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The dimensions of the supporting base of a bridge are as follows: 40 m height and 120 m...

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alejandrogalarce | TA , Undergraduate | (Level 1) Valedictorian

Posted October 18, 2012 at 8:04 AM via web

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The dimensions of the supporting base of a bridge are as follows: 40 m height and 120 m wide.

a) If the arch is an ellipse with the standard form (x-h)2/a2 + (y-k)2/b2 = 1, determine the equation of the arch.

b) If the arch is a parabola with the standard form y-k=a(x-h)2, determine the  equation of the arch.

C)Determine the height of the arch at a point 30 m away from the centre using the equations found above.  The actual height at this point is 30.2 m, determine whether the parabola or ellipse makes the closest approximation. 

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted October 20, 2012 at 2:38 AM (Answer #1)

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Assume the center of the support is at the origin.

(a) In the case of an ellipse where `((x-h)^2)/a^2+(y-k)^2/b^2=1` we have a=60 and b=40. (a is the length of the semi-major axis, or 1/2 the major axis which is 120 while b is the length of the semi-minor axis or 1/2 the minor axis). The center of the ellipse is at the origin so h=k=0.

Then the equation is `x^2/3600+y^2/1600=1`

(b) In the case of a parabola, assume the vertex lies on the y-axis. Then `y=a(x-h)^2+k` where h=0 and k=40. We also have the two intercepts, (-60,0) and (60,0). Using (60,0) to find a we get `0=a(60)^2+40==>a=-1/90`

So `y=-1/90x^2+40`

(c) In the case of the ellipse if x=30 we get

`900/3600+y^2/1600=1==>y^2=1200==>y~~34.64`

For the parabola with x=30 we get

`y=-1/90(30)^2+40==>y=30`

Since 30 is closer to 30.2 than 34.64 is, the parabola is a better model.

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