Expert: Sherry Wallin - 6/16/2011

Question
QUESTION: The curve described by the cable of the suspension bridge is given by y=3hx^3/2L^2-(h/L)x+4h, where x is the distance measured from one end of the bridge, h and L are constant.
> what is the length of the cable
>
> If you could help by step by step that would be grateful

ANSWER: DP~

I will be happy to help you step by step. Please resend question with appropriate parentheses and/or the correct equation. I believe you have a typo and that the correct first term has an x^2 not an x^3. Is it (3hx^2/2l^2)?

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Hi, thanks for responding

yes ! Sorry for the typo !

Its

y= (3hx^2/2L^2)-(h/L)x+4h

how did you know about the typo ?

ANSWER: DP~

There are 3 basics curves that you get for a conic section depending on how you cut/slice through the cone and they are ellipses, parabolas, and hyperbolas. The circle is the special case of an ellipse where the major and minor axes measure the same and the center is the foci.
Anyway, they can all be written from quadratic (2nd degree) equations but ellipses and hyperbolas have both variable x and y squared and the parabola is the only conic section that has one variable usually y to the first power and x squared. Bridge suspension cables are stable only when the horizontal and vertical forces acting on its main cables are in equilibrium (equal). This equilibrium occurs when the main cables are in the shape of parabolas.   In order to find the length of the cable I need to find the length of the curve and to do that I need to know where the function begins and ends. Since your problem says the distance is x, we can place the beginning of the parabola at x = 0.  The neat thing about parabolas is that they are symmetric and the vertex is easy to find. Once you have the x-coordinate of the vertex the whole length is twice the length of the beginning to the vertex. The standard equation for a parabola is y = ax^2 + bx + c. In our problem a = 3h/2L^2, b = -h/L, and c = 4h. One way to find the x-coordinate of the vertex is to use the formula x = -b/2a.  
So the x-coordinate of our vertex is x = -(-h/L)/(2*(3h/2L^2)) = L/3. Hence we will integrate from x = 0 to x  = L/3 and then double that value for the whole length of the cable.
The formula to find the arc length is the Integral over (x1,x2) [sqrt(1+ (dy/dx)^2) dx].
Here our x1 = 0 and x2 = L/3
To get the whole length of the suspension cable you will need to integrate the above and double it.
If y = (3hx^2/2L^2)x^2-(h/L)x+4h, then dy/dx = (3h/L^2)x-(h/L)
Int(0,(L/3)) [sqrt(1+[(3h/L^2)x-(h/L)]^2) dx]. Use u-substitution to integrate this integral. Let u = (3h/L^2)x-(h/L) so du = (3h/L^2) dx. Solve for dx getting (L^2/3h)du = dx.  If we are going to use u-substitution we need to change the limits of integration accordingly. When x = 0 then u =  (3h/l^2)*0 = -h/L and when x = L/3 then u = (3h/L^2)*(L/3) - (h/L) = (h/L) - (h/L) = 0. The new limits of integration are now (-h/L,0). Now put this all into the integral: (L^2/3h) Int(-h/L,0) [sqrt(1+u^2) du]. Using a table of integration, our problem fits into the form:
Int [sqrt(k^2+u^2) du] = (u/2)*sqrt(k^2 + u^2) + (k^2/2)*ln(u + sqrt(k^2 + u^2)) + C. In our case k = 1 so we now have
(L^2/3h)*Int(-h/L,0) [sqrt(1^2+u^2) du] = (L^2/3h)[(u/2) *sqrt(1+u^2) + (1/2)*ln(u+sqrt(1+u^2))] evaluated over u = -h/L to u = 0

Note:  (-h/L)^ 2 = h^2/L^2)
(L^2/3h)[(-h/2L)*sqrt(1+h^2/L^2)+ (1/2)*ln(-h/L+sqrt(1+h^2/L^2))- ((0/2)*sqrt(1+0^2)+(1/2)*ln(1+0^2))] =>
(-L/6)*sqrt(1+h^2/L^2) + (L^2/6h)*ln(-h/L+sqrt(1+h^2/L^2)) -0 -0 => Note: 0 times anything is 0 and ln 1 = 0 =>
(-L/6)*sqrt(1+h^2/L^2) + (L^2/6h)*ln(-h/L+sqrt(1+h^2/L^2))

Now don't forget this is half the cable length so we need to double this number to get the whole cable length:
(-L/3)*sqrt(1+h^2/L^2)+(L^2/12h)*ln(-h/L + sqrt(1=h^2/L^2))

Your answer is in terms of h and L. Once you decide what h and L are you will have an actual value for your cable length based on the distance being x.

I've made every effort to explain each step and to not make any typos. Please feel free to ask if you need any further explanation. I used Int (x1,x2) to mean the integral over the interval (x1,x2).

Math Prof




---------- FOLLOW-UP ----------

QUESTION: hi i think you made a typo,

in the last line towards the end should it be + sqrt(1+h^2/L^2)) ? instead of + sqrt(1=h^2/L^2)),

And is this the final answer in terms of x

thank you very much for you help and time

Answer
DP~

Yep the "=" is a typo and should be a "+".
The final answer is in terms of the constants h and L. What I did was I found the length of the curve from x = 0 to the vertex and doubled that value because the other half of the curve from the vertex to the end will have the same length. You told me x was the distance from one end of the bridge to the other, in other words, x is the horizontal distance, sort of like the length of the bridge and you wanted to length of the cable. Remember the x-coordinate of the vertex is half way to the end of the bridge and the x-coordinate is -b/2a, thus the length of the bridge is 2*-b/2a = -b/a. In your case, using your equation for y, y = (3hx^2/2L^2)-(h/L)x+4h,  -b/a = 2L/3 which is entirely in terms of h and L. i.e., x = 2L/3

Math Prof