Differential Equation: Application of D.E.: Population Growth
A bacterial population B is known to have a rate of growth proportional to (B + 25). Between noon and 2PM the population increases to 3000 and between 2PM and 3PM the population is increased by 1000 in culture. (a) Find an expression for the bacterial population B as a function of time. (b) What is the initial bacterial population in the culture? (c) What is the total bacterial population in the culture at 4:15PM?

Differential Equation: Application of D.E.: Population GrowthA bacterial population

Bis known to have a rate of growth proportional to (B+ 25). Between noon and 2PM the population increases to 3000 and between 2PM and 3PM the population is increased by 1000 in culture. (a) Find an expression for the bacterial populationBas a function of time. (b) What is the initial bacterial population in the culture? (c) What is the total bacterial population in the culture at 4:15PM?$\dfrac{dB}{dt} = k(B + 25)$

$\dfrac{dB}{B + 25} = k \, dt$

$\displaystyle \int \dfrac{dB}{B + 25} = k \int dt$

$\ln (B + 25) = kt + C$

At 2:00PM,

t= 2 andB= 3000$\ln 3025 = 2k + C$ ← eq. (1)

At 3:00PM,

t= 3 andB= 4000$\ln 4025 = 3k + C$ ← eq. (2)

From eq. (1) and eq. (2)

$k = 0.2856$

$C = 7.4434$

Hence,

$\ln (B + 25) = 0.2856t + 7.4434$ answer for (a)

At noon,

t= 0$\ln (B + 25) = 7.4434$

$B = 1683$ answer for (b)

At 4:15PM,

t= 4.25$\ln (B + 25) = 0.2856(4.25) + 7.4434$

$B = 5726$ answer for (c)

Another solution (By Calculator - CASIO fx-991ES PLUS):MODE 3 5

AC

$B + 25 = 0\hat{y}$

$B + 25 = 1708$

$B = 1683$ answer for (b)

$B + 25 = 4.25\hat{y}$

$B + 25 = 5751$

$B = 5726$ answer for (c)

Note:

$\hat{y}$ = SHIFT 1 5 5

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