Mixing
Why is the infinitesimal increment of a vector $R$ not necessarily collinear with $R$?
$begingroup$
Under the topic of differentiation of vectors, it says:
an infinitesimal increment dR of a vector R, does not need to be colinear with vector R.
But why? A vector has magnitude and direction. If you make an increase to that vector in a different direction than the original vector, even if it is infinitesimal, the resultant direction changes. So why or how can they say this?
vector-spaces vectors vector-analysis
edited Jan 5 at 9:33
6005
36.1k751125
asked Jan 5 at 9:16
user3600630user3600630
111
$endgroup$
add a comment |
$begingroup$
Under the topic of differentiation of vectors, it says:
an infinitesimal increment dR of a vector R, does not need to be colinear with vector R.
But why? A vector has magnitude and direction. If you make an increase to that vector in a different direction than the original vector, even if it is infinitesimal, the resultant direction changes. So why or how can they say this?
vector-spaces vectors vector-analysis
edited Jan 5 at 9:33
6005
36.1k751125
asked Jan 5 at 9:16
user3600630user3600630
111
$endgroup$
$begingroup$
Why should the vector stay along the same line after an increment $dbf R$?
$endgroup$
– Shubham Johri
Jan 5 at 9:25
add a comment |
$begingroup$
Under the topic of differentiation of vectors, it says:
an infinitesimal increment dR of a vector R, does not need to be colinear with vector R.
But why? A vector has magnitude and direction. If you make an increase to that vector in a different direction than the original vector, even if it is infinitesimal, the resultant direction changes. So why or how can they say this?
vector-spaces vectors vector-analysis
edited Jan 5 at 9:33
6005
36.1k751125
asked Jan 5 at 9:16
user3600630user3600630
111
$endgroup$
Under the topic of differentiation of vectors, it says:
an infinitesimal increment dR of a vector R, does not need to be colinear with vector R.
But why? A vector has magnitude and direction. If you make an increase to that vector in a different direction than the original vector, even if it is infinitesimal, the resultant direction changes. So why or how can they say this?
vector-spaces vectors vector-analysis
vector-spaces vectors vector-analysis
edited Jan 5 at 9:33
6005
36.1k751125
asked Jan 5 at 9:16
user3600630user3600630
111
edited Jan 5 at 9:33
6005
36.1k751125
asked Jan 5 at 9:16
user3600630user3600630
111
edited Jan 5 at 9:33
6005
36.1k751125
edited Jan 5 at 9:33
6005
36.1k751125
edited Jan 5 at 9:33
6005
36.1k751125
36.1k751125
asked Jan 5 at 9:16
user3600630user3600630
111
asked Jan 5 at 9:16
user3600630user3600630
111
asked Jan 5 at 9:16
user3600630user3600630
111
111
$begingroup$
Why should the vector stay along the same line after an increment $dbf R$?
$endgroup$
– Shubham Johri
Jan 5 at 9:25
add a comment |
$begingroup$
Why should the vector stay along the same line after an increment $dbf R$?
$endgroup$
– Shubham Johri
Jan 5 at 9:25
$begingroup$
Why should the vector stay along the same line after an increment $dbf R$?
$endgroup$
– Shubham Johri
Jan 5 at 9:25
$begingroup$
Why should the vector stay along the same line after an increment $dbf R$?
$endgroup$
– Shubham Johri
Jan 5 at 9:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The picture of $R$ and $dR$ is like this:
Here, $R$ is the original vector. $dR$ is the change in $R$ to get the new, slightly different vector.
Notice that $R$ and $dR$ are not collinear -- they could be pointing in completely different directions.
answered Jan 5 at 9:31
60056005
36.1k751125
$endgroup$
add a comment |
$begingroup$
Note that there are two ways to alter a vector: by changing its magnitude or by changing its direction. Thus, the differential increment $dbf R$ could be collinear with $bf R$ and change its magnitude, not its direction, or it could be non-collinear with $bf R$ and bring about a change in its direction and optionally, its magnitude.
answered Jan 5 at 9:22
Shubham JohriShubham Johri
4,920717
$endgroup$
add a comment |
$begingroup$
Perhaps the following will help you understand. It is how I, definitely not a mathematician, first grasped this.
