Saturday, 24 October 2009

A Cool Tool, Derivatives Without Calculus


After the AP Calculus exam is over and there are still a few weeks of school left, I like to show my students some interesting things that are related to calculus ideas, but have no actual calculus in them. One of those ideas is finding derivatives of polynomials using synthetic division, and another is the idea of how the Polar of a point gives the tangent line to a point on a quadratic curve without calculus.

A brief explanation for those who are not aware of this method. a standard type of problem in calculus is to take a conic, such as an ellipse, and use implicit differentiation to find the tangent line to a point on the curve. For example we might take and find the tangent at some point such as (3/sqrt(10),6/sqrt(10)). It is a nice example because most capable students can solve it for y and do the derivative of the positive half of the ellipse and find the derivative, but it is gets a little messy. Then they can do it by the implicit method and see that (assuming all was done correctly) they get the same result.. the calculus they are learning seems to agree with the stuff they already feel they know.

Now for the Polar method. It seems that if you take the equation, and replace one of the two variables in each square with the value of the point, you get the value of the tangent. For example, the tangent to is found by writing the equation . The image at top shows the ellipse and its tangent.

It seems a little harder to do with a parabola because one of the terms is not a square, but you can replace the y with (y+c)/2 (where c is the y-value at the point of tangency we seek. For example the tangent to the curve y=x2 at the point (2,4) is given by (y+4)/2 = 2x. When simplified to y=4x-4, most semi-capable calculus students will recognize that it is correct.

I was recently wondering about this idea. I had always thought it only applied to quadratics, and had never seen it applied beyond that level. I began to wonder what would happen if you applied it to a cubic, so I picked an easy one, y=x3 at the point (1,1)to see what would happen. Replacing y with (y+1)/2 and one x with 1, and another with (x+1)/2 we get (y+1)/2= (1)(x)(1+x)/2, which simplifies to y=x2 + x -1 . The two graphs are shown below.


Ok, its tangent, but not a linear tangent... that was revealing in itself.. then it hit me that the tangent to y=x2 + x -1 at (1,1) has a slope of three, so if I applied the polar approach one more time, I would get the familiar tangent line we found in calculus...

I don't yet understand how these descending degree equations tangent to the same point are related, and plan to play with them some more... if you understand a little more than I what is happening here, and how it relates to the traditional calculus, a tutorial would be appreciated. I did follow up and check it out with x3 + 3xy + y2 = 4 and the same process works for a mixed implicit function (at least it did with this one).

Enough for now, but more ahead.
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