Chapter 13.4, Problem 48PPSE

Solution

i.

To find: The dependent and independent variable.

  $x=$ Number of days from equinox

  $y=$ Duration of the sunlight.

Given:

The duration of sunlight is $12\text{ h }9\text{ min}$ and the variance of longest and shortest from equinox is $1\text{ h }55\text{ min}$ .

Concept:

1. The variable which depends upon the other factor is known as the dependent variable and the variable that does not depend upon any variable is known as independent variable.

Calculation:

Suppose the dependent variable is $x$ and independent variable is $y$ .

From the problem, it is understood that duration of the sunlight is the independent variable and the number of days from equinox is the dependent variable.

Conclusion:

Hence, the variable $x$ is the number of days from equinox and the variable $y$ is duration of the sunlight.

ii.

To find: The amplitude and the period of the function.

Amplitude = $\frac{23}{12}$

Period = $365$

Given:

The duration of sunlight is $12\text{ h }9\text{ min}$ and the variance of longest and shortest from equinox is $1\text{ h }55\text{ min}$ .

Calculation:

The variance between the longest and the shortest day is $1\text{ h }55\text{ min}$ . So, the amplitude $a$ is,

  $\begin{array}{c}a=1+\frac{55}{60}\\ =\frac{115}{60}\\ =\frac{23}{12}\end{array}$

The number of days in a year is $365$ . So, the period is $365$ .

Conclusion:

Hence, the amplitude is $\frac{23}{12}$ and the period is $365$ .

iii.

To find: The function that relate the number of days from equinox with the variance between the longest and shortest days.

  $y=\frac{23}{12}\sin \frac{2\pi }{365}x$

Given:

The duration of sunlight is $12\text{ h }9\text{ min}$ and the variance of longest and shortest from equinox is $1\text{ h }55\text{ min}$ .

Concept:

Use the sine function $y=a\sin bx$ to obtain the required function. Here $a=$ amplitude and $b=\frac{2\pi }{\text{period}}$ .

Calculation:

The period of the function is $365$ . So, the value of $b$ is,

  $b=\frac{2\pi }{365}$

It is understood that $a=\frac{23}{12}$ .

Substitute the known values in the following function.

  $y=\frac{23}{12}\sin \frac{2\pi }{365}x$

Conclusion:

Hence, the required function is $y=\frac{23}{12}\sin \frac{2\pi }{365}x$ .

iv.

To find: How much sunlight is less on February 14 compared to the March 21.

  $1.1\text{ h}$ less

Given:

The duration of sunlight is $12\text{ h }9\text{ min}$ and the variance of longest and shortest from equinox is $1\text{ h }55\text{ min}$ .

Calculation:

It is understood that March 21 equivalent to the $x=0$ and the February 14 is equivalent to the $x=-35$ .So,

Substitute $x=-35$ in the following function:

  $\begin{array}{c}y=\frac{23}{12}\sin \frac{2\pi }{365}x\\ =\frac{23}{12}\sin \frac{2\pi }{365}\left(-35\right)\\ =-1.1\text{ h}\end{array}$

Conclusion:

Hence, there is $1.1\text{ h}$ less sunlight on February 14 compared to the March 21.