College Math Teaching

October 7, 2021

A “weird” implicit graph

Filed under: calculus, implicit differentiation, pedagogy — oldgote @ 12:46 am

I was preparing some implicit differentiation exercises and decided to give this one:

If sin^2(y) + cos^2(x) =1 find {dy \over dx} That is fairly straight forward, no? But there is a bit more here than meets the eye, as I quickly found out. I graphed this on Desmos and:

What in the world? Then I pondered for a minute or two and then it hit me:

sin^2(y) = 1-cos^2(x) \rightarrow sin^2(y) = sin^2(x) \rightarrow \pm(y \pm 2k \pi ) = \pm (x +2 k \pi) which leads to families of lines with either slope 1 or slope negative 1 and y intercepts multiples of \pi

Now, just blindly doing the problem we get 2sin(x)cos(x) = 2 {dy \over dx} cos(y)sin(y) which leads to: {sin(x)cos(x) \over sin(y)cos(y)} = {dy \over dx} = \pm {\sqrt{1-cos^2(y)} \sqrt{1-sin^2(x)} \over \sqrt{1-cos^2(y)} \sqrt{1-sin^2(x)}} = \pm 1 by both the original equation and the circle identity.

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