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

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 $\frac{d x}{d t} = 2$ units/min and $\frac{d y}{d t} = - 1$ units/min. Using the initial positions, then we see that $x (t) = 4 + 2 t$ 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

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

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

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

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 $a (t)^{2} = 80 + 5 t^{2}$ , we get $2 a (t) a^{'} (t) = 10 t$ . Now we can plug in $a (t) = 10$ and $t = 2$ , which gives $a^{'} (2) = 1 unit per min$ .