Suppose the function is in the form $vec{F}(vec{r}(t),vec{v}(t),t)$
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Suppose the function is in the form $vec{F}(vec{r}(t),vec{v}(t),t)$
I want to know that What type of function is ?
$F:mathbb{R}^3tomathbb{R^3}$ or ....
calculus
asked Nov 15 at 19:50
justin77
494412
|
show 1 more comment
up vote
2
down vote
favorite
Suppose the function is in the form $vec{F}(vec{r}(t),vec{v}(t),t)$
I want to know that What type of function is ?
$F:mathbb{R}^3tomathbb{R^3}$ or ....
calculus
asked Nov 15 at 19:50
justin77
494412
Is $vec{r}$ the abstract position vector or just it's representation?
– Botond
Nov 15 at 19:53
@Botond $vec{r}(t)$ is vector .
– justin77
Nov 15 at 19:55
I know that. But what kind of vector is it? An "abstract thing" or just a triplet of real numbers?
– Botond
Nov 15 at 19:56
@Botond a triplet of real numbers?
– justin77
Nov 15 at 19:57
Is it an element of $mathbb{R}^3$?
– Botond
Nov 15 at 19:58
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose the function is in the form $vec{F}(vec{r}(t),vec{v}(t),t)$
I want to know that What type of function is ?
$F:mathbb{R}^3tomathbb{R^3}$ or ....
calculus
asked Nov 15 at 19:50
justin77
494412
Suppose the function is in the form $vec{F}(vec{r}(t),vec{v}(t),t)$
I want to know that What type of function is ?
$F:mathbb{R}^3tomathbb{R^3}$ or ....
calculus
calculus
asked Nov 15 at 19:50
justin77
494412
asked Nov 15 at 19:50
justin77
494412
asked Nov 15 at 19:50
justin77
494412
asked Nov 15 at 19:50
justin77
494412
asked Nov 15 at 19:50
justin77
494412
494412
Is $vec{r}$ the abstract position vector or just it's representation?
– Botond
Nov 15 at 19:53
@Botond $vec{r}(t)$ is vector .
– justin77
Nov 15 at 19:55
I know that. But what kind of vector is it? An "abstract thing" or just a triplet of real numbers?
– Botond
Nov 15 at 19:56
@Botond a triplet of real numbers?
– justin77
Nov 15 at 19:57
Is it an element of $mathbb{R}^3$?
– Botond
Nov 15 at 19:58
|
show 1 more comment
Is $vec{r}$ the abstract position vector or just it's representation?
– Botond
Nov 15 at 19:53
@Botond $vec{r}(t)$ is vector .
– justin77
Nov 15 at 19:55
I know that. But what kind of vector is it? An "abstract thing" or just a triplet of real numbers?
– Botond
Nov 15 at 19:56
@Botond a triplet of real numbers?
– justin77
Nov 15 at 19:57
Is it an element of $mathbb{R}^3$?
– Botond
Nov 15 at 19:58
Is $vec{r}$ the abstract position vector or just it's representation?
– Botond
Nov 15 at 19:53
Is $vec{r}$ the abstract position vector or just it's representation?
– Botond
Nov 15 at 19:53
@Botond $vec{r}(t)$ is vector .
– justin77
Nov 15 at 19:55
@Botond $vec{r}(t)$ is vector .
– justin77
Nov 15 at 19:55
I know that. But what kind of vector is it? An "abstract thing" or just a triplet of real numbers?
– Botond
Nov 15 at 19:56
I know that. But what kind of vector is it? An "abstract thing" or just a triplet of real numbers?
– Botond
Nov 15 at 19:56
@Botond a triplet of real numbers?
– justin77
Nov 15 at 19:57
@Botond a triplet of real numbers?
– justin77
Nov 15 at 19:57
Is it an element of $mathbb{R}^3$?
– Botond
Nov 15 at 19:58
Is it an element of $mathbb{R}^3$?
– Botond
Nov 15 at 19:58
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
First, let's assume that $vec F$ is a real-valued vector field with three scalar components, $F_x$, $F_y$, and $F_z$, $vec r$ is the position vector in $mathbb{R}^3$, $vec v$ is the velocity vector in $mathbb{R}^3$, and $t$ represents time (a scalar)$.
Then, $vec F(vec r(t), vec v(t), t))$ is a mapping from $mathbb{R}^7$ to $mathbb{R}^3$. For example, in Cartesian coordinates we have
$$begin{align}
vec F(vec r(t), vec v(t), t))&=hat x F_x(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\&
+hat y F_y(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\
&+hat z F_z(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)
end{align}$$
which explicitly shows the dependence of $vec F$ on $7$ variables.
answered Nov 15 at 20:01
Mark Viola
129k1273170
Generally speaking, $vec{r} in mathbb{R}^3$ is not always true (and it's true for the other vectors as well), because it's just a representation.
