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Proof involving central acceleration and vector products

  1. Sep 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]r:R\rightarrow { V }_{ 3 }[/itex] is a twice-differentiable curve with central acceleration, that is, [itex]\ddot { r } [/itex] is parallel with [itex]r[/itex].
    a. Prove [itex]N=r\times \dot { r } [/itex] is constant
    b. Assuming [itex]N\neq 0[/itex], prove that [itex]r[/itex] lies in the plane through the origin with normal [itex]N[/itex].

    2. Relevant equations


    3. The attempt at a solution
    a. [itex]\frac { d }{ dt } N=\frac { d }{ dt } r\times \dot { r } =r\times \ddot { r } +\dot { r } \times \dot { r } =\overrightarrow { 0 } [/itex] because [itex]r[/itex] is parallel with [itex]\ddot { r } [/itex]

    b. [itex]\dot { r } [/itex] is in the same plane as [itex]r[/itex], then the equation of the plane through the origin is [itex]\left< x,y,z \right> \cdot r\times \dot { r } =0[/itex]. If [itex]r=\left< x,y,z \right> [/itex], then [itex]r\cdot r\times \dot { r } =r\times r\cdot \dot { r } =0[/itex] which checks out

    I'm really not sure if I'm right
     
  2. jcsd
  3. Sep 8, 2016 #2

    LCKurtz

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    It looks correct.
     
  4. Sep 9, 2016 #3
    Awesome, I can turn in my homework at ease now. Thanks.
     
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