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Word Problems with Parabolas 
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#1
Feb2212, 08:13 PM

P: 139

1. The problem statement, all variables and given/known data
Roads are often designed with parabolic surfaces to allow rain to drain off. A particular road is that is 32 feet wide is .4 feet higher in in the center then on the sides. a) Find an equation if the parabola that models the road surface. (Assume that the orgin is at the center of the road.) b) How far from the center of the road is the road surface .1 feet lower then in the middle? 2. Relevant equations None 3. The attempt at a solution x^2=4py x^2= 4(.4)y x^2=1.6y ^part (a) Correct? x^2=1.6(.3) x^2=.48 x = .693 feet away from the center ^Part (b) Correct? I kind of have a basis but I am a little wary about the answers. 


#2
Feb2212, 08:52 PM

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The form x^2=4py is fine. If the origin is the center of the road then a point at the center of the road is x=0, y=0 and x is the distance from the center of the road and y is the elevation of the road. What should y be if x is 16?



#3
Feb2212, 08:54 PM

P: 139

y=.4
but what next? 


#4
Feb2212, 08:56 PM

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Word Problems with Parabolas



#5
Feb2212, 09:48 PM

P: 139

so like x^2=4py
x^2=4p(.4) (1/.4)x^2=4p 4p=1/.4 p=1/1.6 x^2=4(1/1.6)y x^2=6.4y is that right? becuase the first time i thought that (0, .4) was the focus, and p=.4 from the focus... 


#6
Feb2212, 10:00 PM

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#7
Feb2212, 10:03 PM

P: 139

p=160
so the equation for the graph would be x^2=640y? and plugging in .3 for y would get me answer to part (b) 13.856 feet from center? 


#8
Feb2212, 10:08 PM

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#9
Feb2212, 10:33 PM

P: 139

oh, i was thinking of like where is the road only .3 feet or whatever,
do i plug .1 in? Solved it for .1 and got 8 


#10
Feb2212, 10:37 PM

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#11
Feb2212, 10:41 PM

P: 139

What do you mean you don't like them? I wasn't trying to be rude or anything.
You do plug in .1 


#12
Feb2212, 10:48 PM

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#13
Feb2312, 05:27 PM

P: 139

x^2=640y is upside down?
I don't understand. 


#14
Feb2312, 05:33 PM

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#15
Feb2312, 05:58 PM

P: 139

ohhhhhhhhhhhhhhhhhhhhh, thank you, that makes a lot more sense, how would you show the work so that the number comes out to x^2=640y?
Would you use .4 when making the equation becuase it is 'upside down' now? But would it matter because the answer comes out to the same thing no matter what? 


#16
Feb2312, 06:53 PM

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#17
Feb2312, 08:10 PM

P: 139

Can you help me with how the graph would look?
It's not for my homework, its just so I can understand it. How would the parabola open and where? Also to get part (b) would you plug in .1 for y and solve? Becuase when I just plugged in .1 it would give me a negative answer 


#18
Feb2312, 08:45 PM

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This means you want a parabola that opens down. If the cross section of the road looked like a parabola that opened up, water would collect in the middle of the road, which isn't good. Do as Dick suggested many posts ago by putting the vertex of the parabola at the origin. The left and right edges of the road will be at (16, .4) and (16, .4). If you substitute the x and y values of either point into x^{2} = 4py, you should be able to solve for p, which will be a positive number. 


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