Rate of change of the magnitude of the displacement

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Homework Help Overview

The discussion revolves around the rate of change of the magnitude of displacement between two points moving in the xy-plane. The original poster presents two approaches to find the distance between the points and its rate of change, questioning which method is correct.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster describes two methods for calculating the rate of change of distance: one using the distance formula and the other using vector differentiation. They seek clarification on which approach yields the correct result.

Discussion Status

Some participants have provided insights into the nature of the calculations, with one suggesting that the first approach is preferable for finding the rate of change of distance. However, there is no explicit consensus on which method is definitively correct.

Contextual Notes

The discussion includes considerations of radial velocity in polar coordinates and the implications of motion, such as a point circling an origin while maintaining a fixed distance.

Ali 2
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If the displacement was given by [tex]\overrightarrow x (t)[/tex]
( i.e vector valued function ) .

Then the velocity is [tex]\overrightarrow v = \frac {d \overrightarrow x}{dt}[/tex]

The speed is the magnitude of v .


But ..

What is the derivative of the magintude of the displacement [tex]\frac {d \|\overrightarrow x\|}{dt}[/tex] ?




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 [tex]x=t^2[/tex]

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 :

[tex]r = \sqrt { x^2 + y^2 } = \sqrt { t^2 + t^4 }[/tex]

Thus simply we differentiate r with respect to t ::

[tex]\frac {dr}{dt} = \frac { 2t^3 + t } { \sqrt { 1 + t^2 }}[/tex]


The second ::

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

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

-------------------------------
Notice the first one is :

[tex]\frac {d \| \overrightarrow r \|}{dt}[/tex]

AND the second is


[tex]\left \| \frac { d \overrightarrow r } {dt} \right \|[/tex]


WHICH ONE IS THE RIGHT ANWER ?
 
Last edited:
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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.
 
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.
 
And that is what I thought about ..

Thanks :smile: ,
 

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