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Word Problems with Parabolas

by darshanpatel
Tags: parabolas, word
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darshanpatel
#1
Feb22-12, 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.
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Dick
#2
Feb22-12, 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?
darshanpatel
#3
Feb22-12, 08:54 PM
P: 139
y=.4

but what next?

Dick
#4
Feb22-12, 08:56 PM
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Word Problems with Parabolas

Quote Quote by darshanpatel View Post
y=.4

but what next?
Use that to solve for p. p isn't equal to 0.4.
darshanpatel
#5
Feb22-12, 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...
Dick
#6
Feb22-12, 10:00 PM
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Quote Quote by darshanpatel View Post
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...
(1/.4)x^2=4p. x=16. What's p? It's not 1.6 either.
darshanpatel
#7
Feb22-12, 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?
Dick
#8
Feb22-12, 10:08 PM
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Quote Quote by darshanpatel View Post
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?
Yes for the first part. .4 is how far the road is below the center at x=16. The problem is asking "How far from the center of the road is the road surface .1 feet lower then in the middle?". I wouldn't plug in .3 for y.
darshanpatel
#9
Feb22-12, 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
Dick
#10
Feb22-12, 10:37 PM
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Quote Quote by darshanpatel View Post
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
I really don't like responses like "do i plug .1 in?". Do you or don't you? Sketch a picture of the road surface and tell me.
darshanpatel
#11
Feb22-12, 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
Dick
#12
Feb22-12, 10:48 PM
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Quote Quote by darshanpatel View Post
What do you mean you don't like them? I wasn't trying to be rude or anything.
You do plug in .1
I didn't think you were trying to be rude or anything. I just like hearing "You do plug in .1" better than the "?" thing. Yes, you are right. And for the equation of the road surface they might like x^2=(-640)y better than x^2=640y. x^2=640y is sort of 'upside down'.
darshanpatel
#13
Feb23-12, 05:27 PM
P: 139
x^2=640y is upside down?

I don't understand.
Mark44
#14
Feb23-12, 05:33 PM
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Quote Quote by darshanpatel View Post
x^2=640y is upside down?

I don't understand.
It's upside down relative to the road surface. The parabola above opens up - you want a parabola that opens down, so that water will drain off the road. That's what Dick meant by "upside down."
darshanpatel
#15
Feb23-12, 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?
Mark44
#16
Feb23-12, 06:53 PM
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Quote Quote by darshanpatel View Post
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?
Start the work you did earlier with x2 = -4py (with p > 0) to get a parabola that opens down.
Quote Quote by darshanpatel View Post

But would it matter because the answer comes out to the same thing no matter what?
darshanpatel
#17
Feb23-12, 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
Mark44
#18
Feb23-12, 08:45 PM
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P: 21,397
Quote Quote by darshanpatel View Post
Can you help me with how the graph would look?
How do you think the graph should look? You want the water to drain off the road, so it should have the high point at the center of the road's cross section.

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 x2 = -4py, you should be able to solve for p, which will be a positive number.
Quote Quote by darshanpatel View Post

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
If the high point of the road is at (0, 0), every other point will have y values that are negative. You should get two x-values when y = -.1.


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