Using d\rr on More General Paths

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Section 8.8 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 8.8. The Vector Differential in Rectangular Coordinates.

The arbitrary infinitesimal displacement vector in Cartesian coordinates is:

\begin{equation} d\rr=dx\,\xhat + dy\,\yhat +dz\,\zhat\tag{8.8.1} \end{equation}

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 (8.8.1), choosing to keep our calculations either in terms of the variable \(x\) or in terms of the variable \(y\text{,}\) but not both.

\begin{align*} d\rr \amp = dx\,\xhat + dy\,\yhat + dz\,\zhat\\ \amp = dx\, (\xhat + \yhat) \\ \Rightarrow \vert d\rr \vert \amp = \vert dx\, (\xhat + \yhat)\vert\\ \amp = dx\, \sqrt{(\xhat + \yhat)\cdot(\xhat + \yhat)}\\ \amp = \sqrt{2} \end{align*}

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.

\begin{align*} LENGTH \amp = \int \vert d\rr \vert\\ \amp = \sqrt{2}\, \vert dx\vert\\ \amp = \sqrt{2}\, dx \end{align*}