Module 3: Practice Problems
Instructions: Work through each problem on paper first, then reveal the solution to check your work. These 10 problems cover exponential growth/decay, logistic models, mixing problems, and Newton's law of cooling.
Problem 1: Radioactive Decay
A substance has a half-life of 12 hours. If you start with 200 mg, how much remains after 30 hours?
Solution
k = -ln(2)/12 ≈ -0.05776. P(30) = 200e-0.05776(30) = 200e-1.7328 ≈ 200(0.1768) ≈ 35.36 mg.
Problem 2: Doubling Time
A bacterial culture grows from 1000 to 4000 in 6 hours. Find the doubling time.
Solution
4000 = 1000e6k ⇒ e6k = 4 ⇒ k = ln(4)/6 ≈ 0.2310. Doubling time = ln(2)/k = ln(2)/0.2310 ≈ 3 hours.
Problem 3: Continuously Compounded Interest
How long does it take $8,000 to grow to $20,000 at 5% continuously compounded?
Solution
20000 = 8000e0.05t ⇒ e0.05t = 2.5 ⇒ t = ln(2.5)/0.05 ≈ 18.33 years.
Problem 4: Logistic Growth
A population satisfies dP/dt = 0.2P(1 - P/5000) with P(0) = 500. Find P(t) and P(10).
Solution
A = (5000 - 500)/500 = 9. P(t) = 5000/(1 + 9e-0.2t). P(10) = 5000/(1 + 9e-2) = 5000/(1 + 9(0.1353)) = 5000/(1 + 1.218) = 5000/2.218 ≈ 2254.
Problem 5: Logistic Harvesting
For dP/dt = 0.6P(1 - P/3000) - h, find the maximum sustainable harvest rate.
Solution
hmax = rK/4 = 0.6(3000)/4 = 450 per unit time.
Problem 6: Mixing (Equal Flow Rates)
A 300-L tank of pure water receives brine at 5 g/L, flowing in at 3 L/min. Outflow is also 3 L/min. Find Q(t) and the amount after 1 hour.
Solution
dQ/dt = 15 - Q/100. Q(t) = 1500(1 - e-t/100). Q(60) = 1500(1 - e-0.6) ≈ 1500(1 - 0.5488) = 1500(0.4512) ≈ 676.8 g.
Problem 7: Mixing (Flushing)
A 500-L tank has 200 g of salt. Pure water enters at 8 L/min, mixed solution exits at 8 L/min. When does the salt drop below 10 g?
Solution
dQ/dt = -8Q/500 = -Q/62.5. Q(t) = 200e-t/62.5. Set Q = 10: 10 = 200e-t/62.5 ⇒ e-t/62.5 = 0.05 ⇒ t = -62.5 ln(0.05) ≈ 62.5(2.996) ≈ 187.2 min.
Problem 8: Newton's Cooling
An object at 95°C cools to 60°C in 15 minutes in a 20°C room. Find k and the temperature at t = 30 min.
Solution
60 = 20 + 75e15k ⇒ 40 = 75e15k ⇒ k = ln(40/75)/15 = ln(8/15)/15 ≈ -0.04186. T(30) = 20 + 75e-0.04186(30) = 20 + 75e-1.256 ≈ 20 + 75(0.2844) ≈ 41.3°C.
Problem 9: Time of Death
A body is found at 32°C at 10 PM in a 22°C room. One hour later it is 30°C. Normal body temp is 37°C. Estimate the time of death.
Solution
At t = 0 (10 PM): T = 32. T(1) = 30. Tenv = 22. T(t) = 22 + 10ekt. 30 = 22 + 10ek ⇒ ek = 0.8 ⇒ k = ln(0.8) ≈ -0.2231.
Set t = 0 at time of death: 37 → 32 in time t1. 32 = 22 + 15ekt1 ⇒ ekt1 = 10/15 = 2/3. t1 = ln(2/3)/(-0.2231) ≈ 1.82 hours. Death was about 1.8 hours before 10 PM, around 8:11 PM.
Problem 10: Carbon-14 Dating
A fossil contains 15% of its original C-14 (half-life 5730 years). How old is it?
Solution
k = -ln(2)/5730. 0.15 = ekt ⇒ t = ln(0.15)/k = -ln(0.15) · 5730/ln(2) = 1.897 · 5730/0.6931 ≈ 15,683 years.