Formula for flux accross a plane curve.
$begingroup$
If $vec{r}(t) = x(t) i + y(t) j$ be a simple curve C in the domain of a continuous vector field $vec{F} = F_1 i + F_2 j + F_3 k$. If $hat{n}$ is the outward pointing unit normal vector to the curve C then the flux accross the plane curve C is given by
$= int_{C} vec{F} . hat{n} dS$
Now if is given in my book that
$hat{n} = hat{T} times hat{k}$ ( if curve is moving in counter clockwise direction)
And
$hat{n} = hat{k} times hat{T}$ (if curve is moving in clockwise direction)
$( hat{T}$ is unit tangent vector to the curve$.)$
I didn't understand the direction of $hat{n}$. I know, it is a unit vector normal to the curve, but why it is represented by cross product of $hat{T}$ and $hat{k} ?$
vector-analysis
asked Dec 26 '18 at 4:10
MathsaddictMathsaddict
3619
$endgroup$
add a comment |
$begingroup$
If $vec{r}(t) = x(t) i + y(t) j$ be a simple curve C in the domain of a continuous vector field $vec{F} = F_1 i + F_2 j + F_3 k$. If $hat{n}$ is the outward pointing unit normal vector to the curve C then the flux accross the plane curve C is given by
$= int_{C} vec{F} . hat{n} dS$
Now if is given in my book that
$hat{n} = hat{T} times hat{k}$ ( if curve is moving in counter clockwise direction)
And
$hat{n} = hat{k} times hat{T}$ (if curve is moving in clockwise direction)
$( hat{T}$ is unit tangent vector to the curve$.)$
I didn't understand the direction of $hat{n}$. I know, it is a unit vector normal to the curve, but why it is represented by cross product of $hat{T}$ and $hat{k} ?$
vector-analysis
asked Dec 26 '18 at 4:10
MathsaddictMathsaddict
3619
$endgroup$
$begingroup$
I dont' have your book. But most likely the north vector in this case is pointing outward normal to $T$ (and $k$)
$endgroup$
– IAmNoOne
Dec 26 '18 at 4:35
$begingroup$
It seems that $hat n$ is the unit outward normal vector to the curve in $xy$ plane
$endgroup$
– Shubham Johri
Dec 26 '18 at 5:03
$begingroup$
$hat{n} = hat{T} times hat{k}$ represents $hat{n}$ is normal to $hat{k}$, but k is already normal to $xy$ plane !
$endgroup$
– Mathsaddict
Dec 26 '18 at 5:13
1
$begingroup$
Since $hat{n}$ lies in the $xy-$plane, it's orthogonal to $hat{k}$ and of course it's orthogonal to $T,$ so it's $pmhat{k}times T$
$endgroup$
– saulspatz
Dec 26 '18 at 6:10
add a comment |
$begingroup$
If $vec{r}(t) = x(t) i + y(t) j$ be a simple curve C in the domain of a continuous vector field $vec{F} = F_1 i + F_2 j + F_3 k$. If $hat{n}$ is the outward pointing unit normal vector to the curve C then the flux accross the plane curve C is given by
$= int_{C} vec{F} . hat{n} dS$
Now if is given in my book that
$hat{n} = hat{T} times hat{k}$ ( if curve is moving in counter clockwise direction)
And
$hat{n} = hat{k} times hat{T}$ (if curve is moving in clockwise direction)
$( hat{T}$ is unit tangent vector to the curve$.)$
I didn't understand the direction of $hat{n}$. I know, it is a unit vector normal to the curve, but why it is represented by cross product of $hat{T}$ and $hat{k} ?$
vector-analysis
asked Dec 26 '18 at 4:10
MathsaddictMathsaddict
3619
$endgroup$
If $vec{r}(t) = x(t) i + y(t) j$ be a simple curve C in the domain of a continuous vector field $vec{F} = F_1 i + F_2 j + F_3 k$. If $hat{n}$ is the outward pointing unit normal vector to the curve C then the flux accross the plane curve C is given by
$= int_{C} vec{F} . hat{n} dS$
Now if is given in my book that
$hat{n} = hat{T} times hat{k}$ ( if curve is moving in counter clockwise direction)
And
$hat{n} = hat{k} times hat{T}$ (if curve is moving in clockwise direction)
$( hat{T}$ is unit tangent vector to the curve$.)$
I didn't understand the direction of $hat{n}$. I know, it is a unit vector normal to the curve, but why it is represented by cross product of $hat{T}$ and $hat{k} ?$
vector-analysis
vector-analysis
asked Dec 26 '18 at 4:10
MathsaddictMathsaddict
3619
asked Dec 26 '18 at 4:10
MathsaddictMathsaddict
3619
asked Dec 26 '18 at 4:10
MathsaddictMathsaddict
3619
asked Dec 26 '18 at 4:10
MathsaddictMathsaddict
3619
asked Dec 26 '18 at 4:10
MathsaddictMathsaddict
3619
3619
$begingroup$
I dont' have your book. But most likely the north vector in this case is pointing outward normal to $T$ (and $k$)
$endgroup$
– IAmNoOne
Dec 26 '18 at 4:35
$begingroup$
It seems that $hat n$ is the unit outward normal vector to the curve in $xy$ plane
$endgroup$
– Shubham Johri
Dec 26 '18 at 5:03
$begingroup$
$hat{n} = hat{T} times hat{k}$ represents $hat{n}$ is normal to $hat{k}$, but k is already normal to $xy$ plane !
