Unit 3 practice test: Trigonometric and Polar Functions

$\tan\theta$$= \frac{y}{x}$$= \frac{3}{-3}$$= -1$, and $\tan^{-1}(-1) = -\frac{\pi}{4}$. But U is in Quadrant II ($x < 0$, $y > 0$), and $\tan^{-1}$ only returns angles in Quadrants I and IV. Add $\pi$: $\theta = \frac{3\pi}{4}$.

• Correct: $\frac{3\pi}{4}$: Right: Quadrant II, reference angle $\frac{\pi}{4}$.
• $-\frac{\pi}{4}$: That is $\tan^{-1}(-1)$, which points into Quadrant IV, and it is not in $[0, 2\pi)$.
• $\frac{7\pi}{4}$: $\frac{7\pi}{4}$ points into Quadrant IV; with $r > 0$ that names $(3, -3)$.
• $\frac{\pi}{4}$: $\frac{\pi}{4}$ points into Quadrant I, where both coordinates are positive.

$r = \sqrt{1 + 3} = 2$. $\tan\theta = -\sqrt{3}$ with $x > 0$ and $y < 0$: Quadrant IV, reference angle $\frac{\pi}{3}$, so $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.

• Correct: $\left(2, \frac{5\pi}{3}\right)$: Right: $r = 2$, Quadrant IV.
• $\left(2, \frac{2\pi}{3}\right)$: $\tan\frac{2\pi}{3} = -\sqrt{3}$ too, but that ray is in Quadrant II; W has $x > 0$.
• $\left(4, \frac{5\pi}{3}\right)$: $4$ is $r^2$; take the square root.
• $\left(2, \frac{\pi}{3}\right)$: $\frac{\pi}{3}$ is in Quadrant I; W has a negative $y$.

Z is $4$ units from the pole, on the negative $x$-axis, which is the ray $\theta = \pi$.

• Correct: $(4, \pi)$: Right: distance $4$, direction $\pi$.
• $(4, 0)$: Angle $0$ points along the positive $x$-axis, to $(4, 0)$.
• $(-4, \pi)$: $r = -4$ at angle $\pi$ goes back through the pole to $(4, 0)$; the question also asks for $r > 0$.
• $\left(4, \frac{\pi}{2}\right)$: Angle $\frac{\pi}{2}$ points straight up, to $(0, 4)$.

The two input intervals have the same length, $\frac{\pi}{6}$, and the average rate on the later one is larger. Increasing average rates over equal intervals mean the graph of $d$ is concave up: the spot moves along the shore faster and faster as the beam turns.

• Correct: The rate of change of $d$ is increasing, so the graph of $d$ is concave up: Right: equal intervals, larger rate later, so concave up.
• The rate of change of $d$ is decreasing, so the graph of $d$ is concave down: The rate on the later interval is larger ($\frac{16\sqrt{3}}{\pi} > \frac{8\sqrt{3}}{\pi}$), so the rates increase.
• $d$ is decreasing, because the rates are not equal: Both rates are positive, so $d$ is increasing on the interval.
• $d$ is linear, because both rates are positive: A linear function has equal rates on equal intervals; these differ.

$d(\theta) = 4\tan\theta$ has a vertical asymptote at $\theta = \frac{\pi}{2}$ because $\cos\frac{\pi}{2} = 0$. From the left $\tan\theta$ grows without bound, so $\lim_{\theta \to \pi/2^-} d(\theta)$$= \infty$. In context, a beam almost parallel to the shore meets it very far away.

• Correct: $\lim_{\theta \to \pi/2^-} d(\theta)$$= \infty$: the beam turns nearly parallel to the shore, so the spot lands farther and farther away: Right: an asymptote at $\frac{\pi}{2}$, approached from the left, gives outputs growing without bound.
• $\lim_{\theta \to \pi/2^-} d(\theta)$$= 4$: the spot approaches the lighthouse's distance from the shore: $d\left(\frac{\pi}{4}\right) = 4$ is one table value; $d$ keeps growing past it.
• $\lim_{\theta \to \pi/2^-} d(\theta)$$= 0$: the beam points back at the nearest point of the shore: $d(0) = 0$ is the start; as $\theta$ grows toward $\frac{\pi}{2}$ the distance grows.
• $\lim_{\theta \to \pi/2^-} d(\theta)$$= -\infty$: the spot moves to the other side of the nearest point: From the left of $\frac{\pi}{2}$, $\sin\theta$ and $\cos\theta$ are both positive, so $d$ is positive.

From the table, $d\left(\frac{\pi}{3}\right) = 4\sqrt{3}$ and $d\left(\frac{\pi}{6}\right) = \frac{4\sqrt{3}}{3}$, so the ratio is $\frac{4\sqrt{3}}{4\sqrt{3}/3} = 3$. Tangent is not linear: doubling the input does not double the output.

• Correct: No: $d\left(\frac{\pi}{3}\right) = 4\sqrt{3}$, which is $3$ times $d\left(\frac{\pi}{6}\right) = \frac{4\sqrt{3}}{3}$: Right: the ratio of the table values is $3$, not $2$.
• Yes: $d$ is proportional to $\theta$, so doubling $\theta$ doubles $d$: $d$ would be proportional to $\theta$ only if its graph were a line through the origin; the rates in the table are not constant.
• No: $d\left(\frac{\pi}{3}\right)$ is $4$ times $d\left(\frac{\pi}{6}\right)$, because $\tan$ squares the angle: The ratio from the table is $3$; tangent does not square its input.
• Yes: $\tan\left(2\cdot\frac{\pi}{6}\right) = 2\tan\left(\frac{\pi}{6}\right)$: $\tan\frac{\pi}{3} = \sqrt{3}$ and $2\tan\frac{\pi}{6} = \frac{2\sqrt{3}}{3}$, which differ.