A cat is trying to catch a mouse in the non-negative quadrant \[N=\{(x_1,x_2)\in \mathbb{R}^2: x_1,x_2\geq 0\}.\]At time $t=0$ the cat is at $(1,1)$ and the mouse is at $(0,0)$. The cat moves with speed $\sqrt{2}$ such that the position $c(t)=(c_1(t),c_2(t))$ is continuous, and differentiable except at finitely many points; while the mouse moves with speed $1$ such that its position $m(t)=(m_1(t),m_2(t))$ is also continuous, and differentiable except at finitely many points. Thus $c(0)=(1,1)$ and $m(0)=(0,0)$; $c(t)$ and $m(t)$ are continuous functions of $t$ such that $c(t),m(t)\in N$ for all $t\geq 0$; the derivatives $c'(t)=(c'_1(t),c'_2(t))$ and $m'(t)=(m'_1(t),m'_2(t))$ each exist for all but finitely many $t$ and \[(c'_1(t)^2+(c'_2(t))^2=2 \qquad (m'_1(t)^2+(m'_2(t))^2=1,\]whenever the respective derivative exists. At each time $t$ the cat knows both the mouse's position $m(t)$ and velocity $m'(t)$. Show that, no matter how the mouse moves, the cat can catch it by time $t=1$; that is, show that the cat can move such that $c(\tau)=m(\tau)$ for some $\tau\in[0,1]$.