Question

AP微积分的问题：急！望高手指导！！！多谢啦，好答案一定加分~promise！

1. The rate at which a purification process can remove contaminants from a tank of water is proportional to the amount of contaminant remaining. If 20% of the contaminant can be removed during the first minute of the process and 98% must be removed to make the water safe, approximately how long will be decontamination process take? (Barron P606 45) Answer: 18min
2. The hypotenuse AB of a right triangle ABC is 5 feet, and one leg, AC, is decreasing at the rate of 2 feet per second. The rate, in aquare feet per second, at which the area is changing when AC=3 is ? Answer: -7/4

Answer 1

1. 依题意，清除污染物的速率和残留污染物浓度成正比。设污染物浓度随时间t（分钟）变化的函数为s = s(t)，则清除速度为v = v(t) = ks(t)，k为常量。
,开始的浓度设为1单位，那么时刻t时，被清除掉的污染物为v(u)du积分从0到t，同时也是1-s(t)。故而这两者相等。这个式子两边对t求导，有
v(t)=-s'(t)。
而v(t) = ks(t)，所以
s(t) + (1/k)s'(t) = 0
解得
s(t) = C exp(-kt)
而根据初始条件，s(0) = 1，于是C = 1，而第一分钟结束时，污染物剩余1-20% = 0.8，所以
s(1) = 0.8 = exp(-k)，所以k = -ln(0.8)
鉴于我们最终只能余下2%的污染物，故而
0.02 = exp(-kt)
即
t = ln 0.02 / (-k) = ln 0.02 / ln 0.8 = 17.53min。

2. 设AC = b(t)，则根据勾股定理，BC = sqrt(25 - b(t) * b(t))。由题意我们知道，b'(t) = -2英尺/秒。
设三角形的面积为S(t) =1/2 * b(t) sqrt(25 - b(t) * b(t))。对时间t求导，
S'(t) = 1/2 b'(t) ( sqrt(25 - b(t)^2) - b^2 / (sqrt(25 - b(t)^2)) )。
当b(t) = 3时，S'(t) = - (4 - 9 / 4) = - 7/4.