\right)\right]_0^{2\pi} = 8$.
On a calculator-active question, enter the integral directly: fnInt gives $8.000$.
A vector-valued function $\vec{r}(t) = \langle x(t), y(t) \rangle$ is a parametric curve written as one object: the position vector. Its derivatives are taken component by component: $$\vec{v}(t) = \vec{r}\,'(t) = \langle x'(t), y'(t) \rangle, \qquad \vec{a}(t) = \vec{v}\,'(t) = \langle x''(t), y''(t) \rangle.$$ The speed is the magnitude of velocity: $|\vec{v}(t)| = \sqrt{(x')^2 + (y')^2}$. The slope of the path is still $\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$.
• $\vec{v}(t) = \langle 3t^2 - 3,\ 2t \rangle$, $\vec{a}(t) = \langle 6t,\ 2 \rangle$.
• At $t = 2$: $\vec{v}(2) = \langle 9, 4 \rangle$, speed $= \sqrt{97}$, $\vec{a}(2) = \langle 12, 2 \rangle$.
• $\vec{v}(2) \cdot \vec{a}(2) = 108 + 8 = 116 \gt 0$, so the speed is increasing at $t = 2$.
$$\int_a^b \vec{v}(t)\,dt = \left\langle \int_a^b x'(t)\,dt,\ \int_a^b y'(t)\,dt \right\rangle, \qquad \vec{r}(b) = \vec{r}(a) + \int_a^b \vec{v}(t)\,dt.$$ The definite integral of velocity is the displacement vector; adding the initial position gives the new position. An indefinite integral carries a constant vector $\vec{C} = \langle C_1, C_2 \rangle$.
$\int_0^{\pi/2} \vec{v}(t)\,dt = \left\langle [\sin t]_0^{\pi/2},\ [t^2]_0^{\pi/2} \right\rangle = \left\langle 1,\ \frac{\pi^2}{4} \right\rangle$, so $\vec{r}\left(\frac{\pi}{2}\right) = \left\langle 2,\ \frac{\pi^2}{4} \right\rangle$.
• Moving right/left: sign of $x'(t)$. Moving up/down: sign of $y'(t)$.
• At rest: $x'(t) = 0$ and $y'(t) = 0$ at the same $t$.
• Speed increasing: $\vec{v} \cdot \vec{a} \gt 0$ (decreasing if $\lt 0$). Equivalently, differentiate the speed $\sqrt{(x')^2 + (y')^2}$.
• Position: $\vec{r}(b) = \vec{r}(a) + \int_a^b \vec{v}(t)\,dt$.
• Total distance traveled: $\int_a^b |\vec{v}(t)|\,dt = \int_a^b \sqrt{(x')^2 + (y')^2}\,dt$ — the same integral as Topic 9.3.
$x'(t) = 0$ at $t = 0, 2$; $y'(t) = 0$ at $t = 2$. The particle is at rest only at $t = 2$. For $0 \lt t \lt 2$, $x'(t) \lt 0$ and $y'(t) \gt 0$: the particle moves left and up.
A point with polar coordinates $(r, \theta)$ has rectangular coordinates $x = r\cos\theta$, $y = r\sin\theta$, and $r^2 = x^2 + y^2$. For a polar curve $r = f(\theta)$ this is a parametric curve with parameter $\theta$: $$x(\theta) = f(\theta)\cos\theta, \quad y(\theta) = f(\theta)\sin\theta, \qquad \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}.$$
$\frac{dy}{d\theta} = 2(0) + 2(1) = 2$, $\frac{dx}{d\theta} = 2(1) - 2(0) = 2$, so $\frac{dy}{dx} = 1$.
Interpreting $\frac{dr}{d\theta}$: when $r \gt 0$, $\frac{dr}{d\theta} \gt 0$ means the curve is moving away from the origin and $\frac{dr}{d\theta} \lt 0$ means it is moving toward the origin. When $r \lt 0$, these conclusions reverse.
