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Condition for constant magnitude of vector

Short answer

Since |a|² = a·a, its derivative is 2a·da/dt. If |a| is constant, derivative zero so a·da/dt = 0. If a·da/dt = 0, then |a|² is constant, so magnitude constant. Thus a·da/dt = 0 is necessary and sufficient for constant magnitude.

Prove that the necessary and sufficient condition for the vector function a of scalar variable 't' to have constant magnitude is a · da/dt = 0.

Let a(t) be a vector function of the scalar t. Its magnitude |a| is constant exactly when |a|² = a · a is constant (|a| ≥ 0).

Derivative of a · a

Using the product rule for dot products:

d/dt (a · a) = (da/dt) · a + a · (da/dt) = 2 a · (da/dt).

Necessary condition

If |a| is constant, then a · a = |a|² is constant, so d/dt(a · a) = 0. Thus 2 a · da/dt = 0, i.e.

a · da/dt = 0.

Sufficient condition

Let a · da/dt = 0. Then d/dt(a · a) = 2a · da/dt = 0, so a · a = |a|² does not change with t. Since |a| ≥ 0, |a| is constant.

Hence |a| is constant if and only if a · da/dt = 0.

Geometric meaning: a vector of constant length can only turn, so its derivative is perpendicular to it (for example, the position vector of a point moving on a circle or sphere centred at the origin).

✓ Answer checked by Claude Opus 5.5 (high effort)

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