How would I got about setting up and solving this problem?

Question

At 1:00pm, Ship A is 30 miles due south of Ship B, and is sailing north at a rate of 15 mi/hr. If Shipe B is sailing west at a rate of 10 mi/hr, find the time at which the distance d is between the ships is minimal.

Answer 1

The 2 speeds form 2 sides of a triangle.
The total distance is the square root of:
(30 -15t)^2 + (10t)^2
A little shuffling of this gives: 25*(13t^2 -36t +36)

Solving the quadratic gives greatest distance. My personal preference- find when dx/dt = 0 (calculus)
That means: 26t - 36 = 0 or t= 36/26,
or 1 and 10/26, or 1.385 hours, or 1 hour 23 minutes 5 seconds...

Final time = 2:23:05 PM

Answer 2

I'll put ship B at the origin, (0,0) at t = 0.

Then ship A is at (0, -30) at t = 0.

Then, since A is going north at 15 mph, A's position at any time would be:

(0, -30 + 15t)

And since ship B's heading west (negative x) at 10 mph, B's position at any time would be:

(-10t, 0)

The distance between them would be:

d(t) = √((0 + 10t)² + (-30 + 15t - 0)²)

But to avoid the nasty chain rule, you could just minimize the square of the distance:

d²(t) = 100t² + 900 - 900t + 225t²
d²(t) = 325t² - 900t + 900
d²'(t) = 650t - 900 = 0
650t = 900
t = 18/13

So they will be closest after 1 5/13 hours, if I didn't mess something up. (Please check!)

Since I agree with Alan, I think that's gonna be right.

5/13 of an hour is 23 minutes, 4.6 seconds.

So, since they started at 1pm, they will be closest at

2:23:04.6 pm.

Answer 3

ok. i drew something. draw exactly what you said, with dashed lines reresenting the way the ships are traveling.
since the ships are traveling away from each other, the distance keep increasing. therefore, the time is 1 PM at which the ships are the less amount of distance away from each other
UNLESS
you are talking about going back in time.
In this case, do the reverse. It creates a nice triangle before, in 12 noon. Solve for the hypotenuse, and you get about 18 miles. In this case, the ans is 12 noon.