How Can We Prove that a Curve with Perpendicular Derivative Lies on a Sphere?

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A curve with a position vector r(t) that is always perpendicular to its tangent vector r'(t) indicates that the curve maintains a constant length. This is established through the relationship that if the dot product r(t) · r'(t) equals zero, then the derivative of the squared magnitude of r(t) is zero, implying constant distance from the origin. Consequently, the curve must lie on the surface of a sphere centered at the origin. The discussion highlights the challenge of proving this property mathematically, emphasizing the importance of understanding vector derivatives and their geometric implications. The conclusion drawn is that the curve's perpendicularity to its derivative confirms its spherical nature.
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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
 
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If \vec r \cdot \frac {d \vec r}{dt} = 0 then \frac {d}{dt} r^2 = 0.
 
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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