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?
Best Answer
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.