IBDP Maths AI: Topic : AHL 1.15: Eigenvalues and eigenvectors: IB style Questions HL Paper 1

Detailed Solution

 The system is given in the form:

\[
\frac{dx}{dt} = ax + by, \quad \frac{dy}{dt} = cx + dy
\]

We’re given the eigenvalues are \(\lambda = -1\) and \(\lambda = 2\), with corresponding eigenvectors \(\begin{pmatrix} 3 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\). The phase portrait is provided, showing lines through the origin \((0, 0)\) parallel to the eigenvectors, and we need to analyze the system in the first quadrant (\(x, y \geq 0\)). The tasks are to determine the direction of motion along the eigenvectors, we will sketch trajectories in the three regions defined by the eigenvectors, and find the minimum number of new \(Y\) animals to add to prevent extinction, given initial populations \(x = 252\) and \(y = 60\).

Step 1: Determine the Coefficients \(a, b, c, d\)
The system can be written in matrix form as:
\[
\begin{pmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}
\]
The matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) has eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = 2\), with eigenvectors \(\begin{pmatrix} 3 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\), respectively.

Eigenvalue \(\lambda_1 = -1\), Eigenvector \(\begin{pmatrix} 3 \\ 1 \end{pmatrix}\):
\[
(A – \lambda_1 I) \mathbf{v}_1 = 0 \implies \begin{pmatrix} a – (-1) & b \\ c & d – (-1) \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]
\[
\begin{pmatrix} a + 1 & b \\ c & d + 1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]
\[
(a + 1) \cdot 3 + b \cdot 1 = 0 \implies 3(a + 1) + b = 0 \implies 3a + 3 + b = 0 \implies b = -3a – 3 \quad (1)
\]
\[
c \cdot 3 + (d + 1) \cdot 1 = 0 \implies 3c + d + 1 = 0 \implies d = -3c – 1 \quad (2)
\]

Eigenvalue \(\lambda_2 = 2\), Eigenvector \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\):
\[
(A – \lambda_2 I) \mathbf{v}_2 = 0 \implies \begin{pmatrix} a – 2 & b \\ c & d – 2 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]
\[
(a – 2) \cdot 1 + b \cdot 3 = 0 \implies a – 2 + 3b = 0 \implies a + 3b = 2 \quad (3)
\]
\[
c \cdot 1 + (d – 2) \cdot 3 = 0 \implies c + 3(d – 2) = 0 \implies c + 3d – 6 = 0 \implies c = -3d + 6 \quad (4)
\]

Solve the System of Equations:
From (1): \(b = -3a – 3\)
Substitute into (3):
\[
a + 3(-3a – 3) = 2 \implies a – 9a – 9 = 2 \implies -8a – 9 = 2 \implies -8a = 11 \implies a = -\frac{11}{8}
\]
\[
b = -3\left(-\frac{11}{8}\right) – 3 = \frac{33}{8} – \frac{24}{8} = \frac{9}{8}
\]
From (4): \(c = -3d + 6\)
Substitute into (2):
\[
d = -3(-3d + 6) – 1 \implies d = 9d – 18 – 1 \implies d – 9d = -19 \implies -8d = -19 \implies d = \frac{19}{8}
\]
\[
c = -3\left(\frac{19}{8}\right) + 6 = -\frac{57}{8} + \frac{48}{8} = -\frac{9}{8}
\]
So, the matrix is:
\[
A = \begin{pmatrix} -\frac{11}{8} & \frac{9}{8} \\ -\frac{9}{8} & \frac{19}{8} \end{pmatrix}
\]

Verify:
Trace of \(A\): \(a + d = -\frac{11}{8} + \frac{19}{8} = \frac{8}{8} = 1 = \lambda_1 + \lambda_2 = -1 + 2\)
 Determinant of \(A\): \(ad – bc = \left(-\frac{11}{8}\right)\left(\frac{19}{8}\right) – \left(\frac{9}{8}\right)\left(-\frac{9}{8}\right) = -\frac{209}{64} + \frac{81}{64} = -\frac{128}{64} = -2 = \lambda_1 \lambda_2 = (-1)(2)\)
The matrix is correct.

