Birthday Polynomials

A few days ago, JD2718 wrote a post about “Birthday Triangles” — having students create three coordinates out of their birthdate and then analyzing the triangle that these coordinates make.

Even though it’s just a bit of fun, and you could have students work with any sets of points you give them, there is something great to be said for students creating their own problems that they feel ownership of. As JD2718 writes:

Best evidence, (and mind this, please) almost every class, when they first plot their own birthday triangle, there is one or two sad looking kiddies (it’s not come to tears, but I’ve seen the quivering lip) who thinks their own triangle is ugly. “Nooo” I say “Yours is obtuuuse. Does anyone else have an obtuse triangle that looks as nice as Anna’s?” (it’s usually a girl)

I thought that this idea could work in calculus too, creating “Birthday Polynomials.” My first thought was exactly JD2718’s: take the three birthday coordinates and find a quadratic that would fit them. But that would be a precalc assignment. (With bonus question: With what birthdays could you not create a quadratic?)

But I wanted more. I wanted to come up with something awesome. Something calculus. Something that would knock my students’ socks off. I initially thought something like this… if I was born on April 21, 1978, the birthday polynomial could look like: f(x)=4x^3+21x^2+19x+78 [1].And questions could be: where are the local maxima and minima? where is it concave up and concave down? where is it increasing and decreasing? And of course you could do things with integration too…

But there’s something unsatisfying about that type of question. It’s nice, but I want to wow! my students. I want to knock their socks off. Show them something elegant and unexpected. So I thought…

I want them to create a polynomial using their birthdate which would have an inflection point that was their age.

I was planning on using this amazing property [if there’s a cubic equation that hits the x-axis three times, then there’s a point of inflection, and it will be the average of these three x-intercepts] they would have to discover.

So if my birthday were January 25, 1980 (it is not), and we evaluated this polynomial on March 30, 1980 (after I celebrated my birthday), a birthday polynomial might look like this:

f(x)=(x+1980*3)(x-2008*3)(x)+1x+25.

f(x)=(x+year born*3)(x-2008*3)(x)+month born*x + date born

[Note that the month and date play no role when finding the point of inflection… they are red herrings.]

But there are many annoying problems with this… First of all, that 3 is annoying. Second of all, that 2008 gives some of the fun away. I guess multiplying it by 3 and writing it like 6024 would help disguise it, but not much. Third of all, if I worked the problem on January 5th (or anytime in the year before January 25th), it would get my age wrong by a year. Fourth of all, it’s not elegant.

I’ve spent a little time tweeking it, and thinking of ways to rework it… but I haven’t anything elegant or clever yet. For now, it’s going to have to go on the backburner. Spring break is over and school is starting tomorrow and I have too much on my plate.

I’ll post an actual, good, interesting way to come up with a birthday polynomial with some amazing property (that somehow magically spews out your age, perhaps) when I have time…

[1] Of course I had to do a google search on “birthday polynomial” to make sure I wasn’t reinventing the wheel. One calculus teacher in Texas did something similar.

Advertisements

2 comments

  1. inflection points at (mm,dd) and (yy,yy)?

    I don’t know, there’s lots of variations. I’ll stick with my precalc parabolas and my algebra triangles.

    (until I teach calc, I guess)

    Jonathan

  2. Not quite! I want to use the birthdate of someone to create a polynomial which will–when you find the x-coordinate of the inflection point–be the student’s age. So the inflection point will be at (age, something).

    So for the January 25, 1980 example, the x-coordinate of the inflection point of f(x)=(x+1980*3)(x-2008*3)(x)+1x+25 will be 28!

    Inflection points (potentially) occur when the second derivative is equal to 0. So f”(x)=6x-1980*3*2+2008*3*2=6x-6(age). So the x-coordinate is the age.

    (The birthdate and month are red herrings because when you take the second derivative it equals zero.)

    Sorry this isn’t clear… I know it’s not… It was late and I was ready to pass out and this week is going to be tough and I wouldn’t have made time for it.

    PS. I love the “given the points, find the parabola that fits it” thing.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s