- #1
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 ?
( 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 ?
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