### On my next Calculus exam ;-)

Also from the Italian maturità,

Nel piano riferito a coordinate cartesiane, ortogonali e monometriche, si tracci il grafico…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.Gdella funzion_{f}f(x) = logx(logaritmo naturale)

- Sia
Ail punto d'intersezione con l'asseydella tangente aGin un suo punto_{f}P. SiaBil punto d'intersezione con l'asseydella parallela perPall'assex. Si dimostri che, qualsiasi siaP, il segmentoABha lunghezza costante. Vale la stessa proprietà per il graficoGdella funzione_{g}g(x) = logcon_{a}xareale positivo diverso da 1?

…okay, okay I'll give it to them in English:

Consider the graphFor the solution, click on "Read more".Gof the function_{f}f(x) = logxon 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.)

- Let
Abe theyintercept of the line tangent toGat a point_{f}P. LetBbe theyintercept of the line parallel to thexaxis atP. Show that, whateverPmay be, the line segmentABhas constant length. Does this remain true for the graphGof the function_{g}G(x) = logwhere_{a}xais a real positive number other than 1?

For

*f*, the line

*AB*has length 1, regardless of the choice of

*P*, because:

- Write
*P*= (*p*_{1},*p*_{2}); - by definition of
*f*,*p*_{2}= ln*p*_{1};(in the image,*p*_{1}= 2 so*p*_{2}= 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/*p*_{1}; - the equation of the line tangent to ln
*x*at*P*is thus*y*= 1/*p*_{1}(*x*-*p*_{1}) + ln*p*_{1}; - the
*y*intercept*A*is thus -1 + ln*p*_{1}≈-0.307 (substitute*x*= 0); - the equation of the line parallel to the
*x*axis at*P*is*y*=*p*_{2}= ln*p*_{1}; - the
*y*intercept*B*is thus ln*p*_{1}; and - the length of
*AB*is the vertical distance between those two points, |(-1 + ln*p*_{1}) - ln*p*_{1}|= 1.

*g*, the solution is almost the same; the slope of the tangent line is 1/(

*p*

_{1}ln

*a*) instead of just 1/

*p*, so the distance is 1/ln

*a*.

(In the image,

*a*= e

^{2}, so the slope of the tangent line is 1/(2 ln e

^{2}) = 1/4, and the length of

*AB*is 1/ln e

^{2}= 1/2.)

(Images generated using sage.)

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

## 2 comments:

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

wouldbe more than a few students who would think you'd lost your mind.I hadn't thought of that. Most of the students wouldn't be interested, but a few might. Thanks.

Post a Comment