$endgroup$
– Mathsaddict
Dec 26 '18 at 5:13
1
$begingroup$
Since $hat{n}$ lies in the $xy-$plane, it's orthogonal to $hat{k}$ and of course it's orthogonal to $T,$ so it's $pmhat{k}times T$
$endgroup$
– saulspatz
Dec 26 '18 at 6:10
add a comment |
$begingroup$
I dont' have your book. But most likely the north vector in this case is pointing outward normal to $T$ (and $k$)
$endgroup$
– IAmNoOne
Dec 26 '18 at 4:35
$begingroup$
It seems that $hat n$ is the unit outward normal vector to the curve in $xy$ plane
$endgroup$
– Shubham Johri
Dec 26 '18 at 5:03
$begingroup$
$hat{n} = hat{T} times hat{k}$ represents $hat{n}$ is normal to $hat{k}$, but k is already normal to $xy$ plane !
$endgroup$
– Mathsaddict
Dec 26 '18 at 5:13
1
$begingroup$
Since $hat{n}$ lies in the $xy-$plane, it's orthogonal to $hat{k}$ and of course it's orthogonal to $T,$ so it's $pmhat{k}times T$
$endgroup$
– saulspatz
Dec 26 '18 at 6:10
$begingroup$
I dont' have your book. But most likely the north vector in this case is pointing outward normal to $T$ (and $k$)
$endgroup$
– IAmNoOne
Dec 26 '18 at 4:35
$begingroup$
I dont' have your book. But most likely the north vector in this case is pointing outward normal to $T$ (and $k$)
$endgroup$
– IAmNoOne
Dec 26 '18 at 4:35
$begingroup$
It seems that $hat n$ is the unit outward normal vector to the curve in $xy$ plane
$endgroup$
– Shubham Johri
Dec 26 '18 at 5:03
$begingroup$
It seems that $hat n$ is the unit outward normal vector to the curve in $xy$ plane
$endgroup$
– Shubham Johri
Dec 26 '18 at 5:03
$begingroup$
$hat{n} = hat{T} times hat{k}$ represents $hat{n}$ is normal to $hat{k}$, but k is already normal to $xy$ plane !
$endgroup$
– Mathsaddict
Dec 26 '18 at 5:13
$begingroup$
$hat{n} = hat{T} times hat{k}$ represents $hat{n}$ is normal to $hat{k}$, but k is already normal to $xy$ plane !
$endgroup$
– Mathsaddict
Dec 26 '18 at 5:13
1
1
$begingroup$
Since $hat{n}$ lies in the $xy-$plane, it's orthogonal to $hat{k}$ and of course it's orthogonal to $T,$ so it's $pmhat{k}times T$
$endgroup$
– saulspatz
Dec 26 '18 at 6:10
$begingroup$
Since $hat{n}$ lies in the $xy-$plane, it's orthogonal to $hat{k}$ and of course it's orthogonal to $T,$ so it's $pmhat{k}times T$
$endgroup$
– saulspatz
Dec 26 '18 at 6:10
add a comment |
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$begingroup$
I dont' have your book. But most likely the north vector in this case is pointing outward normal to $T$ (and $k$)
$endgroup$
– IAmNoOne
Dec 26 '18 at 4:35
$begingroup$
It seems that $hat n$ is the unit outward normal vector to the curve in $xy$ plane
$endgroup$
– Shubham Johri
Dec 26 '18 at 5:03
$begingroup$
$hat{n} = hat{T} times hat{k}$ represents $hat{n}$ is normal to $hat{k}$, but k is already normal to $xy$ plane !
$endgroup$
– Mathsaddict
Dec 26 '18 at 5:13
1
$begingroup$
Since $hat{n}$ lies in the $xy-$plane, it's orthogonal to $hat{k}$ and of course it's orthogonal to $T,$ so it's $pmhat{k}times T$
$endgroup$
– saulspatz
Dec 26 '18 at 6:10