Help with real life calculus problem

[Dion]

It's been ten years since I graduated from college, and now I'm faced with a real life calculus problem. Hope one of you can help me. I don't need any theories, etc, just the solution =)

I have a roll of paper. The paper is 7 microns thick. The inner diameter of the roll is 100 mm. If the length of the paper is 1000 meters, what will the outer diameter of the roll be?

I need the formula itself, as the paper thickness, inner diameter, and legnth varies.

Thanks!

 

[Dion]

oh yeah, one other thing:

How many revolutions are there around the core? The paper is wrapped around a core (whose outer diameter is the inner diameter mentioned above.

 

[shiner]

<< Help with real life calculus problem >>


I don't care what you say, there is no such thing.

 

[Gusfo0]

Maybe you should 'Ask Jeeves' for a calculus teacher.

 

[piku]

Dude, if it involves things measured to an exact measurement, and that measurement is in microns, its not real life.

 

[Dion]

Do you know how thick ordinary bond paper is? How about tracing paper? It's on the order of 10 to 20 microns.

 

[Dion]

Any you measure thickness with an instrument called a caliper. It's a purely mechanical device, and yes it measures in terms of thousandths of an mm, ergo, microns.

 

[Dion]

hmm, I have a feeling I'm posting in the wrong forum

 

[Vincent]

The first revolution is the circumference of a circle with the same diameter as the core. The second revolution is the circumference of a circle with the same diameter as the core + 7 microns. Add up the lengths of all the revolutions and set this equal to 1000 meters (make sure you make the units uniform--you have microns, milimeters, and meters). Then you'll need to figure out a way to solve for the number of revolutions. From the number of revolutions you can find the outer diameter of the roll.

 

[Dion]

Exactly! That's where the calculus part comes in.

 

[jahawkin]

Do you need an answer specific to this case or do you need a formula that works in the general case??

 

[arcain]

<< The second revolution is the circumference of a circle with the same diameter as the core + 7 microns >>



That should be diameter + 14 microns.

Since I'm obviously bored beyond belief.. mind you I have taken a calc class in maybe a year . And I'm going to simplify the problem a little. Instead of a continuous tight spiral around the center (which brings horrible images of polar coordinates and lots of integration), I'm going to work with distinct rings.

n = number of revolutions/wraps/rings/whatever..
r = radius
r_(n+1) means r subscript (n+1)

r_(n+1) = r_n + 7

well that was really easy.

r_(n+2) = (r_n + 7) + 7

r_0 = 100mm

r_n = 100mm + (n * 7 microns) (* Eq. #1 *)

now we need to find the number or revolutions.

c = circumference

c_n = 2 * pi * r_n = 2 * pi * (100mm + (n * 7 microns)) = 2 * pi * 100mm + 2 * pi * n * 7 microns

1000 meters = sigma of n := 0 to x of (2* pi * 100mm + 2 * pi * n * 7microns)
= sigma of n := 0 to x of (2 * pi * 100mm) + (2 * pi * 7microns) * sigma of n := 0 to x of (n)
= x * 2 pi * 100mm + 2 * pi * 7microns * ((x+1) * (x/2))

There you go. Convert units, solve for x (looks like it's quadratic). x will equal the number of rings. Then you can find the outer radius by plugging into equation #1. And diamter = 2 * radius.

If you want to use varying dimensions, just substitute accordingly. I don't have a calculator so you'll have to do the last few steps.

And.. uh.. it's been a while.. I may have forgotten some properties of summations, so I wouldn't build a space station based on those equations until they're double checked. I am not responsible for any application of this work (which may be incorrect).

 

[TonTo]

yeah.....
i was just about to say that

 

[mchammer187]

i did not feel like reading arcains approach but i think this approach might work.

.7 microns * length of the roll so 7 microns * 1000 meters = surface are of a disk with a hole with diameter 100mm and real diameter x meters.

it woud just be 7microns * 1000 meters = area of ring

area of the ring= intergral of ( r * dr d(theta)) for theta = 0 to 2Pi and r= 50 mm to x and solve for x
i think this will work an is simpler

so for the general formula

thickness * length = intergral of (r * dr *d(theta) from theta = 0 to 2 Pi and r= diameter of the whole/2 to x

 

[MichaelD]

Craaaaaaaaaaaaaaap! That made my head hurt! I failed Algebra I. Calculus? I can spell it....that's about it.

 

[Shy]

My grade in Calculus II: 25

Want my help?

-Shy

 

[mchammer187]

Calc A: A
Calc B: A-
Calc C: B+

doh steady progression down
good thing i'm done with calc

off to differential equations

 

[Rhodent]

Differential Equations is a real Bi@tch. Good luck and thank the lord I am done with all of it. I believe both those nice methods will work. my Ti89 would prob work it mostly itself butdon't have access to it right now. I think useing the area of the roll would be easier. But I wonder if it can remain accurate because as the paper wraps around the core wouldn't it compress on the inside and create less area? But then I also guess in theory the outer edge would stretch to compensate and leave an overall area equal to when it was stretched.

 

[eLiu]

*looks*

*feels sick*

Ohh d@mn...calc ab/bc (I/II) is gonna be...err...uhhh...HARD? DOH.

-eric

 

[mchammer187]

to find the revolutions find take the diameter from above subtract the inner diameter and divide by 2

then divide this by the thickness of the paper

 

[Rhosin]

Dion do u work for a paper or pulp industry? usually things are expressed using area or gauge instead of microns