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 \(-4\).