# Cable Arc Length Problem

1. Feb 24, 2007

### Mandanesss

1. The problem statement, all variables and given/known data

A cable hangs between two poles of equal height and 39 feet apart.
At a point on the ground directly under the cable and
x feet from the point on the ground halfway between the poles
the height of the cable in feet is
h(x)=10 +(0.4)( x^{1.5})
The cable weighs 12.5 pounds per linear foot.
Find the weight of the cable.

2. Relevant equations

The Arc Length Formula

3. The attempt at a solution

I need to find the length of the cable to determine the weight. However, since the problem mentions a variable, x, is the "from the point on the ground halfway between the poles," x must be 0. So the bounds of integration are from 0 to 39/2. After finding h'(x) and plugging it into the arc length formula and evaluating the integral from 0 to 39/2, I multiplied my answer by 2 to get the full length of the arc. Once I got this I found the weight to be some absurd number around 1000.

So where did I go wrong (assuming my differentiation, integration and calculations are correct)?

2. Feb 25, 2007

### Schrodinger's Dog

Actually the value you get is possibly correct? I don't know of the top of my head, but since 39 feet of straight cable is 487.5lbs. A figure of around 1000lbs isn't that surprising.

Last edited: Feb 25, 2007
3. Feb 25, 2007

### arildno

It has been given in pounds&feet. Hence, it is impossible for a poor Norwegian to ascertain whether the values are absurd or not.

4. Feb 25, 2007

### Schrodinger's Dog

1 feet = 0.3048 meters

39ft=11.8872 metres

1 pounds = 0.45359237 kilograms

1000lbs=453.59237 Kilos

That should help

Last edited: Feb 25, 2007
5. Feb 25, 2007

### arildno

No it doesn't.
Who would change nice natural numbers like "1" into stupid decimals like that??
I won't accept it. :grumpy:

6. Feb 25, 2007

### Schrodinger's Dog

In that case don't think about feet at all, call them arbitrary unit a. The answer still works out the same either way, the units are pretty irrelevant, unless you want to convert the answer into metric.

I'm glad I learnt imperial and metric, it makes it all easy to understand, if not a bit more complicated.

By the by OP, what equation did you use exactly? Can you show us some working? You may have messed up in your calculations?

7. Feb 25, 2007

### HallsofIvy

I get that the chain is about 108 feet long and weighs around 1300 pounds. Why do you consider that to be "absurd"?

8. Feb 25, 2007

### Schrodinger's Dog

$$\int_0^\frac{39}{2} 1+\sqrt{(h'(x))^2}$$

If I multiply that by 2 and then by 12.5 I get ~1005 but I figure I screwed up somewhere. Anyway whatever your answer is if it's in the region of 1000+ it's in the right ball park I think, my error aside. Anyone know where I went wrong?

Last edited: Feb 25, 2007
9. Feb 25, 2007

### HallsofIvy

Okay, but why in the world would you do that? Arclength is $\int \sqrt{1+ h'^2}dx$! Typo?

10. Feb 25, 2007

### Schrodinger's Dog

Yes, sorry. Thanks.

Last edited: Feb 25, 2007
11. Feb 25, 2007

### Mandanesss

Ok so here's what I did (sorry it's all written in text form!)
First I found h'(x) = .6(x^.5)
Then I plugged it into the Arc Length Formula giving me:
L = integral from 0 to 39/2 of sqrt(1+.36x) dx.
Using substitution I found the integral to be 50/27((8.02)^(3/2)-1)). Then I multiplied that decimal by 2 and then multiplied by 12.5 to get 1097.79.
Does that look right to you guys?

12. Feb 25, 2007

### Mandanesss

Thanks for everyone's help. The answer is ~1005 lbs. Hmm...that's a pretty hefty cable! Thanks again!

13. Feb 27, 2007

### Schrodinger's Dog

Actually I didn't do the integral I just typed it in as it was and let a bit of maths software do it.

Working it out properly I get:- $$\int\sqrt{1+.36x}dx=>50/27(0.36x+1)^{\frac{3}{2}$$

thus

$$\int_0^\frac{39}{2}\sqrt{(1+.36x)}\simeq42.06$$

This is the length of half the chain so multiply by 2 and then 12.5 for each foot. Gives in lbs:-

$$\simeq1051.50lbs$$

Bloody maths program obviously didn't work the integral properly, I'm sure that's right, if someone can check it to make sure I'd appreciate it.

Sorry for the misinformation.

Last edited: Feb 27, 2007
14. Feb 27, 2007

### HallsofIvy

You are aware, are you not, that you are integrating over only half the chain?

15. Feb 27, 2007

### Schrodinger's Dog

I multiplied the answer by 2 and then 12.5, I've edited the result to make it clear what I've done.

Last edited: Feb 27, 2007
16. Feb 27, 2007

### Mandanesss

I am 100% positive the answer is ~1005 lbs.

17. Feb 27, 2007

### Schrodinger's Dog

Actually it's my mistake I made the stupid assumption that the lower value was 0 which of course looking at the equation it isn't. but it is in fact 50/27. If I take 42.05985707897 the result of

$$\frac{50}{27}(1+.036*\frac{39}{2})^{\frac{3}{2}}$$

from 50/27 where x=0

I get 40.20800522712

40.20800522712*2*12.5=

1005.200130678

or

~1005.

well anyway.

Last edited: Feb 27, 2007