Let's say we have a point somewhere on the unit circle, rotating at angular velocity $omega$, crossing the positive $x$ axis at time $t = 0$. (You can basically ignore $omega$, or assume it is equal to $1$ radian per $1$ time unit; $omega = 1$. I use it because it is easier to first grasp the derivatives when the parametric unit is time; it is just more intuitive this way.)
We can define its location using a pair of functions, one for each coordinate,
$$bbox{begin{cases}
x(t) = cos omega t \
y(t) = sin omega t \
end{cases}} tag{1}label{NA1}$$
or as a vector-valued function,
$$bbox{mathbf{p}(t) = left [ begin{matrix} cosomega t \ sinomega t end{matrix} right ]} tag{2}label{NA2}$$
These are completely equivalent representations.
Do note that notations vary. I personally usually use $vec{p}$ for vectors, and $mathbf{p}$ for matrices; and a parameter list (like $(t)$ above) following a scalar, vector, or matrix makes it a scalar-valued, vector-valued, or matrix-valued function. I find it clearest to me. Others use different notations, so always check what kind of notation is used first, before delving into any equations.
If we take the derivatives of $eqref{NA1}$, we obtain the tangent; i.e. the velocity, or the speed and direction, of where the point is headed at any specific moment in time:
$$bbox{ begin{cases}
displaystyle frac{d x(t)}{d t} = dot{x}(t) = -omega sin omega t \
displaystyle frac{d y(t)}{d t} = dot{y}(t) = omega cos omega t \
end{cases} } tag{3}label{NA3}$$
In vector calculus notation, doing the same to $eqref{NA2}$, yields the exact same result, of course:
$$bbox{ frac{d mathbf{p}(t)}{d t} = left [ begin{matrix}
-omega sin omega t \
omega cos omega t end{matrix} right ] } tag{4}label{NA4}$$
These obviously also correspond to the "infinitesimal increments" OP mentioned, via the definition of a scalar derivative. Contrast $eqref{NA1}$ and $eqref{NA3}$ to the definition of a derivative:
$$bbox{ frac{d x(t)}{d t} = lim_{tau to 0} frac{x(t + tau) - x(t)}{tau} }$$
Sidetrack: To obtain the acceleration of the point, we do the second derivative,
$$bbox{ begin{cases}
frac{d^2 x(t)}{d t^2} = ddot{x}(t) = -omega^2 cos omega t \
frac{d^2 y(t)}{d t^2} = ddot{y}(t) = -omega^2 sin omega t \
end{cases} } quad iff quad bbox{ frac{d^2 mathbf{p}(t)}{d t^2} =
left [ begin{matrix} -omega^2 cos omega t \ -omega^2 sin omega t end{matrix} right ]}$$
As an example, if $mathbf{p}(t)$ describes the location of a spacecraft at time $t$, then its derivative describes its velocity vector, and its second derivative the acceleration.
The next step is to realize that this applies no matter how the location of the point is parametrized. The parameter does not have to be time; it can be anything. You can even use more than one parameter (although then we talk about partial derivatives with respect to one specific parameter, or about gradient, instead of derivatives).
That means that there is no need for the derivative, or the infinitesimal change in the vector, to be collinear -- or indeed depend in any way! -- with the vector value itself: they describe completely different things. In the case of a point moving as a function of time, they describe the position (the vector itself) and its velocity (the derivative of the vector), respectively.
If we think of a game where the vector represents the ethical/moral alignment of your character, its derivative represents the effects of their actions and choices on their alignment or ethical/moral stance.
They are obviously related; but represent different, related things.
If we look at $eqref{NA2}$ and $eqref{NA4}$, we find that in the circular case
$$bbox{mathbf{p}(t) perp frac{d mathbf{p}(t)}{d t}}$$
because
$$bbox{mathbf{p}(t) cdot frac{d mathbf{p}(t)}{d t} = -omega (cos omega t)(sin omega t) + omega (sin omega t)(cos omega t) = 0}$$
i.e., the derivative is always perpendicular to the position vector.
answered Jan 5 at 15:57
Nominal AnimalNominal Animal
6,9552517
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The picture of $R$ and $dR$ is like this:
Here, $R$ is the original vector. $dR$ is the change in $R$ to get the new, slightly different vector.
