2010 H2 Mathematics Paper 1 Question 7

Differential Equations (DEs)

Answers

10ln⁡2≈6.93 min{10 \ln 2 \approx 6.93 \textrm{ min}}
θ{\theta} approaches 20{20} for large values of t{t }

Full solutions

dθdt=k(20−θ)\frac{\mathrm{d}\theta}{\mathrm{d}t} = k (20 - \theta)
When θ=10,{\theta = 10, } dθdt=1{\frac{\mathrm{d}\theta}{\mathrm{d}t}=1}
1=k(20−10)k=110\begin{align*} 1 &= k (20-10) \\ k &= \frac{1}{10} \end{align*}
∫120−θ  dθ=∫110  dt\int \frac{1}{20 - \theta} \; \mathrm{d}\theta = \int \frac{1}{10} \; \mathrm{d}t
−ln⁡∣20−θ∣=110t+cln⁡∣20−θ∣=−110t−c∣20−θ∣=e−110t−c=e−110t⋅e−c20−θ=±e−ce−110t=Ae−110tθ=20−Ae−110t\begin{align*} -\ln | 20 - \theta | &= \frac{1}{10}t + c \\ \ln | 20 - \theta | &= -\frac{1}{10}t - c \\ | 20 - \theta | &= \mathrm{e}^{-\frac{1}{10}t - c} \\ & = \mathrm{e}^{-\frac{1}{10}t} \cdot \mathrm{e}^{-c} \\ 20 - \theta &= \pm \mathrm{e}^{-c} \mathrm{e}^{-\frac{1}{10}t} \\ &= A \mathrm{e}^{-\frac{1}{10}t} \\ \theta &= 20 - A \mathrm{e}^{-\frac{1}{10}t} \\ \end{align*}
When t=0,{t=0, } θ=10{\theta = 10}
10=20−Ae−110(0)A=10\begin{align*} 10 &= 20 - A \mathrm{e}^{-\frac{1}{10}(0)} \\ A &= 10 \end{align*}
θ=20−10e−110t  ■\theta = 20 - 10 \mathrm{e}^{-\frac{1}{10}t} \; \blacksquare
When θ=15,{\theta = 15,}
15=20−10e−110t10e−110t=5e−110t=12−110t=ln⁡12t=−10ln⁡12=10ln⁡2 min  ■\begin{align*} 15 &= 20 - 10 \mathrm{e}^{-\frac{1}{10}t} \\ 10 \mathrm{e}^{-\frac{1}{10}t} &= 5 \\ \mathrm{e}^{-\frac{1}{10}t} &= \frac{1}{2} \\ -\frac{1}{10}t &= \ln \frac{1}{2} \\ t &= -10 \ln \frac{1}{2} \\ &= 10 \ln 2 \textrm{ min} \; \blacksquare \end{align*}
As t→∞,{t \to \infty, } e−110t→0.{\mathrm{e}^{-\frac{1}{10}t} \to 0.} Hence θ→20{\theta \to 20}
θ{\theta} approaches 20{20} for large values of t  ■{t \; \blacksquare}
Graph of temperature against time