Part (a) 4 marks
Since the growth rate is proportional to N N
d N d t = k N \frac{dN}{dt} = kN Separating variables and integrating,
∫ 1 N d N = ∫ k d t \int\frac{1}{N}\, dN = \int k\, dt ⇒ ln N = k t + c \Rightarrow \ln N = kt + c ⇒ N = A e k t , where A = e c \Rightarrow N = Ae^{kt},\ \text{where}\ A = e^{c} At t = 0 t = 0 N = 100 N = 100
⇒ A = 100 \Rightarrow A = 100 At t = 2 t = 2 N = 300 N = 300
300 = 100 e 2 k 300 = 100e^{2k} ⇒ e 2 k = 3 \Rightarrow e^{2k} = 3 ⇒ k = 1 2 ln 3 \Rightarrow k = \frac{1}{2}\ln 3 So N = 100 e 1 2 t ln 3 = 100 × 3 t / 2 N = 100e^{\frac{1}{2}t\ln 3} = 100 \times 3^{t/2}
To find when N = 1000 N = 1000
1000 = 100 × 3 t / 2 1000 = 100 \times 3^{t/2} ⇒ 3 t / 2 = 10 \Rightarrow 3^{t/2} = 10 ⇒ t 2 ln 3 = ln 10 \Rightarrow \frac{t}{2}\ln 3 = \ln 10 ⇒ t = 2 ln 10 ln 3 \Rightarrow t = \frac{2\ln 10}{\ln 3}
To check your answer: t = 2 ln 10 ln 3 ≈ 4.192 t = \frac{2\ln 10}{\ln 3} \approx 4.192 t t 100 × 3 t / 2 = 100 × 3 2.096 ≈ 1000 100 \times 3^{t/2} = 100 \times 3^{2.096} \approx 1000
∴ \therefore k = 1 2 ln 3 k = \frac{1}{2}\ln 3 1000 1000 t = 2 ln 10 ln 3 ≈ 4.19 t = \frac{2\ln 10}{\ln 3} \approx 4.19
Part (b) 5 marks
Now the growth rate is reduced by d d
d N d t = k N − d \frac{dN}{dt} = kN - d Separating variables and integrating,
∫ 1 k N − d d N = ∫ d t \int\frac{1}{kN - d}\, dN = \int dt ⇒ 1 k ln ∣ k N − d ∣ = t + c \Rightarrow \frac{1}{k}\ln|kN - d| = t + c At t = 0 t = 0 N = 100 N = 100
0 = 1 k ln ∣ 100 k − d ∣ + c 0 = \frac{1}{k}\ln|100k - d| + c ⇒ c = − 1 k ln ∣ 100 k − d ∣ \Rightarrow c = - \frac{1}{k}\ln|100k - d| Substituting back,
t = 1 k ln ∣ k N − d ∣ − 1 k ln ∣ 100 k − d ∣ t = \frac{1}{k}\ln|kN - d| - \frac{1}{k}\ln|100k - d| = 1 k ln ∣ k N − d 100 k − d ∣ = \frac{1}{k}\ln\left| \frac{kN - d}{100k - d} \right| Now we make N N
k t = ln ∣ k N − d 100 k − d ∣ kt = \ln\left| \frac{kN - d}{100k - d} \right| ⇒ e k t = k N − d 100 k − d \Rightarrow e^{kt} = \frac{kN - d}{100k - d} ⇒ k N − d = ( 100 k − d ) e k t \Rightarrow kN - d = (100k - d)e^{kt} ⇒ N = ( 100 − d k ) e k t + d k \Rightarrow N = \left( 100 - \frac{d}{k} \right)e^{kt} + \frac{d}{k}
To check your answer: at t = 0 t = 0 N = ( 100 − d k ) ( 1 ) + d k = 100 N = \left( 100 - \frac{d}{k} \right)(1) + \frac{d}{k} = 100 d N d t = k ( 100 − d k ) e k t = k N − d \frac{dN}{dt} = k\left( 100 - \frac{d}{k} \right)e^{kt} = kN - d
∴ \therefore t = 1 k ln ∣ k N − d 100 k − d ∣ t = \frac{1}{k}\ln\left| \frac{kN - d}{100k - d} \right| N = ( 100 − d k ) e k t + d k N = \left( 100 - \frac{d}{k} \right)e^{kt} + \frac{d}{k} k = 1 2 ln 3 k = \frac{1}{2}\ln 3
Part (c)(i) 1 mark
Explanation: writing N ( t ) = ( 100 − d k ) e k t + d k N(t) = \left( 100 - \frac{d}{k} \right)e^{kt} + \frac{d}{k} e k t e^{kt} 100 − d k 100 - \frac{d}{k} d > 100 k d > 100k d N d t = k ( 100 − d k ) e k t \frac{dN}{dt} = k\left( 100 - \frac{d}{k} \right)e^{kt} t t N N d ≤ 100 k d \leq 100k N N 100 k = 50 ln 3 100k = 50\ln 3 d > 50 ln 3 d > 50\ln 3
∴ \therefore d > 50 ln 3 ( ≈ 54.9 ) d > 50\ln 3\ ( \approx 54.9)
Part (c)(ii) 2 marks
Since d = 58 > 50 ln 3 ≈ 54.93 d = 58 > 50\ln 3 \approx 54.93 N = 0 N = 0
0 = ( 100 − 58 k ) e k t + 58 k 0 = \left( 100 - \frac{58}{k} \right)e^{kt} + \frac{58}{k} ⇒ e k t = 58 58 − 100 k \Rightarrow e^{kt} = \frac{58}{58 - 100k} ⇒ t = 1 k ln ( 58 58 − 100 k ) \Rightarrow t = \frac{1}{k}\ln\left( \frac{58}{58 - 100k} \right) Substituting k = 1 2 ln 3 k = \frac{1}{2}\ln 3
t = 2 ln 3 ln ( 58 58 − 50 ln 3 ) t = \frac{2}{\ln 3}\ln\left( \frac{58}{58 - 50\ln 3} \right)
To check your answer: 58 − 50 ln 3 ≈ 3.069 58 - 50\ln 3 \approx 3.069 58 3.069 ≈ 18.90 \frac{58}{3.069} \approx 18.90 t ≈ 2 1.0986 ln ( 18.90 ) ≈ 5.350 t \approx \frac{2}{1.0986}\ln(18.90) \approx 5.350 t t N ( t ) N(t) d = 58 d = 58 N ≈ 0 N \approx 0
∴ \therefore t = 2 ln 3 ln ( 58 58 − 50 ln 3 ) ≈ 5.35 t = \frac{2}{\ln 3}\ln\left( \frac{58}{58 - 50\ln 3} \right) \approx 5.35
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