b67e25…: “Next to a still town”

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Suppose $\vec {p} : \R \to \R ^2$ and $\vec {q} : \R \to \R ^2$ are vector-valued functions.

Further assume $\vec {p}(0) = \vector {1,2}$ and $\vec {p}'(0) = \vector {0,1}$ and $\vec {q}(0) = \vector {-3,-5}$ and $\vec {q}'(0) = \vector {1,0}$ .

What is $\dd {t} \left ( \vec {p}(t) \dotp \vec {q}(t) \right )$ when $t = 0$ ? It is $\answer {-4}$ .

By the product rule for dot products, $\dd {t} \left ( \vec {p}(t) \dotp \vec {q}(t) \right )$ equals $\vec {p}(t) \dotp \vec {q}'(t) + \vec {p}'(t) \dotp \vec {q}(t)$ .

At $t = 0$ , this means we wish to compute $\vec {p}(0) \dotp \vec {q}'(0) + \vec {p}'(0) \dotp \vec {q}(0)$ .

In other words, we wish to compute $\vector {1,2} \dotp \vector {1,0} + \vector {0,1} \dotp \vector {-3,-5}$ .

But $\vector {1,2} \dotp \vector {1,0}$ is $1$ .

And $\vector {0,1} \dotp \vector {-3,-5}$ is $-5$ .

So the desired sum is $-4$ .