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


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