Notice that $R$ and $dR$ are not collinear -- they could be pointing in completely different directions.
answered Jan 5 at 9:31
60056005
36.1k751125
$endgroup$
add a comment |
$begingroup$
The picture of $R$ and $dR$ is like this:
Here, $R$ is the original vector. $dR$ is the change in $R$ to get the new, slightly different vector.
Notice that $R$ and $dR$ are not collinear -- they could be pointing in completely different directions.
answered Jan 5 at 9:31
60056005
36.1k751125
$endgroup$
add a comment |
$begingroup$
The picture of $R$ and $dR$ is like this:
Here, $R$ is the original vector. $dR$ is the change in $R$ to get the new, slightly different vector.
Notice that $R$ and $dR$ are not collinear -- they could be pointing in completely different directions.
answered Jan 5 at 9:31
60056005
36.1k751125
$endgroup$
The picture of $R$ and $dR$ is like this:
Here, $R$ is the original vector. $dR$ is the change in $R$ to get the new, slightly different vector.
Notice that $R$ and $dR$ are not collinear -- they could be pointing in completely different directions.
answered Jan 5 at 9:31
60056005
36.1k751125
answered Jan 5 at 9:31
60056005
36.1k751125
answered Jan 5 at 9:31
60056005
36.1k751125
answered Jan 5 at 9:31
60056005
36.1k751125
36.1k751125
add a comment |
add a comment |
$begingroup$
Note that there are two ways to alter a vector: by changing its magnitude or by changing its direction. Thus, the differential increment $dbf R$ could be collinear with $bf R$ and change its magnitude, not its direction, or it could be non-collinear with $bf R$ and bring about a change in its direction and optionally, its magnitude.
answered Jan 5 at 9:22
Shubham JohriShubham Johri
4,920717
$endgroup$
add a comment |
$begingroup$
Note that there are two ways to alter a vector: by changing its magnitude or by changing its direction. Thus, the differential increment $dbf R$ could be collinear with $bf R$ and change its magnitude, not its direction, or it could be non-collinear with $bf R$ and bring about a change in its direction and optionally, its magnitude.
answered Jan 5 at 9:22
Shubham JohriShubham Johri
4,920717
$endgroup$
add a comment |
$begingroup$
Note that there are two ways to alter a vector: by changing its magnitude or by changing its direction. Thus, the differential increment $dbf R$ could be collinear with $bf R$ and change its magnitude, not its direction, or it could be non-collinear with $bf R$ and bring about a change in its direction and optionally, its magnitude.
answered Jan 5 at 9:22
Shubham JohriShubham Johri
4,920717
$endgroup$
Note that there are two ways to alter a vector: by changing its magnitude or by changing its direction. Thus, the differential increment $dbf R$ could be collinear with $bf R$ and change its magnitude, not its direction, or it could be non-collinear with $bf R$ and bring about a change in its direction and optionally, its magnitude.
answered Jan 5 at 9:22
Shubham JohriShubham Johri
4,920717
answered Jan 5 at 9:22
Shubham JohriShubham Johri
4,920717
answered Jan 5 at 9:22
Shubham JohriShubham Johri
4,920717
answered Jan 5 at 9:22
Shubham JohriShubham Johri
4,920717
4,920717
add a comment |
add a comment |
$begingroup$
Perhaps the following will help you understand. It is how I, definitely not a mathematician, first grasped this.
Let's say we have a point somewhere on the unit circle, rotating at angular velocity $omega$, crossing the positive $x$ axis at time $t = 0$. (You can basically ignore $omega$, or assume it is equal to $1$ radian per $1$ time unit; $omega = 1$. I use it because it is easier to first grasp the derivatives when the parametric unit is time; it is just more intuitive this way.)
We can define its location using a pair of functions, one for each coordinate,
$$bbox{begin{cases}
x(t) = cos omega t \
y(t) = sin omega t \
end{cases}} tag{1}label{NA1}$$
or as a vector-valued function,
$$bbox{mathbf{p}(t) = left [ begin{matrix} cosomega t \ sinomega t end{matrix} right ]} tag{2}label{NA2}$$
These are completely equivalent representations.