– Botond
Nov 15 at 20:11
@Botond In general, you are correct. Please note that I stated assumptions and then made a conclusion. That said, in most applications in physics and engineering, $vec r$ represents the position vector in $mathbb{R}^3$. If we don't make that assumption, and instead assume that $vec rin mathbb{R}^m$ and $vec vinmathbb{R}^n$, and $vec F in mathbb{R}^p$, then the mapping is from $mathbb{R}^{m+n+1}$ to $mathbb{R}^p$. But I suspect that the confusion lies in the fact that $vec v(t)=vec r'(t)$ and the OP questions whether there are truly $7$ independent variables or just $4$..
– Mark Viola
Nov 15 at 20:50
I like this answer. If I may make a distinction (because this used to confuse me with "total" derivatives i.e. chain rule stuff) that the composition function $vec{F}(vec{r}(t), vec{v}(t), t)$ is a single variable function while the outside function $vec{F}$ depends on 7 variables
– DWade64
Nov 15 at 21:21
@MarkViola I was not talking about the dimension, but about the concept of the "position vector". The $mathbb{R}^3$ "form$ is just a representation, like in the linear operator - matrix case.
– Botond
Nov 15 at 21:27
@botond I don't understand your point. Just take $vec r$ to be three variables, without regard to physical interpretation if you like.
– Mark Viola
Nov 15 at 21:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
First, let's assume that $vec F$ is a real-valued vector field with three scalar components, $F_x$, $F_y$, and $F_z$, $vec r$ is the position vector in $mathbb{R}^3$, $vec v$ is the velocity vector in $mathbb{R}^3$, and $t$ represents time (a scalar)$.
Then, $vec F(vec r(t), vec v(t), t))$ is a mapping from $mathbb{R}^7$ to $mathbb{R}^3$. For example, in Cartesian coordinates we have
$$begin{align}
vec F(vec r(t), vec v(t), t))&=hat x F_x(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\&
+hat y F_y(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\
&+hat z F_z(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)
end{align}$$
which explicitly shows the dependence of $vec F$ on $7$ variables.
answered Nov 15 at 20:01
Mark Viola
129k1273170
Generally speaking, $vec{r} in mathbb{R}^3$ is not always true (and it's true for the other vectors as well), because it's just a representation.
– Botond
Nov 15 at 20:11
@Botond In general, you are correct. Please note that I stated assumptions and then made a conclusion. That said, in most applications in physics and engineering, $vec r$ represents the position vector in $mathbb{R}^3$. If we don't make that assumption, and instead assume that $vec rin mathbb{R}^m$ and $vec vinmathbb{R}^n$, and $vec F in mathbb{R}^p$, then the mapping is from $mathbb{R}^{m+n+1}$ to $mathbb{R}^p$. But I suspect that the confusion lies in the fact that $vec v(t)=vec r'(t)$ and the OP questions whether there are truly $7$ independent variables or just $4$..
– Mark Viola
Nov 15 at 20:50
I like this answer. If I may make a distinction (because this used to confuse me with "total" derivatives i.e. chain rule stuff) that the composition function $vec{F}(vec{r}(t), vec{v}(t), t)$ is a single variable function while the outside function $vec{F}$ depends on 7 variables
– DWade64
Nov 15 at 21:21
@MarkViola I was not talking about the dimension, but about the concept of the "position vector". The $mathbb{R}^3$ "form$ is just a representation, like in the linear operator - matrix case.
– Botond
Nov 15 at 21:27
@botond I don't understand your point. Just take $vec r$ to be three variables, without regard to physical interpretation if you like.
– Mark Viola
Nov 15 at 21:30
add a comment |
up vote
1
down vote
First, let's assume that $vec F$ is a real-valued vector field with three scalar components, $F_x$, $F_y$, and $F_z$, $vec r$ is the position vector in $mathbb{R}^3$, $vec v$ is the velocity vector in $mathbb{R}^3$, and $t$ represents time (a scalar)$.
Then, $vec F(vec r(t), vec v(t), t))$ is a mapping from $mathbb{R}^7$ to $mathbb{R}^3$. For example, in Cartesian coordinates we have
$$begin{align}
vec F(vec r(t), vec v(t), t))&=hat x F_x(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\&
+hat y F_y(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\
&+hat z F_z(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)
end{align}$$
which explicitly shows the dependence of $vec F$ on $7$ variables.
answered Nov 15 at 20:01
Mark Viola
129k1273170
Generally speaking, $vec{r} in mathbb{R}^3$ is not always true (and it's true for the other vectors as well), because it's just a representation.
