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Section 6.2 Calculus AP/Dual, Revised Β©2018

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1 Section 6.2 Calculus AP/Dual, Revised Β©2018 viet.dang@humbleisd.net
Growth and Decay Section 6.2 Calculus AP/Dual, Revised Β©2018 5/8/ :56 PM Β§6.2: Growth and Decay

2 Review Solve the differential equation, π’…π’š 𝒅𝒙 =πŸ’βˆ’π’™ 5/8/2019 12:56 PM
Β§6.2: Growth and Decay

3 Algebra Review Directly Proportional: π’š=π’Œπ’™ or π’š 𝒙 =π’Œ
Inversely Proportional: π’šπ’™=π’Œ or π’š= π’Œ 𝒙 5/8/ :56 PM Β§6.2: Growth and Decay

4 Example 1 The rate of change of 𝑷 with respect to 𝒕 is proportional to πŸπŸŽβˆ’π’•. Write the equation of the function. 5/8/ :56 PM Β§6.2: Growth and Decay

5 Example 2 The rate of change of 𝑡 with respect to 𝒔 is proportional to πŸπŸ“πŸŽβˆ’π‘΅. Write the equation of the function. 5/8/ :56 PM Β§6.2: Growth and Decay

6 Example 2 The rate of change of 𝑡 with respect to 𝒔 is proportional to πŸπŸ“πŸŽβˆ’π‘΅. Write the equation of the function. 5/8/ :56 PM Β§6.2: Growth and Decay

7 Example 3 The rate of change of π’š is inversely proportional to π’š
5/8/ :56 PM Β§6.2: Growth and Decay

8 Your Turn The rate of change of π’š is proportional to the square root of π’š. Find the function. 5/8/ :56 PM Β§6.2: Growth and Decay

9 Growth and Decay 5/8/ :56 PM Β§6.2: Growth and Decay

10 Growth and Decay 5/8/ :56 PM Β§6.2: Growth and Decay

11 Growth and Decay In many applications, the rate of change of a variable is proportional to the value of π’š If π’…π’š 𝒅𝒕 =π’Œπ’š: The equation, π’š=π‘ͺ 𝒆 π’Œπ’• π‘ͺ is the initial value of π’š 𝑲 is the proportional constant Exponential Growth occurs when π’Œ>𝟎 Exponential Decay occurs when π’Œ<𝟎 5/8/ :56 PM Β§6.2: Growth and Decay

12 Equation π‘ͺ is the initial value of π’š 𝒆 is the natural base
𝑲 is the proportional constant Exponential Growth occurs when π’Œ > 𝟎 Exponential Decay occurs when π’Œ < 𝟎 5/8/ :56 PM Β§6.2: Growth and Decay

13 …is the same as… Growth and Decay 5/8/2019 12:56 PM

14 Steps Separate the variables Integrate
SOLVE for the initial amount (π‘ͺ) Then, solve for other missing parts (π’Œ typically) Solve for 𝒄 or π’š 5/8/ :56 PM Β§6.2: Growth and Decay

15 Example 4 The weight of a population of yeast is given by a differentiable function π’š, where π’š 𝒕 is measured in grams and 𝒕 is measured in minutes. The weight of the yeast population increases according to the equation π’…π’š 𝒅𝒕 =π’Œπ’š, where π’Œ is a constant. At time 𝒕=𝟎, the weight of the yeast population is 144 grams and is increasing at the rate of 24 grams per minute. Write an expression for π’š 𝒕 . 5/8/ :56 PM Β§6.2: Growth and Decay

16 Example 4 The weight of a population of yeast is given by a differentiable function π’š, where π’š 𝒕 is measured in grams and 𝒕 is measured in minutes. The weight of the yeast population increases according to the equation π’…π’š 𝒅𝒕 =π’Œπ’š, where π’Œ is a constant. At time 𝒕=𝟎, the weight of the yeast population is 144 grams and is increasing at the rate of 24 grams per minute. Write an expression for π’š 𝒕 . 5/8/ :56 PM Β§6.2: Growth and Decay

