✔ 最佳答案
dN 800e^（-0.1t）
--- = -----------------------------------
dt ｛1+4e^（-0.1t）｝^2
Integrate with respect to t, using substitution method.
Let U = 1 + 4e^(-0.1t)
dU =- -0.4e^(-0.1t)dt
-dU/0.4 = e^(-0.1t)dt
- 800
N = ------------ dU
0.4U^2
-2000
N = --------- dU
U^2
N = -2000U^(-2)dU
-2000
N = --------- + C (Note that C is a constant)
-1U
2000
N = --------- + C
U
2000
N = ------------------ + C
1 + 4e^(-0.1t)
When t = 0, N = 400, Find C
2000
400 = ------------------------- + C
1 + 4e^(-0.1(0))
2000
400 = ------------------ + C
1 + 4(1)
C = 0
2000
N = ------------------------ [N is a function of t]
1 + 4e^(-0.1t)
2000
1200 = -----------------------
1 + 4e^(-0.1t)
1 + 4e^(-0.1t) = 1.66666
e^(-0.1t) = 0.166666
t = 18 min (approximate)
When t is very large, the term 4e^(-0.1t) approaches 0, N approaches 2000
N = 2000/(1+0) = 2000
When t = 17 min., N = 1156
When t = 18 min., N = 1204
Please check your numeric values in the formula. 18 min. does not agree with your answer 10 min.
1.細菌數目增至原來數目的兩倍 所需的時間 = 18 min。[兩倍 = 1200. Right?]
2.經一段很長時間後，樣本內的細菌數目 = 2000
You can differentiate the function to see if you get back the original dN/dt. (Using quotient rule) ~ That is something you can do!
Would you translate the other question into English. I don’t quite understand it. Please define all the variables N, t, K. N represents what? t ? K ?
What does v to do with the question?
2011-05-09 10:53:08 補充：
Ques. 1
v = 2+3te^（-0.02t）
dV/dt = (3t)d/dt(e^(-0.02t)) + e^(-0.02t）d/dt(3t)
dV/dt = (3t)(-0.02)e^(-0.02t) + 3e^(-0.02t）
dV/dt = -0.06te^(-0.02t) + 3e^(-0.02t）
dV/dt = (3-0.06t)e^(-0.02t)
dV/dt = (3-3/50)e^(-0.02t)
