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Q52.

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Precalculus Mathematics for Calculus

Found in: Page 408

Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

**Find (a) the reference number for each value of $t$ and (b) the terminal point determined by $t$ .**
$t = \frac{31 π}{6}$

a. The reference number $\bar{t} = \frac{π}{6}$ .

b. The terminal point for $t = \frac{31 π}{6}$ is $P (- \frac{\sqrt{3}}{2}, \frac{1}{2})$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part a. Step 1. State the definition of the reference number.

The reference number $\bar{t}$ associated with $t$ , a real number is the shortest distance along the unit circle between the terminal point determined by $t$ and the $x$ -axis.

Part a. Step 2. Determine the quadrant where the terminal point lies.

If $0 < t < \frac{π}{2}$ , then $t$ lies in quadrant I and the reference number $\bar{t} = t$ .

If $\frac{π}{2} < t < π$ , then $t$ lies in quadrant II and the reference number $\bar{t} = π - t$ .

If $π < t < \frac{3 π}{2}$ , then $t$ lies in quadrant III and the reference number $\bar{t} = t - π$ .

If $\frac{3 π}{2} < t < 2 π$ , then $t$ lies in quadrant IV and the reference number $\bar{t} = 2 π - t$ .

Here, $t = \frac{31 π}{6}$ , which lies in quadrant II.

Part a. Step 3. Find the reference number.

Since $t$ lies in quadrant II, therefore, the reference number will be $\bar{t} = t - 5 π$ . Here substitute $\frac{31 π}{6}$ for $t$ and simplify for $\bar{t}$ .

$\bar{t} = t - 5 π = \frac{31 π}{6} - 5 π = \frac{π}{6}$

Therefore, the reference number $\bar{t} = \frac{π}{6}$ .

Part b. Step 1. State the definition of the terminal point on the unit circle.

Starting at the point $(1, 0)$ , if $t \geq 0$ , then $t$ is the distance along the unit circle in clockwise direction and if $t < 0$ , then $|t|$ is the distance along the unit circle in clockwise direction. Here, $t$ is a real number and this distance generates a point $P (x, y)$ on the unit circle. The point $P (x, y)$ obtained in this way is called the terminal point determined by the real number $t$ .

Part b. Step 2. State the terminal points for some special values of t for reference.

The table below gives the terminal points for some special values of $t$ .

$t$	Terminal point determined by $t$
0	$(1, 0)$
$\frac{π}{6}$	$(\frac{\sqrt{3}}{2}, \frac{1}{2})$
$\frac{π}{4}$	$(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$
$\frac{π}{3}$	$(\frac{1}{2}, \frac{\sqrt{3}}{2})$
$\frac{π}{2}$	$(0, 1)$

Part b. Step 3. Simplify and state the conclusion.

Let $P (x, y)$ be the terminal point for $t = \frac{31 π}{6}$ .

Since the reference number for $t = \frac{31 π}{6}$ is $\bar{t} = \frac{π}{6}$ and the corresponding terminal point for $\bar{t} = \frac{π}{6}$ is $(\frac{\sqrt{3}}{2}, \frac{1}{2})$ in the first quadrant, note that $t = \frac{31 π}{6}$ lies in the quadrant II, therefore, its terminal point will have negative $x$ -coordinate and positive y-coordinate. Therefore, the terminal point for $t = \frac{31 π}{6}$ is $P (- \frac{\sqrt{3}}{2}, \frac{1}{2})$ .

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