AP Calculus AB Unit 8A: Areas Between Curves — practice questions — page 3

Explore 66 free Unit 8A: Areas Between Curves questions in AP Calculus AB Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 8A: Areas Between Curves questions and exercises with answers

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If $\omega$ is a non-real cube root of unity, find the value of $1 + \omega + \omega^2$.Read the answer and worked solution →Which statement about a null vector is correct?Read the answer and worked solution →The projection of vector $\vec A$ on vector $\vec B$ is given byRead the answer and worked solution →When subtracting vectors, $\vec A-\vec B$ is evaluated asRead the answer and worked solution →For nonzero vector $\mathbf{A}$, which expression gives a vector in the same direction as $\mathbf{A}$ but of magnitude $1$?Read the answer and worked solution →Let $\mathbf{A}=\langle 1,2,3\rangle$, $\mathbf{B}=\langle 2,0,1\rangle$, and $\mathbf{C}=\langle 0,1,1\rangle$. Then $\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})$ equalsRead the answer and worked solution →

Worked practice questions

If $\omega$ is a non-real cube root of unity, find the value of $1 + \omega + \omega^2$.

$0$
$3$
$-1$
$1$

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Correct answer: A. $0$

The three cube roots of unity are $1, \omega, \omega^2$, and they are roots of $z^3 - 1 = (z-1)(z^2 + z + 1) = 0$. Vieta's formulas on $z^2 + z + 1$ give $\omega + \omega^2 = -1$, so $1 + \omega + \omega^2 = 0$. Equivalently, the sum of *all* $n$-th roots of unity is zero for $n \ge 2$.

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Which statement about a null vector is correct?

Its direction is fixed along the positive x-axis
Its dot product with any vector is always zero
Adding it to a vector reverses the vector
Its cross product with any vector gives a scalar

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Correct answer: B. Its dot product with any vector is always zero

A null vector has zero magnitude and arbitrary direction. The notes state that its dot product with any vector is zero, and its cross product with any vector is also a null vector. Adding or subtracting it leaves the original vector unchanged.

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The projection of vector $\vec A$ on vector $\vec B$ is given by

$\dfrac{\vec A\times\vec B}{|\vec B|}$
$\dfrac{\vec A\cdot\vec B}{|\vec B|}$
$\dfrac{|\vec A|}{\vec B}$
$\dfrac{\vec A\cdot\vec B}{|\vec A|}$

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Correct answer: B. $\dfrac{\vec A\cdot\vec B}{|\vec B|}$

The notes explicitly give the projection of $\vec A$ on $\vec B$ as $\dfrac{\vec A\cdot\vec B}{|\vec B|}$. This is the scalar projection along $\vec B$.

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AP Calculus AB Unit 8A: Areas Between Curves — practice questions — page 3

Unit 8A: Areas Between Curves questions and exercises with answers

Worked practice questions

If $\omega$ is a non-real cube root of unity, find the value of $1 + \omega + \omega^2$.

Which statement about a null vector is correct?

The projection of vector $\vec A$ on vector $\vec B$ is given by

How to practise

Put it into practice

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