If the gateway of arch is represented by x2
-5x+4. Then its zeroes are
Answers
Answer:
y = ax2 + bx + c
We have 3 points and 3 unknowns, so we just substitute them in:
Substituting (0,0) gives:
0 = 0 + 0 + c
So c = 0.
Next, substitute (−3, −4) and (3, −4) to give 2 equations in 2 unknowns:
−4 = 9a − 3b (1)
−4 = 9a + 3b (2)
Adding (1) and (2) gives:
−8 = 18a
So a = −4/9
Substituting this back into either (1) or (2) gives:
b = 0
So the parabola we are looking for is:
{y}=-{\frac{{4}}{{9}}}{x}^{2}
transparent screen shot from there over the image using PaintShop Pro, which is similar to Photoshop).
Francesca says:
22 May 2016 at 10:38 pm [Comment permalink]
Thanks a lot!
Francesca says:
23 May 2016 at 6:56 am [Comment permalink]
What is the computer algebra system that you used to solve the simultaneous equations for the hyperbolic case?
Murray says:
23 May 2016 at 8:29 am [Comment permalink]
@Francesca: I used Scientific Notebook to solve that system. You could probably use Wolfram|Alpha to get the solution as well.
BTW, it was a "hyperbolic cosine" (or "catenary") case, which is different from a "hyperbola".
Francesca says:
14 Jun 2016 at 9:24 am [Comment permalink]
What further calculations can you do once you have the function for the arch?
Murray says:
15 Jun 2016 at 9:32 am [Comment permalink]
@Francesca: Probably the main reason for finding such a function would be to find height of the arch at various points along the ground underneath it.
One practical example could be if you needed to erect scaffolding under one section (for maintenance, or whatever), then how high would it need to be at various points.
Husna says:
16 Jun 2016 at 4:00 am [Comment permalink]
Hello Murray:) it's an interesting sharing. Thank you very much. May I request for your email address as there's some question that I would love to ask in detail.
Thanks in advance!
Murray says:
16 Jun 2016 at 9:35 am [Comment permalink]
@Husna: If your question is about parabolas or catenaries, it's probably best to ask here, then everyone can benefit.
If not, perhaps one of the categories in the IntMath Forum would be appropriate for your question. Once again, other readers would benefit from the discussion.
Me says:
24 Jun 2016 at 10:09 am [Comment permalink]
@Murray:I tried to proof the shape of the arch by using the same way as yours, using my own picture. Huhu But I failed to draw the parabolic shape. The size of the parabola is smaller than the arch, I mean very small..may I know, why is this happening
Murray says:
24 Jun 2016 at 10:53 am [Comment permalink]
Hello again, Husna. What are you using to draw the parabola? What image editing software are you using?
Me says:
26 Jun 2016 at 3:29 pm [Comment permalink]
@murray Oh so I have to use image editing software?sorry didnt know about itim just using geogebra..so sorry
Star says:
1 Jul 2016 at 10:45 am [Comment permalink]
Hey murray.how could we derive to the flattened catenary formula manually without using software?tq
Murray says:
1 Jul 2016 at 4:49 pm [Comment permalink]
@Star: I don't think there's a "nice" algebraic way to solve it. I believe you would need to do it numerically.
Star says:
2 Jul 2016 at 3:21 pm [Comment permalink]
@Murray: Would you mind to show it how
Murray says:
3 Jul 2016 at 9:04 am [Comment permalink]
Well, the 2 equations in 2 unknowns from the known values (-3, -4) and (-2, -1.42) are:
-{4}=-{A}\frac{{e}^{{3}\//{B}}+{e}^{-{3}\//{B}}}{2}+{A}
-{1.42}=-{A}\frac{{e}^{{2}\//{B}}+{e}^{-{2}\//{B}}}{2}+{A}
Multiplying both through by 2 gives:
-{8}=-{A}{\left({e}^{{3}\//{B}}+{e}^{-{3}\//{B}}\right)}+{2}{A}\ldots{\left[{1}\right]}
-{2.84}=-{A}{\left({e}^{{2}\//{B}}+{e}^{-{2}\//{B}}\right)}+{2}{A}\ldots{\left[{2}\right]}
Subtract [1] from [2]:
{5.16}=-{A}{\left({e}^{{2}\//{B}}+{e}^{-{2}\//{B}}-{e}^{{3}\//{B}}-{e}^{-{3}\//{B}}\right)}
You still have 2 unknowns, but could assume {A}={1} (for now) and solve
{e}^{{2}\//{B}}+{e}^{-{2}\//{B}}-{e}^{{3}\//{B}}-{e}^{-{3}\//{B}}+{5.16}={0}
!!
Hope this will help you ......
