How High Must a Parabolic Arch Be for Minimum Clearance Over a Stream?

  • Thread starter ahmadmz
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In summary, a parabolic arch spanning a stream 200 feet wide must be above the stream with a minimum clearance of 40 feet to avoid interference.
  • #1
ahmadmz
62
0

Homework Statement



A parabolic arch spans a stream 200 feet wide. How high
must the arch be above the stream to give a minimum
clearance of 40 feet over a channel in the center that is
120 feet wide?

Homework Equations





The Attempt at a Solution



They are asking for the k value right?
If I place the parabola starting from x=0,y=0 then the vertex is at (100,40+n).
When x=40,y=40. x=160,y=40 I want to find out what y is when x=100.

So I used the equation (x-h)^2 = a(y-k)

(x-100)^2 = a(y-(40+n)) After using x=40 ,y=40 I get n=-3600/a And now I have no idea what to do .. Am I doing this right?
 
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  • #2
I think so. Why not just put x=0 and y=0 into your equation and get another relation between 'a' and 'n'?
 
  • #3
I think you can just reason it out, no need for all those crazy equations and what have you

Basically you're going to have a parabola that crosses the x-axis at x=-100 and x=100

You need at least y=40 at x=60 and -60, that ultimately what it's asking, right? And it's basically of the form Y=-Ax^2+H (I just centered it at 0 for ease)

H is what we're looking for, and A is also unknown, however we know if y=40, then x=60, and if y=0, x=100

I went ahead and finished this out to find A and H and checked it and it was right, see if you can follow.
 
  • #4
Ok I used the equation y=-ax^2+h with the values x=60,y=40 and x=100,y=0.
Then solved the system and got a = -1/160 and h = 62.5 (the answer for the question).
And the equation of the parabola is y=-1/160x^2+62.5

Thanks for the help :). This is the first time I've done a question like this and I had trouble approaching it. I'm self studying these days and this shows i got a long way to go.
 
  • #5
ahmadmz said:
Thanks for the help :). This is the first time I've done a question like this and I had trouble approaching it. I'm self studying these days and this shows i got a long way to go.
I feel your pain, I been self-studying for months now. I'm proud of you :-] Keep at it!
 
  • #6
I performed a process somewhat like what was described in #3 and #4, but obtained a somewhat different result. The standard form of a parabola was still necessary.

Do you know of any other interesting problems like this; using parabolas or other conic sections? Interesting and varied problems are often difficult to find. Also, I'm curious; was the question in post #1 from PreCalculus, or was it from Intermediate Algebra (I suspect it is from PreCalculus). Once in a while, I restudy College Algebra or Intermediate Algebra, and the applied situation exercises are often interesting but I just do not find enough of them.
 
  • #7
What result did you get symbolipoint? The book says the answer is 62.5 which i got.
I am studying conics right now and I'll let you know if I find any more similar problems. This is from a intermediate algebra book. You are correct that most books do not have many problems like these. They are indeed interesting :)
 
  • #8
ahmadmz said:
What result did you get symbolipoint? The book says the answer is 62.5 which i got.
I am studying conics right now and I'll let you know if I find any more similar problems. This is from a intermediate algebra book. You are correct that most books do not have many problems like these. They are indeed interesting :)

I just wrote a lengthy response and the forum cut me off.

In short, [tex] \[
f(x) = - \frac{1}{{90}}x^2 + 40 + 71{\textstyle{1 \over 9}}
\]
[/tex]

I used an "untranslated" parabola, and then a "translated" parabola. Too difficult to rewrite all the details NOW. One used x=60, the other relied on x=100.
 
  • #9
Hmm I think you made a mistake somewhere. Can someone confirm the answer please? What did you get blochwave?
 
  • #10
blochwave said:
I think you can just reason it out, no need for all those crazy equations and what have you

Basically you're going to have a parabola that crosses the x-axis at x=-100 and x=100

You need at least y=40 at x=60 and -60, that ultimately what it's asking, right? And it's basically of the form Y=-Ax^2+H (I just centered it at 0 for ease)

H is what we're looking for, and A is also unknown, however we know if y=40, then x=60, and if y=0, x=100

I went ahead and finished this out to find A and H and checked it and it was right, see if you can follow.

Blockwave understood the problem description. I may have obtained the "wrong" answer because I did not fully understand the problem description. Yet, he seems to have taken most of the approach that I took. Are we both misunderstanding something?
 

Related to How High Must a Parabolic Arch Be for Minimum Clearance Over a Stream?

1. How does a parabolic arch work?

A parabolic arch is a curved structure that is able to support a large amount of weight by distributing it evenly along its length. This is achieved through the shape of the arch, which follows the curve of a parabola, allowing for a more efficient distribution of weight and stress.

2. What materials are commonly used to build a parabolic arch?

The most commonly used materials for building a parabolic arch are stone, brick, and concrete. These materials are able to withstand the weight and stress of the arch and can be easily shaped into the desired curve.

3. What are the advantages of using a parabolic arch over other types of arches?

One of the main advantages of a parabolic arch is its ability to support a large amount of weight while maintaining its structural integrity. This makes it ideal for building bridges, tunnels, and other structures that require a strong and durable support system.

4. How is the shape of a parabolic arch determined?

The shape of a parabolic arch is determined by the mathematical curve of a parabola. This curve is created by plotting points on a graph and connecting them with a smooth, continuous line. The height and width of the arch can be adjusted to fit the specific needs of the structure being built.

5. Are there any famous examples of parabolic arches?

Yes, there are many famous examples of parabolic arches, including the Gateway Arch in St. Louis, Missouri; the Sydney Harbour Bridge in Australia; and the Pont du Gard in France. These structures showcase the strength and beauty of parabolic arches and their ability to withstand the test of time.

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