# Homework Help: Rate of change of the magnitude of the displacement

1. Jun 18, 2007

### Ali 2

If the displacement was given by $$\overrightarrow x (t)$$
( i.e vector valued function ) .

Then the velocity is $$\overrightarrow v = \frac {d \overrightarrow x}{dt}$$

The speed is the magnitude of v .

But ..

What is the derivative of the magintude of the displacement $$\frac {d \|\overrightarrow x\|}{dt}$$ ?

To Clerify my question more ..

Suppose on the xy - plane , there are two point started moving from the origin , the first one in the y - axis direction and its displacement at time t is y =t
the other point moves in the x - axis direction , and its displacement is $$x=t^2$$

If we want to find the rate of change of the distnace between them as a function of time .. there are 2 approaches ..

The First : :

We can say that .. the distnace between them is :

$$r = \sqrt { x^2 + y^2 } = \sqrt { t^2 + t^4 }$$

Thus simply we differentiate r with respect to t ::

$$\frac {dr}{dt} = \frac { 2t^3 + t } { \sqrt { 1 + t^2 }}$$

The second ::

Consider .. the vector $$\overrightarrow x = t^2 \mathbf i$$ and $$\overrightarrow y = t \mathbf j$$ ..
Thus , $$\overrightarrow r = \overrightarrow x - \overrightarrow y = t^2 \mathbf i - t \mathbf j$$
The velocity is
$$\frac { d \overrightarrow r } {dt} = 2t \mathbf i - \mathbf j$$

Thus the rate of change of the distance between them is
$$\left \| \frac { d \overrightarrow r } {dt} \right \| = \sqrt { 4t^2 + 1 }$$

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Notice the first one is :

$$\frac {d \| \overrightarrow r \|}{dt}$$

AND the second is

$$\left \| \frac { d \overrightarrow r } {dt} \right \|$$

WHICH ONE IS THE RIGHT ANWER ?

Last edited: Jun 18, 2007
2. Jun 18, 2007

### Dick

d|r|/dt, in the case where r the distance from a point to a distinguished origin, is called a 'radial velocity' and is a component (in polar coordinates) of the velocity dr/dt.

3. Jun 18, 2007

### Dick

If you want to find the 'rate of change of the distance between them' use the first approach. E.g. if a point is circling the origin |dr/dt| is nonzero, yet the distance is fixed.

4. Jun 18, 2007

### Ali 2

And that is what I thought about ..

Thanks ,