Single curve (Topic 9.8): polar area is built from thin circular sectors of area $\frac{1}{2}r^2\,d\theta$, so the region swept from $\theta = \alpha$ to $\theta = \beta$ is $$A = \frac{1}{2}\int_\alpha^\beta [r(\theta)]^2\,d\theta.$$ Find the $\theta$-interval on which the region is traced exactly once — solve $r(\theta) = 0$ to find where loops and petals start and end.
For a region inside both curves, split the $\theta$-interval at the intersection and, on each piece, use the curve that is closer to the origin.
1. Speed increasing/decreasing: "Because $\vec{v}(1) \cdot \vec{a}(1) = -10 \lt 0$, the speed of the particle is decreasing at $t = 1$." A sign of one acceleration component is not sufficient.
2. Position: write $x(3) = x(0) + \int_0^3 x'(t)\,dt$ before the value; the initial condition is part of the setup point.
3. Distance traveled: write $\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ explicitly, then the calculator value to three decimal places.
4. Polar distance from the origin: "Since $r(\theta) \gt 0$ and $r'(\theta) \lt 0$ at $\theta = \theta_0$, the curve is getting closer to the origin." Both signs must be cited.
5. Polar area: show $\frac{1}{2}\int_\alpha^\beta \left(R^2 - r^2\right)d\theta$ with intersection angles found algebraically or stored to full calculator precision.
Unit 9 Parametric Equations, Polar Coordinates & Vector-Valued Functions: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Calculus BC — Unit 10: Infinite Sequences and Series BC ONLY
Unit 10 asks a question that no earlier unit could answer: what does it mean to add infinitely many numbers, and when does the answer exist? The first half of the unit (Topics 10.1–10.10) builds a toolbox of convergence tests; the second half (Topics 10.11–10.15) uses that toolbox to replace functions such as $e^x$, $\sin x$ and $\ln(1 + x)$ by polynomials and power series, and to measure how large the error of that replacement can be. The whole unit is tested only on AP Calculus BC and carries 15–20% of the BC exam score — the largest single unit on the exam (College Board AP Central, AP Calculus BC course page, 2026-27). Topics 10.1–10.15 below follow the current Course and Exam Description; a Taylor-polynomial / error-bound or interval-of-convergence free-response question appears on almost every BC exam.
Sequence vs. series: a sequence $\{a_n\}$ is an ordered list of numbers; a series $\sum_{n=1}^{\infty} a_n$ is the attempt to add them. The series is given meaning through its sequence of partial sums $$S_n = a_1 + a_2 + \cdots + a_n.$$ If $\lim_{n\to\infty} S_n = S$ exists (as a finite number), the series converges and its sum is $S$; otherwise it diverges.
$S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1}$, so $S_n \to 1$ and the series converges to $1$.
If instead the partial sums are given, e.g. $S_n = \frac{4n+1}{n+3}$, then the sum is $\lim S_n = 4$, while the first term is $a_1 = S_1 = \frac{5}{4}$.
A geometric series has a constant ratio $r$ between consecutive terms: $$\sum_{n=0}^{\infty} a r^n = a + ar + ar^2 + \cdots = \frac{a}{1 - r} \quad \text{if } |r| \lt 1,$$ and it diverges if $|r| \ge 1$ (and $a \ne 0$). The numerator is always the first term actually written, whatever the starting index.
Alternating ratio: $\sum_{n=0}^{\infty} \frac{(-1)^n 3^{n+1}}{4^n} = \sum_{n=0}^{\infty} 3\left(-\frac{3}{4}\right)^n = \frac{3}{1 + 3/4} = \frac{12}{7}$.
Statement: if $\lim_{n\to\infty} a_n \ne 0$ (or the limit does not exist), then $\sum a_n$ diverges. The reason: if the series converged, then $a_n = S_n - S_{n-1} \to S - S = 0$.
Inconclusive case: $\lim \frac{1}{n} = 0$, yet the harmonic series $\sum \frac{1}{n}$ diverges; $\lim \frac{1}{n^2} = 0$, and $\sum \frac{1}{n^2}$ converges. A zero limit decides nothing.
