Chapter 6.6, Problem 6CFU

a.

To determine

To find: the amplitude of a sinusoidal function that models the monthly temperature.

Answer to Problem 6CFU

12.5

Explanation of Solution

Given information:

The table representing the average monthly temperatures for a city of Washington is shown below.

Jan	Feb	Mar	April	May	Jun	Jul	Aug	Sept	Oct	Nov	Dec
41 $°$	44 $°$	47 $°$	50 $°$	56 $°$	61 $°$	65 $°$	66 $°$	61 $°$	54 $°$	46 $°$	42 $°$

Calculation:

The amplitude of the sinusoidal function that models the monthly temperature can be evaluated as follows:

  $\begin{array}{c}\text{Amplitude}=\left|\frac{\text{Highest temperature}-\text{Lowest temperature}}{2}\right|\\ =\left|\frac{66-41}{2}\right|\left[\text{Put the values}\right]\\ =\left|\frac{25}{2}\right|\\ =12.5\end{array}$

Thus the amplitude for the given function is 12.5.

b.

To determine

To find: the vertical shift of a sinusoidal function that models the monthly temperature.

Answer to Problem 6CFU

53.5

Explanation of Solution

Given information:

The table representing the average monthly temperatures for a city of Washington is shown below.

Jan	Feb	Mar	April	May	Jun	Jul	Aug	Sept	Oct	Nov	Dec
41 $°$	44 $°$	47 $°$	50 $°$	56 $°$	61 $°$	65 $°$	66 $°$	61 $°$	54 $°$	46 $°$	42 $°$

Calculation:

The vertical shift of the sinusoidal function that models the monthly temperature can be evaluated as follows:

  $\begin{array}{c}\text{Vertical shift}=\left|\frac{\text{Lowest temperature}+\text{}\text{Amplitude}}{2}\right|\\ =\left|41+12.5\right|\left[\text{Put the values}\right]\\ =53.5\end{array}$

Thus the vertical shift for the given function is 53.5.

c.

To determine

To find: the period of a sinusoidal function that models the monthly temperature.

Answer to Problem 6CFU

12 months

Explanation of Solution

Given information:

The table representing the average monthly temperatures for a city of Washington is shown below.

Jan	Feb	Mar	April	May	Jun	Jul	Aug	Sept	Oct	Nov	Dec
41 $°$	44 $°$	47 $°$	50 $°$	56 $°$	61 $°$	65 $°$	66 $°$	61 $°$	54 $°$	46 $°$	42 $°$

Calculation:

Period of a function is defined as the time taken by the function to repeat itself. Thus the period of a sinusoidal function that models the monthly temperature is 12 months.

d.

To determine

To find: the sinusoidal function that models the monthly temperature using $t=1$ to represent January.

Answer to Problem 6CFU

  $y=12.5\sin \left(\frac{\pi }{6}t-\frac{2\pi }{3}\right)+53.5$

Explanation of Solution

Given information:

The table representing the average monthly temperatures for a city of Washington is shown below.

Jan	Feb	Mar	April	May	Jun	Jul	Aug	Sept	Oct	Nov	Dec
41 $°$	44 $°$	47 $°$	50 $°$	56 $°$	61 $°$	65 $°$	66 $°$	61 $°$	54 $°$	46 $°$	42 $°$

The values evaluated from part a. to part d. are as follows:

  $\begin{array}{c}\text{Amplitude}\left(\pm A\right)=12.5\\ \text{Period}=12\\ \text{Vertical shift }\left(h\right)=53.5\\ t=1\\ y=41\end{array}$

Calculation:

The sinusoidal function can be represented as $y=A\sin \left(kt+c\right)+h$ , where $A$ is the coefficient of the absolute value of amplitude, $k$ is equal to $\frac{2\pi }{\text{Period}}$ , $c$ is equals to $-\left(\text{Phase shift}\right)\left(k\right)$ and $h$ is the vertical shift.

The value of $A$ can be evaluated as follows:

  $\begin{array}{c}\text{Amplitude}=\left|\pm A\right|\\ A=\pm \left|\text{Amplitude}\right|\left[\text{Put the value of amplitude}\right]\\ =\pm 12.5\end{array}$

The value of $k$ can be evaluated as follows:

  $\begin{array}{c}k=\frac{2\pi }{\text{Period}}\\ \text{=}\frac{2\pi }{\text{12}}\left[\text{Put the value of period}\right]\\ =\frac{\pi }{6}\end{array}$

Now evaluate the value of $c$ as shown below.

