An architect's design for a building includes some large pillars with cross sections in the shape of hyperbolas. The curves are modeled by x^2/.25 - y^2/9 = 1. If the pillars are 4 meters tall, find the width of the top of each pillar.
+118703
When y=2
4x2=1+4/9
4×x2=1+49⇒{x=−√136x=√136}⇒{x=−0.6009252125773315x=0.6009252125773315}
width at the top = √133 metres
+26397 An architect's design for a building includes some large pillars with cross sections in the shape of hyperbolas. The curves are modeled by . If the pillars are 4 meters tall, find the width of the top of each pillar.
x20.25−y29=1we need xx20.25=1+y29x2=0.25∗(1+y29)|±√x1,2=±0.5∗√(1+y29)x1,2=±0.53∗√y2+9x1,2=±16∗√y2+9x1=−16∗√y2+9x2=16∗√y2+9
width=x2−x1=16∗√y2+9−(−16∗√y2+9)width=2∗16∗√y2+9width=13∗√y2+9|y=4 meterwidth=13∗√16+9width=13∗√25width=13∗5width=53width=1.¯6 m
+130477 We have
(x^2 / .25) - (y^2 / 9 ) = 1
And, let's assume that the base of the pillar lies on the line y = -2
Then, since the pillars are 4 m high, at the top of the pillar, y = 2
So we have
x^2 / .25 - 2^9 / 9 = 1
x^2 / .25 - 4/9 = 1 add 4/9 to each side
x^2 / .25 = 1 + 4/9
x^2 / .25 = 13 / 9 multiply by .25 on both sides
x^2 = (.25 * 13) / 9
x^2 = 13/36
x = ±√(13) / 6
And these are the two x values on the hyperbola when y = 2
So.....the distance between these two x values is the width of the cross-section at the top of the pillar = (2)√(13)/6 = (1/3)√(13) = aboiut 1.2019 m
Here's a graph of the situation.........https://www.desmos.com/calculator/ts3acz4ffr
+118703
When y=2
4x2=1+4/9
4×x2=1+49⇒{x=−√136x=√136}⇒{x=−0.6009252125773315x=0.6009252125773315}
width at the top = √133 metres
