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

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SUMMARY

A curve with the property that its position vector r(t) is always perpendicular to its tangent vector r'(t) lies on a sphere centered at the origin. This is established by the fact that the dot product r(t) · r'(t) equals zero, indicating that the derivative of the squared length of the position vector, d/dt(r^2), is constant. Consequently, the magnitude of r(t) remains unchanged, confirming that the curve maintains a constant distance from the origin, characteristic of a spherical path.

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sharpstones
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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 [itex]\vec r \cdot \frac {d \vec r}{dt} = 0[/itex] then [itex]\frac {d}{dt} r^2 = 0[/itex].
 
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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