Tangent Perspectives

I assigned AP Calculus BC 1975 problem #7 to my class a couple weeks ago.  I got a 100% legitimate answer I didn’t expect from a student, so I thought I’d share.  It’s what can happen when you encourage students to follow their instincts.

Paraphrasing, the students first had to find an equation of a line through the origin tangent to the graph of y=ln(x).  Most had no problems concluding that this was \displaystyle y=\frac{x}{e}.

The next part asked if the tangent was above or below y=ln(x).  In class, we had discussed why the position of tangent lines was dependent on the underlying function’s concavity, so I fully expected successful solutions to end up at \displaystyle y''=\frac{-1}{x^2} which is negative for all x\neq 0 therefore making the original curve concave down and the tangent line above.  Most successful solutions did this, but one was different.

Paraphrasing M’s work, he concluded that if the tangent line was entirely on one side, then \displaystyle g(x)=\frac{x}{e}-ln(x) must have an extremum.  From there, \displaystyle g'(x)=\frac{1}{e}-\frac{1}{x}=0 confirms the tangency point at x=e from earlier, but this time as a critical point on g.  From here, he concluded that \displaystyle g''(x)=\frac{1}{x^2}>0 for all x\neq 0 making his critical point a global minimum.  From the construction of g, the tangent line then had to be above y=ln(x).

Admittedly, M’s algebra work took a bit longer, but what impressed me was his completely different visualization of the problem.  I’m betting he didn’t remember the down-concavity-means-tangent-above factoid from class, so he had to invent his own approach.  And he did this by turning a concavity problem into an optimization problem.  Nice.

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