– Botond
Nov 15 at 20:11
@Botond In general, you are correct. Please note that I stated assumptions and then made a conclusion. That said, in most applications in physics and engineering, $vec r$ represents the position vector in $mathbb{R}^3$. If we don't make that assumption, and instead assume that $vec rin mathbb{R}^m$ and $vec vinmathbb{R}^n$, and $vec F in mathbb{R}^p$, then the mapping is from $mathbb{R}^{m+n+1}$ to $mathbb{R}^p$. But I suspect that the confusion lies in the fact that $vec v(t)=vec r'(t)$ and the OP questions whether there are truly $7$ independent variables or just $4$..
– Mark Viola
Nov 15 at 20:50
I like this answer. If I may make a distinction (because this used to confuse me with "total" derivatives i.e. chain rule stuff) that the composition function $vec{F}(vec{r}(t), vec{v}(t), t)$ is a single variable function while the outside function $vec{F}$ depends on 7 variables
– DWade64
Nov 15 at 21:21
@MarkViola I was not talking about the dimension, but about the concept of the "position vector". The $mathbb{R}^3$ "form$ is just a representation, like in the linear operator - matrix case.
– Botond
Nov 15 at 21:27
@botond I don't understand your point. Just take $vec r$ to be three variables, without regard to physical interpretation if you like.
– Mark Viola
Nov 15 at 21:30
add a comment |
up vote
1
down vote
up vote
1
down vote
First, let's assume that $vec F$ is a real-valued vector field with three scalar components, $F_x$, $F_y$, and $F_z$, $vec r$ is the position vector in $mathbb{R}^3$, $vec v$ is the velocity vector in $mathbb{R}^3$, and $t$ represents time (a scalar)$.
Then, $vec F(vec r(t), vec v(t), t))$ is a mapping from $mathbb{R}^7$ to $mathbb{R}^3$. For example, in Cartesian coordinates we have
$$begin{align}
vec F(vec r(t), vec v(t), t))&=hat x F_x(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\&
+hat y F_y(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\
&+hat z F_z(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)
end{align}$$
which explicitly shows the dependence of $vec F$ on $7$ variables.
answered Nov 15 at 20:01
Mark Viola
129k1273170
First, let's assume that $vec F$ is a real-valued vector field with three scalar components, $F_x$, $F_y$, and $F_z$, $vec r$ is the position vector in $mathbb{R}^3$, $vec v$ is the velocity vector in $mathbb{R}^3$, and $t$ represents time (a scalar)$.
Then, $vec F(vec r(t), vec v(t), t))$ is a mapping from $mathbb{R}^7$ to $mathbb{R}^3$. For example, in Cartesian coordinates we have
$$begin{align}
vec F(vec r(t), vec v(t), t))&=hat x F_x(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\&
+hat y F_y(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)\\
&+hat z F_z(x(t),y(t),z(t), v_x(t),v_y(t),v_z(t),t)
end{align}$$
which explicitly shows the dependence of $vec F$ on $7$ variables.
answered Nov 15 at 20:01
Mark Viola
129k1273170
answered Nov 15 at 20:01
Mark Viola
129k1273170
answered Nov 15 at 20:01
Mark Viola
129k1273170
answered Nov 15 at 20:01
Mark Viola
129k1273170
129k1273170
Generally speaking, $vec{r} in mathbb{R}^3$ is not always true (and it's true for the other vectors as well), because it's just a representation.
– Botond
Nov 15 at 20:11
@Botond In general, you are correct. Please note that I stated assumptions and then made a conclusion. That said, in most applications in physics and engineering, $vec r$ represents the position vector in $mathbb{R}^3$. If we don't make that assumption, and instead assume that $vec rin mathbb{R}^m$ and $vec vinmathbb{R}^n$, and $vec F in mathbb{R}^p$, then the mapping is from $mathbb{R}^{m+n+1}$ to $mathbb{R}^p$. But I suspect that the confusion lies in the fact that $vec v(t)=vec r'(t)$ and the OP questions whether there are truly $7$ independent variables or just $4$..
– Mark Viola
Nov 15 at 20:50
I like this answer. If I may make a distinction (because this used to confuse me with "total" derivatives i.e. chain rule stuff) that the composition function $vec{F}(vec{r}(t), vec{v}(t), t)$ is a single variable function while the outside function $vec{F}$ depends on 7 variables
– DWade64
Nov 15 at 21:21
@MarkViola I was not talking about the dimension, but about the concept of the "position vector". The $mathbb{R}^3$ "form$ is just a representation, like in the linear operator - matrix case.
– Botond
Nov 15 at 21:27
@botond I don't understand your point. Just take $vec r$ to be three variables, without regard to physical interpretation if you like.
– Mark Viola
Nov 15 at 21:30
add a comment |
Generally speaking, $vec{r} in mathbb{R}^3$ is not always true (and it's true for the other vectors as well), because it's just a representation.
