If I am rewriting v in terms of x and t, I write it as dx/dt, but how about v^2?
Feb 27, 2006 #1 Lewis If I am rewriting v in terms of x and t, I write it as dx/dt, but how about v^2?
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Feb 27, 2006 #2 mr_coffee 1,629 1 urnt? Are you talking about Reduction of order? Like: http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0021M.gif [Broken] http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0022M.gif [Broken] http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0023M.gif [Broken] Last edited by a moderator: May 2, 2017
urnt? Are you talking about Reduction of order? Like: http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0021M.gif [Broken] http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0022M.gif [Broken] http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0023M.gif [Broken]
Feb 27, 2006 #3 Lewis I don't know, my prof likes to ask questions on things he doesn't teach :\ I mean, in an equation like: F=-mv, you can rewrite it like m [(d^2 x)/(dt^2)] = -m dx/dt But how would you rewrite F=-m(v^2) ?
I don't know, my prof likes to ask questions on things he doesn't teach :\ I mean, in an equation like: F=-mv, you can rewrite it like m [(d^2 x)/(dt^2)] = -m dx/dt But how would you rewrite F=-m(v^2) ?
Feb 27, 2006 #4 FrogPad 805 0 [tex] (v)^2 = v*v[/tex] [tex] (\frac{dx}{dt})^2 = \frac{dx}{dt} * \frac{dx}{dt} [/tex] Let's say [itex] x=t^2+t^4 [/itex] then [itex] v = \frac{dx}{dt}[/itex]. So we then have [itex] F = mv^2 = m \times \frac{dx}{dt} \frac{dx}{dt} = 2t+4t^3 \times 2t+4t^3[/itex] assuming F = mv Last edited: Feb 27, 2006
[tex] (v)^2 = v*v[/tex] [tex] (\frac{dx}{dt})^2 = \frac{dx}{dt} * \frac{dx}{dt} [/tex] Let's say [itex] x=t^2+t^4 [/itex] then [itex] v = \frac{dx}{dt}[/itex]. So we then have [itex] F = mv^2 = m \times \frac{dx}{dt} \frac{dx}{dt} = 2t+4t^3 \times 2t+4t^3[/itex] assuming F = mv