Question #263932

A cable hangs in a parabolic arc between two poles 100 feet apart. The poles are 30 feet high and the lowest point on a suspended cable is 5 feet above the ground.

a) find the equation of the arc if the vertex is the lowest point of the cable

b) the opening of a cave 20 feet wide is in the shape of a semi ellipse. If the height of the opening is 15 feet at the center, how high is the opening 5 feet from the center.



1
Expert's answer
2021-11-10T18:24:56-0500

a) Let point (0,5)(0, 5) be the vertex of the parabola. Then the equation of the parabola will be


y=ax2+5,a>0y=ax^2+5, a>0

Given y(50)=y(50)=30.y(-50)=y(50)=30. Substitute


a(50)2+5=30a(50)^2+5=30

a=0.01a=0.01

The equation of the arc is

y=0.01x2+5y=0.01x^2+5

b)

The equation of the ellipse is


(xh)2a2+(yk)2b2=1\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1

Substitute


(x0)2(20/2)2+(y0)2152=1\dfrac{(x-0)^2}{(20/2)^2}+\dfrac{(y-0)^2}{15^2}=1

The equation of the given semi ellipse is

x2100+y2225=1,y0\dfrac{x^2}{100}+\dfrac{y^2}{225}=1, y\geq0

If x1=5x_1=5


(5)2100+y2225=1,y0\dfrac{(5)^2}{100}+\dfrac{y^2}{225}=1, y\geq0

y=225(11/4)y=\sqrt{225(1-1/4)}

y=7.53y=7.5\sqrt{3}

The opening 5 feet from the center is 7.53(13)7.5\sqrt{3}(\approx13) feet high.



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