– Botond
Nov 15 at 20:11
@Botond In general, you are correct. Please note that I stated assumptions and then made a conclusion. That said, in most applications in physics and engineering, $vec r$ represents the position vector in $mathbb{R}^3$. If we don't make that assumption, and instead assume that $vec rin mathbb{R}^m$ and $vec vinmathbb{R}^n$, and $vec F in mathbb{R}^p$, then the mapping is from $mathbb{R}^{m+n+1}$ to $mathbb{R}^p$. But I suspect that the confusion lies in the fact that $vec v(t)=vec r'(t)$ and the OP questions whether there are truly $7$ independent variables or just $4$..
– Mark Viola
Nov 15 at 20:50
I like this answer. If I may make a distinction (because this used to confuse me with "total" derivatives i.e. chain rule stuff) that the composition function $vec{F}(vec{r}(t), vec{v}(t), t)$ is a single variable function while the outside function $vec{F}$ depends on 7 variables
– DWade64
Nov 15 at 21:21
@MarkViola I was not talking about the dimension, but about the concept of the "position vector". The $mathbb{R}^3$ "form$ is just a representation, like in the linear operator - matrix case.
– Botond
Nov 15 at 21:27
@botond I don't understand your point. Just take $vec r$ to be three variables, without regard to physical interpretation if you like.
– Mark Viola
Nov 15 at 21:30
Generally speaking, $vec{r} in mathbb{R}^3$ is not always true (and it's true for the other vectors as well), because it's just a representation.
– Botond
Nov 15 at 20:11
Generally speaking, $vec{r} in mathbb{R}^3$ is not always true (and it's true for the other vectors as well), because it's just a representation.
– Botond
Nov 15 at 20:11
@Botond In general, you are correct. Please note that I stated assumptions and then made a conclusion. That said, in most applications in physics and engineering, $vec r$ represents the position vector in $mathbb{R}^3$. If we don't make that assumption, and instead assume that $vec rin mathbb{R}^m$ and $vec vinmathbb{R}^n$, and $vec F in mathbb{R}^p$, then the mapping is from $mathbb{R}^{m+n+1}$ to $mathbb{R}^p$. But I suspect that the confusion lies in the fact that $vec v(t)=vec r'(t)$ and the OP questions whether there are truly $7$ independent variables or just $4$..
– Mark Viola
Nov 15 at 20:50
@Botond In general, you are correct. Please note that I stated assumptions and then made a conclusion. That said, in most applications in physics and engineering, $vec r$ represents the position vector in $mathbb{R}^3$. If we don't make that assumption, and instead assume that $vec rin mathbb{R}^m$ and $vec vinmathbb{R}^n$, and $vec F in mathbb{R}^p$, then the mapping is from $mathbb{R}^{m+n+1}$ to $mathbb{R}^p$. But I suspect that the confusion lies in the fact that $vec v(t)=vec r'(t)$ and the OP questions whether there are truly $7$ independent variables or just $4$..
– Mark Viola
Nov 15 at 20:50
I like this answer. If I may make a distinction (because this used to confuse me with "total" derivatives i.e. chain rule stuff) that the composition function $vec{F}(vec{r}(t), vec{v}(t), t)$ is a single variable function while the outside function $vec{F}$ depends on 7 variables
– DWade64
Nov 15 at 21:21
I like this answer. If I may make a distinction (because this used to confuse me with "total" derivatives i.e. chain rule stuff) that the composition function $vec{F}(vec{r}(t), vec{v}(t), t)$ is a single variable function while the outside function $vec{F}$ depends on 7 variables
– DWade64
Nov 15 at 21:21
@MarkViola I was not talking about the dimension, but about the concept of the "position vector". The $mathbb{R}^3$ "form$ is just a representation, like in the linear operator - matrix case.
– Botond
Nov 15 at 21:27
@MarkViola I was not talking about the dimension, but about the concept of the "position vector". The $mathbb{R}^3$ "form$ is just a representation, like in the linear operator - matrix case.
– Botond
Nov 15 at 21:27
@botond I don't understand your point. Just take $vec r$ to be three variables, without regard to physical interpretation if you like.
– Mark Viola
Nov 15 at 21:30
@botond I don't understand your point. Just take $vec r$ to be three variables, without regard to physical interpretation if you like.
– Mark Viola
Nov 15 at 21:30
add a comment |
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Is $vec{r}$ the abstract position vector or just it's representation?
– Botond
Nov 15 at 19:53
@Botond $vec{r}(t)$ is vector .
– justin77
Nov 15 at 19:55
I know that. But what kind of vector is it? An "abstract thing" or just a triplet of real numbers?
– Botond
Nov 15 at 19:56
@Botond a triplet of real numbers?
– justin77
Nov 15 at 19:57
Is it an element of $mathbb{R}^3$?
– Botond
Nov 15 at 19:58