Step 2: General Solution
The system is:
\[
\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix}
\]
The general solution is:
\[
\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 = c_1 e^{-t} \begin{pmatrix} 3 \\ 1 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} 1 \\ 3 \end{pmatrix}
\]
\[
x(t) = 3 c_1 e^{-t} + c_2 e^{2t}, \quad y(t) = c_1 e^{-t} + 3 c_2 e^{2t}
\]

Eigenvector Lines:
\(\lambda_1 = -1\), \(\mathbf{v}_1 = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\): Line \(y = \frac{1}{3}x\).
\(\lambda_2 = 2\), \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\): Line \(y = 3x\).

These lines divide the first quadrant into three regions:
Region I: Above \(y = 3x\) (\(y > 3x\)).
Region II: Between \(y = 3x\) and \(y = \frac{1}{3}x\) (\(3x > y > \frac{1}{3}x\)).
Region III: Below \(y = \frac{1}{3}x\) (\(y < \frac{1}{3}x\)).

Part (a)(i): Show the Direction of Motion Along the Eigenvectors
Along \(\mathbf{v}_1 = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\), \(\lambda_1 = -1\):
The solution along this eigenvector is:
\[
\begin{pmatrix} x \\ y \end{pmatrix} = c_1 e^{-t} \begin{pmatrix} 3 \\ 1 \end{pmatrix}
\]
As \(t \to \infty\), \(e^{-t} \to 0\), so the solution approaches the origin \((0, 0)\). As \(t \to -\infty\), \(e^{-t} \to \infty\), so the solution moves away from the origin. Thus, the direction of motion is **toward the origin** along this line in both directions from the origin.

Along \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\), \(\lambda_2 = 2\):
\[
\begin{pmatrix} x \\ y \end{pmatrix} = c_2 e^{2t} \begin{pmatrix} 1 \\ 3 \end{pmatrix}
\]
As \(t \to \infty\), \(e^{2t} \to \infty\), so the solution moves away from the origin. As \(t \to -\infty\), \(e^{2t} \to 0\), so the solution approaches the origin. Thus, the direction of motion is **away from the origin** along this line in both directions.

Answer for (a)(i):
 Along \(y = \frac{1}{3}x\), the direction is toward the origin.
Along \(y = 3x\), the direction is **away from the origin**.

Part (a)(ii): Sketch One Trajectory in Each of the Three Regions
The origin is a saddle point (one positive eigenvalue, one negative), so trajectories will generally be influenced by the unstable direction (\(\lambda_2 = 2\)) as \(t \to \infty\).

Determine the Behavior:
As \(t \to \infty\), the \(e^{2t}\) term dominates, so trajectories will tend to align with the eigenvector \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) (line \(y = 3x\)), moving away from the origin.
As \(t \to -\infty\), the \(e^{-t}\) term dominates, so trajectories approach the origin along the direction of \(\begin{pmatrix} 3 \\ 1 \end{pmatrix}\) (line \(y = \frac{1}{3}x\)).

Region I (\(y > 3x\)):
Choose a point, e.g., \((1, 4)\):
\[
y = 4 > 3 \times 1 = 3
\]
Solve for \(c_1, c_2\):
\[
x(0) = 3 c_1 + c_2 = 1 \quad (1)
\]
\[
y(0) = c_1 + 3 c_2 = 4 \quad (2)
\]
\[
(2) – 3 \times (1): (c_1 + 3 c_2) – 3 (3 c_1 + c_2) = 4 – 3 \implies c_1 + 3 c_2 – 9 c_1 – 3 c_2 = 1 \implies -8 c_1 = 1 \implies c_1 = -\frac{1}{8}
\]
\[
3 \left(-\frac{1}{8}\right) + c_2 = 1 \implies -\frac{3}{8} + c_2 = 1 \implies c_2 = 1 + \frac{3}{8} = \frac{11}{8}
\]
\[
x(t) = 3 \left(-\frac{1}{8}\right) e^{-t} + \frac{11}{8} e^{2t} = -\frac{3}{8} e^{-t} + \frac{11}{8} e^{2t}
\]
\[
y(t) = -\frac{1}{8} e^{-t} + 3 \left(\frac{11}{8}\right) e^{2t} = -\frac{1}{8} e^{-t} + \frac{33}{8} e^{2t}
\]
As \(t \to \infty\), \(x \to \infty\), \(y \to \infty\), and \(\frac{y}{x} \to \frac{\frac{33}{8} e^{2t}}{\frac{11}{8} e^{2t}} = 3\), so the trajectory approaches the line \(y = 3x\).
As \(t \to -\infty\), the \(e^{-t}\) terms dominate, but \(c_1\) is negative, so the trajectory comes from the third quadrant, crosses into the first quadrant, and then diverges along \(y = 3x\).

