# The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after 5/2 hours. - Mathematics and Statistics

Sum

The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after 5/2 hours.

(Given: sqrt2 = 1.414)

#### Solution

Let ‘x’ be the number of bacteria present at time ‘t’.

∴ dx/dt ∝ x

∴ dx/dt = kx, where k is the constant of proportionality.

∴ dx/x = k  dt

Integrating on both sides, we get

int dx/x = k int dt

∴ log x = kt + c  …(i)

When t = 0, x = 1000

∴ log (1000) = k(0) + c

∴ c = log (1000)

∴ log x = kt + log (1000) …(ii)[From (i)]

When t = 1, x = 2000

∴ log (2000) = k(1) + log (1000)

∴ log (2000) - log (1000) = k

∴ k = log (2000/1000) = log 2 ...(iii)

When t = 5 /2 , we get

log x = 5/2 k + log(1000)  …[From (ii)]

∴ log x = (5/ 2)  log 2 + log (1000)  ...[From (iii)

= log  (2^(5/2)) + log (1000)

= log (4sqrt2)+ log (1000)

= log (4000sqrt2)

= log (4000 x1.414)

∴ log x = log (5656)

∴ x = 5656

Thus, there will be 5656 bacteria after 5/2 hours.

Concept: Application of Differential Equations
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#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Differential Equation and Applications
Exercise 8.6 | Q 3 | Page 170
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