Ex 9.3

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.3, 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 โ y ๐๐/๐๐ = ky ๐๐ฆ/๐ฆ = kdt Integrating both sides โซ1โใ๐๐ฆ/๐ฆ=๐ใ โซ1โ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 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 โ log 1,00,000 = 2k log (1,10,000/1,00,000) = 2k ๐/๐ log (๐๐/๐๐) = k Putting value of k in (2) log y = kt + log 1,00,000 log y = ๐/๐ log (๐๐/๐๐) 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 โ log 1,00,000 = 1/2 log (11/10) t log ((๐,๐๐,๐๐๐)/(๐,๐๐,๐๐๐)) = ๐/๐ log (๐๐/๐๐) t log 2 = 1/2 log (11/10) t t = (๐ ๐ฅ๐จ๐ โก๐)/๐ฅ๐จ๐ โกใ (๐๐/๐๐)ใ

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.