Velocity and Acceleration
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Velocity and acceleration
As we already saw, a vector function \(\mathbf{r}:[a,b] \to \mathbb{R}^3\):
\[\mathbf{r}(t) = \left\langle x1(t), y(t), z(t)\right\rangle\]
for \(a \le t \le b\) can be interpreted as a position vector of a moving point in \(\mathbb{R}^3\) at the time \(t\). As the point moves through the space, it traces a trajectory, which is a curve in \(\mathbb{R}^3\). We say that the vector function \(\mathbf{r}\) is a parametrization of the curve.
We also already know that the derivative \(\mathbf{r}'(t)\) can be interpreted as the velocity vector of the moving point – it will tell us how fast and in which direction is the position changing. The magnitude of the velocity vector \(\left\lvert\mathbf{r}'(t) \right\rvert\) is the speed of the motion.
The second derivative \(\mathbf{r}''(t)\) will tell us how the velocity changes. It can be interpreted as the acceleration vector.
Try it yourself
Multiple parametrizations
Different vector functions can correspond to the same trajectory in \(\mathbb{R}^3\). We can think about it as two objects traveling along the same curve, but at different times and at possibly different speeds (or even in the opposite direction).
We will only consider so called regular parametrizations, which are vector functions that are continuous on \([a,b]\), twice differentiable on \((a,b)\), and \(\left\lvert\mathbf{r}'(t) \right\rvert \neq 0\) for all \(t\in (a,b)\).
The following video explores relations between two regular parametrizations of the same curve.
Tangential and normal components of acceleration
When the velocity of a motion change, there are two things that can be changing:
the speed, or magnitude of the velocity
the direction of the motion
The following video looks at how an acceleration vector can cause these two changes.
Try it yourself
Curvature
In the previous video we saw that the curvature is defined as
\[\kappa(t) = \frac{\left\lvert\text{normal component of } \mathbf{r}''(t) \right\rvert}{\left\lvert\mathbf{r}'(t) \right\rvert^2}\]
and can be calculated as
\[\kappa(t) = \frac{\left\lvert\mathbf{T}'(t) \right\rvert}{\left\lvert\mathbf{r}'(t) \right\rvert}\]
Most of the time, calculating the derivative of the unit tangent vector is hard. In this video we will derive a formula for curvature that is usually easier to use: