Enjoying the View of a Sinusoidal Canyon, Pt. 1

If there is a tangent line to the rim of the canyon going through your location, and this tangent line is less steep than any other tangent line to the rim, then there will be points on the rim that are invisible to you. Thus as long as you can see the points on the rim with the steepest tangent lines, you will be able to see entire canyon; these points are of course the inflection points of the curve.

We calculate that

$\begin{aligned} f'(x) &= \dfrac{\pi}{2} \cos \left( \pi \left(x - \dfrac{1}{2} \right) \right) \\ \\ f''(x) &= - \dfrac{\pi^2}{2} \sin \left( \pi \left(x - \dfrac{1}{2} \right) \right) \end{aligned}$

From your position on the bridge, the important inflection points are the four closest, one in each quadrant; these are easily seen to be $(\pm \dfrac{1}{2}, \pm 1)$ .

Suppose you are moving upwards and looking to the right. As long as you can see the point $(\dfrac{1}{2},1)$ , you can see the rest of the canyon.

The slope of the tangent line at that point is $f'\left(\dfrac{1}{2}\right) = \dfrac{\pi}{2}$ , so the equation of the tangent line is $y - 1 = \dfrac{\pi}{2} \left( x - \dfrac{1}{2} \right)$ .

The point at which that tangent line crosses the bridge, and thus the highest we can move and still see the whole canyon, is the y-intercept, $1 + \dfrac{\pi}{2} \left( 0 - \dfrac{1}{2} \right) = \boxed{1 - \dfrac{\pi}{4} }$ .

As for the helicopter tour, if we draw all four tangent lines from the points $(\pm \dfrac{1}{2}, \pm 1)$ , they will border an area in the shape of a rhombus from which you will be able to see the entire canyon; the vertices of this rhombus are the x- and y-intercepts of the tangent lines.

The x-intercept of the previously found tangent line is $\dfrac{1}{2} - \dfrac{2}{\pi}$ , so the area of the rhombus will be $2 \left| \left( \dfrac{1}{2} - \dfrac{2}{\pi} \right) \left( 1 - \dfrac{\pi}{4} \right) \right| \approx \boxed{0.0586377}$

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