Q. 92 - Precalculus Enhanced with Graphing Utilities

Question

Bridge Clearance A one-lane highway runs through a tunnel    in the shape of one-half a sine curve cycle. The opening is 28 feet wide at road level and is 15 feet tall at its highest point.

(a) Find an equation for the sine curve that fits the opening. Place the origin at the left end of the sine curve.
(b) If the road is 14 feet wide with 7 -foot shoulders on each side, what is the height of the tunnel at the edge of the road?

Source: en.wikipedia.org/wiki/Interstate_Highway_standards and Ohio Revised Code

Step-by-Step Solution

Answer

(a) The required sin function $y=15\sin \left(\frac{\mathrm{\pi }}{28}t\right)$

(b) The height of the tunnel at the edge of road is $\frac{15}{\sqrt{2}}$units

Step 1.(a) Find an equation for the sine curve that fits the opening. Place the origin at the left end of the sine curve.

The equation for the sine curve that fits the opening can be found out if we consider the curve with left end as the origin and half cycle as the half curve of sine function above the x-axis as follows-: 

Now, the half-curve begins at $x=0$ and ends at $x=28$, so the period of the function will be $T=56$ as the full cycle will end at $x=56$. Therefore,

$$\begin{gathered} \omega =\frac{2\mathrm{\pi }}{T} \\ =\frac{2\mathrm{\pi }}{56} \\ =\frac{\mathrm{\pi }}{28} \end{gathered}$$

Also, the amplitude of the function will be $A=15$ because the maximum value of this sine function at $x=0$ is 15 units.

Since the sine function is of the form $y=A\sin \left(\omega t\right)$, we get the required sine function as-:

$y=15\sin \left(\frac{\mathrm{\pi }}{28}t\right)$

Step 2.(b) If the road is 14 feet wide with 7 -foot shoulders on each side, what is the height of the tunnel at the edge of the road?

If the road is 14 feet wide, the road will be 7 feet away from the origin point from where the sine curve starts.

So, we need to find the value of function at $x=7$ to find the height of tunnel at the edge of road which is-:

$$\begin{gathered} y=15\sin \left(\frac{\mathrm{\pi }}{28}\left(7\right)\right) \\ =15\sin \left(\frac{\mathrm{\pi }}{4}\right) \\ =15\left(\frac{1}{\sqrt{2}}\right) \\ =\frac{15}{\sqrt{2}} \end{gathered}$$

Hence,the height of the tunnel at the edge of road is $\frac{15}{\sqrt{2}}$units