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Q52.

Expert-verifiedFound in: Page 408

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 $P\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$.

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.

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

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

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

If $\frac{3\pi}{2}<t<2\pi $, 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.

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 $\overline{t}=\frac{\pi}{6}$.

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 *$t$*.

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

$t$ | Terminal point determined by $t$ |

0 | $\left(1,0\right)$ |

$\frac{\pi}{6}$ | $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ |

$\frac{\pi}{4}$ | $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ |

$\frac{\pi}{3}$ | $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ |

$\frac{\pi}{2}$ | $\left(0,1\right)$ |

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 $P\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$.

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