The document discusses differential equations and their use in modeling various real-world scenarios. It provides examples of first-order differential equations that describe how the population of rabbits changes over time, how the volume of water leaks from a tank, and how the temperature of an object decreases. It also gives an example of using a differential equation to model the rate of drug absorption by the kidneys and solves for the quantity of drug remaining after a given time.

1. Objective: to form and solve first order differential equations DIFFERENTIAL EQUATIONS

2. DIFFERENTIAL EQUATIONS Here are some first order differential equations. A differential equation is simply an equation that contains one or more derivatives.

3. DIFFERENTIAL EQUATIONS A second order differential equation. A third order differential equation.

4. Exposure to Iodine-131 can cause hypothyroidism, thyroid nodules and cancer. Iodine-131 is a short-lived radioactive isotope. It enters the air from nuclear power plants.

5. The quantity of Iodine-131 in the environment reduces by half every 8.1 days.

6. m = mass of Iodine-131 in grams t = time elapsed in days

10. The rate of decay of Iodine-131 is proportional to its current mass.

19. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

20. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

21. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

22. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

23. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

24. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25:

25. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25:

26. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25:

27. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25: When t = 2, x = 10:

28. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25: When t = 2, x = 10:

29. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25: When t = 2, x = 10:

30. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25: When t = 2, x = 10:

31. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25: When t = 2, x = 10:

32. The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it. When t = 0, x = 25: When t = 2, x = 10:

33. A hot-air balloon can reach a maximum height of 1.25 km and the rate at which it gains height decreases as it climbs according to the formula: where h is the height in km and t is the time in hours since lift-off. How long does the balloon take to reach 1 km?

34. A hot-air balloon can reach a maximum height of 1.25 km and the rate at which it gains height decreases as it climbs according to the formula: where h is the height in km and t is the time in hours since lift-off. How long does the balloon take to reach 1 km?

36. When t =0 hours, h = 0 km so:

37. When t =0 hours, h = 0 km so:

38. When t =0 hours, h = 0 km so: When h = 1 km:

39. When t =0 hours, h = 0 km so: When h = 1 km:

40. When t =0 hours, h = 0 km so: When h = 1 km:

41. When t =0 hours, h = 0 km so: When h = 1 km: The balloon takes 6 minutes to rise to 1 km

42. When t =0 hours, h = 0 km so: When h = 1 km: The balloon takes 6 minutes to rise to 1 km