Originally posted by iamatigerIt 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.
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.
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Originally posted by joe shmoUsing the result from above:
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.
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 iamatigerWe can conclude that she's thin, but I wouldn't say that she's exactly long....hahah
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 iamatigerSo 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?
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?
- Joined
- 26 Apr 03
- Moves
- 26771
Originally posted by joe shmoIt 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.
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?
