# Thread: Calculus II Problem - some bridge problem!!

1. Can someone please help me out with this problem.. the towers of a suspension bridge are 800ft apart and rise 160ft above road. The cable between the towers has the shape of a parabloa and the cable just touches the sides of the road midway between the towers.

i) what is the height of the cable 100ft from a tower?
ii) what is the length of the cable between two towers?
iii) use this problem to explain why mathematicians can be an evolving discipline, interrelated with human culture, and its connection with other disciplines. ** Sorry for my bad paint drawing Thanks a bunch  2.

3. It is the general policy of this web site not to do students' homework for them. We will help you out but you have to show some effort.

Just to get you started, do you know what the equation of a parabola is? Think about how you could solve the problem starting with that equation.  4. The second part of the problem is the difficult bit. Originally Posted by jgezau
ii) what is the length of the cable between two towers?
Youâll need to integrate an expression of the form . I can show you how to do this integral, anyhow.  The first integral on the RHS can tackled using the substitution ; youâll get .

For the second integral on the RHS, use integration by parts. and you can bring the last bit over to the LHS in equation . Hence   5. Just wondering jane, how did you come up with the equation to represnt the problem? I tried my best but have no clue.

ax'2 + bx +c

I know set the vertex at (0,0) and that when x =(-400,400) y = 160 but how do I use that infomation to make a quad equation. Once again thank you and sorry for asking so many questions, but I just want to review as much as I can before returning to school  6. Originally Posted by 11rdc11
Just wondering jane, how did you come up with the equation to represnt the problem? I tried my best but have no clue.

ax'2 + bx +c
If you set your origin at the vertex of your parabola, the equation of your parabola should just be in the form .

The arc length of a curve from to is .  7. Originally Posted by 11rdc11
Just wondering jane, how did you come up with the equation to represnt the problem? I tried my best but have no clue.

ax'2 + bx +c

I know set the vertex at (0,0) and that when x =(-400,400) y = 160 but how do I use that infomation to make a quad equation. Once again thank you and sorry for asking so many questions, but I just want to review as much as I can before returning to school
If you start with and set both y and x = 0 then c is 0. Similarly you can plug in y=160, x=400 and y=160, x=-400 and work from there.

If you google parabolic arc length you will find sites like this which may be helpful
http://home.att.net/~numericana/answ...s.htm#parabola  8. Originally Posted by Harold14370 Originally Posted by 11rdc11
Just wondering jane, how did you come up with the equation to represnt the problem? I tried my best but have no clue.

ax'2 + bx +c

I know set the vertex at (0,0) and that when x =(-400,400) y = 160 but how do I use that infomation to make a quad equation. Once again thank you and sorry for asking so many questions, but I just want to review as much as I can before returning to school
If you start with and set both y and x = 0 then c is 0. Similarly you can plug in y=160, x=400 and y=160, x=-400 and work from there.

If you google parabolic arc length you will find sites like this which may be helpful
http://home.att.net/~numericana/answ...s.htm#parabola

How can you determine the shape of the parabola? Couldn't the cone used to create the parabola be a very long cone?

Like the dunce cap I used to have to wear at school. Ha-ha.

Sincerely,

William McCormick  9. Originally Posted by William McCormick
How can you determine the shape of the parabola?
Well, in the real world it wouldnt be a parabola. If you hang a cable freely between two points, the shape of the cable will actually be a catenary instead. But the problem is assuming the shape to be a parabola, so we are just going along with it. In fact, a catenary would have made calculating the cable length in part (ii) much easier, for if then and   i) 90 feet
ii) 878.586 feet

Have fun!!!   11. You can determine the shape of the parabola by the height and distance between the towers and the fact that it touches the bridge between supports.

Jane, you are right that a free hanging uniform cable or chain will assume the shape of a catenary, but I think I recall that the weight of the hanging bridge will make it parabolic. This would assume the bridge is much heavier than the cable and the weight is distributed uniformly.  12. Originally Posted by Harold14370
Jane, you are right that a free hanging uniform cable or chain will assume the shape of a catenary, but I think I recall that the weight of the hanging bridge will make it parabolic. This would assume the bridge is much heavier than the cable and the weight is distributed uniformly.
Oh, so it is. I didnt realize that the cable was used to hold up the bridge; I thought it was just a freely suspending cable. Well, of course if it is not suspended freely, then it wont be a catenary.   13. Originally Posted by Harold14370 Originally Posted by 11rdc11
Just wondering jane, how did you come up with the equation to represnt the problem? I tried my best but have no clue.

ax'2 + bx +c

I know set the vertex at (0,0) and that when x =(-400,400) y = 160 but how do I use that infomation to make a quad equation. Once again thank you and sorry for asking so many questions, but I just want to review as much as I can before returning to school
If you start with and set both y and x = 0 then c is 0. Similarly you can plug in y=160, x=400 and y=160, x=-400 and work from there.

If you google parabolic arc length you will find sites like this which may be helpful
http://home.att.net/~numericana/answ...s.htm#parabola
Thanks now it makes sense. And thanks for the tidbit about the cable being a parabola and not hyperbolic.  Bookmarks
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