2021 H2 Mathematics Paper 1 Question 10

Differential Equations (DEs)

Answers

(a)

t=2ln⁡10ln⁡3≈4.19 min{t = \frac{2 \ln 10}{\ln 3} \approx 4.19 \textrm{ min}}

(b)

N=2ln⁡3(d+(50ln⁡3−d)(3)t2)N = \frac{2}{\ln 3} \left( d + (50 \ln 3 - d) \left( 3 \right)^{\frac{t}{2}} \right)
(ci)
d>50ln⁡3{d > 50 \ln 3}
(cii)
5.35 min{5.35 \textrm{ min}}

Full solutions

(a)

dNdt=kN  ■1NdNdt=k∫1N  dθ=∫k  dtln⁡N=kt+cN=ekt+cN=Aekt  ■\begin{gather*} \frac{\mathrm{d}N}{\mathrm{d}t} = kN \; \blacksquare \\ \frac{1}{N} \frac{\mathrm{d}N}{\mathrm{d}t} = k \\ \int \frac{1}{N} \; \mathrm{d}\theta = \int k \; \mathrm{d}t \\ \ln N = kt + c \\ N = \mathrm{e}^{kt+c} \\ N = A\mathrm{e}^{kt} \; \blacksquare \\ \end{gather*}
When t=0,N=100{t=0, N = 100}
100=Aek(0)A=100\begin{gather*} 100 = A \mathrm{e}^{k(0)} \\ A = 100 \end{gather*}
When t=2,N=300{t=2, N = 300}
300=100ek(2)2k=ln⁡3k=12ln⁡3  ■\begin{gather*} 300 = 100 \mathrm{e}^{k(2)} \\ 2k = \ln 3 \\ k = \frac{1}{2} \ln 3 \; \blacksquare \end{gather*}
N=100et2ln⁡3N = 100 \mathrm{e}^{\frac{t}{2} \ln 3}
When N=1000{N = 1000}
1000=100et2ln⁡3t2ln⁡3=ln⁡10t=2ln⁡10ln⁡3 min  ■\begin{align*} 1000 &= 100 \mathrm{e}^{\frac{t}{2} \ln 3} \\ \frac{t}{2} \ln 3 &= \ln 10 \\ t &= \frac{2 \ln 10}{\ln 3} \textrm{ min} \; \blacksquare \end{align*}

(b)

dNdt=kN−d  ■1kN−ddNdt=1∫1kN−d  dθ=∫1  dt1kln⁡∣kN−d∣=t+Ct=1kln⁡∣kN−d∣−C  ■∣kN−d∣=ekt+kC∣kN−d∣=ekCektkN−d=A2ekt  ■\begin{gather*} \frac{\mathrm{d}N}{\mathrm{d}t} = kN - d \; \blacksquare \\ \frac{1}{kN-d} \frac{\mathrm{d}N}{\mathrm{d}t} = 1 \\ \int \frac{1}{kN-d} \; \mathrm{d}\theta = \int 1 \; \mathrm{d}t \\ \frac{1}{k} \ln | kN-d | = t + C \\ t = \frac{1}{k} \ln | kN-d | - C \; \blacksquare \\ |kN-d| = \mathrm{e}^{kt+kC} \\ |kN-d| = \mathrm{e}^{kC}\mathrm{e}^{kt} \\ kN - d = A_2 \mathrm{e}^{kt} \; \blacksquare \\ \end{gather*}
When t=0,T=100{t=0, T=100}
100k−d=A2ek(0)A2=100k−dA2=50ln⁡3−d\begin{gather*} 100k - d = A_2 \mathrm{e}^{k(0)} \\ A_2 = 100k-d \\ A_2 = 50 \ln 3 - d \end{gather*}
kN−d=(50ln⁡3−d)ekt(12ln⁡3)N=d+(50ln⁡3−d)e(12ln⁡3)t(12ln⁡3)N=d+(50ln⁡3−d)eln⁡3t2(12ln⁡3)N=d+(50ln⁡3−d)(3)t2N=2ln⁡3(d+(50ln⁡3−d)(3)t2)  ■\begin{gather*} kN - d = (50 \ln 3 - d) \mathrm{e}^{kt} \\ \left( \frac{1}{2} \ln 3 \right) N = d + (50 \ln 3 - d) \mathrm{e}^{\left( \frac{1}{2} \ln 3 \right)t} \\ \left( \frac{1}{2} \ln 3 \right) N = d + (50 \ln 3 - d) \mathrm{e}^{\ln 3^{\frac{t}{2}}} \\ \left( \frac{1}{2} \ln 3 \right) N = d + (50 \ln 3 - d) \left( 3 \right)^{\frac{t}{2}} \\ N = \frac{2}{\ln 3} \left( d + (50 \ln 3 - d) \left( 3 \right)^{\frac{t}{2}} \right) \; \blacksquare \\ \end{gather*}
(ci)
For the bacteria to decrease,
50ln⁡3−d<0d>50ln⁡3  ■\begin{gather*} 50 \ln 3 - d < 0 \\ d > 50 \ln 3 \; \blacksquare \end{gather*}
(cii)
When N=0,{N=0,}
0=2ln⁡3(58+(50ln⁡3−58)(3)t2)3t2=−5850ln⁡3−58t2=ln⁡−5850ln⁡3−58ln⁡3t=2ln⁡−5850ln⁡3−58ln⁡3t=5.35 min (3 sf)  ■\begin{gather*} 0 = \frac{2}{\ln 3} \left( 58 + (50 \ln 3 - 58) \left( 3 \right)^{\frac{t}{2}} \right) \\ 3^{\frac{t}{2}} = \frac{-58}{50\ln 3 - 58} \\ \frac{t}{2} = \frac{\ln \frac{-58}{50\ln 3 - 58}}{\ln 3}\\ t = \frac{2\ln \frac{-58}{50\ln 3 - 58}}{\ln 3}\\ t = 5.35 \textrm{ min} \textrm{ (3 sf)} \; \blacksquare \end{gather*}