In this introductory lecture/SWBQ, students are given a picture as a guide. They then write down an algebraic expression for the vector differential in rectangular coordinates for coordinate equals constant paths.
This activity can be done as a mini-lecture/SWBQ as an introduction to Vector Differential--Curvilinear where students find the vector differential in cylindrical and spherical coordinates..
1. Integration Sequence | Vector Differential--Curvilinear > >
Vector Differential: Rectangular Coordinates:Find the general form for \(d\vec{r}\) in rectangular coordinates by determining \(d\vec{r}\) along the specific paths in the figure below.
- Path 1: \[d\vec{r}=\hspace{35em}\]
- Path 2: \[d\vec{r}=\hspace{35em}\]
- Path 3: \[d\vec{r}=\hspace{35em}\]
If all three coordinates are allowed to change simultaneously, by an infinitesimal amount, we could write this \(d\vec{r}\) for any path as:
\[d\vec{r}=\hspace{35em}\]
This is the general line element in rectangular coordinates.
This activity allows students to derive formulas for \(d\vec{r}\) in rectangular coordinates, using purely geometric reasoning. The formula forms the basis of a unified view of all of vector calculus, so this activity is essential. For more information on this unified view, see our publications, especially: Using differentials to bridge the vector calculus gap
Using a picture as a guide, students write down an algebraic expression for the vector differential in rectangular coordinates.
Begin by drawing a curve (like a particle trajectory, but avoid "time" in the language) and an origin on the board. Show the position vector \(\vec{r}\) that points from the origin to a point on the curve and the position vector \(\vec{r}+d\vec{r}\) to a nearby point. Show the vector \(d\vec{r}\) and explain that it is tangent to the curve.
We often do this rectangular case, paths 1-3, for the students, as a mini-lecture to get them started quickly before doing activity Vector Differential--Curvilinear.
The only wrap-up needed is to make sure that all students have (and understand the geometry of!) the correct formulas for \(d\vec{r}\).