17 Example 4 The weight of a population of yeast is given by a differentiable function π’š, where π’š 𝒕 is measured in grams and 𝒕 is measured in minutes. The weight of the yeast population increases according to the equation π’…π’š 𝒅𝒕 =π’Œπ’š, where π’Œ is a constant. At time 𝒕=𝟎, the weight of the yeast population is 144 grams and is increasing at the rate of 24 grams per minute. Write an expression for π’š 𝒕 . 5/8/ :56 PM Β§6.2: Growth and Decay

18 Example 4 The weight of a population of yeast is given by a differentiable function π’š, where π’š 𝒕 is measured in grams and 𝒕 is measured in minutes. The weight of the yeast population increases according to the equation π’…π’š 𝒅𝒕 =π’Œπ’š, where π’Œ is a constant. At time 𝒕=𝟎, the weight of the yeast population is 144 grams and is increasing at the rate of 24 grams per minute. Write an expression for π’š 𝒕 . 5/8/ :56 PM Β§6.2: Growth and Decay

19 Example 5 During a certain epidemic, the number of people that are infected at any time increases at a rate proportional to the number of people that are infected at that time. If 1,000 people are infected when the epidemic is first discovered, and 1,200 are infected 7 days later, then write the equation where there are 12 days after the epidemic is first discovered. 5/8/ :56 PM Β§6.2: Growth and Decay

20 Example 5 During a certain epidemic, the number of people that are infected at any time increases at a rate proportional to the number of people that are infected at that time. If 1,000 people are infected when the epidemic is first discovered, and 1,200 are infected 7 days later, then write the equation where there are 12 days after the epidemic is first discovered. 5/8/ :56 PM Β§6.2: Growth and Decay

21 Example 5 During a certain epidemic, the number of people that are infected at any time increases at a rate proportional to the number of people that are infected at that time. If 1,000 people are infected when the epidemic is first discovered, and 1,200 are infected 7 days later, then write the equation where there are 12 days after the epidemic is first discovered. 5/8/ :56 PM Β§6.2: Growth and Decay

22 Example 5 During a certain epidemic, the number of people that are infected at any time increases at a rate proportional to the number of people that are infected at that time. If 1,000 people are infected when the epidemic is first discovered, and 1,200 are infected 7 days later, then write the equation where there are 12 days after the epidemic is first discovered. 5/8/ :56 PM Β§6.2: Growth and Decay

23 Example 5 During a certain epidemic, the number of people that are infected at any time increases at a rate proportional to the number of people that are infected at that time. If 1,000 people are infected when the epidemic is first discovered, and 1,200 are infected 7 days later, then write the equation where there are 12 days after the epidemic is first discovered. 5/8/ :56 PM Β§6.2: Growth and Decay

24 Calculator Instruction
[2nd] [+] [7] [1] [2] to ensure clarity ALWAYS in radian mode, NEVER in degree mode Round decimal answers to 4 DECIMAL places. DO NOT round the third number of a decimal answer. 5/8/ :56 PM Β§6.2: Growth and Decay

25 Example 6 Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure. Write a differential equation that states this fact. Evaluate the proportionality constant if, at the time of zero, the pressure is 35 psi and decreasing at a rate of .28 psi/min. Solve the differential equation subject to the initial condition in 𝒂. Sketch the graph of the function without a calculator. Show its behavior a long time after the tire is punctured. What would the pressure be 10 minutes after the tire was punctured? The car is safe to drive as long as the tire pressure is 12 psi or greater. For how long after the puncture will the car be safe to drive? 5/8/ :56 PM Β§6.2: Growth and Decay

26 Example 6a Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure. Write a differential equation that states this fact. Evaluate the proportionality constant if, at the time of zero, the pressure is 35 psi and decreasing at a rate of .28 psi/min. 5/8/ :56 PM Β§6.2: Growth and Decay

