On my next Calculus exam ;-)

Also from the Italian maturità,

Nel piano riferito a coordinate cartesiane, ortogonali e monometriche, si tracci il grafico G_f della funzion f(x) = log x (logaritmo naturale)

1. Sia A il punto d'intersezione con l'asse y della tangente a G_f 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 G_g della funzione g(x) = log_a 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 G_f 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 G_f 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 G_g of the function G(x) = log_a 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 = (p₁, p₂);
• 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. :-)