Question involving cylinders

Originally posted by iamatiger
I think, because when you increase the radius of a disk by one atom, you are adding a whole ring of atoms round the edge to do that, whereas when the length of the string is increased by one atom, only one atom is added at the end. So the rate of change of surface area with respect to rate of change of radius is misleading.

It probably depends on how efficiency is defined, which I really didn't specify. Sure changing the radius is more efficient for changing surface area in theory, but the addition of material in a practical way changes the notion of what efficiency is.

Originally posted by joe shmo
It probably depends on how efficiency is defined, which I really didn't specify. Sure changing the radius is more efficient for changing surface area in theory, but the addition of material in a practical way changes the notion of what efficiency is.

Using the result from above:

s = 2.pi.r^2 + 2.pi.K/r

ds/dr = 4.pi.r - 2.pi.K/r^2

ds = 4.pi.r.dr - 2.pi.K.dr/r^2

At large r ds is positive when dr is positive.
at small r, the righthand term dominates, and ds is positive (and larger) when dr is negative.

Can we conclude that your girlfriend is long and thin?

Originally posted by iamatiger
Using the result from above:

s = 2.pi.r^2 + 2.pi.K/r

ds/dr = 4.pi.r - 2.pi.K/r^2

ds = 4.pi.r.dr - 2.pi.K.dr/r^2

At large r ds is positive when dr is positive.
at small r, the righthand term dominates, and ds is positive (and larger) when dr is negative.

Can we conclude that your girlfriend is long and thin?

We can conclude that she's thin, but I wouldn't say that she's exactly long....hahah

Originally posted by iamatiger
Using the result from above:

s = 2.pi.r^2 + 2.pi.K/r

ds/dr = 4.pi.r - 2.pi.K/r^2

ds = 4.pi.r.dr - 2.pi.K.dr/r^2

At large r ds is positive when dr is positive.
at small r, the righthand term dominates, and ds is positive (and larger) when dr is negative.

Can we conclude that your girlfriend is long and thin?

So at small radii that are becoming still smaller... the differential of surface area is becoming larger in the positive sense. Ok, It seems that I agree with your result from above, However I'm not exactly sure what new point (if any intended) you were getting at?

Originally posted by joe shmo
So at small radii that are becoming still smaller... the differential of surface area is becoming larger in the positive sense. Ok, It seems that I agree with your result from above, However I'm not exactly sure what new point (if any intended) you were getting at?

It was just to make it clear that surface area is large at either extreme, but changes quicker at larger because 1/r^2 changes more rapidly than r. This shows that you were exactly right about when it changes fastest.

Originally posted by joe shmo
Actually, I was laying in bed with my girlfriend, she said she was cold and I was not...we began to discuss why tha might be. I focused in on muscle mass, vein structure ect...

I can't be positive, of course, but I think you missed your cue... 😉

Originally posted by Soothfast
I can't be positive, of course, but I think you missed your cue... 😉

Yes I'm very aware that I missed my cue, but thank you very much! 🙂