I dont know what to do with these Differentiation/Intergration questions

1) The number of particles at time t of a certain radioactive substance is N. The substance is decaying such a way that dN/dt = -N/3
Given that the time t=0 the number of particles is N=0 find the time when the particles remaining is 0.5N0.

2)The mass M at time t of the leaves of a certain plant varies according to the differential equation dM/dt = M - M^2
a)Given at time t=0, M=0.5, find an expression for M in terms of t.
b) Find the value for M when t=ln2.
c) Explain what happens to the value of M as t increases.

3) The rate of increase of the radius r kilometres of an oil slick is given by dr/dt=k/r^2 , where k is a positive constant. When the slick was observed the radius was 3km. Two days later it was 5 km. Find to the nearest day when the radius will be 6.

Reply 1

*bobo*

2)a)int. (1/(M-M^2)) dM= int. 1 dt

=> int. ((1/M + 1/(1-M)) dm = int. 1 dt
ln M - ln l 1-M l = t + c
ln l (M/ (1-M) l = t +c
M/ (1-M)= Ae^t

M=0.5, t=0

1=A

so M/(1-M)= e^t

Reply 2

*bobo*

b) if t=ln 2
e^t= 2

M/(1-M)=2
M=2/3

Reply 3

dvs

1.
dN/dt = -N/3
(1/N) dN = - (1/3) dt
ln(N) = -t/3 + C
N = A e^(-t/3), where A = e^c

Now when t=0, N=N_0. So:
N_0 = A e^(0) => A = N_0

Hence:
N = N_0 e^(-t/3)

So if N = 0.5 N_0, then:
N_0 e^(-t/3) = 0.5 N_0
t = 3 ln(2)

2c.
M/(1-M)= e^t
=> M = e^t/(1+e^t) = 1/(1 + e^(-t)), after we divide top and bottom by e^t.

Now as t gets very big, e^t gets very big, so e^(-t) = 1/e^t gets very small, i.e. e^(-t) -> 0 as t gets big. Hence M -> 1.

3.
dr/dt = k/r^2
r^2 dr = k dt
r^3 = 3kt + C

When t=0, r=3. So C=27. And when t=2, r=5. So k=49/3. Hence if r=6, then:
6^3 = 49t + 27
=> t = 4 to the nearest day