Do note that notations vary. I personally usually use $vec{p}$ for vectors, and $mathbf{p}$ for matrices; and a parameter list (like $(t)$ above) following a scalar, vector, or matrix makes it a scalar-valued, vector-valued, or matrix-valued function. I find it clearest to me. Others use different notations, so always check what kind of notation is used first, before delving into any equations.
If we take the derivatives of $eqref{NA1}$, we obtain the tangent; i.e. the velocity, or the speed and direction, of where the point is headed at any specific moment in time:
$$bbox{ begin{cases}
displaystyle frac{d x(t)}{d t} = dot{x}(t) = -omega sin omega t \
displaystyle frac{d y(t)}{d t} = dot{y}(t) = omega cos omega t \
end{cases} } tag{3}label{NA3}$$
In vector calculus notation, doing the same to $eqref{NA2}$, yields the exact same result, of course:
$$bbox{ frac{d mathbf{p}(t)}{d t} = left [ begin{matrix}
-omega sin omega t \
omega cos omega t end{matrix} right ] } tag{4}label{NA4}$$
These obviously also correspond to the "infinitesimal increments" OP mentioned, via the definition of a scalar derivative. Contrast $eqref{NA1}$ and $eqref{NA3}$ to the definition of a derivative:
$$bbox{ frac{d x(t)}{d t} = lim_{tau to 0} frac{x(t + tau) - x(t)}{tau} }$$
Sidetrack: To obtain the acceleration of the point, we do the second derivative,
$$bbox{ begin{cases}
frac{d^2 x(t)}{d t^2} = ddot{x}(t) = -omega^2 cos omega t \
frac{d^2 y(t)}{d t^2} = ddot{y}(t) = -omega^2 sin omega t \
end{cases} } quad iff quad bbox{ frac{d^2 mathbf{p}(t)}{d t^2} =
left [ begin{matrix} -omega^2 cos omega t \ -omega^2 sin omega t end{matrix} right ]}$$
As an example, if $mathbf{p}(t)$ describes the location of a spacecraft at time $t$, then its derivative describes its velocity vector, and its second derivative the acceleration.
The next step is to realize that this applies no matter how the location of the point is parametrized. The parameter does not have to be time; it can be anything. You can even use more than one parameter (although then we talk about partial derivatives with respect to one specific parameter, or about gradient, instead of derivatives).
That means that there is no need for the derivative, or the infinitesimal change in the vector, to be collinear -- or indeed depend in any way! -- with the vector value itself: they describe completely different things. In the case of a point moving as a function of time, they describe the position (the vector itself) and its velocity (the derivative of the vector), respectively.
If we think of a game where the vector represents the ethical/moral alignment of your character, its derivative represents the effects of their actions and choices on their alignment or ethical/moral stance.
They are obviously related; but represent different, related things.
If we look at $eqref{NA2}$ and $eqref{NA4}$, we find that in the circular case
$$bbox{mathbf{p}(t) perp frac{d mathbf{p}(t)}{d t}}$$
because
$$bbox{mathbf{p}(t) cdot frac{d mathbf{p}(t)}{d t} = -omega (cos omega t)(sin omega t) + omega (sin omega t)(cos omega t) = 0}$$
i.e., the derivative is always perpendicular to the position vector.
answered Jan 5 at 15:57
Nominal AnimalNominal Animal
6,9552517
$endgroup$
add a comment |
$begingroup$
Perhaps the following will help you understand. It is how I, definitely not a mathematician, first grasped this.
Let's say we have a point somewhere on the unit circle, rotating at angular velocity $omega$, crossing the positive $x$ axis at time $t = 0$. (You can basically ignore $omega$, or assume it is equal to $1$ radian per $1$ time unit; $omega = 1$. I use it because it is easier to first grasp the derivatives when the parametric unit is time; it is just more intuitive this way.)