Integral Test (10.4): let $a_n = f(n)$, where $f$ is positive, continuous and decreasing on $[k, \infty)$. Then $\sum_{n=k}^{\infty} a_n$ and $\int_k^{\infty} f(x)\,dx$ either both converge or both diverge. The test gives convergence, not the value of the sum.
Examples: $\sum \frac{1}{n\sqrt{n}}$ ($p = \frac{3}{2}$) converges; $\sum \frac{1}{\sqrt{n}}$ ($p = \frac{1}{2}$) diverges; $\sum \frac{1}{n^{2p-1}}$ converges exactly when $2p - 1 \gt 1$, i.e. $p \gt 1$.
Both tests apply to series with positive terms.
Direct Comparison: if $0 \lt a_n \le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges ("smaller than convergent"). If $0 \lt b_n \le a_n$ and $\sum b_n$ diverges, then $\sum a_n$ diverges ("larger than divergent").
Limit Comparison: if $\lim_{n\to\infty} \frac{a_n}{b_n} = L$ with $0 \lt L \lt \infty$, then $\sum a_n$ and $\sum b_n$ both converge or both diverge.
• $\sum \frac{1}{n^2 + 5}$: $\frac{1}{n^2 + 5} \lt \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges ($p = 2$), so the series converges by Direct Comparison.
• $\sum \frac{3n + 1}{n^3 + 2}$: choose $b_n = \frac{1}{n^2}$ (keep the dominant powers: $\frac{3n}{n^3}$). $\lim \frac{a_n}{b_n} = \lim \frac{3n^3 + n^2}{n^3 + 2} = 3$, finite and positive, and $\sum \frac{1}{n^2}$ converges, so the series converges by Limit Comparison.
Alternating Series Test (AST): a series $\sum (-1)^n b_n$ (or $\sum (-1)^{n+1} b_n$) converges if (i) $b_n \gt 0$, (ii) $b_{n+1} \le b_n$ for all $n$ (eventually), and (iii) $\lim_{n\to\infty} b_n = 0$.
• $\sum \frac{(-1)^n}{\ln n}$ ($n \ge 2$): $\frac{1}{\ln n}$ is positive, decreasing and tends to $0$, so the series converges by the AST.
• $\sum (-1)^n \frac{n}{n+1}$: $b_n \to 1 \ne 0$, so the AST does not apply and the series diverges by the nth term test.
Ratio Test: let $L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$. If $L \lt 1$, $\sum a_n$ converges absolutely; if $L \gt 1$ (or $L = \infty$), it diverges; if $L = 1$, the test is inconclusive. It is the test of choice when $a_n$ contains $n!$ or $c^n$.
• $\sum \frac{n^2}{3^n}$: $\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{n^2}\cdot\frac{1}{3} \to \frac{1}{3} \lt 1$, so the series converges.
• $\sum \frac{(2n)!}{(n!)^2\,5^n}$: $\frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)}{(n+1)^2}\cdot\frac{1}{5} \to \frac{4}{5} \lt 1$, so the series converges.
• Any p-series gives $L = 1$: the Ratio Test cannot decide $\sum \frac{1}{n}$ or $\sum \frac{1}{n^2}$.
A series is absolutely convergent if $\sum |a_n|$ converges; absolute convergence implies convergence. It is conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges.
Examples: $\sum \frac{(-1)^n}{\sqrt[3]{n}}$ is conditionally convergent ($\sum \frac{1}{n^{1/3}}$ diverges, AST holds). $\sum \frac{\sin n}{n^2}$ is absolutely convergent ($\frac{|\sin n|}{n^2} \le \frac{1}{n^2}$). $\sum (-1)^n \frac{n}{n^2 + 1}$ is conditionally convergent (Limit Comparison of $|a_n|$ with $\frac{1}{n}$ gives divergence; AST gives convergence).
If $\sum (-1)^{n+1} b_n$ satisfies the AST and $S$ is its sum, then the error of the partial sum $S_n$ is at most the first omitted term: $$|S - S_n| \le b_{n+1}.$$ Moreover, $S$ lies between consecutive partial sums $S_n$ and $S_{n+1}$.
