How fast is the point moving? Using calculus, its velocity, $ v$, is defined as the derivative of $ y$ with respect to time:

$\displaystyle v = { dy(t) \over dt } .$ (8.1)

Using numerical computations, $ v$ is approximately equal to $ \Delta y / \Delta t$, in which $ \Delta t$ denotes a small change in time and

$\displaystyle \Delta y = y(t + \Delta t) - y(t).$ (8.2)

In other words, $ \Delta y$ is the change in $ y$ from the start to the end of the time change. The velocity $ v$ can be used to estimate the change in $ y$ over $ \Delta t$ as

$\displaystyle \Delta y \approx v \Delta t .$ (8.3)

The approximation quality improves as $ \Delta t$ becomes smaller and $ v$ itself varies less during the time from $ t$ to $ t + \Delta t$.

We can write $ v(t)$ to indicate that velocity may change over time. The position can be calculated for any time $ t$ from the velocity using integration as8.1

$\displaystyle y(t) = y(0) + \int_0^t v(s) ds ,$ (8.4)

which assumes that $ y$ was known at the starting time $ t = 0$. If $ v(t)$ is constant for all time, represented as $ v$, then $ y(t) = y(0) + vt$. The integral in (8.4) accounts for $ v(t)$ being allowed to vary.

Steven M LaValle 2016-12-31