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Ex 9.4, 22 - In a culture, bacteria count is 1,00,000. Number

Ex 9.4, 22 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 22 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.4, 22 - Chapter 9 Class 12 Differential Equations - Part 4


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Ex 9.4, 22 In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours, In how many hours will the count reach 2,00,000 , if the rate of growth of bacteria is proportional to the number present? Let the number of bacteria at time t be y Given that rate of growth of bacteria is proportional to the number present ๐‘‘๐‘ฆ/๐‘‘๐‘ก โˆ y ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ky ๐‘‘๐‘ฆ/๐‘ฆ = kdt Integrating both sides โˆซ1โ–’ใ€–๐‘‘๐‘ฆ/๐‘ฆ=๐‘˜ใ€— โˆซ1โ–’๐‘‘๐‘ก log y = kt + C Now, according to question The bacteria count is 1,00,000. The number is increased by 10% in 2 hours, In how many hours will the count reach 2,00,000 Putting t = 0 and y = 1,00,000 in (1) log 1,00,000 = k ร— 0 + C C = log 1,00,000 โ€ฆ(1) Putting value of C in (1) log y = kt + C log y = kt + log 1,00,000 Now, Putting t = 2 and y = 1,00,000 in (2) log 1,10,000 = 2k + log 1,00,000 log 1,10,000 โˆ’ log 1,00,000 = 2k log (1,10,000/1,00,000) = 2k 1/2 log (11/10) = k โ€ฆ(2) Putting value of k in (2) log y = kt + log 1,00,000 log y = 1/2 log (11/10) t + log 1,00,000 Now, If Bacterial = 2,00,000, we have to find t Putting y = 2,00,000 in (3) log 2,00,000 = 1/2 log (11/10) t + log (1,00,000) log 2,00,000 โˆ’ log 1,00,000 = 1/2 log (11/10) t log (2,00,000/1,00,000) = 1/2 log (11/10) t log 2 = 1/2 log (11/10) t t = (๐Ÿ ๐ฅ๐จ๐ โก๐Ÿ)/๐ฅ๐จ๐ โกใ€– (๐Ÿ๐Ÿ/๐Ÿ๐ŸŽ)ใ€— โ€ฆ(3)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.