Botanists are studying how excessive logging is affecting the population of forest trees in a country in South-East Asia. The population of trees in a forest after tt years is denoted by xx, in thousands. One simple model proposed is that the population increases at a rate proportional to its population size. It is known that trees are chopped off at a constant rate of 3 thousand trees per year and the population stays constant when x=px = p.

  1. (i)

    Find a differential equation involving xx and tt.

  2. (ii)

    The initial population of trees in a forest is 5 thousand trees. Solve the differential equation in part (i) and show that x=(5−p)e3pt+px = (5 - p)\mathrm{e}^{\frac{3}{p}t} + p.

  3. (iii)

    Given that the population of tree decreases in a finite time, state an inequality satisfied by pp.

    Sketch the graph of xx against tt for this case, showing clearly the coordinates of the graph's point of intersection with the horizontal axis. Hence explain the meaning of this axial intercept in the context of the question.

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Worked solution

4(i)

1. Formulate a differential equation with an unknown constant of proportionality.

Let the rate of change of the population with respect to time be dxdt\frac{\mathrm{d}x}{\mathrm{d}t} and the constant of proportionality be kk.
The net rate of change is the difference between the growth rate (kxkx) and the rate at which trees are chopped off (3 thousand/year):
dxdt=kx−3\frac{\mathrm{d}x}{\mathrm{d}t} = kx - 3

2. Use the given steady-state condition to find the proportionality constant $k$.

The population stays constant when x=px = p, which means the rate of change dxdt=0\frac{\mathrm{d}x}{\mathrm{d}t} = 0.
Substituting these values into the equation:
0=kp−30 = kp - 3
k=3pk = \frac{3}{p}

3. State the final differential equation.

Substitute k=3pk = \frac{3}{p} back into the initial equation:
dxdt=3px−3\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3}{p}x - 3
Alternatively, this can be factored as:
dxdt=3p(x−p)\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3}{p}(x - p)

Answer

dxdt=3px−3\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3}{p}x - 3

4(ii)

1. Establish the differential equation from part (i).

From the model, the rate of increase is proportional to the population size minus a constant chopping rate: dxdt=kx−3\frac{\mathrm{d}x}{\mathrm{d}t} = kx - 3
Since the population stays constant when x=px = p, dxdt=0\frac{\mathrm{d}x}{\mathrm{d}t} = 0.
0=kp−3  ⟹  k=3p0 = kp - 3 \implies k = \frac{3}{p}
Thus, dxdt=3px−3=3p(x−p)\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3}{p}x - 3 = \frac{3}{p}(x - p)

2. Separate variables to prepare for integration.

∫1x−p dx=∫3p dt\int \frac{1}{x - p} \, \mathrm{d}x = \int \frac{3}{p} \, \mathrm{d}t

3. Integrate both sides to find the general solution.

ln⁡∣x−p∣=3pt+c\ln |x - p| = \frac{3}{p}t + c

4. Express the equation in exponential form.

∣x−p∣=e3pt+c=ec⋅e3pt|x - p| = \mathrm{e}^{\frac{3}{p}t + c} = \mathrm{e}^c \cdot \mathrm{e}^{\frac{3}{p}t}
x−p=Ae3ptx - p = A\mathrm{e}^{\frac{3}{p}t} (where A=±ecA = \pm \mathrm{e}^c)

5. Apply the initial condition to find $A$ and rearrange to show the given expression.

Given t=0t = 0 when x=5x = 5:
5−p=Ae0  ⟹  A=5−p5 - p = A\mathrm{e}^{0} \implies A = 5 - p
Substitute AA back into the equation:
x−p=(5−p)e3ptx - p = (5 - p)\mathrm{e}^{\frac{3}{p}t}
x=(5−p)e3pt+px = (5 - p)\mathrm{e}^{\frac{3}{p}t} + p (Shown)

Answer

x=(5−p)e3pt+px = (5 - p)\mathrm{e}^{\frac{3}{p}t} + p

4(iii)

1. State the inequality for $p$ that results in a decreasing tree population.

For the population x(t)=(5−p)e3pt+px(t) = (5 - p)\mathrm{e}^{\frac{3}{p}t} + p to decrease over time, the term involving tt must be decreasing.
Since p>0p > 0 (a constant population size is strictly positive), the exponential function e3pt\mathrm{e}^{\frac{3}{p}t} is increasing.
Thus, its coefficient must be negative for the overall function to decrease:
5−p<0  ⟹  p>55 - p < 0 \implies p > 5

2. Find the coordinates of the $t$-intercept algebraically.

At the tt-axis intercept, x=0x = 0.
0=(5−p)e3pt+p0 = (5 - p)\mathrm{e}^{\frac{3}{p}t} + p
(p−5)e3pt=p(p - 5)\mathrm{e}^{\frac{3}{p}t} = p
e3pt=pp−5\mathrm{e}^{\frac{3}{p}t} = \frac{p}{p - 5}
3pt=ln⁡(pp−5)\frac{3}{p}t = \ln\left(\frac{p}{p - 5}\right)
t=p3ln⁡(pp−5)t = \frac{p}{3}\ln\left(\frac{p}{p - 5}\right)
The coordinates are (p3ln⁡(pp−5),0)\left(\frac{p}{3}\ln\left(\frac{p}{p-5}\right), 0\right).

3. Describe the key features for the graph of $x$ against $t$.

The graph is a concave-downwards exponential curve starting at the vertical axis intercept (0,5)(0, 5) and crossing the horizontal axis at the computed intercept.

Graph sketching details:

  • Vertical axis labelled xx, horizontal axis labelled tt.
  • Point (0,5)(0, 5) marked on the xx-axis.
  • Point (p3ln⁡(pp−5),0)\left(\frac{p}{3}\ln\left(\frac{p}{p-5}\right), 0\right) marked on the tt-axis.
  • A smooth curve passing through both points, steadily decreasing from left to right.

4. Explain the meaning of the axial intercept.

In the context of the model, xx represents the population in thousands and tt represents time in years. The tt-axial intercept represents the time taken for the population of trees in the forest to reach exactly zero (the point of extinction).

Answer

p>5p > 5

Coordinates of intercept: (p3ln⁡(pp−5),0)\left(\frac{p}{3}\ln\left(\frac{p}{p-5}\right), 0\right)

Meaning: The exact time, in years, when the population of trees in the forest reaches zero (i.e., becomes completely wiped out).

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