AP Calculus AB

Conducting Body: College Board
Primary Audience: High school students
Core Topics: Limits, derivatives, and integrals
Question Types: Multiple Choice and Free Response
Exam Duration: 3 hours 15 minutes
Frequency: Annually in May
Purpose: To earn potential college credit or placement

The AP Calculus AB exam is a standardized test from the College Board that assesses a student's understanding of single-variable calculus. It is typically taken by high school students who have completed a corresponding AP course, and a strong score can earn them college credit or placement out of introductory calculus courses at many universities. The curriculum is equivalent to a first-semester college calculus course, covering topics like limits, derivatives, and integrals.

The exam is divided into two main sections: multiple choice questions and free response questions. Each of these sections is further split into a part where a graphing calculator is permitted and a part where it is not. The challenge of the exam comes from its emphasis on conceptual understanding over rote memorization. Students must not only compute derivatives and integrals correctly but also interpret their meaning in various contexts and justify their methods using definitions and theorems. The free response questions, in particular, demand clear, logical written explanations, and managing time across the different question types is a common difficulty for many students.

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From a mathematics paper rather than AP Calculus AB, worked the way every question on this site is. AP Calculus AB questions are being added.

JEE Advanced 2023

Let $P$ be the plane $\sqrt{3}x + 2y + 3z = 16$ and let $S = \{\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} : \alpha^{2} + \beta^{2} + \gamma^{2} = 1$ and the distance of $(\alpha, \beta, \gamma)$ from the plane $P$ is $\frac{7}{2}\}$ . Let $\vec{u}, \vec{v}$ and $\vec{w}$ be three distinct vectors in $S$ such that $|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}|$ . Let $V$ be the volume of the parallelepiped determined by vectors $\vec{u}, \vec{v}$ and $\vec{w}$ . Then the value of $\frac{80}{\sqrt{3}}V$ is

Three Dimensional Geometry Let

P

be the plane

\sqrt{3}x + 2y + 3z = 16

and let

S = \{\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} : \alpha^{2} + \beta^{2} + \gamma^{2} = 1

and the distance of

(\alpha, \beta, \gamma)

from the plane

P

\frac{7}{2}\}

. Let

\vec{u}, \vec{v}

and

\vec{w}

be three distinct vectors in

S

such that

|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}|

. Let

V

be the volume of the parallelepiped determined by vectors

\vec{u}, \vec{v}

and

\vec{w}

. Then the value of

\frac{80}{\sqrt{3}}V

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Let P be the plane √3x + 2y + 3z = 16 and let S = {αî + βĵ + γk̂ : α² + β² + γ² = 1 and the distance of (α, β, γ) from the plane P is 7/2}. Let u⃗, v⃗ and w⃗ be three distinct vectors in S such that |u⃗ - v⃗| = |v⃗ - w⃗| = |w⃗ - u⃗|. Let V be the volume of the parallelepiped determined by vectors u⃗, v⃗ and w⃗. Then the value of 80/√3V is

$\text{Plane } P:\quad \sqrt{3}\,x + 2y + 3z = 16$

$S = \{\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} : \alpha^2+\beta^2+\gamma^2 = 1,\ d\big((\alpha,\beta,\gamma),P\big) = \tfrac{7}{2}\}$

$\vec{u},\vec{v},\vec{w} \in S \text{ distinct},\quad |\vec{u}-\vec{v}| = |\vec{v}-\vec{w}| = |\vec{w}-\vec{u}|$

$\text{To find: } \dfrac{80}{\sqrt{3}}\,V,\qquad V = |[\vec{u}\ \vec{v}\ \vec{w}]|$

$\text{Plan:}\quad \vec{n}\cdot\vec{x} = 2 \Rightarrow S \text{ is a circle}$

$\text{centre } c,\ \text{radius } r$

triangle uvw equilateral ⇒ A; V = 2Ah

$\vec{n} = (\sqrt{3},\ 2,\ 3),\quad |\vec{n}| = \sqrt{3+4+9} = 4$

$d = \dfrac{|\vec{n}\cdot(\alpha,\beta,\gamma) - 16|}{|\vec{n}|} = \dfrac{7}{2}$

$|\vec{n}\cdot\vec{x} - 16| = 14 \Rightarrow \vec{n}\cdot\vec{x} = 30 \text{ or } 2$

$|\vec{n}\cdot\vec{x}| \le |\vec{n}|\,|\vec{x}| = 4 \Rightarrow \vec{n}\cdot\vec{x} = 2$

$S = \{\vec{x} : |\vec{x}| = 1,\ \vec{n}\cdot\vec{x} = 2\} = \text{circle}$

$c = \dfrac{2}{|\vec{n}|^2}\,\vec{n} = \left(\dfrac{\sqrt{3}}{8},\ \dfrac{1}{4},\ \dfrac{3}{8}\right)$

$h = \dfrac{2}{|\vec{n}|} = \dfrac{1}{2},\quad r = \sqrt{1 - h^2} = \dfrac{\sqrt{3}}{2}$

|u⃗-v⃗| = |v⃗-w⃗| = |w⃗-u⃗| ⇒ triangle uvw equilateral

$s = r\sqrt{3} = \dfrac{\sqrt{3}}{2}\cdot\sqrt{3} = \dfrac{3}{2}$

$A = \dfrac{\sqrt{3}}{4}\,s^2 = \dfrac{\sqrt{3}}{4}\cdot\dfrac{9}{4} = \dfrac{9\sqrt{3}}{16}$

V = 6,V_{Ouvw} V_(Ouvw) = 1/3 A h ⇒ V = 6·1/3Ah = 2Ah

$V = 2\cdot\dfrac{9\sqrt{3}}{16}\cdot\dfrac{1}{2} = \dfrac{9\sqrt{3}}{16}$

$\dfrac{80}{\sqrt{3}}\,V = \dfrac{80}{\sqrt{3}}\cdot\dfrac{9\sqrt{3}}{16} = 45$

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