In established forest fires, the proportion of the total area of the forest which has been destroyed is denoted by $x$, and the rate of change of $x$ with respect to time, $t$ hours, is called the destruction rate. Investigations show that the destruction rate is directly proportional to the product of the proportion destroyed and the proportion not destroyed. A particular fire is initially noticed when one half of the forest is destroyed, and it is found that the destruction rate at this time is such that, if it remained constant thereafter, the forest would be destroyed completely in a further 24 hours.
Show that&nbsp;$12\frac{\mathrm{d}x}{\mathrm{d}t}=x(1-x)$ .[3]
Deduce that approximately 73% of the forest is destroyed 12 hours after it is first noticed.[6]

In established forest fires, the proportion of the total area of the forest which has been destroyed is denoted by $x$, and the rate of change of $x$ with respect to time, $t$ hours, is called the destruction rate. Investigations show that the destruction rate is directly proportional to the product of the proportion destroyed and the proportion not destroyed. A particular fire is initially noticed when one half of the forest is destroyed, and it is found that the destruction rate at this time is such that, if it remained constant thereafter, the forest would be destroyed completely in a further 24 hours.

Show that $12\frac{\mathrm{d}x}{\mathrm{d}t}=x(1-x)$ .

Deduce that approximately 73% of the forest is destroyed 12 hours after it is first noticed.

Differential Equations - General

Differential Equations

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

IJC P2 Q5 (b) Modified

Let $x$ = proportion of forest destroyed, $(1-x)$ = proportion not destroyed.\n\nThe rate of destruction is proportional to $x(1-x)$:\n\n$\frac{\mathrm{d}x}{\mathrm{d}t} \propto x(1-x)$\n\n$\Rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = kx(1-x)$ where $k$ is a positive constant ... (1)

**Finding k:**\n\nAt $t = 0$, $x = \frac{1}{2}$\n\nDestruction rate = $\frac{\text{proportion of forest}}{\text{time}} = \frac{\frac{1}{2}}{24 \text{ hours}} = \frac{1}{48}$ forest/hour\n\n$\left.\frac{\mathrm{d}x}{\mathrm{d}t}\right|_{x=\frac{1}{2}} = \frac{1}{48}$\n\nSubstitute $x = \frac{1}{2}$, $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{48}$ into (1):\n\n$\frac{1}{48} = k\left(\frac{1}{2}\right)\left(1 - \frac{1}{2}\right)$\n\n$\frac{1}{4}k = \frac{1}{48}$\n\n$k = \frac{1}{12}$\n\n$\therefore \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{12}x(1-x)$

**Solving the differential equation:**\n\n$\frac{1}{x(1-x)}\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{12}$\n\n$\displaystyle\displaystyle\int \frac{A}{x} + \frac{B}{1-x} \, \mathrm{d}x = \frac{1}{12}\displaystyle\displaystyle\int 1 \, dt$\n\nUsing partial fractions: $A = 1$, $B = 1$\n\n$\displaystyle\displaystyle\int \frac{1}{x} + \frac{1}{1-x} \, \mathrm{d}x = \frac{1}{12}t + C$\n\n$\ln|x| + \frac{\ln|1-x|}{-1} = \frac{1}{12}t + C$\n\n$\ln\left|\frac{x}{1-x}\right| = \frac{1}{12}t + C$\n\n$\frac{x}{1-x} = \pm e^C \cdot e^{\frac{t}{12}} = Ae^{\frac{t}{12}}$

**Using initial condition** $t = 0$, $x = \frac{1}{2}$:\n\n$\frac{\frac{1}{2}}{1 - \frac{1}{2}} = Ae^0$\n\n$\frac{\frac{1}{2}}{\frac{1}{2}} = A$\n\n$A = 1$\n\n$\therefore \frac{x}{1-x} = e^{\frac{t}{12}}$

**When $t = 12$:**\n\n$\frac{x}{1-x} = e^1 = e$\n\n$x = e - ex$\n\n$(1 + e)x = e$\n\n$x = \frac{e}{1 + e}$\n\n$= 0.73105$\n\n$\approx 73\%$ (shown)

IJC P2 Q5 (b) Modified

IJC P2 Q5 (b) Modified

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