Question

Particle A moves along the positive horizontal axis, and particle B along the graph of \(f(x)=-\sqrt{3}x,x\le 0\). At a certain time, A is at the point (5,0) and moving with speed 3 units/sec; and B is at a distance of 3 units from the origin and moving with speed 4 units/sec. At what rate is the distance between A and B changing?

Answer 1

Nope, too hard.

Answer 2

Don't be a quitter inky haha

Answer 3

Are the coordinates from which they start given????

Answer 4

I posted the question text in full, that's all the information I have.

Answer 5

I think the first thing to do is find what points on the line \(y=-\sqrt3 x\) are a distance of 3 from the origin.

Answer 6

Well, it's restricted to \(x\le0\) so it's just the one point, haven't solved for the exact point yet.

Answer 7

\(x^2+y^2=9\) and \(y=-\sqrt{3}x\), so if we solve that system we get... an imaginary answer. But we can actually change that to be \(y=\sqrt{3}x\) and then flip the sign later, and we get x=-3/2, and y is \(3\sqrt{3}/2\), right?

Answer 8

That looks right.

Answer 9

So we know the exact position but I still have no idea how to figure out a formula for the distance between A and B so as to have something to derive to determine a rate of change.

Answer 10

Now we want to find a parameterization for the lines.

Answer 11

So for particle A:\[\binom{5}{0}+t\binom{3}{0}\]

Answer 12

And for particle B it's more complicated.

Answer 13

Hrmm this seems almost like a vector problem, finding the y and x components of the vector B

Answer 14

Almost.

Answer 15

But we haven't done anything like that in this book so I don't think that's how he wants me to do it

Answer 16

Well, let's find the distance between the two point right now.

Answer 17

Right now the distance between them is 7

Answer 18

Can we find distance after a small time 'dt' and then find the distance again, and then find the rate of change of distance????

Answer 19

That would be the basic approach.

Answer 20

I guess we'll have to use that only because we have no other info as to where the particles are starting, do the have uniform velocity.....

Answer 21

is the answer 83/14 ????

Answer 22

I don't have the answer, it's not in the book.

Answer 23

We would want to find a relation between the time t and and the distance between the two particles where t=0 corresponds to where the points are located now.

Answer 24

Yep. I know that's what we need to do I just have no idea how to do it.

Answer 25

You could always just find the distance when A, B are just a wee bit farther along, and estimate from that XP

Answer 26

Haha I want to learn the right method for this, though. It's in a chapter about Differentiation techniques, sandwiched in between two problems on the Chain Rule. There's gotta be an easier way to do this I'm just missing it.

Answer 27

I'm far too tired to think.

Answer 28

It's a question of differentials. Consider the situation after time 'dt', the new positions of the particles would be:
A : (5 + 3 dt , 0)
B : (-1.5 - 2 dt , 3^(3/2)/2 + 2(3^.5) dt)
Now, find the distance between them.
You can neglect (dt)^2, since dt is very small.
The rate of change of distance is change in distance / dt.
Now, limit of the ratio as dt->0 should give you the answer, i guess, by the L'Hospitals Rule.

Answer 29

Alright, I did it by hand, after t they are at positions \((-3.5, 3.5\sqrt{3})\) and (8,0), and the distance between them is 13. So the distance changed by 6. The answer isn't 6, though, is it?

Answer 30

Probably not.

Answer 31

After time t, the positions should be in terms of t.....

Answer 32

Okay I actually didn't do after time t, I did after one second. After one second the distance increases by 6.

Answer 33

I guess we have to find the rate of change of distance at a particular instant, so we should use
d/dt (distance between A and B)

Answer 34

Thats where your differentiation will come in.....

Answer 35

we know the angle is 120 because tan(theta) = sqrt3
use law of cosines to find distance (c) in terms of lengths a and b
Note: cos(120) = -1/2
\[c = \sqrt{a^{2}+b^{2}+ab}\]
Now the goal is to find rate distance is changing or dc/dt
\[\frac{dc}{dt} = \frac{da}{dt}*\frac{dc}{da}+\frac{db}{dt}*\frac{dc}{db}\]
da/dt = 3
db/dt = 4
\[\frac{dc}{da}=\frac{2a+b}{2\sqrt{a^{2}+b^{2}+ab}}\]
\[\frac{dc}{db}=\frac{a+2b}{2\sqrt{a^{2}+b^{2}+ab}}\]
only thing left is plug in values for a,b which are the given lengths
a = 5
b = 3
this will give the rate the distance is changing at that moment in time
i believe Ashu was correct...83/14

hope this made sense