27 Example 6b Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure. B. Solve the differential equation subject to the initial condition in 𝒂. Initial pressure is 35 psi. 5/8/ :56 PM Β§6.2: Growth and Decay

28 Example 6b Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure. B. Solve the differential equation subject to the initial condition in 𝒂. Initial pressure is 35 psi. 5/8/ :56 PM Β§6.2: Growth and Decay

29 Example 6c Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure. C. Sketch the graph of the function without a calculator. Show its behavior a long time after the tire is punctured. 5/8/ :56 PM Β§6.2: Growth and Decay

30 Example 6d Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure. D. What would the pressure be 10 minutes after the tire was punctured? 5/8/ :56 PM Β§6.2: Growth and Decay

31 Example 6e Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure. E. The car is safe to drive as long as the tire pressure is 12 psi or greater. For how long after the puncture will the car be safe to drive? 5/8/ :56 PM Β§6.2: Growth and Decay

32 AP Multiple Choice Practice Question 1 (non-calculator)
The rate of change of atmospheric pressure 𝑷 with respect to the altitude 𝒉 is proportional to 𝑷, provided that the temperature is constant. Which of the following differential equations represents this problem? (A) 𝒅𝑷 𝒅𝒉 =π‘·βˆ’πŸŽ.πŸπŸ“ (B) 𝒅𝑷 𝒅𝒉 =βˆ’πŸŽ.πŸπŸ“π‘· (C) 𝒅𝑷 𝒅𝒉 = 𝒆 βˆ’πŸŽ.πŸπŸ“π’‰ (D) 𝒅𝑷 𝒅𝒉 = 𝒆 βˆ’πŸŽ.πŸπŸ“π‘· 5/8/ :56 PM Β§6.2: Growth and Decay

33 AP Multiple Choice Practice Question 1 (non-calculator)
The rate of change of atmospheric pressure 𝑷 with respect to the altitude 𝒉 is proportional to 𝑷, provided that the temperature is constant. Which of the following differential equations represents this problem? Vocabulary Connections and Process Answer and Justifications 5/8/ :56 PM Β§6.2: Growth and Decay

34 AP Multiple Choice Practice Question 2 (non-calculator)
The number of employees in an office who have heard a rumor at time 𝒕 hours is modeled by the function 𝑷, the solution to a differential equation. At noon, 50 of the office’s 500 employees have heard the rumor. Also at noon, 𝑷 is increasing at a rate of 18 employees per hour. Which of the following could be a differential equation? (A) 𝒅𝑷 𝒅𝒕 = 𝟏 πŸ‘πŸ”πŸŽ 𝑷 πŸ“πŸŽπŸŽβˆ’π‘· (B) 𝒅𝑷 𝒅𝒕 = 𝟏 πŸπŸπŸ“πŸŽ 𝑷 πŸ“πŸŽπŸŽβˆ’π‘· (C) 𝒅𝑷 𝒅𝒕 = 𝟏 πŸπŸ’πŸ“πŸŽ 𝑷 πŸ“πŸŽπŸŽβˆ’π‘· (D) 𝒅𝑷 𝒅𝒕 = πŸ“ πŸ’πŸ• 𝑷 πŸ“πŸŽπŸŽβˆ’π‘· 5/8/ :56 PM Β§6.2: Growth and Decay

35 AP Multiple Choice Practice Question 2 (non-calculator)
The number of employees in an office who have heard a rumor at time 𝒕 hours is modeled by the function 𝑷, the solution to a differential equation. At noon, 50 of the office’s 500 employees have heard the rumor. Also at noon, 𝑷 is increasing at a rate of 18 employees per hour. Which of the following could be a differential equation? Vocabulary Connections and Process Answer and Justifications 5/8/ :56 PM Β§6.2: Growth and Decay

36 Assignment Worksheet 5/8/ :56 PM Β§6.2: Growth and Decay


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