Expert: Scotto - 4/23/2009

Question
Ship A is 15 miles east of O and moving west at 20 mi/h, while ship B is 60 mi south of O and moving north at 15mi/h.

Use analytical way of proving each.

A. Is the distance between the ships increasing or decreasing after 1 hour? Justify your answer.

B. Is the distance between the ships increasing or decreasing after 3 hours?  Justify your answer.

C. When are the ships closest to each other?

Answer
Ship A equation: 15-20t miles to the east where t is hours.
Ship B equation: 60-15t miles to the south where t is hours.

A. Ship A will barely be west of the point and
ship B will be a whole lot closer to the point,
so the distance between the ships will be less.

When t=0, the distance between the ships is √(15²+60²) = √(225+3600)
= √3825, which is almost 62 apart.
When t=1, the distance is....
well, A is 15-20 miles east, which is 5 miles west and
ship B is 60 miles south and has moved 15 north, so is 45 south.
this distance is √(5²+45²) = √(25+2025) = √2050,
which is between 45 and 46 miles.
As the equations demonstrate, the ships are closer after 1 hour
and appear to be moving closer as time goes on.

B. After three hours, A is now 15-20(3) = -45 east = 45 west.
B is now 60-15(3) = 5 miles south.
The distance between them is √(45²+5²) = { didn't we just do this? }
√2050, which is between 45 and 46 and appear to be moving farther apart as time goes on.

C. The distance between them is the equation
d(t) = (15-20t)² + (60-15t)².
We can see that d'(t) = 2(15-20t)(-20) + 2(60-15t)(-15).
Simplified, that is d'(t) = -600 + 800t -1800 + 450t.
Settting this to 0 gives -2400 + 1250t = 0.
We can see that 275t = 2400, to t = 2400/1250.
That's 48/25, or 1 23/25, or 1.92.
The places can be found by the equations.
Put t = 1.92  into the two equations.
That would be A: 15-20(1.92) = 11.6 miles east,
B: 60-15(1.92) = 31.2 miles south.