  $\begin{array}{c}y=A\sin \left(kt+c\right)+h\\ 41=12.5\sin \left(\frac{\pi }{6}\left(1\right)+c\right)+53.5\left[\text{Put the value of }y,A,k,tandh\right]\\ 41-53.5=12.5\sin \left(\frac{\pi }{6}+c\right)\left[\text{Subtract both sides by 53}\text{.3}\right]\\ -12.5=12.5\sin \left(\frac{\pi }{6}+c\right)\\ -1=\sin \left(\frac{\pi }{6}+c\right)\left[\text{Divide both sides by 12}\text{.5}\right]\\ -\frac{\pi }{2}=\frac{\pi }{6}+c\left[\text{Taking inverse of function on each side}\right]\\ \frac{-2\pi }{3}=c\left[\text{Subtract }\frac{\pi }{6}\text{from both sides}\right]\end{array}$

Now substitute the value of $A,k\text{ and }c$ in the sine function $y=A\sin \left(kt+c\right)+h$ to evaluate the required sinusoidal function for January.

  $\begin{array}{l}y=A\sin \left(kt+c\right)+h\\ y=12.5\sin \left(\frac{\pi }{6}t-\frac{2\pi }{3}\right)+53.5\end{array}$

Thus the sinusoidal function that models the monthly temperature to represent January is $y=12.5\sin \left(\frac{\pi }{6}t-\frac{2\pi }{3}\right)+53.5$ .

e.

To determine

To find: the average monthly temperature in February according to your model and compare this value with the actual average.

Answer to Problem 6CFU

  $43°$

It is one degree lesser than the actual average.

Explanation of Solution

Given information:

The table representing the average monthly temperatures for a city of Washington is shown below.

Jan	Feb	Mar	April	May	Jun	Jul	Aug	Sept	Oct	Nov	Dec
41 $°$	44 $°$	47 $°$	50 $°$	56 $°$	61 $°$	65 $°$	66 $°$	61 $°$	54 $°$	46 $°$	42 $°$

The sinusoidal function (evaluated in part a.) that models the monthly temperature is as follows:

  $y=12.5\sin \left(\frac{\pi }{6}t-\frac{2\pi }{3}\right)+53.5$

Calculation:

Substitute $t=2$ in the evaluated sinusoidal function and evaluate the average monthly temperature as follows:

  $\begin{array}{c}y=12.5\sin \left(\frac{\pi }{6}\left(2\right)-\frac{2\pi }{3}\right)+53.5\\ =12.5\sin \left(\frac{\pi }{3}-\frac{2\pi }{3}\right)+53.5\\ =12.5\sin \left(-\frac{\pi }{3}\right)+53.5\\ 12.5\left(\frac{-\sqrt{3}}{2}\right)+53.5\\ =-10.8+53.5\\ =42.7\\ \approx 43°\end{array}$

The average monthly temperature of February is $43°$ by using the sinusoidal function. The value is 1 degree lesser than the actual value i.e. $44°$ .

f.

To determine

To find: the average monthly temperature in October according to your model and compare this value with the actual average.

Answer to Problem 6CFU

  $53.5°$

It is approximately same to the actual average.

Explanation of Solution

Given information:

The table representing the average monthly temperatures for a city of Washington is shown below.

Jan	Feb	Mar	April	May	Jun	Jul	Aug	Sept	Oct	Nov	Dec
41 $°$	44 $°$	47 $°$	50 $°$	56 $°$	61 $°$	65 $°$	66 $°$	61 $°$	54 $°$	46 $°$	42 $°$

The sinusoidal function (evaluated in part a.) that models the monthly temperature is as follows:

  $y=12.5\sin \left(\frac{\pi }{6}t-\frac{2\pi }{3}\right)+53.5$

Calculation:

Substitute $t=10$ in the evaluated sinusoidal function and evaluate the average monthly temperature as follows:

  $\begin{array}{c}y=12.5\sin \left(\frac{\pi }{6}\left(10\right)-\frac{2\pi }{3}\right)+53.5\\ =12.5\sin \left(\frac{5\pi }{3}-\frac{2\pi }{3}\right)+53.5\\ =12.5\sin \left(\pi \right)+53.5\\ 12.5\left(0\right)+53.5\\ =53.5\\ \approx 54°\end{array}$

The average monthly temperature of October is $53.5°$ by using the sinusoidal function. The value is approximately same as compared to the actual value i.e. $54°$ .