Answer :
To determine the maximum height at which an eight-foot-wide rectangular truck can safely pass under the elliptical arch, we follow these steps:
1. Identify the parameters of the ellipse:
- The width of the elliptical arch, which is the span of the arch at ground level, is 50 feet. This width represents the major axis, so the semi-major axis [tex]\( a \)[/tex] is half of the width:
[tex]\[ a = \frac{50}{2} = 25 \text{ feet} \][/tex]
- The height of the elliptical arch at the center is 15 feet. This height represents the semi-minor axis [tex]\( b \)[/tex]:
[tex]\[ b = 15 \text{ feet} \][/tex]
2. Determine the half-width of the truck:
- Since the truck is 8 feet wide, half of this width [tex]\( x \)[/tex] is:
[tex]\[ x = \frac{8}{2} = 4 \text{ feet} \][/tex]
3. Use the equation of the ellipse to find the height [tex]\( y \)[/tex] at [tex]\( x = 4 \)[/tex]:
- The standard equation of an ellipse centered at the origin is:
[tex]\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \][/tex]
- Plugging in the known values [tex]\( x = 4 \)[/tex], [tex]\( a = 25 \)[/tex], and [tex]\( b = 15 \)[/tex]:
[tex]\[ \frac{4^2}{25^2} + \frac{y^2}{15^2} = 1 \][/tex]
- First, calculate [tex]\( \frac{4^2}{25^2} \)[/tex]:
[tex]\[ \frac{16}{625} = \frac{16}{625} \][/tex]
- Subtract this value from 1 to isolate [tex]\( \frac{y^2}{15^2} \)[/tex]:
[tex]\[ 1 - \frac{16}{625} = \frac{625}{625} - \frac{16}{625} = \frac{609}{625} \][/tex]
- Simplify the fraction on the right-hand side:
[tex]\[ \frac{y^2}{15^2} = \frac{609}{625} \][/tex]
- Multiply both sides of the equation by [tex]\( 15^2 \)[/tex] to solve for [tex]\( y^2 \)[/tex]:
[tex]\[ y^2 = \frac{609}{625} \cdot 225 = \frac{609 \times 225}{625} \][/tex]
- Find [tex]\( y \)[/tex] by taking the square root of both sides:
[tex]\[ y = \sqrt{\frac{609 \times 225}{625}} \][/tex]
Evaluating this expression, we get:
[tex]\[ y \approx 14.81 \text{ feet} \][/tex]
Therefore, the maximum height at which the eight-foot-wide truck can safely drive underneath the elliptical bridge is approximately [tex]\( 14.81 \)[/tex] feet.
[tex]\[ h \approx 14.81 \text{ feet} \][/tex]
1. Identify the parameters of the ellipse:
- The width of the elliptical arch, which is the span of the arch at ground level, is 50 feet. This width represents the major axis, so the semi-major axis [tex]\( a \)[/tex] is half of the width:
[tex]\[ a = \frac{50}{2} = 25 \text{ feet} \][/tex]
- The height of the elliptical arch at the center is 15 feet. This height represents the semi-minor axis [tex]\( b \)[/tex]:
[tex]\[ b = 15 \text{ feet} \][/tex]
2. Determine the half-width of the truck:
- Since the truck is 8 feet wide, half of this width [tex]\( x \)[/tex] is:
[tex]\[ x = \frac{8}{2} = 4 \text{ feet} \][/tex]
3. Use the equation of the ellipse to find the height [tex]\( y \)[/tex] at [tex]\( x = 4 \)[/tex]:
- The standard equation of an ellipse centered at the origin is:
[tex]\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \][/tex]
- Plugging in the known values [tex]\( x = 4 \)[/tex], [tex]\( a = 25 \)[/tex], and [tex]\( b = 15 \)[/tex]:
[tex]\[ \frac{4^2}{25^2} + \frac{y^2}{15^2} = 1 \][/tex]
- First, calculate [tex]\( \frac{4^2}{25^2} \)[/tex]:
[tex]\[ \frac{16}{625} = \frac{16}{625} \][/tex]
- Subtract this value from 1 to isolate [tex]\( \frac{y^2}{15^2} \)[/tex]:
[tex]\[ 1 - \frac{16}{625} = \frac{625}{625} - \frac{16}{625} = \frac{609}{625} \][/tex]
- Simplify the fraction on the right-hand side:
[tex]\[ \frac{y^2}{15^2} = \frac{609}{625} \][/tex]
- Multiply both sides of the equation by [tex]\( 15^2 \)[/tex] to solve for [tex]\( y^2 \)[/tex]:
[tex]\[ y^2 = \frac{609}{625} \cdot 225 = \frac{609 \times 225}{625} \][/tex]
- Find [tex]\( y \)[/tex] by taking the square root of both sides:
[tex]\[ y = \sqrt{\frac{609 \times 225}{625}} \][/tex]
Evaluating this expression, we get:
[tex]\[ y \approx 14.81 \text{ feet} \][/tex]
Therefore, the maximum height at which the eight-foot-wide truck can safely drive underneath the elliptical bridge is approximately [tex]\( 14.81 \)[/tex] feet.
[tex]\[ h \approx 14.81 \text{ feet} \][/tex]
