Section 6.8 Using \(d\rr\) on More General Paths
In this section, we use the formula \(d\rr=dx\,\xhat + dy\,\yhat +dz\,\zhat\) in rectangular coordinates to calculate some simple geometric/physical quantities.
Activity 6.8.1. The Vector Differential in Rectangular Coordinates.
The arbitrary infinitesimal displacement vector in Cartesian coordinates is:
Given the unit cube shown below, find \(d\rr\) on path 4, leading from \(a\) to \(c\text{.}\) Then, use this expression to find the length of the line segment between \(a\) and \(c\text{.}\)
Path 4: \(d\rr=\)
``Use what you know!'' What is the value of \(z\) on this path? How is \(x\) related to \(y\text{?}\) How is \(dx\) related to \(dy\text{?}\)
What we know on this path is that \(z=0\) and therefore \(dz=0\text{.}\) We also know that \(x=y\) and therefore \(dx=dy\text{.}\) We can plug this information into Equation (6.8.1), choosing to keep our calculations either in terms of the variable \(x\) or in terms of the variable \(y\text{,}\) but not both.
Notice how using \(d\rr\) automatically generates the geometric factor of \(\sqrt{2}\text{.}\)
We can integrate \(\vert d\rr\vert\) to get the length of the line segment. Since we have expressed \(\vert d\rr\vert\) in terms of \(dx\text{,}\) we must be careful to be consistent and express the limits of integration also in terms of \(x\text{,}\) i.e. for a unit cube, \(0\le x\le 1\text{.}\) Be careful to integrate in a direction so that \(\vert dx\vert\) is a positive direction.