I'll wrap up this discussion of velocity vectors with a list of some helpful properties that might look familiar to you from AP Calculus. The proofs of these will show up in the extension questions for this unit.[br][br]Suppose [math]\vec{p},\vec{r}:\left[a,b\right]\to\mathbb{R}^2[/math] are both differentiable paths. Let [math]t\in\left[a,b\right][/math]. Then the following are true:[br][list][*]If [math]\vec{c}\left(t\right)=\vec{p}\left(t\right)\pm\vec{r}\left(t\right)[/math], then [math]\vec{c}'\left(t\right)=\vec{p}'\left(t\right)\pm\vec{r}'\left(t\right)[/math][br][/*][*]If [math]k\in\mathbb{R}[/math] is a scalar and [math]\vec{c}\left(t\right)=k\vec{p}\left(t\right)[/math], then [math]\vec{c}'\left(t\right)=k\vec{p}'\left(t\right)[/math][br][/*][*]If [math]f:\mathbb{R}\to\mathbb{R}[/math] is a scalar valued function and [math]\vec{c}\left(t\right)=f\left(t\right)\vec{p}\left(t\right)[/math], then [math]\vec{c}'\left(t\right)=f\left(t\right)\vec{p}'\left(t\right)+f'\left(t\right)\vec{p}\left(t\right)[/math][br][/*][*]If [math]f\left(t\right)=\vec{p}\left(t\right)\cdot\vec{r}\left(t\right)[/math], then [math]f'\left(t\right)=\vec{r}\left(t\right)\cdot\vec{p}'\left(t\right)+\vec{r}'\left(t\right)\cdot\vec{p}\left(t\right)[/math][br][/*][*]If [math]\vec{c}\left(t\right)=\vec{p}\left(t\right)\times\vec{r}\left(t\right)[/math], then [math]\vec{c}'\left(t\right)=\vec{p}'\left(t\right)\times\vec{r}\left(t\right)+\vec{p}\left(t\right)\times\vec{r}'\left(t\right)[/math] (A few notes on this one: First of all, note that the path [math]\vec{c}\left(t\right)[/math] is actually a path in [math]\mathbb{R}^3[/math] (what we will call a space curve in a few days). Moreover, while usually in a product rule the order in which you differentiate doesn't matter, because cross product is [i]anti-commutative[/i] the order here really does matter.)[/*][*]If [math]f:\mathbb{R}\to\left[a,b\right][/math] and [math]\vec{c}\left(t\right)=\vec{p}\left(f\left(t\right)\right)[/math], then [math]\vec{c}'\left(t\right)=f'\left(t\right)\vec{p}'\left(f\left(t\right)\right)[/math][br][/*][*]If [math]f\left(t\right)=\left|\left|\vec{p}\left(t\right)\right|\right|[/math], then [math]f'\left(t\right)=\frac{\vec{p}\left(t\right)\cdot\vec{p}'\left(t\right)}{f\left(t\right)}[/math][br][br][br][/*][/list]