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Ali 2
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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 }
-------------------------------
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 ?
( 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 }
-------------------------------
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 ?
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