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Comments by Mary Bridget Kustusch (post-doc, co-teaching "Static Vector Fields" in Fall 2011):

According to Ed Price, when his students were given a non-conservative field first, they assumed they had done the differentiation incorrectly instead of concluding that the field is non-conservative.

To address this issue, we reversed the order of the first two problems to provide the conservative example first. Unfortunately, the conservative example is in cylindrical coordinates and there is only one term in the potential. It went ok in class, but based on some of the questions we were getting both during and after the activity, we have revised the activity even further.

Now, the first example is a conservative field in rectangular coordinates that has a potential with multiple terms. The hope is that this will better facilitate discussion of about the necessity of corroborating witness accounts with each other since each was looking in a different direction. The second example is a non-conservative field in rectangular coordinates. The third is the field of a point charge in cylindrical coordinates (no $\phi$ component). The last example is the same field from the second example in cylindrical coordinates, which requires them to deal with the $\phi$ coordinate and allows for a comparison between the second and fourth problems.

Reflections on teaching the Murder Mystery Method (TD 2/12)

Introducing the MMM

When teaching the MMM, I typically start by drawing a tree diagram with a known example potential (for example, $xy$) at the top, and its gradient on the next line. I then take another set of derivatives, and point out that the mixed partial derivatives must agree. This process is the usual “derivative” test for a conservative vector field: Start with the second row, representing the components of the field, and compare the appropriate derivatives. The vector field is conservative if and only if they agree,

In the past, I would then introduce the MMM simply by saying that, given the middle row, one can move up the diagram instead of down, followed immediately by a description of the MMM, in turn followed by numerous examples.

This year, I tried something different. I used the same example, so $\FF=y\,\ii+x\,\jj$, for which of course moving up the diagram by integrating the components results in perfect agreement; you get $xy$ in both cases. So I hypothesize an “integral” test for conservative vector fields, namely that these integrals must agree.

Fortuitously, I then tried the example $\GG=y\,\ii+(x+y)\,\jj$, for which the integrals do not agree, but which is in fact conservative. Most students predicted incorrectly that this vector field was not conservative, and were then surprised when I wrote down a potential function. This led to a discussion of how to fix the proposed integral test, leading directly to the consistency requirements of the MMM, namely that functions of $n$ variables must show up exactly $n$ times. The extra $y$ in the $y$-component of $\GG$ is a function of just one variable, and therefore can only be “seen” by one “witness”.

I felt that this discussion strongly reinforced the need for the consistency check in the MMM

Choosing an alternate path

When using the derivative test, one shows that a vector field is conservative without however actually finding a potential function. This makes it possible to adopt an alternate strategy for evaluating vector line integrals of a conservative vector field. You can of course just do the integral, or you can find a potential function and evaluate it at the endpoints. But a third option is to choose a simpler path, then do the integral.

When using the MMM, this strategy is rarely warranted, since the potential function has already been found by the time you know that the vector field is conservative. But here is an example using the MMM where you might nonetheless want to use this strategy.

Consider the vector field $\HH=x\,\ii+e^{y^2}\,\jj$, to be integrated along a spiral path starting from the origin and ending at $(1,0)$. When trying to use the MMM, you can't actually do the second integral, so you can't find a potential function. (More precisely, you can only express the potential function as an indefinite integral.) Nonetheless, it is easy to apply the consistency test, since that integral is clearly a function of just one variable ($y$). Thus, you do know that $\HH$ is conservative, but you don't know the potential. So choose the obvious alternate path, the straight line segment between the two points — and watch the exponential term drop out completely, since $dy=0$!

This example is indeed cute, and does serve as a proof-of-concept for the need for this alternate strategy. But it is quite artificial, and perhaps better suited for sharing with colleagues than with students.


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