We can define its location using a pair of functions, one for each coordinate,
$$bbox{begin{cases}
x(t) = cos omega t \
y(t) = sin omega t \
end{cases}} tag{1}label{NA1}$$
or as a vector-valued function,
$$bbox{mathbf{p}(t) = left [ begin{matrix} cosomega t \ sinomega t end{matrix} right ]} tag{2}label{NA2}$$
These are completely equivalent representations.
Do note that notations vary. I personally usually use $vec{p}$ for vectors, and $mathbf{p}$ for matrices; and a parameter list (like $(t)$ above) following a scalar, vector, or matrix makes it a scalar-valued, vector-valued, or matrix-valued function. I find it clearest to me. Others use different notations, so always check what kind of notation is used first, before delving into any equations.
If we take the derivatives of $eqref{NA1}$, we obtain the tangent; i.e. the velocity, or the speed and direction, of where the point is headed at any specific moment in time:
$$bbox{ begin{cases}
displaystyle frac{d x(t)}{d t} = dot{x}(t) = -omega sin omega t \
displaystyle frac{d y(t)}{d t} = dot{y}(t) = omega cos omega t \
end{cases} } tag{3}label{NA3}$$
In vector calculus notation, doing the same to $eqref{NA2}$, yields the exact same result, of course:
$$bbox{ frac{d mathbf{p}(t)}{d t} = left [ begin{matrix}
-omega sin omega t \
omega cos omega t end{matrix} right ] } tag{4}label{NA4}$$
These obviously also correspond to the "infinitesimal increments" OP mentioned, via the definition of a scalar derivative. Contrast $eqref{NA1}$ and $eqref{NA3}$ to the definition of a derivative:
$$bbox{ frac{d x(t)}{d t} = lim_{tau to 0} frac{x(t + tau) - x(t)}{tau} }$$
Sidetrack: To obtain the acceleration of the point, we do the second derivative,
$$bbox{ begin{cases}
frac{d^2 x(t)}{d t^2} = ddot{x}(t) = -omega^2 cos omega t \
frac{d^2 y(t)}{d t^2} = ddot{y}(t) = -omega^2 sin omega t \
end{cases} } quad iff quad bbox{ frac{d^2 mathbf{p}(t)}{d t^2} =
left [ begin{matrix} -omega^2 cos omega t \ -omega^2 sin omega t end{matrix} right ]}$$
As an example, if $mathbf{p}(t)$ describes the location of a spacecraft at time $t$, then its derivative describes its velocity vector, and its second derivative the acceleration.
The next step is to realize that this applies no matter how the location of the point is parametrized. The parameter does not have to be time; it can be anything. You can even use more than one parameter (although then we talk about partial derivatives with respect to one specific parameter, or about gradient, instead of derivatives).
That means that there is no need for the derivative, or the infinitesimal change in the vector, to be collinear -- or indeed depend in any way! -- with the vector value itself: they describe completely different things. In the case of a point moving as a function of time, they describe the position (the vector itself) and its velocity (the derivative of the vector), respectively.
If we think of a game where the vector represents the ethical/moral alignment of your character, its derivative represents the effects of their actions and choices on their alignment or ethical/moral stance.
They are obviously related; but represent different, related things.
If we look at $eqref{NA2}$ and $eqref{NA4}$, we find that in the circular case
$$bbox{mathbf{p}(t) perp frac{d mathbf{p}(t)}{d t}}$$
because
$$bbox{mathbf{p}(t) cdot frac{d mathbf{p}(t)}{d t} = -omega (cos omega t)(sin omega t) + omega (sin omega t)(cos omega t) = 0}$$
i.e., the derivative is always perpendicular to the position vector.
answered Jan 5 at 15:57
Nominal AnimalNominal Animal
6,9552517
$endgroup$
add a comment |
$begingroup$
Perhaps the following will help you understand. It is how I, definitely not a mathematician, first grasped this.
Let's say we have a point somewhere on the unit circle, rotating at angular velocity $omega$, crossing the positive $x$ axis at time $t = 0$. (You can basically ignore $omega$, or assume it is equal to $1$ radian per $1$ time unit; $omega = 1$. I use it because it is easier to first grasp the derivatives when the parametric unit is time; it is just more intuitive this way.)
