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Found in: Page 408

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

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

a. The reference number $\overline{t}=\frac{\pi }{6}$.

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

See the step by step solution

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

The reference number $\overline{t}$ 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, then $t$ lies in quadrant I and the reference number $\overline{t}=t$.

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

If $\pi , then $t$ lies in quadrant III and the reference number $\overline{t}=t-\pi$.

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

Here, $t=\frac{31\pi }{6}$, 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 $\overline{t}=t-5\pi$. Here substitute $\frac{31\pi }{6}$ for $t$ and simplify for $\overline{t}$.

$\overline{t}=t-5\pi \phantom{\rule{0ex}{0ex}}=\frac{31\pi }{6}-5\pi \phantom{\rule{0ex}{0ex}}=\frac{\pi }{6}$

Therefore, the reference number $\overline{t}=\frac{\pi }{6}$.

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

Starting at the point $\left(1,0\right)$, if $t\ge 0$, then $t$ is the distance along the unit circle in clockwise direction and if $t<0$, then $\left|t\right|$ is the distance along the unit circle in clockwise direction. Here, $t$ is a real number and this distance generates a point $P\left(x,y\right)$ on the unit circle. The point $P\left(x,y\right)$ 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 $\left(1,0\right)$ $\frac{\pi }{6}$ $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ $\frac{\pi }{4}$ $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ $\frac{\pi }{3}$ $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ $\frac{\pi }{2}$ $\left(0,1\right)$

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

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

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