Invariable Calculus Project

Here’s one of my favorite calculus projects.  I initially discovered it over 20 years ago in Cohen, et al’s superb Student Research Projects in Calculus.

For x>0, what is true about every triangle formed by the x- and y-axes and any tangent line to \displaystyle y=\frac{1}{x} ?  Prove thy claim.

I’d love to say nothing more than that, but I usually don’t.  The problem sounds vague in its statement, but is pretty simple to solve.  The hidden property is a delightful surprise.  I encourage you to try it out for yourself before reading further.

I just assigned the problem to one of my classes of seniors.  The class is a one-semester introduction to calculus for primarily students who’ve never been in honors and largely aren’t enamored by mathematics.  Most take the class to get an introduction to statistics (fall) and calculus (spring) before likely taking a course in one of these two in college and–for most–never taking another math course.  With that background in mind, I’ve probably scaffolded this iteration of the problem more than I should.  Here’s the assignment I gave them this week.

WARNING!  Partial Solution Alert!  Don’t read further if you want to solve the problem for yourself.

I typically use this project early in my introduction to derivatives and walk students through a little review and data gathering to help them discover the surprising hidden property.  While I don’t expect my students to do this, my default approach to geometric-type problems is to use a dynamic geometry package.  The animation below shows what happens when I varied the point of tangency while tracking the base, height, and area of the resulting triangle.

Well, I hope that animation screams something.  The x– and y-intercepts are the base and height, respectively, of a right triangle.  While those intercepts obviously vary as the point of tangency changes, the area of the triangle always seems to be 4.  It never changes!  If you’ve any geometry sense, something like that just shouldn’t happen.  So, is this a universal property, or is my animation misleading or limited in some way?  That’s a good question, and it requires proof.  Can you prove this apparent property about tangent lines to \displaystyle y=\frac{1}{x}?

FINAL SOLUTION ALERT!  Don’t read further if you want to prove this property for yourself.

For \displaystyle f(x)=\frac{1}{x}, \displaystyle\frac{d}{dx}\left(f(x)\right)=\frac{-1}{x^2}, so an equation for the tangent line to f at any point x=a is

\displaystyle \left(y-\frac{1}{a}\right)=\frac{-1}{a^2}\left(x-a\right).

The x-intercept of this generic line is \left(2a,0\right), and its y-intercept is \displaystyle \left(0,\frac{2}{a}\right).  Therefore, the area of the triangle formed by the x-and y-axes and the tangent line to f at any point x=a is

\displaystyle Area=\frac{1}{2}\cdot base\cdot height = \frac{1}{2}\cdot 2a\cdot\frac{2}{a}=2.

Cool!  The triangle’s area is always 2, completely independent of the point of tangency!

EXTENSION:

Are there any other functions that have a similar property, or is \displaystyle y=\frac{1}{x} alone in the mathematical universe for having constant area triangles?  Well, that’s a problem for another post.

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5 responses to “Invariable Calculus Project

  1. Isn’t a triangle 1/2 base height yielding an area that’s always 2?

    • And a nice extension is to try y=k/x instead of just y=1/x

      • Rocky, of course. I was so focused on the movie and LaTeX that I dropped the \frac{1}{2}. Thanks for the catch. I’m re-editing the post now.

        As to the y=k/x, I’m hoping to upload a lovely extension that derives that equation rather than starting with it. Hopefully, it will be up no later than Monday.

        Thanks for the responses!

  2. Pingback: Invariable Calculus Project II | CAS Musings

  3. Pingback: Non-Calculus approach to Invariable Calculus Project | CAS Musings

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