IB Mathematics AHL 1.15 Eigenvalues and eigenvectors AI HL Paper 1- Exam Style Questions- New Syllabus

(a)

At \(t=0\), \(x=5\) and \(y=2\). First calculate both rates of change:

\(\dfrac{dx}{dt}=2.2(5)-2.6(2)\)

\(\dfrac{dx}{dt}=11-5.2=5.8\).

Also,

\(\dfrac{dy}{dt}=3.4(5)-2.2(2)\)

\(\dfrac{dy}{dt}=17-4.4=12.6\).

Using

\(\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\),

\(\dfrac{dy}{dx}=\dfrac{12.6}{5.8}=\dfrac{63}{29}\).

\(\dfrac{dy}{dx}=2.17241\ldots\)

✅ Answer: \(\dfrac{dy}{dx}\approx2.17\)

(b)

The eigenvalues are \(\pm2i\), which are purely imaginary. Therefore, the trajectories form closed curves—circles or ellipses—centred at the equilibrium point \((0,0)\).

The required trajectory must pass through \((5,2)\). At this point,

\(\dfrac{dx}{dt}=5.8>0\)

and

\(\dfrac{dy}{dt}=12.6>0\).

Therefore, the trajectory is moving to the right and upwards at \((5,2)\). The direction around the ellipse is consequently anticlockwise.

The curve should also have a positive gradient of approximately \(2.17\) at \((5,2)\).

For a more precise sketch, the trajectory through \((5,2)\) can be represented by

\(17x^2-22xy+13y^2=257\),

which is a tilted ellipse centred at the origin.

✅ Answer: Sketch an ellipse centred at \((0,0)\), passing through \((5,2)\), with arrows showing anticlockwise motion and a positive gradient at \((5,2)\).