Implicit differentiation really helps my students understand why we can't just arbitrarily slap a prime on a function when we differentiate, that the variable we differentiate with respect to matters. Discussing the differences between y' and dy/dx, or P' and dP/dt helps facilitate later topics too, like related rates and optimization.

We had our second calculus class of the semester today. We spend part of the time going over results from our fall semester final exam. We also worked on a free response question from an old AP exam, 2000 AB #5 that mirrored another old AP question I have on our fall final. The question we worked today is below.

We had some good conversation about what it means to show the expression given in part a for dy/dx corresponds to the given implicit curve. After doing the implicit differentiation, we ended up with this:

I argued with the students our job wasn't quite done because the expression to the right of the equals sign does not match exactly with the original expression given in part a. I then wrote

And declared part a "done." A student challenged me, saying the y' did not match the left hand side of the equals sign of the original in part a (dy/dx). Great observation. I told the students if we are being precise, we should also go a step further and demonstrate we know y' and dy/dx are equivalent by writing dy/dx on the left. While this might seem like splitting hairs, I want my students to know attention to detail matters.

Part b came and went without much trouble. We found the two points on the curve whose x-coordinate is 1 and constructed the tangent line equations.

Then came part c. As the students were working part c, I went to Wolfram Alpha and created a graph of the implicit curve.

 

 

 

 

 

 

 

 

 

 

 

I wasn't too concerned with the restrictions on the scale or viewing window. As we worked through part c together, the students understood why we needed to set because this would retrieve an undefined result for dy/dx.

The algebra we did looked like this:

The reason x = 0 is circled is because I forgot to verify at the end of the problem that x=0 is not a valid answer. If we try to directly substitute x = 0 into xy² – x³y = 6, we get 0 = 6 which is nonsense. The substitution confirms x cannot be zero on the graph despite the fact we must infer the asymptote on the graph.

Here's another screenshot of the last slide I wrote during class:

I wrote -x⁵ = 24 and then asked how to isolate the x. I then wrote the solution (in black) above. The students were having trouble reconciling that the black value is equivalent to the red value. I referenced even and odd roots as the reason why I could pull the negative out of the radical (going from red to black).

HERE'S WHAT I WISH I HAD WRITTEN ON THE BOARD DURING CLASS...

I could sense the unease in the room around the solution, but I couldn't quite put my finger on it. After reflecting on this, I think I know why the kids seemed to disengage a bit. I have a lot of experience with negatives and roots and so on, so the simplification was not a stretch for me based on my experience. But the kids needed to see the work above, to reference the algebraic rules they know, to understand why it is permissible to move the negative out in front. I struggled to diagnose the students' need for this explanation during class. In hindsight, it explains why the students seemed to disengage and why my spider sense started tingling.

This post is really a reference for myself later, to remind me to think carefully on anticipated errors and what I can do to help students reconcile quantities that are numerically equivalent but not obviously numerically equivalent.