27 June, 2009

On my next Calculus exam ;-)

Also from the Italian maturità,

Nel piano riferito a coordinate cartesiane, ortogonali e monometriche, si tracci il grafico Gf della funzion f(x) = log x (logaritmo naturale)
  1. Sia A il punto d'intersezione con l'asse y della tangente a Gf in un suo punto P. Sia B il punto d'intersezione con l'asse y della parallela per P all'asse x. Si dimostri che, qualsiasi sia P, il segmento AB ha lunghezza costante. Vale la stessa proprietà per il grafico Gg della funzione g(x) = loga x con a reale positivo diverso da 1?
…and that's only one part of one of the problems. Imagine if my students were to receive a question in Italian like this; they'd think I've lost my mind. Little would they know that I lost it long, long ago.

…okay, okay I'll give it to them in English:
Consider the graph Gf of the function f(x) = log x on the Cartesian plane with orthogonal, equally measured coordinates. (Ed: This is the only Cartesian plane taught in the US to my knowledge. I'm not entirely sure how to translate "monometriche" but I think the point is that it isn't logarithmic scale.)
  1. Let A be the y intercept of the line tangent to Gf at a point P. Let B be the y intercept of the line parallel to the x axis at P. Show that, whatever P may be, the line segment AB has constant length. Does this remain true for the graph Gg of the function G(x) = loga x where a is a real positive number other than 1?
For the solution, click on "Read more".

For f, the line AB has length 1, regardless of the choice of P, because:
  • Write P = (p1p2);
  • by definition of f, p2 = ln p1;(in the image, p1 = 2 so p2 = ln 2 ≈ 0.7)
  • the derivative of ln x is 1/x, so that the slope of the line tangent to ln x at P is 1/p1;
  • the equation of the line tangent to ln x at P is thus y = 1/p1 (x - p1) + ln p1;
  • the y intercept A is thus -1 + ln p1 ≈-0.307 (substitute x = 0);
  • the equation of the line parallel to the x axis at P is y = p2 = ln p1;
  • the y intercept B is thus ln p1; and
  • the length of AB is the vertical distance between those two points, |(-1 + ln p1) - ln p1 |= 1.
For g, the solution is almost the same; the slope of the tangent line is 1/(p1 ln a) instead of just 1/p, so the distance is 1/ln a.

(In the image, a = e2, so the slope of the tangent line is 1/(2 ln e2) = 1/4, and the length of AB is 1/ln e2 = 1/2.)

(Images generated using sage.)

Edit: Fixed spelling of "Calculus" in the title. :-)


Brandon said...

Actually I suspect at least some would find it cool, even if only for example problems, if you gave them a math problem in a foreign language and show them that they can do the mathematics even if it's not in English, as long as they can get enough information to identify the problem. The problems would have to be carefully selected so that they could crack it themselves with just a vocabulary list and a nudge here and there. Of course, there probably would be more than a few students who would think you'd lost your mind.

jack perry said...

I hadn't thought of that. Most of the students wouldn't be interested, but a few might. Thanks.