Institute and Museum of History of Science, Florence, ITALY

*Torricellian conundrum
for keen students*

Can you solve the
following problem [imagined and solved by Torricelli] by using modern
integral calculus? Send your solution to the following address:
Correct answers will receive a response. The text of the following definition is taken from Torricelli,
Amongst my inventions is that of infinite hyperbolas, which we will define in the following way. Let the two lines AB and AC be at right angles to each other (although this is not even necessary), and let AB and AC be the asymptotes, that is to say the non-concurrent lines of the hyperbolas. Then let there be a curve FEF etc. of such a type that when the lines DE, GF, wherever they are, are applied to the asymptote AB the dignities at equal degree of the applied lines [i.e. abscissae taken with equal exponent] relate to each other as the dignities at equal degree of the asymptotals [i.e. ordinates taken with equal exponent], but reciprocally. For example, the square ED relates to the square FG (which are dignities of equal degree of the applied lines) as the cube GA relates to the cube AD (which are dignities of equal degree of the asymptotals taken reciprocally). And the figure BFEF prolonged an infinite distance at both ends will be one of the infinite hyperbolas, which will never intersect with either of the asymptotes AB, AC.
XLI Given one of the infinite hyperbolas, with asymptotes AC, AB, and letting AC be the asymptote with greater dignities, if the figure is cut by the line DE, and the parallelogram BEDA is made, there will be the same proportion between the parallelogram BEDA and all of the remaining figure ECD, although infinitely long, as there is between the difference between the exponents and the minor exponent. |

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