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  • $\begingroup$ I don't think it's a mistake seeing as the d^2s in question does not represent (dsds) but rather (dds). This comes from dds=d^2s=dds/r=d^2s/r . So more explicitly the canceling out looks like this: α = (dds)v/r(d*s)dt = dsv/rsdt . However, I may be wrong so please do explain if I am. $\endgroup$ Commented Feb 17, 2016 at 9:22
  • $\begingroup$ @H.Turki Aha, I understand, but how can d then cancel out, when d is not a variable? It is just a symbol meaning change in the ds variable. $\endgroup$ Commented Feb 17, 2016 at 9:26
  • $\begingroup$ I was under the impression that d can cancel out as long as it pertains to the same variable (in this case s) - also in my edited comment I mistakenly wrote dds=d^2s=... when I meant ddθ=d^2θ=... my apologies $\endgroup$ Commented Feb 17, 2016 at 9:29
  • $\begingroup$ Remember that ds is a small change. That is the name we have given to a small change in s. If a change in distance s equals the change in some other distsnce x, then we can say ds=dx. But that doesn't mean that s=x. If this was the case, then v=ds/dt would equal v=s/t, which is not at all the case (the later is rather the average speed over the entire distance s and not the instantaneous speed, which the former is). d's don't cancel out since they are not variables $\endgroup$ Commented Feb 17, 2016 at 9:34
  • $\begingroup$ Ok I see your point :) But now instead of canceling if we replace ds with x (representing a small change in s) we end up with α = dxv/rxdt , and then isn't dx/dt approximately v? or is this a different v as ds/dt is not approximately equal to d(ds)/dt? $\endgroup$ Commented Feb 17, 2016 at 9:58