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Calculus II Problem - some bridge problem!!

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jgezau

Posted: Sat Aug 02, 2008 5:26 pm Post subject: Calculus II Problem - some bridge problem!!

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Can someone please help me out with this problem.. Rolling Eyes

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.

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Thanks a bunch

Harold14370

Posted: Sat Aug 02, 2008 5:58 pm Post subject:

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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.

JaneBennet

Posted: Sat Aug 02, 2008 9:59 pm Post subject: Re: Calculus II Problem - some bridge problem!!

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The second part of the problem is the difficult bit.

jgezau wrote:

ii) what is the length of the cable between two towers?

Youâ€™ll need to integrate an expression of the form $\sqrt{1+k^2x^2}\ (k>0)$ . I can show you how to do this integral, anyhow.

$\displaystyle\int{\sqrt{1+k^2x^2}}\,dx=\int{\frac{1+k^2x^2}{\sqrt{1+k^2x^2}}}\,dx$

$.\hspace{24mm}=\displaystyle\int\frac{dx}{\sqrt{1+k^2x^2}}+\int\frac{k^2x^2}{\sqrt{1+k^2x^2}}\,dx\ \ldots\fbox{1}$

The first integral on the RHS can tackled using the substitution $kx=\tan{u}$ ; youâ€™ll get $\int{\frac{1}{k}\sec{u}}\,dx=\frac{1}{k}\ln|\tan{u}+\sec{u}|=\frac{1}{k}\ln\left|kx+\sqrt{1+k^2x^2}\right|$ .

For the second integral on the RHS, use integration by parts.

$\displaystyle\int x\cdot\frac{k^2x}{\sqrt{1+k^2x^2}}\,dx=x\sqrt{1+k^2x^2}-\int{\sqrt{1+k^2x^2}}\,dx$

and you can bring the last bit over to the LHS in equation [1]. Hence

$\displaystyle\int{\sqrt{1+k^2x^2}}\,dx=\frac{1}{2k}\ln\left|kx+\sqrt{1+k^2x^2}\right|+\frac{1}{2}x\sqrt{1+k^2x^2}+C$
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11rdc11

Posted: Sat Aug 02, 2008 10:42 pm Post subject:

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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

JaneBennet

Posted: Sat Aug 02, 2008 11:03 pm Post subject:

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11rdc11 wrote:

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 y=ax^2

.

The arc length of a curve from x=a

is $\displaystyle\int_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx$ .
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Harold14370

Posted: Sun Aug 03, 2008 2:16 am Post subject:

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11rdc11 wrote:

If you start with y=ax^2+bx+c

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/answer/calculus.htm#parabola

William McCormick

Posted: Sun Aug 03, 2008 9:17 am Post subject:

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Harold14370 wrote:

11rdc11 wrote:

If you start with y=ax^2+bx+c

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.

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JaneBennet

Posted: Sun Aug 03, 2008 12:48 pm Post subject:

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William McCormick wrote:

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. Rolling Eyes

In fact, a catenary would have made calculating the cable length in part (ii) much easier, for if $y=a\cosh\left(\frac{x}{a}\right)$ then

$\displaystyle\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ =\ \cosh\left(\frac{x}{a}\right)$

and

$\displaystyle\int{\cosh\left(\frac{x}{a}\right)}\,dx\ =\ a\sinh\left(\frac{x}{a}\right)+C$
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jgezau

Posted: Sun Aug 03, 2008 1:50 pm Post subject:

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Thanks for the response. I have the answers already

i) 90 feet
ii) 878.586 feet

Have fun!!! Very Happy

Harold14370

Posted: Sun Aug 03, 2008 1:55 pm Post subject:

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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.

JaneBennet

Posted: Sun Aug 03, 2008 3:36 pm Post subject:

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Harold14370 wrote:

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. Smile

Interesting link
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11rdc11

Posted: Sun Aug 03, 2008 6:41 pm Post subject:

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Harold14370 wrote:

11rdc11 wrote:

If you start with y=ax^2+bx+c

Thanks now it makes sense. And thanks for the tidbit about the cable being a parabola and not hyperbolic.

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