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2024 Paper 3 Q6
D: 1500.0 B: 1500.0
differential equations systems of odes linear algebra eigenvalues invariant quantities necessary conditions exponential functions

  1. A particle moves in two-dimensional space. Its position is given by coordinates \((x, y)\) which satisfy \[\frac{\mathrm{d}x}{\mathrm{d}t} = -x + 3y + u\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = x + y + u\] where \(t\) is the time and \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0)\).
    1. By considering \(\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{\mathrm{d}y}{\mathrm{d}t}\), show that if the particle is at the origin \((0,0)\) at some time \(t > 0\), then it is necessary that \(x_0 = y_0\).
    2. Given that \(x_0 = y_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).
  2. A particle whose position in three-dimensional space is given by co-ordinates \((x, y, z)\) moves with time \(t\) such that \[\frac{\mathrm{d}x}{\mathrm{d}t} = 4y - 5z + u\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = x - 2z + u\] \[\frac{\mathrm{d}z}{\mathrm{d}t} = x - 2y + u\] where \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0,\, z_0)\).
    1. Show that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(y_0\) is the mean of \(x_0\) and \(z_0\).
    2. Show further that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(x_0 = y_0 = z_0\).
    3. Given that \(x_0 = y_0 = z_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).

Solution:

    1. \(\,\) \begin{align*} && \frac{\d x}{\d t} - \frac{\d y}{\d t} &= -2x + 2y \\ &&&= -2(x-y) \\ \Rightarrow && \dot{z} &= -2z \tag{\(z = x-y\)} \\ \Rightarrow && z &= Ae^{-2t} \\ z = 0, t : && A &= 0 \\ \Rightarrow && z &= 0 \quad \forall t \\ \Rightarrow && x &= y \quad \forall t \\ \Rightarrow && x_0 &= y_0 \end{align*}
    2. Since \(x = y\) for all \(t\) our equation can we written as \(\frac{\d x}{\d t} = 2x + u\). This has solution \(x = Ae^{2t} - \frac{u}{2}\) we also have \begin{align*} t = 0: && x_0 &= A - \frac{u}{2} \\ t = T: && 0 &= Ae^{2T} - \frac{u}{2} \\ \Rightarrow && x_0 &= \frac{u}{2} - e^{-2T}\frac{u}{2} \\ \Rightarrow && u &= \frac{2x_0}{1-e^{-2T}} \end{align*}
    1. Let \(w = y - \frac12(x+z)\) then \begin{align*} && \dot{w} &= \dot{y} - \tfrac12(\dot{x}+\dot{z}) \\ &&&= (x-2z+u) - \tfrac12(4y-5z+u+x-2y+u) \\ &&&= -y + \tfrac12(x+z) \\ &&&= -w \\ \Rightarrow && w &= Ae^{-t} \\ \text{at origin}: && w &= 0 \\ \Rightarrow && y &= \tfrac12(x+z) \end{align*}
    2. We now have \(\dot{x} = 2x+2z-5z+u = 2x-3z+u\) and \(\dot{z} = x - x- z +u = - z+u\) so in particular \((x - z)' = 2(x-z)\) or \(x-z = Ae^{2t}\) and since we hit the origin \(x = z\) for all \(t\) and so \(y = x = z\) for all \(t\).
    3. Notice we now have \(\dot{x} = -x + u\) or \(x = Ae^{-t} + u\) \begin{align*} t = 0: && x_0 &= A + u \\ t = T: && 0 &= Ae^{-T} + u \\ \Rightarrow && x_0 &= u-e^{T} u \\ \Rightarrow && u &= \frac{x_0}{1-e^{T}} \end{align*}

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1989 Paper 2 Q10
D: 1600.0 B: 1500.0
3x3 matrices line intersection planes vectors parametric simultaneous equations linear algebra geometric proof

State carefully the conditions which the fixed vectors \(\mathbf{a,b,u}\) and \(\mathbf{v}\) must satisfy in order to ensure that the line \(\mathbf{r=a+}\lambda\mathbf{u}\) intersects the line \(\mathbf{r=b+\mu}\mathbf{v}\) in exactly one point. Find the two values of the fixed scalar \(b\) for which the planes with equations \[ \left.\begin{array}{c} x+y+bz=b+2\\ bx+by+z=2b+1 \end{array}\right\} \tag{*} \] do not intersect in a line. For other values of \(b\), express the line of intersection of the two planes in the form \(\mathbf{r=a}+\lambda\mathbf{u},\) where \(\mathbf{a\cdot u}=0\). Find the conditions which \(b\) and the fixed scalars \(c\) and \(d\) must satisfy to ensure that there is exactly one point on the line \[ \mathbf{r=}\left(\begin{array}{c} 0\\ 0\\ c \end{array}\right)+\mu\left(\begin{array}{c} 1\\ d\\ 0 \end{array}\right) \] whose coordinates satisfy both equations \((*)\).

Solution: There are two requirements (assuming they are lines not fixed points): 1. They cannot be parallel, ie \(\mathbf{u} \neq \lambda \mathbf{v}\) for any \(\lambda\) 2. They must lie in the same plane, ie \((\mathbf{b}-\mathbf{a})\cdot (\mathbf{u} \times \mathbf{v}) = 0\) The planes will not intersect in a line if they are either parallel and separate or parallel and the same. If \(b = 1\) or \(b=-1\) the planes are parallel. \begin{align*} && (x+y) + b z &= b+ 2\\ &&b(x+y) + z &= 2b + 1 \\ \Rightarrow && (1-b^2)z &= 2b+1 - b^2 -2b \\ &&&= 1-b^2 \\ \Rightarrow && z &= 1 \\ && x+ y &= 2 \\ \end{align*} Therefore our line is \(\mathbf{r} = \begin{pmatrix} 1+t \\ 1-t \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \) We must have: \(d \neq -1\) to ensure that the lines aren't parallel. We must also have: \begin{align*} 0 &= \left ( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} -\begin{pmatrix} 0 \\ 0 \\ c \end{pmatrix}\right) \cdot \left ( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ d \\ 0 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 \\ 1 \\ 1-c \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ d+1 \end{pmatrix} \\ &= (1-c)(d+1) \end{align*} So \(c =1\)

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