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Spherical vector problem

  • #1
My friends and I have been trying to work this one out all night, but to no avail.
If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r'(t), show that the curve lies on a sphere with center the origin.

We know the dot product of r(t) and r'(t) = 0 or that r(t) cross r'(t) equals the multiplication of their magnitudes but to go about showing that it is a sphere because of this is causing a great deal of difficulty. Any help would be appreciated
 

Answers and Replies

  • #2
Tide
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If [itex]\vec r \cdot \frac {d \vec r}{dt} = 0[/itex] then [itex]\frac {d}{dt} r^2 = 0[/itex].
 
  • #3
HallsofIvy
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Or, to put what Tide said in different words, if the derivative of a vector is always perpendicular to the vector, the vector has constant length.
 

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