• $S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}$ approximated by $S_4$: $|S - S_4| \le b_5 = \frac{1}{125}$.
• How many terms guarantee $|S - S_n| \lt 0.001$ for $\sum \frac{(-1)^{n+1}}{n^2}$? Need $\frac{1}{(n+1)^2} \lt 0.001$, i.e. $(n+1)^2 \gt 1000$. Since $31^2 = 961$ and $32^2 = 1024$, $n + 1 = 32$ and $n = 31$.
The $n$th-degree Taylor polynomial for $f$ about $x = c$ matches $f$ and its first $n$ derivatives at $c$: $$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!}(x - c)^k = f(c) + f'(c)(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \cdots + \frac{f^{(n)}(c)}{n!}(x - c)^n.$$ When $c = 0$ it is called a Maclaurin polynomial.
$P_2(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{64}(x - 4)^2$, and $P_2(4.2) = 2 + 0.05 - 0.000625 = 2.049375$ (true value $\sqrt{4.2} \approx 2.049390$).
If $|f^{(n+1)}(t)| \le M$ for every $t$ between $c$ and $x$, then $$\left|f(x) - P_n(x)\right| \le \frac{M}{(n+1)!}\,|x - c|^{n+1}.$$ $M$ must bound the $(n+1)$st derivative on the whole interval between $c$ and $x$ — not only at the centre.
Bound $= \frac{8e^{0.2}}{3!}(0.1)^3 \approx 0.00163$. The actual error is $e^{0.2} - 1.22 \approx 0.00140$, below the bound. Note that $M = 8$ (the value at $x = 0$) would give $0.00133$, which is smaller than the actual error — an invalid bound.
A power series $\sum_{n=0}^{\infty} c_n (x - a)^n$ converges on an interval centred at $a$. Its radius of convergence $R$ is found with the Ratio Test: the series converges absolutely when $\lim \left|\frac{c_{n+1}}{c_n}\right| |x - a| \lt 1$. The interval of convergence is $(a - R, a + R)$ plus whichever endpoints give convergent series — each endpoint must be tested separately.
• $x = 7$: $\sum \frac{4^n}{n\,4^n} = \sum \frac{1}{n}$, harmonic, diverges.
• $x = -1$: $\sum \frac{(-4)^n}{n\,4^n} = \sum \frac{(-1)^n}{n}$, converges by the AST.
Interval of convergence: $[-1, 7)$.
A Taylor series is the Taylor polynomial with infinitely many terms, $\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x - c)^n$. The four Maclaurin series named in the CED: $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \qquad \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \qquad \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \quad (\text{all } x),$$ $$\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n \quad (|x| \lt 1).$$
Reading derivatives from a series: for $e^{x^2} = \sum \frac{x^{2n}}{n!}$, the coefficient of $x^6$ is $\frac{1}{3!} = \frac{1}{6}$, so $f^{(6)}(0) = 6! \cdot \frac{1}{6} = 120$.
Inside its interval of convergence, a power series may be differentiated and integrated term by term; the radius stays the same, but endpoint behaviour can change.
• $\frac{x}{1 - x^3} = x\sum (x^3)^n = \sum_{n=0}^{\infty} x^{3n+1}$ for $|x| \lt 1$.
• Differentiate $\frac{1}{1 - x} = \sum x^n$: $\frac{1}{(1 - x)^2} = \sum_{n=1}^{\infty} n x^{n-1}$, so $\sum_{n=1}^{\infty} n x^n = \frac{x}{(1 - x)^2}$; at $x = \frac{1}{2}$, $\sum \frac{n}{2^n} = 2$.
• Integrate $\frac{1}{1 + t} = \sum (-1)^n t^n$ from $0$ to $x$: $\ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}$, so $\sum \frac{(-1)^{n+1}}{n\,2^n} = \ln\frac{3}{2}$.
• Integrate $\frac{1}{1 + t^2}$: $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$.
