67fd37…: “Beyond the lonely expression”

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Let $\vec {p}(t) = \vector {t+1, t^2 + 2t + 1, t+1}$ .

Find $t$ so that $\vec {p}(t)$ is orthogonal to its derivative $\vec {p}'(t)$ .

$t = \answer {-1}.$

Calculate $\vec {p}'(t) = \vector {1, 2t + 2, 1}$ .

Consequently $\vec {p}(t) \dotp \vec {p}'(t) = \vector {t+1, t^2 + 2t + 1, t+1} \dotp \vector {1, 2t + 2, 1}$ .

Therefore $\vec {p}(t) \dotp \vec {p}'(t) = (t+1) + (t^2 + 2t + 1)(2t+2) + (t+1)$ .

Simplifying yields that $\vec {p}(t) \dotp \vec {p}'(t) = 2 \, t^{3} + 6 \, t^{2} + 8 \, t + 4$ which factors as $2 \, {\left (t^{2} + 2 \, t + 2\right )} {\left (t + 1\right )}$ .

Note that $t^{2} + 2 \, t + 2$ has no real roots.

So the only way that $\vec {p}(t) \dotp \vec {p}'(t)$ vanishes is when $t+1 = 0$ , meaning $t = -1$ .