What is the magnitude of displacement?
I am studying arclength for a path $c(t)=(x(t),y(t),z(t))$. I understand that we integrate the derivative of speed etc.
My question is, geometrically, what is $|c(t)|$? $c(t)$ is a formula of displacement, so my naive interpretation is that taking its magnitude would yield distance travelled, but this clearly isn't the case.
As a follow-up, what is the difference between the rate of change of displacement and the rate of change of distance?
$\endgroup$ 12 Answers
$\begingroup$If you look at the formula for $|c(t)|$, then it's equal to$$ \sqrt{x(t)^2+y(t)^2+z(t)^2} $$which measures distance from the origin at time $t$ according to the distance formula (this corresponds to the intuition that displacement is like a "signed/vector form of distance"). Rate of change of distance from the origin would be $\frac{d}{dt} |c(t)|$ and would be a scalar, whereas $c'(t)$ as rate of change of distance would correspond to velocity and is a vector.
$\endgroup$ $\begingroup$"My naive interpretation is that $|\overrightarrow{\boldsymbol{c}}(t)|$ is distance travelled."
- $|\overrightarrow{\boldsymbol{c}}(t)|$ is distance from the reference point, and is an instantaneous quantity ($t$ represents a moment in time).
- In contrast, distance travelled is measured over a time interval, say, $[t_1,t_2]$, and equals$$\int_{t_1}^{t_2}\left|\overrightarrow{\boldsymbol{v}}(t)\right|\,\mathrm{d}t\\=\int_{t_1}^{t_2}\left|\frac{\mathrm{d}\overrightarrow{\boldsymbol{c}}(t)}{\,\mathrm{d}t}\right|\,\mathrm{d}t\\=\boxed{\int_{t_1}^{t_2}\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2}\,\mathrm{d}t}\\\neq\sqrt{x^2+y^2+z^2}\\=|\overrightarrow{\boldsymbol{c}}(t)|.$$
