c1f544…: “Far from the clever square”

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Suppose $\vec {p} : \R \to \R ^2$ is the vector-valued function given by the rule $\vec {p}(t) = \vector {\cos t, \sin t}.$ Let $\vec {v}(t) = \vec {p}'(t)$ be the derivative, and $\vec {a}(t) = \vec {p}''(t)$ be the second derivative.

What is $\vec {a}(t) \dotp \vec {v}(t)$ ? It is $\answer {0}$ .

We compute $\vec {v}(t) = \vector {-\sin \left (t\right ),\,\cos \left (t\right )}$ .

We further compute $\vector {-\cos \left (t\right ),\,-\sin \left (t\right )}$ .

Therefore $\vec {a}(t) \dotp \vec {v}(t)$ equals $\vector {-\cos \left (t\right ),\,-\sin \left (t\right )} \dotp \vector {-\sin \left (t\right ),\,\cos \left (t\right )}$ .

But this is $\cos \left (t\right ) \cdot \sin \left (t\right ) - \sin \left (t\right ) \cos \left (t\right )$ which is zero.

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