Find the best answers to your questions with the help of IDNLearn.com's knowledgeable users. Get accurate answers to your questions from our community of experts who are always ready to provide timely and relevant solutions.
Sagot :
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]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for visiting IDNLearn.com. We’re here to provide clear and concise answers, so visit us again soon.