Sec.6.5 - Where to Sit at the Movies
Applied Project in Sec.6.5, Calculus by Stewart
Chinese version 電影院要坐哪?
A movie theater has a screen that is positioned $3$m off the floor and is $10$m high. The first row of seats is placed $3$m from the screen and the rows are set $1$m apart. The floor of the seating area is inclined at an angle of $\alpha=20^{\circ}$ above the horizontal and the distance up the incline that you sit is $x$. The theater has $21$ rows of seats, so $0 \leq x \leq 20$. Suppose you decide that the best place to sit is in the row where the angle $\theta$ subtended by the screen at your eyes is a maximum. Let’s also suppose that your eye are $1.2$m above the floor.
Question 1:
Show that $\theta = \cos^{-1}\left(\frac{a^2+b^2-100}{2ab}\right)$, where $$ a^2 = (3+x\cos{\alpha})^2+(11.8-x\sin{\alpha})^2 $$ and $$ b^2 = (3+x\cos{\alpha})^2+(x\sin{\alpha} - 1.8)^2 $$
Question 2:
Use a graph of $\theta$ as a function of $x$ to estimate the value of $x$ that maximizes $\theta$. In which row should you sit? What is the viewing angle $\theta$ in this row? What is the angle $\theta$ subtended by the screen at your eyes?
Answer:
$x$ | $\theta$ | $x$ | $\theta$ |
---|---|---|---|
0 | 0.781414 | 11 | 0.688473 |
1 | 0.885175 | 12 | 0.654426 |
2 | 0.932324 | 13 | 0.622381 |
3 | 0.943356 | 14 | 0.592377 |
4 | 0.932085 | 15 | 0.564380 |
5 | 0.907358 | 16 | 0.538310 |
6 | 0.874860 | 17 | 0.514061 |
7 | 0.838292 | 18 | 0.491514 |
8 | 0.800070 | 19 | 0.470546 |
9 | 0.761763 | 20 | 0.451038 |
10 | 0.724364 |
The maximum of $\theta$ is 0.943356, you should sit at third row.
Question 3:
Use your computer algebra system to differentiate $\theta$ and find a numerical value for the root of the equation $\frac{d\theta}{dx} = 0$. Does this value confirm your result in Problem 2?
Answer:
According to the graph of $y = \frac{d\theta}{dx}$
we found that there is a root at $\approx 2.84556$
Question 4:
Use the graph of $\theta$ to estimate the average value of $\theta$ on the interval $0 \leq x \leq 20$. Then use your CAS to compute the average value. Compare with the maximum and minimum values of $\theta$.
Answer:
average value:
$$ avg = \frac{1}{20}\int_{0}^{20} \theta(x) dx \approx 0.71941 $$
There exists three critical points at $x=0$, 2.84556 and 20, the maximum value is approximately 0.94336, the minimum value is approximately 0.45594