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?

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 [tex]12\dot t (n.m)[/tex] and ship B has moved [tex]8\dot t (n.m)[/tex]. 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:

[tex]\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}[/tex]

b)

We want to find [tex]\frac{ds}{dt}[/tex] for t=0 and t=1

[tex]\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots[/tex]

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 = [tex]s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0[/tex]

Since function [tex]f(x)=199-288x+208x^2[/tex] is quadratic, concave up and has no real roots, we know that [tex]199-288x+208x^2>0[/tex] 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

[tex]\lim_{t \to \infty} \frac{ds}{dt}<\infty[/tex]

So, lets check this limit:

[tex]\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}<\infty[/tex]

Notice that:

[tex]4\sqrt{13}=\sqrt{12^2+5^2}[/tex]=√(speed of ship A² + speed of ship B²)