Parametric Mimima of Two Moving Points At noon, one ship (A) was 100km directly north of another ship (B). Ship A was sailing south at 30 kph and B was sailing east at 15kph. After how many hours will the two ships be nearest each other?
The Answer is 2.67
If we turn the pythagorean theorem into a parametric equation we get; 
2D(dD/dt)^2 = (2x(dx/dT))^2 +(2y(dy/dT))^2

Initial Distances are the following
D = 100
x = 0
y = 100

We also get the following rates:
dx/dT = 15kpH
dy/dT = -30kpH

If we substitute all of it into the equation we get; -15kPh for dD/dT then I realize. I have to set dD/dT = 0 in order to get minima; just like when i set y' to 0 in other problems. I know im missing another equation? method? Any hint? 
 A: There are two ships, suppose that Ship B is at the origin.
Then the equation for the position is Ship  is $p_B(t) = (15t,0)$, where $t$ is
the elapsed time in hours. The equation for the position of Ship A is
$p_A(t) = (0, 100-30t)$.
Then the distance squared between them is described by $s(t) = \|p_A(t)-p_B(t)\|^2$. 
Compute this formula, differentiate with respect to $t$ and determine the
time at which the minimum occurs. Use this to find the closest distance.
Notes: (1) You can pick any initial point, not just the origin, as the distance between the boats will remain the same. (2) We use the distance squared because it is a simpler formula the the distance (no square root) and, not that it matters in this problem, it is differentiable everywhere, the distance is not differentiable at the origin.
A: The answer is after $\frac{8}{3}$ hours. This is because the square of the distance in terms of t is, as in the first answer, $f(t) =(15t)^2+(100-39t)^2=225t^2-6000t+10000$. The minimum is attained if the derivative is zero. $t=\frac{8}{3}$ 
A: As a general result, if you express the relative displacement as a vector $\underline R$ and the relative velocity as a vector $\underline V$, then the objects are closest when the dot product $\underline R.\underline V=0$