We can define its location using a pair of functions, one for each coordinate,
$$bbox{begin{cases}
x(t) = cos omega t \
y(t) = sin omega t \
end{cases}} tag{1}label{NA1}$$
or as a vector-valued function,
$$bbox{mathbf{p}(t) = left [ begin{matrix} cosomega t \ sinomega t end{matrix} right ]} tag{2}label{NA2}$$
These are completely equivalent representations.
Do note that notations vary. I personally usually use $vec{p}$ for vectors, and $mathbf{p}$ for matrices; and a parameter list (like $(t)$ above) following a scalar, vector, or matrix makes it a scalar-valued, vector-valued, or matrix-valued function. I find it clearest to me. Others use different notations, so always check what kind of notation is used first, before delving into any equations.
If we take the derivatives of $eqref{NA1}$, we obtain the tangent; i.e. the velocity, or the speed and direction, of where the point is headed at any specific moment in time:
$$bbox{ begin{cases}
displaystyle frac{d x(t)}{d t} = dot{x}(t) = -omega sin omega t \
displaystyle frac{d y(t)}{d t} = dot{y}(t) = omega cos omega t \
end{cases} } tag{3}label{NA3}$$
In vector calculus notation, doing the same to $eqref{NA2}$, yields the exact same result, of course:
$$bbox{ frac{d mathbf{p}(t)}{d t} = left [ begin{matrix}
-omega sin omega t \
omega cos omega t end{matrix} right ] } tag{4}label{NA4}$$
These obviously also correspond to the "infinitesimal increments" OP mentioned, via the definition of a scalar derivative. Contrast $eqref{NA1}$ and $eqref{NA3}$ to the definition of a derivative:
$$bbox{ frac{d x(t)}{d t} = lim_{tau to 0} frac{x(t + tau) - x(t)}{tau} }$$
Sidetrack: To obtain the acceleration of the point, we do the second derivative,
$$bbox{ begin{cases}
frac{d^2 x(t)}{d t^2} = ddot{x}(t) = -omega^2 cos omega t \
frac{d^2 y(t)}{d t^2} = ddot{y}(t) = -omega^2 sin omega t \
end{cases} } quad iff quad bbox{ frac{d^2 mathbf{p}(t)}{d t^2} =
left [ begin{matrix} -omega^2 cos omega t \ -omega^2 sin omega t end{matrix} right ]}$$
As an example, if $mathbf{p}(t)$ describes the location of a spacecraft at time $t$, then its derivative describes its velocity vector, and its second derivative the acceleration.
The next step is to realize that this applies no matter how the location of the point is parametrized. The parameter does not have to be time; it can be anything. You can even use more than one parameter (although then we talk about partial derivatives with respect to one specific parameter, or about gradient, instead of derivatives).
That means that there is no need for the derivative, or the infinitesimal change in the vector, to be collinear -- or indeed depend in any way! -- with the vector value itself: they describe completely different things. In the case of a point moving as a function of time, they describe the position (the vector itself) and its velocity (the derivative of the vector), respectively.
If we think of a game where the vector represents the ethical/moral alignment of your character, its derivative represents the effects of their actions and choices on their alignment or ethical/moral stance.
They are obviously related; but represent different, related things.
If we look at $eqref{NA2}$ and $eqref{NA4}$, we find that in the circular case
$$bbox{mathbf{p}(t) perp frac{d mathbf{p}(t)}{d t}}$$
because
$$bbox{mathbf{p}(t) cdot frac{d mathbf{p}(t)}{d t} = -omega (cos omega t)(sin omega t) + omega (sin omega t)(cos omega t) = 0}$$
i.e., the derivative is always perpendicular to the position vector.
answered Jan 5 at 15:57
Nominal AnimalNominal Animal
6,9552517
$endgroup$
Perhaps the following will help you understand. It is how I, definitely not a mathematician, first grasped this.
Let's say we have a point somewhere on the unit circle, rotating at angular velocity $omega$, crossing the positive $x$ axis at time $t = 0$. (You can basically ignore $omega$, or assume it is equal to $1$ radian per $1$ time unit; $omega = 1$. I use it because it is easier to first grasp the derivatives when the parametric unit is time; it is just more intuitive this way.)
