1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parabolic arch question

  1. Feb 21, 2008 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    3. 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?
  2. jcsd
  3. Feb 21, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    I think so. Why not just put x=0 and y=0 into your equation and get another relation between 'a' and 'n'?
  4. Feb 21, 2008 #3
    I think you can just reason it out, no need for all those crazy equations and what have you

    Basically you're gonna 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.
  5. Feb 21, 2008 #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.
  6. Feb 21, 2008 #5
    I feel your pain, I been self-studying for months now. I'm proud of ya :-] Keep at it!!!
  7. Feb 22, 2008 #6


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    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.
  8. Feb 22, 2008 #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 :)
  9. Feb 23, 2008 #8


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    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}}

    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.
  10. Feb 24, 2008 #9
    Hmm I think you made a mistake somewhere. Can someone confirm the answer please? What did you get blochwave?
  11. Feb 26, 2008 #10


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Parabolic arch question
  1. Parabolic arch (Replies: 1)

  2. Hyperbola Arch (Replies: 3)