Question

will a truck that is 14 feet wide carrying a load that reaches 10 feet above the ground clear the semielliptical arch on the one - way road that passes under the bridge shown in the figure on the right? it □ clear the arch because the height of the archway of the bridge 7 feet from the center is approximately □ feet. (round to two decimal places as needed.)

Explanation:

Step1: Define the ellipse equation

The semi - elliptical arch has a major axis (the width of the arch at the ground level) of length \(2a = 44\) feet, so \(a=22\) feet. The height of the semi - ellipse at the center (the semi - minor axis) is \(b = 11\) feet (from the figure, the height at the center is 11 ft). The standard equation of an ellipse centered at the origin \((0,0)\) is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\). For the upper half (the semi - ellipse), we can solve for \(y\): \(y=\sqrt{b^{2}(1 - \frac{x^{2}}{a^{2}})}\)

Step2: Substitute \(x = 7\) (since the truck is 14 feet wide, we consider the distance from the center, which is \(\frac{14}{2}=7\) feet)

We know that \(a = 22\), \(b = 11\), and \(x = 7\). Substitute these values into the equation for \(y\):

First, calculate \(\frac{x^{2}}{a^{2}}=\frac{7^{2}}{22^{2}}=\frac{49}{484}\)

Then, \(1-\frac{x^{2}}{a^{2}}=1 - \frac{49}{484}=\frac{484 - 49}{484}=\frac{435}{484}\)

Next, \(b^{2}(1 - \frac{x^{2}}{a^{2}})=11^{2}\times\frac{435}{484}=121\times\frac{435}{484}\)

Simplify \(121\times\frac{435}{484}=\frac{121}{484}\times435=\frac{1}{4}\times435 = 108.75\)

Then, \(y=\sqrt{108.75}\approx10.43\)

Answer:

will; \(10.43\)