3.3 – Applications: Uninhibited and Limited Growth Models THEOREM A function y = f (x) satisfies the equation if and only if for some constant c. 𝑦=𝑐 𝑒 𝑘𝑥 𝑑𝑦 𝑑𝑥 =𝑐∙ 𝑒 𝑘𝑥 ∙𝑘 𝑑𝑦 𝑑𝑥 =𝑐 𝑒 𝑘𝑥 𝑘 𝑑𝑦 𝑑𝑥 =𝑘𝑦

3.3 – Applications: Uninhibited and Limited Growth Models Example: Find the general form of the function that satisfies the equations. 𝑑𝐴 𝑑𝑡 =5𝐴 𝑘=5 𝑦=𝐴 ∴ 𝐴=𝑐 𝑒 5𝑡 ℎ ′ 𝑥 =8 ℎ(𝑥) 𝑘=8 𝑦=ℎ(𝑥) ∴ ℎ(𝑥)=𝑐 𝑒 8𝑥

3.3 – Applications: Uninhibited and Limited Growth Models Example: If the world population is about 6 billion people now and if it grows continuously at an annual rate of 1.7%, what will the population be in 10 years? When will the population double? 𝑑𝑃 𝑑𝑡 =0.017𝑃 𝑃(𝑡)= 𝑃 0 𝑒 𝑘𝑡 𝑃(𝑡)= 𝑃 0 𝑒 𝑘𝑡 𝑃(10)=6 𝑒 0.017(10) 12=6 𝑒 0.017𝑡 𝑃(10)=6 𝑒 0.17 2= 𝑒 0.17 𝑃(10)=7.112 ln 2 = ln 𝑒 0.017𝑡 ln 2 =0.017𝑡 7.112 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑝𝑒𝑜𝑝𝑙𝑒 40.773=𝑡 40.8 𝑦𝑒𝑎𝑟𝑠

3.3 – Applications: Uninhibited and Limited Growth Models The logistic equation is another model for population growth. This is used when there are factors preventing the population from exceeding some limiting value L, such as a limitation on food, living space, or other natural resources. The logistic equation: 𝑃 𝑡 = 𝐿 1+𝑏 𝑒 −𝑘𝑡 where 𝑘>0. A town with a population of 3,500 has an epidemic disease occur. The number of people (N) infected t days after the disease first appears is calculated by: 𝑁 𝑡 = 3500 1+19.9 𝑒 −0.6𝑡

3.3 – Applications: Uninhibited and Limited Growth Models Example: A town with a population of 3,500 has an epidemic disease occur. The number of people (N) infected t days after the disease first appears is calculated by: 𝑁 𝑡 = 3500 1+19.9 𝑒 −0.6𝑡 How many people are initially infected? 𝑁 0 = 3500 1+19.9 𝑒 −0.6(0) 𝑁 0 = 3500 20.9 𝑁 0 =167.46 𝑎𝑏𝑜𝑢𝑡 167 𝑝𝑒𝑜𝑝𝑙𝑒

3.3 – Applications: Uninhibited and Limited Growth Models Example: A town with a population of 3,500 has an epidemic disease occur. The number of people (N) infected t days after the disease first appears is calculated by: 𝑁 𝑡 = 3500 1+19.9 𝑒 −0.6𝑡 How many people are infected after 5 days and 12 days? 𝑁 5 = 3500 1+19.9 𝑒 −0.6(5) 𝑁 12 = 3500 1+19.9 𝑒 −0.6(12) 𝑁 12 =3448.76 𝑁 5 =1758.12 𝑎𝑏𝑜𝑢𝑡 1,758 𝑝𝑒𝑜𝑝𝑙𝑒 𝑎𝑏𝑜𝑢𝑡 3,449 𝑝𝑒𝑜𝑝𝑙𝑒

3.3 – Applications: Uninhibited and Limited Growth Models 𝑁 𝑡 = 3500 1+19.9 𝑒 −0.6𝑡