2. Is it geometric ($a r^n$) or a p-series ($\frac{1}{n^p}$)? ⟹ decide immediately (10.2, 10.5).
3. Factorials or $c^n$? ⟹ Ratio Test (10.8).
4. A rational or algebraic expression in $n$? ⟹ Limit Comparison with $\frac{1}{n^p}$ from the dominant powers (10.6).
5. Alternating signs? ⟹ test $\sum |a_n|$ first; if that diverges, use the AST and call it conditionally convergent (10.7, 10.9).
6. A term $f(n)$ that is easy to integrate, such as $\frac{1}{n(\ln n)^2}$? ⟹ Integral Test (10.4).
(a) Write $P_3(x)$ about $x = 1$: $P_3(x) = 4 - 2(x - 1) + \frac{6}{2!}(x - 1)^2 - \frac{12}{3!}(x - 1)^3 = 4 - 2(x - 1) + 3(x - 1)^2 - 2(x - 1)^3$.
(b) Approximate $f(1.2)$: $P_3(1.2) = 4 - 0.4 + 0.12 - 0.016 = 3.704$.
(c) Lagrange bound: $|f(1.2) - P_3(1.2)| \le \frac{48}{4!}(0.2)^4 = 0.0032$.
(d) Conclusion: $3.7008 \le f(1.2) \le 3.7072$.
Typical scoring: one point for the two lower-order terms, one for the two higher-order terms (factorials included), one for the approximation, one for the form of the Lagrange bound, one for the numerical bound.
(a) Ratio Test: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{\sqrt{n}}{\sqrt{n+1}}\cdot\frac{|x - 2|}{3} \to \frac{|x - 2|}{3} \lt 1$ ⟹ $R = 3$.
(b) Endpoints: at $x = 5$ the series is $\sum \frac{(-1)^n}{\sqrt{n}}$, which converges by the AST ($\frac{1}{\sqrt{n}}$ positive, decreasing, limit $0$); at $x = -1$ it is $\sum \frac{1}{\sqrt{n}}$, a p-series with $p = \frac{1}{2} \le 1$, which diverges. Interval of convergence: $(-1, 5]$.
(c) At $x = 5$: $\sum \left|\frac{(-1)^n}{\sqrt{n}}\right| = \sum \frac{1}{\sqrt{n}}$ diverges, so the convergence is conditional.
(d) $f(3) = \sum \frac{(-1)^n}{\sqrt{n}\,3^n}$ satisfies the AST, so $|f(3) - S_3| \le b_4 = \frac{1}{\sqrt{4}\cdot 3^4} = \frac{1}{162}$.
Typical scoring: one point for setting up the ratio, one for the limit / radius, one for each endpoint with the test named, one for the interval, one for the error bound with the first omitted term.
Unit 10 Infinite Sequences and Series: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Official AP Calculus BC Free-Response Archive (2025–2026)
Bu bölüm College Board'un AP Central'da herkese açık yayımladığı gerçek sınav sorularına dayanır (© College Board; AP Central past exam questions). Soru metni BilgeAP’ye kopyalanmadı. Her kartta kısa bağlam, konu ve ünite eşlemesi, yerel PDF'teki sayfanın bağlantısı ve resmi puanlama kılavuzundan özetlenmiş puan ölçütleri var. "AB and BC" rozetli sorular aynı yıl AP Calculus AB sınavında da ortak soru olarak yer alır.
1. Süre: Part A (FRQ 1–2) 30 dakika, grafik hesap makinesi zorunlu, radyan modu. Part B (FRQ 3–6) 60 dakika, hesap makinesi yok. Tam deneme için 90 dakikalık tek oturum planla.
2. Ekran + kâğıt: Soruyu PDF'ten ekranda oku, Bluebook'taki gibi. Çözümü kâğıda yaz: kurşun kalem ya da siyah veya koyu mavi mürekkepli kalem (resmi yönerge). Her parçayı A, B, C, D diye etiketle.
