Vectors, Polar Coordinates, and Complex Numbers: math test solutions, Grade 12 – download the PDF

1 Vector basics / 3 pts

1. \(\|\mathbf{u}\|=\sqrt{16+9}=5\) and \(\|\mathbf{v}\|=\sqrt{25+144}=13\). (1 pt)
2. \(2\mathbf{u}+\mathbf{v}=\langle-8,6\rangle+\langle5,12\rangle=\langle-3,18\rangle\). (1 pt)
3. \(\mathbf{u}\cdot\mathbf{v}=-20+36=16\). (1 pt)

2 Work and angle / 3 pts

1. \(W=48+48=96\) ft·lb. (1 pt)
2. \(\|\mathbf{F}\|=10\) lb and \(\|\mathbf{d}\|=10\) ft. (1 pt)
3. \(\cos\theta=\dfrac{96}{100}=0.96\), so \(\theta\approx16.3^\circ\). (1 pt)

3 Polar conversions / 4 pts

1. \(x=6\cos330^\circ=3\sqrt3\) and \(y=6\sin330^\circ=-3\): \((3\sqrt3,\,-3)\). (1.5 pts)
2. \(r=\sqrt{18}=3\sqrt2\); quadrant II gives \(\theta=\dfrac{3\pi}{4}\): \(\left(3\sqrt2,\dfrac{3\pi}{4}\right)\). (1.5 pts)
3. \(r^2=8r\sin\theta\) gives \(x^2+y^2=8y\), i.e., \(x^2+(y-4)^2=16\): a circle of center \((0,4)\) and radius 4. (1 pt)

4 A polar curve / 3 pts

1. \(r=2,\ 1,\ 0,\ 4\) respectively. (2 pts: 0.5 pt per value)
2. It is a cardioid, symmetric about the vertical line \(\theta=\dfrac\pi2\) (the \(y\)-axis); its farthest point is 4 units below the pole. (1 pt)

5 Complex numbers in polar form / 4 pts

1. \(z_1z_2=10\operatorname{cis}135^\circ\) (1 pt) \(=-5\sqrt2+5\sqrt2\,i\). (1 pt)
2. \(z_1^3=8\operatorname{cis}300^\circ\) (1 pt) \(=8\cos300^\circ+8i\sin300^\circ=4-4\sqrt3\,i\). (1 pt)

6 Cube roots / 3 pts

\(-64=64\operatorname{cis}180^\circ\), so the modulus is \(\sqrt[3]{64}=4\). (1 pt)

The arguments are \(\dfrac{180^\circ+360^\circ k}{3}=60^\circ,\ 180^\circ,\ 300^\circ\). (1 pt)

\(z_0=2+2\sqrt3\,i\), \(z_1=-4\), \(z_2=2-2\sqrt3\,i\). (1 pt)