Let $x$ be the distance from the origin to the particle $A$ and let $y$ be the distance from the origin to the particle $B$ . Also Let $a$ be the distance between $A$ and $B$ . Notice that $x,y$ and $a$ are functions of time and we shall thus write them as $x(t),y(t),a(t)$ respectively.

By the given information, we have ${\frac {dx}{dt}}=2$ units/min and ${\frac {dy}{dt}}=-1$ units/min. Using the initial positions, then we see that $x(t)=4+2t$ and $y(t)=8-t$ . We can relate these two functions by Pythagoras' theorem: we have $x(t)^{2}+y(t)^{2}=a(t)^{2}$ , and therefore

$a(t)^{2}=x(t)^{2}+y(t)^{2}=(4+2t)^{2}+(8-t)^{2}=16+16t+4t^{2}+64-16t+t^{2}=80+5t^{2}.$ Now, we will first determine the time $t$ at which the distance between the particles is $10$ units. We do this by setting $a(t)=10$ and solving for $t$ . This gives

$100=80+5t^{2}$ and therefore $t=2$ (we can ignore the negative root since time cannot be negative).

Next, we want to find $a'(t)$ when $a(t)=10$ . By differentiating both sides of the equation $a(t)^{2}=80+5t^{2}$ , we get $2a(t)a'(t)=10t$ . Now we can plug in $a(t)=10$ and $t=2$ , which gives $\color {blue}a'(2)=1{\text{ unit per min}}$ .