Law of natural growth or decay

1. The number \( N \) of bacteria in a culture grew at a rate proportional to \( N \). The value of \( N \) was initially 100 and increased to 332 in one hour. What would be the value of \( N \) after \( 1\frac{1}{2} \) hours?

Solution: By the Law of natural growth,

\[ N = ce^{kt} \quad \text{...(1)} \]

Given that \( N = 100 \) when \( t = 0 \) hours

From (1), we get

\[ 100 = ce^0 \Rightarrow c = 100 \]

Also given that \( N = 332 \) when \( t = 1 \) hour

From (1), we get

\[ 332 = 100e^k \Rightarrow k = \ln \left( \frac{332}{100} \right) \]

If \( t = \frac{3}{2} \) hours, then \( N = ? \)

From (1), we get

\[ N = 100 e^{ \frac{3}{2} \ln \left( \frac{332}{100} \right) } \] \[ N = 100 e^{ \ln \left( \frac{332}{100} \right)^{3/2} } \] \[ N = 100 \left( \frac{332}{100} \right)^{3/2} \Rightarrow N = 604.9 \] \[ N \approx 605 \]

2. Bacteria in a culture grows exponentially so that the initial number has doubled in 3 hours. i) How many times the initial number will be present after 9 hours? ii) When will it be tripled?

Solution: Let \( x \) be the amount of bacteria in a culture at time \( t \), by the law of natural growth

\[ x = ce^{kt} \quad \text{...(1)} \]

Let \( x_0 \) be the initial number of bacteria in the culture, i.e., \( x = x_0 \) when \( t = 0 \) hours.

From (1), we get

\[ x_0 = ce^0 \Rightarrow c = x_0 \]

Given that \( x = 2x_0 \) when \( t = 3 \) hours.

From (1), we get

\[ 2x_0 = x_0 e^{3k} \Rightarrow 2 = e^{3k} \Rightarrow \ln 2 = 3k \Rightarrow k = \frac{1}{3} \ln 2 \]

(i) If \( t = 9 \) hours, then \( x = ? \)

From (1), we get

\[ x = x_0 e^{9(1/3) \ln 2} = x_0 e^{\ln 8} = 8x_0 \]

∴ 8 times the initial number will be present after 9 hours.

(ii) If \( x = 3x_0 \), then \( t = ? \)

From (1),

\[ 3x_0 = x_0 e^{\frac{t}{3} \ln 2} \] \[ \Rightarrow \ln 3 = \frac{t}{3} \ln 2 \] \[ \Rightarrow t = 3 \times \frac{\ln 3}{\ln 2} \] \[ \Rightarrow t = 4.75 \text{ hours} \]

∴ The initial number will be tripled in 4.75 hours.

3. Radium disintegrates at a rate proportional to its mass. When the mass is 10 mg, the rate of disintegration is 0.051 mg/day. How long will it take for the mass to be reduced to 5 mg?

Solution:Given that \( x = 10 \) mg when \( t = 0 \) days

From (1), we get:

\[ 10 = C e^0 \]

\[ \Rightarrow C = 10 \]

Given that 0.051 mg disintegrates after one day.

i.e., Remaining \( x = 10 - 0.051 \) after \( t = 1 \) day

\[ \Rightarrow x = 9.949 \]

From (1), we get \[ \Rightarrow 9.949 = 10 e^{-k(1)} \]

\[ \frac{9.949}{10} = e^{-k} \]

\[ -k = \ln(0.9949) \]

If x=5 mg (remaining) the t= ?

From (1), we get \[ 5 = 10 e^{t \ln(0.9949)} \]

\[ \frac{1}{2} = e^{t \ln(0.9949)} \]

\[ \ln(\frac{1}{2}) = t \ln(0.9949) \]

\[ t = \frac{\ln(1/2)}{\ln(0.9949)} \]

\[ t = 135.56 \text{ days} \]