Since A is a unit vector, $\mathbf{A}·\mathbf{A}\=1$. Using the result in Example 1.3.8(b), differentiate both sides to get

$0\=\frac{d}{\mathrm{dt}}\left(\mathbf{A}·\mathbf{A}\right)\=2 \mathbf{A}·\mathbf{A}\prime$

By the orthogonality property of the dot product, the vectors A and $\mathbf{A}\prime$ are necessarily orthogonal.

The following calculations are totally self-sufficient.

$\mathbf{A}\=\left[\begin{array}{c}\cos \(t\)\\ \sin \(t\)\end{array}\right]\Rightarrow \mathbf{A}\prime \=\left[\begin{array}{c}-\sin \(t\)\\ \cos \(t\)\end{array}\right]\Rightarrow \mathbf{A}·\mathbf{A}\prime \=-\cos \left(t\right)\sin \left(t\right)\+\sin \left(t\right)\cos \left(t\right)\=0$ 

Within the Student MultivariateCalculus package, differentiation automatically maps onto the components of vectors.

The value of $t$ in the graph in Figure 1.3.9(a) is controlled by the slider to its right. As $t$ varies, the tip of vector A (in red) traces out the blue curve. The derivative $\mathbf{A}\prime$ (in green) is drawn at the tip of A so that the orthogonality of A and $\mathbf{A}\prime$ is more readily observed.