Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour, a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day.

a. Start counting time with t =0 at noon and express the distance s between the ships as a function of t.
b. How rapidly was the distance between the ships changing at noon? One hour later?
c. The visibility that day was 5 nautical miles. Did the ships ever sight each other?
d. Graph s and ds/dt together as functions oft for -1 <= t<= 3, using different colors if possible. Compare the graphs and reconcile what you see with your answers in parts (b) and (c).
e. The graph of ds/dtlooks as if it might have a horizontal asymptote in the first quadrant. This in turn suggests that ds/dt approaches a limiting value as t →[infinity] . What is this value? What is its relation to the ships' individual speeds?

a) b.) -12knots, 8 knots c) No e)

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved and ship B has moved . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

b)

We want to find for t=0 and t=1

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other =

Since function is quadratic, concave up and has no real roots, we know that for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

So, lets check this limit:

Notice that:

=√(speed of ship A² + speed of ship B²)

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: