Question

Two ships, A and B, are sailing away from point O along routes such that the angle AOB is 120 degrees. How fast is the distance between them changing when OA=8km, OB=6km, ship A is traveling at a rate of 20km/h and ship B at a rate of 30km/h? (use Cosine Law)

Answer 1

i got 148, is that correct?

Answer 2

that is c^2, take \(\sqrt {148}\)

Answer 3

oops, forgot the square root.

Answer 4

Next is a mess!! hehehe

Answer 5

A mess?! What do I do next?

Answer 6

I am sorry, I don't know how to take it easier. Let me think more. It's hard.

Answer 7

okay, take your time..

Answer 8

@SithsAndGiggles

Answer 9

$$
c^2=a^2+b^2-2ab\cos \theta\\
{d\over dt}c^2={d\over dt}{a^2+b^2-2ab\cos \theta}\\
2c{dc(t)\over dt}=2a{da\over dt}+2b {db\over dt}-2a {da\over dt}b\cos\theta -2b{db\over dt}a\cos \theta +2ab\sin \theta {d\theta (t)\over dt}
$$
We want to know \(\large {dc\over dt}\).

We know all the variables except \(\large {d\theta \over dt}\)

Answer 10

where did you get the sine in the differentiation equation?

Answer 11

Using http://en.wikipedia.org/wiki/Angular_velocity
We can find \(\large {d\theta \over dt}\)
$$
{d\theta \over dt}={\left |{da\over dt}\right |\sin \theta '\over |a|}={\left |{db\over dt}\right |\sin \theta '\over |b|}
$$
Where \(\theta '\) is the angle c makes with a or b make with c

Answer 12

The sine came form differentiating the cosine term because \(\theta=\theta(t) \) is a function of time
$$
{d\over dt}\cos\theta =-\sin \theta
$$

Answer 13

I didn't learn that yet...

Answer 14

So the solution will be
$$
2c{dc(t)\over dt}=2a{da\over dt}+2b {db\over dt}-2a {da\over dt}b\cos\theta -2b{db\over dt}a\cos \theta +2ab\sin \theta {d\theta (t)\over dt}\\
{dc(t)\over dt}={1\over c}{da\over dt} \left ( a-ab\cos\theta +ab\sin\theta{|\sin \theta '| \over a}\right)+{1\over c } {db\over dt} \left ( b-ba\cos \theta\right )\\
$$
Where
$$
a=8~km\\
b=6~km\\
c=\sqrt{148}~km\\
\theta = 120^\circ\\
{da\over dt}=30~km/hr\\
{db\over dt}=20~km/hr\\
$$
You'll just need to reapply the cosine law to get \(\theta '\)