HMMT 二月 2001 · CALC 赛 · 第 9 题

英文解析

9. On the planet Lemniscate, the people use the elliptic table of elements, a far moreadvanced version of our periodic table. They’re not very good at calculus, though, so they’veasked for your help. They know that Kr is somewhat radioactive and deteriorates into Pl, avery unstable element that deteriorates to form the stable element As. They started with ablock of Kr of size 10 and nothing else. (Their units don’t translate into English, sorry.) andnothing else. At time t , they let x ( t ) be the amount of Kr, y ( t ) the amount of Pl, and z ( t )
   ′ ′
   the amount of As. They know that x ( t ) = − x , and that, in the absence of Kr, y ( t ) = − 2 y .
   Your job is to find at what time t the quantity of Pl will be largest. You should assume that the entire amount of Kr that deteriorates has turned into Pl.
   Solution: This problem is long-winded since it’s giving an autonomous linear system ofdifferential equations without using any such language (and it includes a number of subtle
   ′ ′ − treferences). The system we have is x = − x , y = x − 2 y . It’s not hard to see that x = 10 esatisfies the first equation and the initial condition. Plugging this into the second equation
   2 tand using the integrating factor e (or using eigenvalues and eigenvectors to solve the systemdirectly, though I don’t want to begin to explain what that means) lets us solve for y . More
   ′ − t 2 tprecisely, we want to solve y + 2 y = 10 e . Multiply by e and simplify the left hand side
   2 t ′ t 2 t tto get ( ye ) = 10 e . Integrating both sides with respect to t then yields ye = 10 e + C , or − t − 2 t ′
   y = 10 e + Ce . Since y (0) = 0, we find C = − 10. Now to maximize y , we solve y ( t ) = 0,
   − t − 2 tor − 10 e + 20 e = 0, or t = ln 2 .
   +1
   ∫
   332 998 1664 691
   2 u + u +4 u sin u