Chapter 6.6, Problem 35AYU

Hours of Daylight According to the Old Farmer’s Almanac, in Miami. Florida, the number of hours of sunlight on the summer solstice of 2015 was $13.75$, and the number of hours of sunlight on the winter solstice was $10.52$.

(a) Find a sinusoidal function of the form $y=A\sin (\omega x-\varphi )+B$ that models the data.

(b) Use the function found in part $\left(a\right)$ to predict the number of hours of sunlight on April 1, the 91st day of the year.

(c) Draw a graph of the function found in part $\left(a\right)$.

(d) Look up the number of hours of sunlight for April 1 in the Old Farmer’s Almanac, and compare the actual hours of daylight to the results found in part $\left(b\right)$.

To determine

To find:

a. Find a sinusoidal function of the form $y=A\sin \left(\omega x-\varphi \right)+B$ that models the data.

Answer to Problem 35AYU

Solution:

a. $y=1.615\sin \left(\frac{2\pi x}{365}-1.39\right)+12.135$

Explanation of Solution

Given:

According to the Old Farmer’s Almanac, in Miami, Florida, the number of hours of sunlight on the summer solstice of 2015 was $13.75$, and the number of hours of sunlight on the winter solstice was $10.52$.

Calculation:

a. Find a sinusoidal function of the form $y=A\sin \left(\omega x-\varphi \right)+B$ that models the data.

Determine A, the amplitude of the function.

$\text{Amplitude }=\frac{ \text{largest data value }-\text{ smallest data value} }{2}=\frac{13.75-10.52}{2}=1.615$

Determine B, the vertical shift of the function.

$\text{Vertical shift }=\frac{ \text{largest data value }+\text{ smallest data value}}{2}=\frac{13.75\text{ + }10.52}{2}=12.135$

Determine $\omega$, It is easier to find the horizontal stretch factor first. Since the period of the function is $T=365$. Because $T=\frac{2\pi }{\omega }=365$. Therefore $\omega =\frac{2\pi }{365}$.

Now superimpose the graph of $y=1.615\sin \left(\frac{2\pi }{365}x\right)+12.135$

Determine the horizontal shift of the function by using the period of the data.

$\left(0,91.25\right),\left(91.25,182.5\right),\left(182.5,273.75\right),\left(273.75,365\right)$

Since sine curve is increasing in the interval $\left(0,91.25\right)$ and decreasing in the interval $\left(91.25,182.5\right)$. So a local maximum point occurs at $x=91.25$.

Since the maximum occurs at $x=172$. So we must shift the graph of the function $172-91.25=80.75$ units to the right by replacing $x$ by $x-80.75$

$y=1.615\sin \left(\frac{2\pi }{365}(x-80.75\right)+12.135$

$y=1.615\sin \left(\frac{2\pi x}{365}-1.39\right)+12.135$

To determine

To find:

b. Use the function found in part (a) to predict the number of hours of sunlight on April 1, the 91st day of the year.

Answer to Problem 35AYU

Solution:

b. 12hours 25 minutes

Explanation of Solution

Given:

According to the Old Farmer’s Almanac, in Miami, Florida, the number of hours of sunlight on the summer solstice of 2015 was $13.75$, and the number of hours of sunlight on the winter solstice was $10.52$.

Calculation:

b. Use the function found in part (a) to predict the number of hours of sunlight on April 1, the 91st day of the year.

$y=1.615\sin \left(\frac{2\pi \times 91}{365}-1.39\right)+12.135$

$y=1.615\sin \left(1.5656-1.39\right)+12.135$

$y=1.615\sin \left(0.1757\right)+12.135$

$y=1.615\times 0.1748+12.135$

$y=12.417$

So we predict that there will be about $12.42\text{ hours }=12\text{ hours }25\text{ minutes}$

To determine

To find:

c. Draw a graph of the function found in part (a).

Answer to Problem 35AYU

Solution:

c.

Explanation of Solution

Given:

According to the Old Farmer’s Almanac, in Miami, Florida, the number of hours of sunlight on the summer solstice of 2015 was $13.75$, and the number of hours of sunlight on the winter solstice was $10.52$.

Calculation:

c. Draw a graph of the function found in part (a).

To determine

To find:

d. Look up the number of hours of sunlight for April 1 in the Old Farmer’s Almanac, and compare the actual hours of daylight to the results found in part (b).

Answer to Problem 35AYU

Solution:

d. The actual hours of sunlight on April 1, 2015, were $12.43$ hours. This is close to the predicted amount of $12.42$ hours.

Explanation of Solution

Given:

According to the Old Farmer’s Almanac, in Miami, Florida, the number of hours of sunlight on the summer solstice of 2015 was $13.75$, and the number of hours of sunlight on the winter solstice was $10.52$.

Calculation:

d. Look up the number of hours of sunlight for April 1 in the Old Farmer’s Almanac, and compare the actual hours of daylight to the results found in part (b).

The actual hours of sunlight on April 1, 2015, were $12.43$ hours. This is close to the predicted amount of $12.42$ hours.