📂Vector Analysis

A Constant-Magnitude Vector-Valued Function is Orthogonal to Its Derivative

Theorem¹

For a vector function $\mathbf{r} : \mathbb{R} \to \mathbb{R}^{n}$, if $|\mathbf{r}(t)| = c$ then the following holds. ($c$ is a constant)

$$\mathbf{r}(t) \perp \mathbf{r}^{\prime}(t) \quad \forall t$$

Explanation

An example can be given for uniform circular motion with a constant radius. In this case, the velocity vector and the acceleration vector are always perpendicular to each other.

Proof

By the property of the dot product,

$$\mathbf{r} \cdot \mathbf{r} = | \mathbf{r} |^{2} = c^{2}$$

Differentiating both sides by $t$ yields the following:

$$2 \mathbf{r} \cdot \mathbf{r}^{\prime} = 0 \implies \mathbf{r} \cdot \mathbf{r}^{\prime} = 0$$

Thus, the two vectors are orthogonal.

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