Question

Let$\mathit{S}\mathbf{\left(}\mathit{t}\mathbf{\right)}$ be the length of the ${\mathit{t}}^{\mathbf{t}\mathbf{h}}$day of the year$\mathbf{2013}$ in Mumbai (formerly known as Bombay), India (measured in hours, from sunrise to sunset). We are given the following values of$\mathit{S}\mathbf{\left(}\mathit{t}\mathbf{\right)}$ :

$\mathbf{\left[}\begin{array}{cc}\mathbf{t}& \mathbf{S}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\\ \mathbf{47}& \mathbf{11}\mathbf{.5}\\ \mathbf{74}& \mathbf{12}\\ \mathbf{273}& \mathbf{12}\end{array}\mathbf{\right]}$

For example$\mathit{S}\mathbf{\left(}\mathbf{47}\mathbf{\right)}\mathbf{=}\mathbf{11}\mathbf{.5}$, means that the timefrom sunrise to sunset on February16is11hours and30minutes. For locations close to the equator, the function$\mathit{S}\mathbf{\left(}\mathit{t}\mathbf{\right)}$ is well approximated by a trigonometric functionof the form

$\mathit{S}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\pi }\mathbf{t}}{\mathbf{365}}\mathbf{\right)}\mathbf{+}\mathit{c}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\pi }\mathbf{t}}{\mathbf{365}}\mathbf{\right)}$

(The period is 365 days, or 1 year.) Find this approximationfor Mumbai, and graph your solution. Accordingto this model, how long is the longest day of the year inMumbai?

Short answer

In Mumbai, the longest day of the year will be$13.3$hours long.

And the required graph is:

Step-by-step solution

Consider the labeled diagram. 

Near the equator location, the length of the ${\mathit{t}}^{\mathbf{t}\mathbf{h}}$day of the year is given by the equation,

.$\mathit{S}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\pi }\mathbf{t}}{\mathbf{365}}\mathbf{\right)}\mathbf{+}\mathit{c}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\pi }\mathbf{t}}{\mathbf{365}}\mathbf{\right)}$

Consider the standard equation and the table.

$S\left(t\right)=a+b\cos \left(\frac{2\pi t}{365}\right)+c\sin \left(\frac{2\pi t}{365}\right)$

$\left[\begin{array}{l}tS\left(t\right)\\ 4711.5\\ 7412\\ 27312\end{array}\right]$

Substitute the values in the equation.

$\begin{array}{l}11.5=a+b\cos \left(\frac{2\pi \times 47}{365}\right)+c\sin \left(\frac{2\pi \times 47}{365}\right)\\ \Rightarrow a+0.690b+0.724c=11.5\\ 12=a+b\cos \left(\frac{2\pi \times 74}{365}\right)+c\sin \left(\frac{2\pi \times 74}{365}\right)\\ \Rightarrow a+0.293b+0.956c=12\\ 12=a+b\cos \left(\frac{2\pi \times 273}{365}\right)+c\sin \left(\frac{2\pi \times 273}{365}\right)\\ \Rightarrow a-0.0129b-0.999c=12\end{array}$

Represent the equations in terms of matrix. 

The matrix form of the equations is,

$\left[\begin{array}{cccc}1& 0.693& 0.724& 11.5\\ 1& 0.293& 0.956& 12\\ 1& -0.0129& -0.999& 12\end{array}\right]$

Reduce the matrix using Row Echelon form,

$\left[\begin{array}{cccc}1& 0& 0& 12.17\\ 0& 1& 0& -1.15\\ 0& 0& 1& 0.18\end{array}\right]$

The values are, $a=12.17,b=-1.15,c=0.18$

Substitute the values in the formula for computing the length of the day. 

The equations is,

$\begin{array}{l}S\left(t\right)=a+b\cos \left(\frac{2\pi t}{365}\right)+c\sin \left(\frac{2\pi t}{365}\right)\\ \Rightarrow S\left(t\right)=12.17+(-1.15)\cos \left(\frac{2\pi t}{365}\right)+0.18\sin \left(\frac{2\pi t}{365}\right)\\ \therefore S\left(t\right)=12.17-1.15\cos \left(\frac{2\pi t}{365}\right)+0.18\sin \left(\frac{2\pi t}{365}\right)\end{array}$

The graph of the obtained equation is shown below:

The longest day will have,

$\begin{array}{c}S\left(t\right)=12.17-1.15\times (-1)+0.18\left(0\right)\\ \therefore S\left(t\right)=13.32\text{hours}\end{array}$

Hence, the longest day is about $13.32\text{hours}$.