ca0e53…: “After a light rectangle”

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Suppose $\vec {p} : \R \to \R ^3$ is a vector-valued function; its image is a curve in $\R ^3$.

Let $\vec {v}(t) = \vec {p}'(t)$ be the derivative, and $\vec {a}(t) = \vec {p}''(t)$ be the second derivative.

Suppose for all $t \in \R $ that $|\vec {v}(t)| = 1$.

Then $\vec {a}(t) \dotp \vec {v}(t) = \answer {0}$.

We are told that $\vec {v}(t) \dotp \vec {v}(t) = 1$.

Therefore $\dd {t} \left ( \vec {v}(t) \dotp \vec {v}(t) \right ) = 0$.

By the product rule, $\dd {t} \left ( \vec {v}(t) \dotp \vec {v}(t) \right )$ is $\vec {a}(t) \dotp \vec {v}(t) + \vec {v}(t) \dotp \vec {a}(t)$.

By commutativity of dot product, this means that $\dd {t} \left ( \vec {v}(t) \dotp \vec {v}(t) \right )$ is $2 \cdot \vec {a}(t) \dotp \vec {v}(t)$.

Since $\dd {t} \left ( \vec {v}(t) \dotp \vec {v}(t) \right )$ vanishes, it must be that $\vec {a}(t) \dotp \vec {v}(t)$ vanishes as well.

Therefore we conclude $\vec {a}(t) \dotp \vec {v}(t) = 0$.