3.3 – Applications: Uninhibited and Limited Growth Models Example: What is the rate at which the disease is spreading after 16 days? 𝑁 𝑡 = 3500 1+19.9 𝑒 −0.6𝑡 𝑁′ 𝑡 = 1+19.9 𝑒 −0.6𝑡 ∙0−3500 19.9 𝑒 −0.6𝑡 (−0.6 ) 1+19.9 𝑒 −0.6𝑡 2 𝑁′ 𝑡 = 41790 𝑒 −0.6𝑡 1+19.9 𝑒 −0.6𝑡 2 𝑁′ 16 = 41790 𝑒 −0.6(16) 1+19.9 𝑒 −0.6(16) 2 𝑁′ 16 =2.823 𝑎𝑏𝑜𝑢𝑡 2.8 𝑝𝑒𝑜𝑝𝑙𝑒/𝑑𝑎𝑦

3.4 – Applications: Decay Uninhibited Exponential Growth 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 𝐴 𝑡 =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝐴 0 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑘=𝑟𝑎𝑡𝑒 𝑜𝑓 𝑔𝑟𝑜𝑤𝑡ℎ; 𝑡ℎ𝑒 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑘>0) 𝑡=𝑡𝑖𝑚𝑒 𝑝𝑎𝑠𝑠𝑒𝑑 Uninhibited Exponential Decay 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 𝐴 𝑡 =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝐴 0 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑘=𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑐𝑎𝑦; 𝑡ℎ𝑒 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑘<0) 𝑡=𝑡𝑖𝑚𝑒 𝑝𝑎𝑠𝑠𝑒𝑑 © 2010 Pearson Education, Inc. All rights reserved

3.4 – Applications: Decay Example Find k 2010 The population of the United States was approximately 227 million in 1980 and 282 million in 2000. Estimate the population in the year 2010. 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 Find k 2010 𝐴 𝑡 =227 𝑒 0.010848 (2010−1980) 282=227 𝑒 𝑘(2000−1980) 𝐴 𝑡 =314.3 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 282 227 = 𝑒 20𝑘 𝐹𝑟𝑜𝑚 2010 𝐶𝑒𝑛𝑠𝑢𝑠: 308.7 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑙𝑛 282 227 =20𝑘 𝑘= 𝑙𝑛 282 227 20 =0.010848 © 2010 Pearson Education, Inc. All rights reserved

3.4 – Applications: Decay Example Find k 300 years 7.5 grams or or A radioactive material has a half-life of 700 years. If there were ten grams initially, how much would remain after 300 years? When will the material weigh 7.5 grams? 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 Find k 300 years 7.5 grams 𝐴 𝑡 =10 𝑒 −0.00099 (300) 7.5=10 𝑒 −0.00099 𝑡 1=2 𝑒 𝑘(700) 𝐴 𝑡 =7.43 𝑔𝑟𝑎𝑚𝑠 0.75= 𝑒 −0.00099 𝑡 0.5= 𝑒 700𝑘 or 𝑙𝑛0.75=−0.00099𝑡 𝑙𝑛0.5=700𝑘 𝑘= 𝑙𝑛0.5 700 𝐴 𝑡 =10 𝑒 𝑙𝑛0.5 700 (300) 𝑡=290.6 𝑦𝑒𝑎𝑟𝑠 or 𝑘=−0.000990 𝐴 𝑡 =7.43 𝑔𝑟𝑎𝑚𝑠 7.5=10 𝑒 𝑙𝑛0.5 700 𝑡 0.75= 𝑒 𝑙𝑛0.5 700 𝑡 𝑙𝑛0.75= 𝑙𝑛0.5 700 𝑡 𝑡=290.5 𝑦𝑒𝑎𝑟𝑠 © 2010 Pearson Education, Inc. All rights reserved

3.4 – Applications: Decay Example Find A(5) Rebecca invested $15,000 into an account that pays 10% interest compounded continuously. What is the value of the account after five years? How long must the money be left in the account for it to grow to $110,835.84? 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 Find A(5) Find t if A(t) = $110,835.84 𝐴(5)=15000 𝑒 0.1(5) 110835.84=15000 𝑒 0.1 𝑡 𝐴(5)= 15000𝑒 0.5 110835.84 15000 = 𝑒 0.1 𝑡 𝐴(5)=$24,730.82 ln 110835.84 15000 = ln 𝑒 0.1 𝑡 2=0.1𝑡 20=𝑡 𝑡=20 𝑦𝑒𝑎𝑟𝑠 © 2010 Pearson Education, Inc. All rights reserved

