Let's discuss Mathematics

Say you have a rope which fits snugly around the equator of the Earth. How much longer would it have to be if it were 1 metre above the ground instead?

Taking the derivative of the circumference of a circle (2 pi r) and multiplying by delta r (=1) suggests 2 pi meters, but that doesn't really seem to be a lot.
 
Wrong and right!

Most people expect it would need to be much longer.

EDIT: I posted before I saw dutchfire's reply, thhat's correct as well. You don't need calculus, just the circumference difference between a circle of radius r and radius r+1

i..e.

2pi*(r+1) - 2pi*r
 
Oh, it's 2 for the 1st one isn't it? Cos they both make 1 complete rotation (edit: e.g. if you had two cogs), so if one of them's still, then the moving one rotates twice.

EDIT2: Another way of saying it is that the centre of the moving coin traces a circular locus or radius 2r (where r is the radius of both coins); the circumference of this locus is twice the circumference of the coin, so one circumference's worth of movement would only get you half way round the locus.
 
Yup Mise, most people say 1 at first.
 
If you roll a coin around the circumference of another coin, how many complete rotations has it done when it arrives back at the start point?

Umm - have you guys really thought about this? Unless I'm misunderstanding the question, the answer is indeed 1. I'm not a mathematician... but you don't need to be a mathematician to take 2 coins, draw a mark on each, line up the marks, and then perform the experiment.

If the coins are of different sizes, they will line up in a period represented by the ratio of their diameters. Simple gear stuff - use it all the time at work. For instance, on a clock, the ration of the hour hand gear to the minute gear hand will be 1:60.
 
One gear is stationary.

Try it, you will get the answer 2.
 
Try it, you will get the answer 2
I did. Did you?

One gear is stationary.
Irrelevent.
One gear is stationary from the point of view of the other gear, or they both 'rotate' - either way, it's the same. You're looking at a 1:1 ratio. And that means 1 rotation. Seriously. If you don't believe me, perform the experiment.
 
What purposes does non-Euclidean geometry have other than describing spacetime and eldritch horrors?
 
I performed the experiment. I am correct I'm afraid. I'll try and find a link...
 
What purposes does non-Euclidean geometry have other than describing spacetime and eldritch horrors?

Space-time is pretty important. If you don't believe me Cthulhu will pay you a visit.

It's also interesting.
 
Yes, I know describing spacetime is important. I was just wondering if it had any other uses.
 
I performed the experiment. I am correct I'm afraid. I'll try and find a link...

[hangs-head-in-shame]
You are completely correct. I was wrong. 2 rotations, indeed!

The error in my experiment (and my practical experience) is that I was only noting the location of the hash-mark on the coin (gear) that was rotating around the other [relative rotation = 1] -- neglecting to note the absolute location of the hash mark against a point on my workbench [absolute rotation = 2!!]

... as you wisely pointed out. :blush:

Interesting how something I've done for years is suddenly illuminated in a whole new light! :D
 
I did. Did you?


Irrelevent.
One gear is stationary from the point of view of the other gear, or they both 'rotate' - either way, it's the same. You're looking at a 1:1 ratio. And that means 1 rotation. Seriously. If you don't believe me, perform the experiment.

Here you go (first animation, on the left)

http://mathworld.wolfram.com/Epicycloid.html

EDIT: No need to be ashamed, it is very counter-intuitive.
 
No-one wants to discuss maths :(

Come on guys and gals don't be shy :)
 
Well, there's this Vector Calculus exam I've got Tuesday...

Lagrange multipliers are cool :)
 
I admit I know nothing about Lagrange multipliers ;)

I am familiar with his interpolating polynomial though.
 
http://en.wikipedia.org/wiki/Lagrange_multipliers

When optimizing a function f(x) under a constaint g(x)=c, you have got to find points where the gradient of f is a scalar multiple of the gradient of g. (Of course, there's the usual blabla about f and g having to be continuus and differentiable etc.)

(x being a vector)
 
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