Question
Two joggers, $A$ and $B$, start at either end of a $100$ m track. $A$ runs horizontally towards the other end at $3$ m/s and $B$ runs diagonally at $5$ m/s. The diagonal forms an angle of $30$ degrees with the track. What is the closest the two joggers can get to each other?
Hint: Use Pythagoras' theorem to find the the distance between the two joggers as a function of time.
My working
Let the shortest distance between the joggers be $S$.
$$\begin{aligned} S & = \sqrt{(5t)^2 - (100 - 3t)^2}\\ & = \sqrt{16t^2 + 600t - 10000}\\ \implies \frac {ds} {dt} & = \frac 1 2 (16t^2 + 600t - 10000)^{-\frac 1 2} (32t + 600) \end{aligned}$$
When $$\frac {ds} {dt} = 0,$$ $$32t + 600 = 0.$$
However, this gives a negative value of $t$, which obviously cannot be true.
Where have I gone wrong? I am thinking that my interpretation of the shortest distance probably has some issues. Any intuitive explanations as to what the solution should be will be greatly appreciated :)
The given hint says to apply Pythagoras.
$A$ travels horizontally towards $B's$ original position whereas $B$ travels diagonally. So after time $t$, if $B$ has reached point $C$ and $A$ has reached point $D$,
$BC = 5t, AD = 3t$
Dropping perp from $C$ to $AB$,
$CE = BC/2 = \dfrac{5t}{2}, BE = \dfrac{5 \sqrt3 ~t}{2}$
$DE = 100 - \left(3t + \dfrac{5 \sqrt3 ~t}{2}\right)$
Distance between new positions of $A$ and $B$ is,
$ \displaystyle CD^2 = CE^2 + DE^2 = \frac{25 t^2}{4} + \left(100 - \left(3t + \dfrac{5 \sqrt3 ~t}{2}\right)\right)^2$
Now take derivative and equate to zero to minimize $CD^2$ (or $CD$)
You should get minimum distance at $~t \approx 12.22$