3. Hesap makinesi: Kâğıda yalnızca kurulumu yaz (integral, denklem, türev). Hesap makinesi sözdizimi puan almaz. Ondalıklar üç basamak doğru olmalı (yuvarlanmış ya da kesilmiş). Bkz. komut kartı.
4. Gerekçe: "Justify" geçen her parçada kullandığın teoremin koşulunu yaz (ör. "$R$ is differentiable, so $R$ is continuous").
5. Öz değerlendirme: Süre bitince kartın içindeki "Puanlama ölçütleri" kutusunu aç ve kazandığın her ölçütü işaretle. İşaretler hesabına kaydedilir.
2025 AP Calculus BC
AP Central: 2025 FRQ (9 s.) · 2025 Scoring Guidelines (27 s.) · 2025 Chief Reader Report
FRQ 1 · Invasive plant species in a fruit grove
Source: College Board Official Exam (AP Calculus BC 2025 FRQ #1) · © 2025 College Board
An accumulation model $C(t) = 7.6\arctan(0.2t)$ (acres, $t$ in weeks) with a given rate $C'(t)$; a second function $A(t)$ subtracts a removal integral after week 4.
Skills tested: average value of a function (8.1) · average vs. instantaneous rate of change (4.1) · limit at infinity of a rate (1.15) · derivative of an accumulation function and the Candidates Test for an absolute maximum (6.4, 5.5)
Review first: Unit 8 · Unit 4 · Unit 1 · Unit 6 · Unit 5
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 2 · Polar curve $r = 2\sin^2\theta$ and the semicircle $r = \frac{1}{2}$
Source: College Board Official Exam (AP Calculus BC 2025 FRQ #2) · © 2025 College Board
A polar curve on $0 \le \theta \le \pi$ together with the semicircle $r = \frac12$; a particle moves along the curve with $\frac{d\theta}{dt} = 15$.
Skills tested: differentiating in polar form (9.7) · area between two polar curves (9.9) · Candidates Test on $x(\theta)$ (5.5) · chain rule for a related rate $\frac{dr}{dt}$ (3.1, 4.5)
Review first: Unit 9 · Unit 5 · Unit 3 · Unit 4
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 3 · Reading rate $R(t)$ from a table
Source: College Board Official Exam (AP Calculus BC 2025 FRQ #3) · © 2025 College Board
A table of a differentiable reading rate $R(t)$ (words per minute) at $t = 0, 2, 8, 10$, and a teacher's quadratic rate model $W(t)$.
Skills tested: estimating a derivative from a table with units (2.3, 4.1) · Intermediate Value Theorem (1.16) · trapezoidal sum with unequal subintervals (6.2) · definite integral of a polynomial (6.7)
Review first: Unit 2 · Unit 4 · Unit 1 · Unit 6
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 4 · Graph of $f$ and $g(x) = \int_6^x f(t)\,dt$
Source: College Board Official Exam (AP Calculus BC 2025 FRQ #4) · © 2025 College Board
The graph of $f$ on $-6 \le x \le 12$ consists of two semicircles and one line segment; $g$ is its accumulation function from $6$.
Skills tested: Fundamental Theorem of Calculus with accumulation functions (6.4) · points of inflection from the graph of $g' = f$ (5.9) · areas from geometry (6.2) · Candidates Test for an absolute minimum (5.5)
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 5 · Differential equation $\frac{dy}{dx} = (3-x)y^2$ with $f(1) = -1$
Source: College Board Official Exam (AP Calculus BC 2025 FRQ #5) · © 2025 College Board
A particular solution of a differential equation is analysed with a second derivative, a Taylor polynomial, an error bound and Euler's method.
Skills tested: second derivative from a differential equation (3.6) · Taylor polynomial (10.11) · Lagrange error bound (10.12) · Euler's method (7.5)
Review first: Unit 3 · Unit 7 · Unit 10
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 6 · Taylor series about $x = 4$
Source: College Board Official Exam (AP Calculus BC 2025 FRQ #6) · © 2025 College Board
A Taylor series with general term $\frac{(x-4)^{n+1}}{(n+1)3^n}$; its derivative series turns out to be geometric.
