Chapter 10.4, Problem 42PS

Solution

To find: At what days the sun is ${20}^{\circ }$ north of due east at sunrise.

The day at which the sun is ${20}^{\circ }$ north of due east at sunrise is May $2$ and August $11$ .

Given information:

  $\theta \left(t\right)=31\sin \left(\frac{2\pi }{365}t-1.4\right)$

Calculation:

The given function is given by,

  $\theta \left(t\right)=31\sin \left(\frac{2\pi }{365}t-1.4\right)$

Substituting the value of $\theta \left(t\right)=20$ ,

  $\begin{array}{c}31\sin \left(\frac{2\pi }{365}t-1.4\right)=20\\ \sin \left(\frac{2\pi }{365}t-1.4\right)=\frac{20}{31}\\ \frac{2\pi }{365}t-1.4={\sin }^{-1}\frac{20}{31}\end{array}$

Here,

The value of ${\sin }^{-1}\frac{20}{31}$ can either be $0.7$ or $2.44$ .

When ${\sin }^{-1}\frac{20}{31}=0.7$ ,

  $\begin{array}{c}\frac{2\pi }{365}t-1.4=0.7\\ \frac{2\pi }{365}t=0.7+1.4\\ \frac{2\pi }{365}t=2.1\\ t=2.1\cdot \frac{365}{2\pi }\\ t\approx 122\end{array}$

Thus, the ${122}^{\text{nd}}$ day of the year is May $2$ .

When ${\sin }^{-1}\frac{20}{31}=2.44$ ,

  $\begin{array}{c}\frac{2\pi }{365}t-1.4=2.44\\ \frac{2\pi }{365}t=2.44+1.4\\ \frac{2\pi }{365}t=3.84\\ t=3.84\cdot \frac{365}{2\pi }\\ t\approx 223\end{array}$

Thus, the ${223}^{\text{rd}}$ day of the year is August $11$ .

Therefore, the day at which the sun is ${20}^{\circ }$ north of due east at sunrise is May $2$ and August $11$ .

Simply plot the supplied function and $y=20$ on the graph of $\theta \left(t\right)$ to observe where they intersect to verify the solution.