3.4 – Applications: Decay Example 𝑭𝒊𝒏𝒅 𝑨 𝟎 Jack’s grandfather started an account for him on the day he was born. The account will have $25,000 in it on Jack’s 25th birthday. If the interest rate is 6.2%, how much was the initial investment made by Jack’s grandfather? 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 𝑭𝒊𝒏𝒅 𝑨 𝟎 Find t if A(t) = $110,835.84 25000= 𝐴 0 𝑒 0.062(25) 110835.84=15000 𝑒 0.1 𝑡 25000= 𝐴 0 𝑒 1.55 110835.84 15000 = 𝑒 0.1 𝑡 25000 𝑒 1.55 = 𝐴 0 ln 110835.84 15000 = ln 𝑒 0.1 𝑡 5306.199= 𝐴 0 2=0.1𝑡 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 𝑤𝑎𝑠 $5,306.20 20=𝑡 𝑡=20 𝑦𝑒𝑎𝑟𝑠 © 2010 Pearson Education, Inc. All rights reserved

3.4 – Applications: Decay A bone was found in a dig-site and had lost 91.4% of its carbon-14. The half-life of carbon-14 is 5370 years. How old is the bone? 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 Find carbon-14 in bone Find k 8.6% of carbon-14 100%−91.4% 1=2 𝑒 𝑘(5370) 8.6=10 0𝑒 −0.000129 𝑡 0.5= 𝑒 5370𝑘 8.6% 0.086= 𝑒 −0.000129 𝑡 𝑙𝑛0.5=5370𝑘 𝑙𝑛0.086=−0.000129𝑡 𝑘= 𝑙𝑛0.5 5370 19018.667=𝑡 𝑡=19019 𝑦𝑒𝑎𝑟𝑠 𝑘=−0.000129

3.4 – Applications: Decay Newton’s Law of Cooling 𝐻− 𝐻 𝑠 =( 𝐻 0 − 𝐻 𝑆 ) 𝑒 𝑘𝑡 𝐻=𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑎 ℎ𝑒𝑎𝑡𝑒𝑑 𝑜𝑏𝑗𝑒𝑐𝑡 𝑎𝑡 𝑎𝑛𝑦 𝑔𝑖𝑣𝑒𝑛 𝑡𝑖𝑚𝑒 𝐻 𝑆 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑚𝑒𝑑𝑖𝑢𝑚 𝐻 0 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑒𝑎𝑡𝑒𝑑 𝑜𝑏𝑗𝑒𝑐𝑡 𝑘=𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑘<0) 𝑡=𝑡𝑖𝑚𝑒 𝑝𝑎𝑠𝑠𝑒𝑑 © 2010 Pearson Education, Inc. All rights reserved

3.4 – Applications: Decay Newton’s Law of Cooling Example Find k 𝐻− 𝐻 𝑠 =( 𝐻 0 − 𝐻 𝑆 ) 𝑒 𝑘𝑡 Example A pizza pan is removed at 3:00 PM from an oven whose temperature is fixed at 450 F into a room that is a constant 70 F. After 5 minutes, the pizza pan is at 300 F. How long will it take for the pan to cool to 135 F? Find k t @ 135 F 300−70=(450−70) 𝑒 𝑘5 135−70=(450−70) 𝑒 −0.100418𝑡 230=380 𝑒 𝑘5 65=380 𝑒 −0.100418𝑡 230 380 = 𝑒 𝑘5 65 380 = 𝑒 −0.100418𝑡 𝑙𝑛 23 38 =5𝑘 𝑙𝑛 13 75 =−0.100418𝑡 𝑘= 𝑙𝑛 23 38 5 =−0.100418 𝑡=17.45 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 © 2010 Pearson Education, Inc. All rights reserved