# Science:Math Exam Resources/Courses/MATH100/December 2018/Question 06/Solution 1

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

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

${\displaystyle a(t)^{2}=x(t)^{2}+y(t)^{2}=(4+2t)^{2}+(8-t)^{2}=16+16t+4t^{2}+64-16t+t^{2}=80+5t^{2}.}$

Now, we will first determine the time ${\displaystyle t}$ at which the distance between the particles is ${\displaystyle 10}$ units. We do this by setting ${\displaystyle a(t)=10}$ and solving for ${\displaystyle t}$. This gives

${\displaystyle 100=80+5t^{2}}$

and therefore ${\displaystyle t=2}$ (we can ignore the negative root since time cannot be negative).

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