IB Maths AI HL Phase Portraits Questions

1. State the condition for an equilibrium point: both $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=0$.

2. Set up the system of linear equations: $3x-y=5$ and $2x+y=5$.

3. Add the two equations to eliminate $y$: $5x=10⟹x=2$.

4. Substitute $x=2$ back into the second equation: $2(2)+y=5⟹y=1$.

5. The exact coordinates of the equilibrium point are $(2,1)$.

1. Set up the characteristic equation $det(M-\lambda I)=0$.

2. Evaluate the determinant: $(1-\lambda )(-1-\lambda )-(4)(2)=0$.

3. Expand and simplify the quadratic: $-1-\lambda +\lambda +{\lambda }^2-8=0⟹{\lambda }^2-9=0$.

4. Solve for the eigenvalues: $\lambda =3$ and $\lambda =-3$.

5. Classify the point: Since the eigenvalues are real and have opposite signs, the origin is a saddle point.

1. Extract the coefficient matrix $M=\left(\begin{array}{cc}-1& -1\\ 1& -1\end{array}\right)$.

2. Form the characteristic equation: $(-1-\lambda {)}^2-(-1)(1)=0⟹{\lambda }^2+2\lambda +2=0$.

3. Solve using the quadratic formula: $\lambda =\frac{-2\pm \sqrt{4-8}}{2}=-1\pm i$.

4. Analyse the real part of the complex eigenvalues: The real part is $-1$. Since it is negative, trajectories spiral inwards, so it is a stable spiral.

5. Determine the direction of rotation by testing a coordinate on the positive $x$-axis, e.g., $(1,0)$. Here, $y^{\prime }=1-0=1$, which means the vector points upwards. Thus, the rotation is anti-clockwise.