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

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.

## 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?

## 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?

I think so. Why not just put x=0 and y=0 into your equation and get another relation between 'a' and 'n'?

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.

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.

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!

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.

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 :)

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, $$$f(x) = - \frac{1}{{90}}x^2 + 40 + 71{\textstyle{1 \over 9}}$$$

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.

Hmm I think you made a mistake somewhere. Can someone confirm the answer please? What did you get blochwave?

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?

## 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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