Region II (\(3x > y > \frac{1}{3}x\)):
Choose \((1, 1)\):
\[
3 \times 1 > 1 > \frac{1}{3} \times 1
\]
\[
3 c_1 + c_2 = 1 \quad (1)
\]
\[
c_1 + 3 c_2 = 1 \quad (2)
\]
\[
(1) – 3 \times (2): (3 c_1 + c_2) – 3 (c_1 + 3 c_2) = 1 – 3 \implies 3 c_1 + c_2 – 3 c_1 – 9 c_2 = -2 \implies -8 c_2 = -2 \implies c_2 = \frac{1}{4}
\]
\[
c_1 + 3 \left(\frac{1}{4}\right) = 1 \implies c_1 + \frac{3}{4} = 1 \implies c_1 = \frac{1}{4}
\]
\[
x(t) = 3 \left(\frac{1}{4}\right) e^{-t} + \frac{1}{4} e^{2t} = \frac{3}{4} e^{-t} + \frac{1}{4} e^{2t}
\]
\[
y(t) = \frac{1}{4} e^{-t} + 3 \left(\frac{1}{4}\right) e^{2t} = \frac{1}{4} e^{-t} + \frac{3}{4} e^{2t}
\]
As \(t \to \infty\), the trajectory diverges along \(y = 3x\).
As \(t \to -\infty\), it approaches the origin along \(y = \frac{1}{3}x\).

Region III (\(y < \frac{1}{3}x\)):
Choose \((3, 0.5)\):
\[
\frac{1}{3} \times 3 = 1 > 0.5
\]
\[
3 c_1 + c_2 = 3 \quad (1)
\]
\[
c_1 + 3 c_2 = 0.5 \quad (2)
\]
\[
(1) – 3 \times (2): (3 c_1 + c_2) – 3 (c_1 + 3 c_2) = 3 – 1.5 \implies -8 c_2 = 1.5 \implies c_2 = -\frac{1.5}{8} = -\frac{3}{16}
\]
\[
c_1 + 3 \left(-\frac{3}{16}\right) = 0.5 \implies c_1 – \frac{9}{16} = 0.5 \implies c_1 = 0.5 + \frac{9}{16} = \frac{17}{16}
\]
\[
x(t) = 3 \left(\frac{17}{16}\right) e^{-t} + \left(-\frac{3}{16}\right) e^{2t} = \frac{51}{16} e^{-t} – \frac{3}{16} e^{2t}
\]
\[
y(t) = \frac{17}{16} e^{-t} + 3 \left(-\frac{3}{16}\right) e^{2t} = \frac{17}{16} e^{-t} – \frac{9}{16} e^{2t}
\]
As \(t \to \infty\), the \(e^{2t}\) term dominates, and since \(c_2 < 0\), \(x \to -\infty\), \(y \to -\infty\), so the trajectory moves into the third quadrant along \(y = 3x\).
As \(t \to -\infty\), it approaches the origin along \(y = \frac{1}{3}x\).

 (a)(ii):
Region I: Trajectory starts above \(y = 3x\), comes from the third quadrant, crosses into the first quadrant, and diverges along \(y = 3x\) upward.
Region II: Trajectory starts between the lines, approaches the origin along \(y = \frac{1}{3}x\), then diverges along \(y = 3x\).
Region III: Trajectory starts below \(y = \frac{1}{3}x\), approaches the origin along \(y = \frac{1}{3}x\), then diverges along \(y = 3x\) into the third quadrant.

The desired graph is shown below

Part (b): Minimum Number of New \(Y\) Animals to Prevent Extinction
Given initial populations \(x(0) = 252\), \(y(0) = 60\), we need to add \(\delta\) animals to \(Y\), so the new initial condition is \((252, 60 + \delta)\). We want \(y(t) \geq 0\) for all \(t \geq 0\) to prevent extinction.

