Cable Arc Length Problem

[Mandanesss]

Homework Statement

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.

Homework Equations

The Arc Length Formula

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)?

 

[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.

 

[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.

 

[Schrodinger's Dog]

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

1 feet = 0.3048 meters

39ft=11.8872 metres

1 pounds = 0.45359237 kilograms

1000lbs=453.59237 Kilos

That should help

 

[arildno]

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

 

[Schrodinger's Dog]

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

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 learned 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?

 

[HallsofIvy]

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

 

[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?

 

[HallsofIvy]

I got a lot more from.

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

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.

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

 

[Schrodinger's Dog]

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

Yes, sorry. Thanks.

 

[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?

 

[Mandanesss]

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

 

[Schrodinger's Dog]

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?

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.

 

[HallsofIvy]

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

 

[Schrodinger's Dog]

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

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

 

[Mandanesss]

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

 

[Schrodinger's Dog]

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

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.