Contradicting solutions to a basic movement problem.

[Adgorn]

Homework Statement

Hi, everyone. I came across a basic calculus problem concerning movement of 2 problems, I've attempted to solve it using vector analysis and got 1 answer, and then solved it with differentiation and got a different answer. I'll show a version of it I made which is a bit more simplified to make things easier since even then the same problem occurs:

"Particle A moves along the positive horizontal axis, and Particle B moves along the positive vertical axis. At a certain time, the A is at the point $(5,0)$ and moving with speed 3 units/sec; and B is at the point $(0,3)$ and moving with speed 4 units/sec. At what rate is the distance between A and B changing?"

Homework Equations

Pythagorean theorem, chain rule.

The Attempt at a Solution

Intuitively, this is just a "riverboat problem", the rate of change of the distance between A and B at a certain moment is just their speed relative to each other at that moment, and does not depend on their location. To get this relative speed you just have to add the velocity vectors. I have the feeling that this intuition is the fault in my calculation, although I don't realist why, which would explain why the first method produces a wrong answer:

Method 1, vectors: The particles move away from each other with a horizontal speed of 3 units/sec and a vertical speed of 4 units/sec. Thus, the speed at which they move away from each other is $\sqrt {3^2+4^2}=5$ units/sec.

Method 2, differentiation: Let the position of A be $(a(t),0)$, the position of B be $(0,b(t))$, the distance between A and B be $d(t)$ and the specified time of the problem be $t_0$. Then $a(t_0)=5$, $a'(t_0)=3$, $b(t_0)=3$ and $b'(t_0)=4.$
At any moment $t$, $d(t)^2=a(t)^2+b(t)^2$, so $d(t_0)=\sqrt {25+9}=\sqrt {34}$. Differentiating the 2 sides of the equation yields
$2d(t)d'(t)=2a(t)a'(t)+2b(t)b'(t)$ or $d'(t)=\frac {2a(t)a'(t)+2b(t)b'(t)} {2d(t)}$, inserting the values at $t_0$ yields $d'(t)=\frac {30+24} {2\sqrt {34}}=\frac {27} {\sqrt {34}}$.

The book uses the 2nd method, so I assume something is wrong with the 1st method, probably something very basic that I looked over. The equations of the 2nd method seem to indicate that the relative velocity does depend on the location of the particles, which is something I don't quite understand. Clarification would be very appreciated.

 

[mjc123]

The velocity vector is not in the same direction as the distance vector, so its magnitude doesn't directly represent the rate at which the distance is changing.

 

[Abhishek11235]

The wrong concept is related to direction. In first case,we found relative velocity as seen by first observer. What question says is find relative velocity in the direction of rod. And,indeed you get result. The velocity vector components along the direction of rod are :

$$4 \cos\theta $$ and $$ 3\sin\theta$$. The relative velocity is:

$$4\cos\theta + 3\sin\theta$$. After substituting required values,we get the same answer as in second method

 

[Adgorn]

The velocity vector is not in the same direction as the distance vector, so its magnitude doesn't directly represent the rate at which the distance is changing.

You'll have to forgive me for my slow understanding, I haven't actually got into vector analysis at this point, but I used it here since it matches my (evidently wrong) intuition about the problem. I'll explain my intuition:

Both particles move at a constant velocity in a straight line, so from the point of view of each particle, the other particle must see the other one moving at a constant speed in a straight line. The rate of change of the distance between them is just the (absolute value of) velocity of one particle relative to the other one, say, the velocity of particle B from the point of view or particle A.

If particle B moves $x$ units away from particle A during $t$ seconds, the distance grows by $x$ units during $t$ seconds. Since particle A needs to see particle B moving at a constant speed in a straight line, the rate of change of the distance should be constant at all times, and can't depend on the position of the particles. Yet 2 immediate problems stem from this intuition.

1. The absolute value of the relative velocity of particle B with respect to A, 5 units/sec, is not the rate of change of the distance.

2. The actual rate of change of the distance does depend on the location of the particles, even though they are moving at the same speed and at the same direction. This seems to completely contradict my intuition about inertial reference frames. If the rate of change of the distance is how fast the particles move away from each other, and their frames is inertial, the rate at which they move away from each other needs to be constant too.

Obviously, I'm missing something basic here...

 

[PeroK]

You'll have to forgive me for my slow understanding, I haven't actually got into vector analysis at this point, but I used it here since it matches my (evidently wrong) intuition about the problem. I'll explain my intuition:Both particles move at a constant velocity in a straight line, so from the point of view of each particle, the other particle must see the other one moving at a constant speed in a straight line. The rate of change of the distance between them is just the (absolute value of) velocity of one particle relative to the other one, say, the velocity of particle B from the point of view or particle A.If particle B moves $x$ units away from particle A during $t$ seconds, the distance grows by $x$ units during $t$ seconds. Since particle A needs to see particle B moving at a constant speed in a straight line, the rate of change of the distance should be constant at all times, and can't depend on the position of the particles. Yet 2 immediate problems stem from this intuition.1. The absolute value of the relative velocity of particle B with respect to A, 5 units/sec, is not the rate of change of the distance.2. The actual rate of change of the distance does depend on the location of the particles, even though they are moving at the same speed and at the same direction. This seems to completely contradict my intuition about inertial reference frames. If the rate of change of the distance is how fast the particles move away from each other, and their frames is inertial, the rate at which they move away from each other needs to be constant tobviously, I'm missing something basic here...

Using your vector approach you got a constant answer of $5$ units per second. Let's look at this.At $t =0$ they are at $(0,3)$ and $(5,0)$, which are $\sqrt{34}$ units apart.

At $t =1$ they are at $(0,7)$ and $(8, 0)$, which are $\sqrt{119}$ apart.

If you do the calculation you'll see that these distances do not differ by $5$ units.

 

[Adgorn]

Using your vector approach you got a constant answer of $5$ units per second. Let's look at this.At $t =0$ they are at $(0,3)$ and $(5,0)$, which are $\sqrt{34}$ units apart.

At $t =1$ they are at $(0,7)$ and $(8, 0)$, which are $\sqrt{119}$ apart.

If you do the calculation you'll see that these distances do not differ by $5$ units.

Nevermind, I understood what my problem was. I thought that from the point of view of particle A, particle B was moving directly away from it (i.e. in a straight line that goes through the center of the reference frame), though it didn't, so the rate of change of the distance is not the relative speed of particle B. Thanks for the help.