We can define its location using a pair of functions, one for each coordinate,
$$bbox{begin{cases}
x(t) = cos omega t \
y(t) = sin omega t \
end{cases}} tag{1}label{NA1}$$
or as a vector-valued function,
$$bbox{mathbf{p}(t) = left [ begin{matrix} cosomega t \ sinomega t end{matrix} right ]} tag{2}label{NA2}$$
These are completely equivalent representations.
Do note that notations vary. I personally usually use $vec{p}$ for vectors, and $mathbf{p}$ for matrices; and a parameter list (like $(t)$ above) following a scalar, vector, or matrix makes it a scalar-valued, vector-valued, or matrix-valued function. I find it clearest to me. Others use different notations, so always check what kind of notation is used first, before delving into any equations.
If we take the derivatives of $eqref{NA1}$, we obtain the tangent; i.e. the velocity, or the speed and direction, of where the point is headed at any specific moment in time:
$$bbox{ begin{cases}
displaystyle frac{d x(t)}{d t} = dot{x}(t) = -omega sin omega t \
displaystyle frac{d y(t)}{d t} = dot{y}(t) = omega cos omega t \
end{cases} } tag{3}label{NA3}$$
In vector calculus notation, doing the same to $eqref{NA2}$, yields the exact same result, of course:
$$bbox{ frac{d mathbf{p}(t)}{d t} = left [ begin{matrix}
-omega sin omega t \
omega cos omega t end{matrix} right ] } tag{4}label{NA4}$$
These obviously also correspond to the "infinitesimal increments" OP mentioned, via the definition of a scalar derivative. Contrast $eqref{NA1}$ and $eqref{NA3}$ to the definition of a derivative:
$$bbox{ frac{d x(t)}{d t} = lim_{tau to 0} frac{x(t + tau) - x(t)}{tau} }$$
Sidetrack: To obtain the acceleration of the point, we do the second derivative,
$$bbox{ begin{cases}
frac{d^2 x(t)}{d t^2} = ddot{x}(t) = -omega^2 cos omega t \
frac{d^2 y(t)}{d t^2} = ddot{y}(t) = -omega^2 sin omega t \
end{cases} } quad iff quad bbox{ frac{d^2 mathbf{p}(t)}{d t^2} =
left [ begin{matrix} -omega^2 cos omega t \ -omega^2 sin omega t end{matrix} right ]}$$
As an example, if $mathbf{p}(t)$ describes the location of a spacecraft at time $t$, then its derivative describes its velocity vector, and its second derivative the acceleration.
The next step is to realize that this applies no matter how the location of the point is parametrized. The parameter does not have to be time; it can be anything. You can even use more than one parameter (although then we talk about partial derivatives with respect to one specific parameter, or about gradient, instead of derivatives).
That means that there is no need for the derivative, or the infinitesimal change in the vector, to be collinear -- or indeed depend in any way! -- with the vector value itself: they describe completely different things. In the case of a point moving as a function of time, they describe the position (the vector itself) and its velocity (the derivative of the vector), respectively.
If we think of a game where the vector represents the ethical/moral alignment of your character, its derivative represents the effects of their actions and choices on their alignment or ethical/moral stance.
They are obviously related; but represent different, related things.
If we look at $eqref{NA2}$ and $eqref{NA4}$, we find that in the circular case
$$bbox{mathbf{p}(t) perp frac{d mathbf{p}(t)}{d t}}$$
because
$$bbox{mathbf{p}(t) cdot frac{d mathbf{p}(t)}{d t} = -omega (cos omega t)(sin omega t) + omega (sin omega t)(cos omega t) = 0}$$
i.e., the derivative is always perpendicular to the position vector.
answered Jan 5 at 15:57
Nominal AnimalNominal Animal
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answered Jan 5 at 15:57
Nominal AnimalNominal Animal
6,9552517
answered Jan 5 at 15:57
Nominal AnimalNominal Animal
6,9552517
answered Jan 5 at 15:57
Nominal AnimalNominal Animal
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$begingroup$
Why should the vector stay along the same line after an increment $dbf R$?
$endgroup$
– Shubham Johri
Jan 5 at 9:25