- #1
yuiop
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Hi, this is from a physics subforum of physicsforums:
My calculus is not very good, but the above does not strike me as true and I would like a second opinion.
If it is true, then my calculus is even worse than I thought and I need someone to explain to me how dr/dt=0 always implies that d^2r/dt^2 must be zero too.
P.S. I am one of those "certain members", but I welcome re-education and conversion to the group of "uncertain members".
starthaus said:... given the ODE:
[tex]\frac{d^2r}{dt^2}+A\frac{dr}{dt}+B=0[/tex] for any [tex]a<t<b[/tex]
if you make [tex]\frac{dr}{dt}=0[/tex] for any [tex]a<t<b[/tex]
this means
[tex]\frac{d^2r}{dt^2}=0[/tex] any [tex]a<t<b[/tex]
meaning that:
[tex]B=0[/tex]
So, in general, [tex]\frac{dr}{dt}=0[/tex] is not a solution for the ODE. This is why we try solving the ODE with known methods rather than doing silly things like setting [tex]\frac{dr}{dt}=0[/tex].
Even worse is the naive attempt by certain members of this forum at calculating [tex]\frac{d^2r}{dt^2}[/tex] by inserting [tex]\frac{dr}{dt}=0[/tex] in the above ODE and declaring that [tex]\frac{d^2r}{dt^2}=-B[/tex].
My calculus is not very good, but the above does not strike me as true and I would like a second opinion.
If it is true, then my calculus is even worse than I thought and I need someone to explain to me how dr/dt=0 always implies that d^2r/dt^2 must be zero too.
P.S. I am one of those "certain members", but I welcome re-education and conversion to the group of "uncertain members".
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