Differential equation describing the decay and growth Answers

Memorandum – Question 1
Question
A radioactive substance decomposes at a rate proportional to the amount present. Initially
there are 40 g, and after 15 years, 80% of the original amount remains.

1.1 Formulate and solve the differential equation
Step 1: Formulate the DE
Since the rate of decay is proportional to the amount present,
𝑑𝑀
= −𝑘𝑀
𝑑𝑡

where
• 𝑀(𝑡)= mass after time 𝑡
• 𝑘 > 0= decay constant.

Step 2: Separate variables
1
𝑑𝑀 = −𝑘 𝑑𝑡
𝑀


Step 3: Integrate
1
∫ 𝑑𝑀 = ∫ − 𝑘 𝑑𝑡
𝑀
ln ∣ 𝑀 ∣= −𝑘𝑡 + 𝐶


Step 4: Solve for 𝑴
Exponentiate both sides:
𝑀 = 𝑒 −𝑘𝑡+𝐶
𝑀 = 𝐶1 𝑒 −𝑘𝑡


Step 5: Apply the initial condition
Initially,
𝑀(0) = 40

Hence,
40 = 𝐶1

Therefore,
𝑀(𝑡) = 40𝑒 −𝑘𝑡


Step 6: Determine 𝒌
80% remains after 15 years.
𝑀(15) = 0.80(40) = 32

Substitute:
32 = 40𝑒 −15𝑘
0.8 = 𝑒 −15𝑘

Take logarithms:

, ln(0.8) = −15𝑘
ln(0.8)
𝑘=−
15
𝑘 = 0.01488


Final Model
𝑀(𝑡) = 40𝑒 −0.01488𝑡


Assumptions
• Decay rate is proportional to the amount present.
• No additional material is added.
• Environmental conditions remain constant.
• The decay constant remains constant.

1.2 Amount after 25 years
𝑀(25) = 40𝑒 −0.01488(25)
= 40𝑒 −0.3719
= 40(0.6895)
𝑀(25) = 27.58 g


Memorandum – Question 2
Initially 120 g.
After 10 years, 15% has decomposed.
Therefore,
85% remains


2.1
Step 1
𝑑𝑀
= −𝑘𝑀
𝑑𝑡


Step 2
𝑑𝑀
= −𝑘𝑑𝑡
𝑀


Step 3
ln ∣ 𝑀 ∣= −𝑘𝑡 + 𝐶


Step 4
𝑀 = 𝐶1 𝑒 −𝑘𝑡


Step 5
𝑀(0) = 120

, Therefore,
𝐶1 = 120
𝑀(𝑡) = 120𝑒 −𝑘𝑡


Step 6
After 10 years,
𝑀(10) = 120(0.85) = 102

Substitute:
102 = 120𝑒 −10𝑘
0.85 = 𝑒 −10𝑘
ln(0.85) = −10𝑘
ln(0.85)
𝑘=−
10
𝑘 = 0.01625


Final Model
𝑀(𝑡) = 120𝑒 −0.01625𝑡


2.2 Remaining after 18 years
𝑀(18) = 120𝑒 −0.01625(18)
= 120𝑒 −0.2925
= 120(0.7464)
𝑀(18) = 89.57 g


Memorandum – Question 3
Initially 90 g.
After 6 days, 65% remains.

3.1
Step 1
𝑑𝑀
= −𝑘𝑀
𝑑𝑡


Step 2
𝑑𝑀
= −𝑘𝑑𝑡
𝑀


Step 3
ln ∣ 𝑀 ∣= −𝑘𝑡 + 𝐶


Step 4
𝑀 = 𝐶1 𝑒 −𝑘𝑡