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January 10, 2026, 12:28:08 pm
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nbalakers24

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parabola
« on: June 15, 2009, 10:35:20 pm »
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The pathway of the bridge can be reasonbly modelled by a parabola joining the points on each bank 68m apart and rising 3m to reach its maximum height in the centre.

The bank where the bridge starts is 18m above the flood level of the river.


Find the equation of the parabola?
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GoodGuys

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Re: parabola
« Reply #1 on: June 15, 2009, 10:39:03 pm »
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where did you get this question from???

this sounds like the very exact question in my application sac tomorrow...
My teacher was telling me///


SOMEONE FIND THE FORMULA !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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TrueTears

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Re: parabola
« Reply #2 on: June 15, 2009, 10:42:09 pm »
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Is the bridge (parabola) symmetrical around y axis?

If it is then I can do it. If not then :S
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lacoste

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Re: parabola
« Reply #3 on: June 15, 2009, 10:43:42 pm »
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Yeh, I'm assuming it is.
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Mao

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Re: parabola
« Reply #4 on: June 15, 2009, 10:48:44 pm »
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The question gave three points, (0, 18), (68, 18) and (34,21)

now, y=ax^2 + bx + c, plug the data points in there and solve simultaneously for a, b and c
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TrueTears

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Re: parabola
« Reply #5 on: June 15, 2009, 10:49:27 pm »
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Quote from: lacoste on June 15, 2009, 10:43:42 pm
Yeh, I'm assuming it is.
Okay.

Here is how I would do it:

The 2 points are 68 m apart, assuming its symmetrical around y axis, one point would be (34, 18) the other would be (-34, 18)

[y axis is 18 because its lifted 18 above the x axis, assuming x axis is the flood level of the river]

The coordinate of the turning point of parabola is (0, 21)

Now general formula of parabola is

sub in your 3 points work out a, b and c and you are done.

EDIT: Mao's interpretation is probs correct, so listen to him :). I assumed the parabola was symmetrical around y axis. Same parabola anyways but just translated horizontally by z units, where
« Last Edit: June 15, 2009, 10:53:01 pm by TrueTears »
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lacoste

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Re: parabola
« Reply #6 on: June 15, 2009, 10:55:09 pm »
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Quote from: Mao on June 15, 2009, 10:48:44 pm
The question gave three points, (0, 18), (68, 18) and (34,21)

now, y=ax^2 + bx + c, plug the data points in there and solve simultaneously for a, b and c

ahh man, when I tried to do it i got totally diff to TT and Mao.

Is it an inverted parabola, someone draw a diagram?

Need to fine tune some parabola skills =]
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GoodGuys

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Re: parabola
« Reply #7 on: June 15, 2009, 10:56:06 pm »
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i think it is inverted....

as a bridge would be a negative..
understand? LOL

you get the point.
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Mao

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Re: parabola
« Reply #8 on: June 15, 2009, 10:56:54 pm »
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TT's method is valid. Though for parabolas symmetrical around the y-axis, the general form simplifies to , which is easier to work with.
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TrueTears

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Re: parabola
« Reply #9 on: June 15, 2009, 10:57:44 pm »
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Stroodle

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Re: parabola
« Reply #10 on: June 15, 2009, 11:01:06 pm »
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Or just plug it into turning point form:




so
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physics

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Re: parabola
« Reply #11 on: June 17, 2009, 08:31:33 pm »
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why does it seem that every school has the same analysis task????
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Re: parabola
« Reply #12 on: June 17, 2009, 11:09:48 pm »
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Man I hope this isn't on Friday's SAC.
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