Step 1: Solve for \(c_1, c_2\):
\[
3 c_1 + c_2 = 252 \quad (1)
\]
\[
c_1 + 3 c_2 = 60 + \delta \quad (2)
\]
\[
(1) – 3 \times (2): (3 c_1 + c_2) – 3 (c_1 + 3 c_2) = 252 – 3 (60 + \delta) \implies -8 c_2 = 252 – 180 – 3\delta \implies -8 c_2 = 72 – 3\delta \implies c_2 = \frac{3\delta – 72}{8}
\]
\[
c_1 + 3 \left(\frac{3\delta – 72}{8}\right) = 60 + \delta \implies c_1 + \frac{9\delta – 216}{8} = 60 + \delta
\]
\[
c_1 = 60 + \delta – \frac{9\delta – 216}{8} = \frac{480 + 8\delta – 9\delta + 216}{8} = \frac{696 – \delta}{8}
\]
\[
y(t) = \frac{696 – \delta}{8} e^{-t} + 3 \left(\frac{3\delta – 72}{8}\right) e^{2t} = \frac{696 – \delta}{8} e^{-t} + \frac{9\delta – 216}{8} e^{2t}
\]

Step 2: Ensure \(y(t) \geq 0\):
\[
y(t) = \frac{696 – \delta}{8} e^{-t} + \frac{9\delta – 216}{8} e^{2t}
\]
As \(t \to \infty\), the \(e^{2t}\) term dominates:
– If \(9\delta – 216 > 0 \implies \delta > 24\), then \(y(t) \to \infty\), and \(y(t) > 0\) for large \(t\).
– If \(9\delta – 216 = 0 \implies \delta = 24\), then \(y(t) \to 0\) as \(t \to \infty\).
– If \(9\delta – 216 < 0 \implies \delta < 24\), then \(y(t) \to -\infty\), which leads to extinction.

Set \(\delta = 24\):
\[
y(t) = \frac{696 – 24}{8} e^{-t} = \frac{672}{8} e^{-t} = 84 e^{-t}
\]
This is always positive but approaches 0, which may not be considered “preventing extinction” in a practical sense. We need \(y(t)\) to remain positive and not approach 0, so \(\delta > 24\).

Try \(\delta = 25\):
\[
c_1 = \frac{696 – 25}{8} = \frac{671}{8}, \quad c_2 = \frac{3 \times 25 – 72}{8} = \frac{75 – 72}{8} = \frac{3}{8}
\]
\[
y(t) = \frac{671}{8} e^{-t} + \frac{9 \times 25 – 216}{8} e^{2t} = \frac{671}{8} e^{-t} + \frac{225 – 216}{8} e^{2t} = \frac{671}{8} e^{-t} + \frac{9}{8} e^{2t}
\]
Find the minimum of \(y(t)\):
\[
\frac{dy}{dt} = -\frac{671}{8} e^{-t} + \frac{9}{8} \cdot 2 e^{2t} = 0 \implies -\frac{671}{8} e^{-t} + \frac{18}{8} e^{2t} = 0 \implies 18 e^{2t} = 671 e^{-t}
\]
\[
e^{3t} = \frac{671}{18} \approx 37.2778 \implies 3t = \ln\left(\frac{671}{18}\right) \implies t = \frac{1}{3} \ln\left(\frac{671}{18}\right) \approx 1.211
\]
\[
e^{-t} \approx e^{-1.211} \approx 0.2978, \quad e^{2t} \approx (37.2778)^{2/3} \approx 11.149
\]
\[
y_{\text{min}} \approx \frac{671}{8} \cdot 0.2978 + \frac{9}{8} \cdot 11.149 \approx 24.99 + 12.54 \approx 37.53 > 0
\]
Since \(y(t) > 0\) for all \(t\), \(\delta = 25\) ensures \(Y\) does not go extinct.

The minimum number of new \(Y\) animals to add is 25.

………………………….Markscheme………………………………….

Solution: –

(a) (i)&(ii)

(b) for Y not to die out $y>\frac{1}{3}x$

as $x=252, y>84$

(minimum number of new animals is) 25