Suppose [math]\mathbf{r}(t)=f(t)\mathbf{i}+g(t)\mathbf{j}+h(t)\mathbf{k}[/math] is the position vector of a particle moving along a curve in space where [math]f[/math], [math]g[/math], and [math]h[/math] are differentiable functions of [math]t[/math]. Then the difference between the particle's position at time [math]t[/math] and time [math]t + \Delta t[/math] is [math]\Delta r(t) = \mathbf{r}(t+ \Delta t) – \mathbf{r}(t)[/math]. As [math]\Delta t \to 0[/math], [math]\mathbf{r}(t)[/math] approaches [math]\mathbf{r}'(t)[/math].[br][br]In the interactive figure, explore how [math]\Delta \mathbf{r}(t)[/math] becomes the tangent vector [math]\mathbf{r}'(t)[/math] as [math]\Delta t \to 0[/math].

[i]Developed for use with Thomas' Calculus, published by Pearson.[/i]