Skills tested: ratio test and interval of convergence with endpoint analysis (10.13, 10.9, 10.6) · differentiating a power series (10.15) · sum of a geometric series (10.2)
Review first: Unit 10
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
2026 AP Calculus BC
AP Central: 2026 FRQ (9 s.). 2026 puanlama kılavuzu: 2026 Scoring Guidelines (23 s.). Ölçütler bu belgeden özetlendi (16 Ekim 2026 kontrolü).
FRQ 1 · Male and female birds arriving at a nesting area
Source: College Board Official Exam (AP Calculus BC 2026 FRQ #1) · © 2026 College Board
A table of the male arrival rate $M(t)$ (birds per day) over 30 days and a piecewise sinusoidal female arrival rate $F(t)$.
Skills tested: estimating a derivative from a table with units (2.3, 4.1) · midpoint Riemann sum and interpretation (6.2, 8.3) · definite integral of a rate on the calculator (8.3) · Intermediate Value Theorem (1.16)
Review first: Unit 2 · Unit 4 · Unit 6 · Unit 8 · Unit 1
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 2 · Polar region $S$: area, tangent slope, average distance
Source: College Board Official Exam (AP Calculus BC 2026 FRQ #2) · © 2026 College Board
The polar curve $r = 3 + 2\sin(2\theta) + \cos(2\theta)$ on $0 \le \theta \le \pi$; area, slope of a tangent line, a critical point of $r$ and an average distance from the origin.
Skills tested: area of a polar region (9.8) · slope in polar form (9.7) · First Derivative Test on $r(\theta)$ (5.4) · average value (8.1)
Review first: Unit 9 · Unit 5 · Unit 8
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 3 · Cooling pie: $\frac{dH}{dt} = -\frac{1}{15}(H - 20)$, $H(0) = 75$
Source: College Board Official Exam (AP Calculus BC 2026 FRQ #3) · © 2026 College Board
Newton's law of cooling with a proposed slope field, a tangent line approximation and an exact solution.
Skills tested: slope fields (7.3) · tangent line approximation and concavity (4.6, 5.6) · separation of variables with an initial condition (7.6, 7.7)
Review first: Unit 7 · Unit 4 · Unit 5
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 4 · Graph of $f'$ on $[-4, 4]$ with $f(2) = 3$
Source: College Board Official Exam (AP Calculus BC 2026 FRQ #4) · © 2026 College Board
The graph of the derivative $f'$ is given; $g(x) = f(x) - \ln x$ is defined for $x > 0$.
Skills tested: derivative of $\ln x$ and difference rule (2.7) · inflection points from the graph of $f'$ (5.6, 5.9) · intervals where $f$ is increasing and concave down (5.3, 5.6) · absolute extrema using the area under $f'$ (5.5, 8.3)
Review first: Unit 2 · Unit 5 · Unit 8
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 5 · Region under $f(x) = \sqrt[3]{x-1}$ and the curve $g(x) = e^{-2x+4}$
Source: College Board Official Exam (AP Calculus BC 2026 FRQ #5) · © 2026 College Board
Region $R$ is bounded by the graph of $f$, the $x$-axis and $x = 2$; the curves meet at $(2, 1)$.
Skills tested: antiderivative by substitution (6.9) · disc method about the $x$-axis (8.9) · arc length and perimeter (8.13) · improper integral with limit notation (6.13)
Puanlama ölçütleri: önce kâğıtta çöz, sonra aç
FRQ 6 · Maclaurin series $\sum_{n=0}^{\infty} 2\left(-\frac{x}{5}\right)^n$
Source: College Board Official Exam (AP Calculus BC 2026 FRQ #6) · © 2026 College Board
An alternating geometric Maclaurin series for $g$ (valid for $|x| < 5$), its derivative $f = g'$, and $h(x) = 25g(x) - 2e^x$.
Skills tested: sum of a geometric series (10.2) · differentiating a power series (10.15) · alternating series error bound (10.10) · Maclaurin series for $e^x$ and combining series (10.14)
Review first: Unit 10