Alternative way to define the Normal Vector
$begingroup$
My calculus book defines the Normal vector in terms of the Tangent vector like so:
$overrightarrow{r}(t) =$ the original vector function
$$overrightarrow{T}(t) = frac{overrightarrow{r'}(t)}{|overrightarrow{r'}(t)|}$$
$$overrightarrow{N}(t) = frac{overrightarrow{T'}(t)}{|overrightarrow{T'}(t)|}$$
Since $overrightarrow{T}(t)$ is just $overrightarrow{r'}(t)$ normalized, can I say that
$$overrightarrow{N}(t) = frac{overrightarrow{r''}(t)}{|overrightarrow{r''}(t)|}$$
without changing anything?
calculus vectors
asked Dec 12 '18 at 16:35
J AtkinJ Atkin
1205
$endgroup$
add a comment |
$begingroup$
My calculus book defines the Normal vector in terms of the Tangent vector like so:
$overrightarrow{r}(t) =$ the original vector function
$$overrightarrow{T}(t) = frac{overrightarrow{r'}(t)}{|overrightarrow{r'}(t)|}$$
$$overrightarrow{N}(t) = frac{overrightarrow{T'}(t)}{|overrightarrow{T'}(t)|}$$
Since $overrightarrow{T}(t)$ is just $overrightarrow{r'}(t)$ normalized, can I say that
$$overrightarrow{N}(t) = frac{overrightarrow{r''}(t)}{|overrightarrow{r''}(t)|}$$
without changing anything?
calculus vectors
asked Dec 12 '18 at 16:35
J AtkinJ Atkin
1205
$endgroup$
$begingroup$
Is this because the quotient could be a function oft? So, in the specific case where the first derivative ofris a constant, I would be correct?
$endgroup$
– J Atkin
Dec 12 '18 at 16:55
add a comment |
$begingroup$
My calculus book defines the Normal vector in terms of the Tangent vector like so:
$overrightarrow{r}(t) =$ the original vector function
$$overrightarrow{T}(t) = frac{overrightarrow{r'}(t)}{|overrightarrow{r'}(t)|}$$
$$overrightarrow{N}(t) = frac{overrightarrow{T'}(t)}{|overrightarrow{T'}(t)|}$$
Since $overrightarrow{T}(t)$ is just $overrightarrow{r'}(t)$ normalized, can I say that
$$overrightarrow{N}(t) = frac{overrightarrow{r''}(t)}{|overrightarrow{r''}(t)|}$$
without changing anything?
calculus vectors
asked Dec 12 '18 at 16:35
J AtkinJ Atkin
1205
$endgroup$
My calculus book defines the Normal vector in terms of the Tangent vector like so:
$overrightarrow{r}(t) =$ the original vector function
$$overrightarrow{T}(t) = frac{overrightarrow{r'}(t)}{|overrightarrow{r'}(t)|}$$
$$overrightarrow{N}(t) = frac{overrightarrow{T'}(t)}{|overrightarrow{T'}(t)|}$$
Since $overrightarrow{T}(t)$ is just $overrightarrow{r'}(t)$ normalized, can I say that
$$overrightarrow{N}(t) = frac{overrightarrow{r''}(t)}{|overrightarrow{r''}(t)|}$$
without changing anything?
calculus vectors
calculus vectors
asked Dec 12 '18 at 16:35
J AtkinJ Atkin
1205
asked Dec 12 '18 at 16:35
J AtkinJ Atkin
1205
edited Dec 12 '18 at 16:44
J Atkin
asked Dec 12 '18 at 16:35
J AtkinJ Atkin
1205
asked Dec 12 '18 at 16:35
J AtkinJ Atkin
1205
asked Dec 12 '18 at 16:35
J AtkinJ Atkin
1205
1205
$begingroup$
Is this because the quotient could be a function oft? So, in the specific case where the first derivative ofris a constant, I would be correct?
$endgroup$
– J Atkin
Dec 12 '18 at 16:55
add a comment |
$begingroup$
Is this because the quotient could be a function oft? So, in the specific case where the first derivative ofris a constant, I would be correct?
$endgroup$
– J Atkin
Dec 12 '18 at 16:55
$begingroup$
Is this because the quotient could be a function of
t? So, in the specific case where the first derivative of r is a constant, I would be correct?$endgroup$
– J Atkin
Dec 12 '18 at 16:55
$begingroup$
Is this because the quotient could be a function of
t? So, in the specific case where the first derivative of r is a constant, I would be correct?$endgroup$
– J Atkin
Dec 12 '18 at 16:55
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The specific reason why this does not work in the simple way you attempted
is that $r''$ includes both the normal acceleration (which is in the direction you want) and the tangential acceleration (which is perpendicular to the direction you want).
You can separate out these two components by projecting $r''$ onto the tangential direction (that is, onto the direction of $r'$). This gives the tangential component of $r''$:
$$ frac{r' cdot r''}{lVert r'rVert^2} r'.$$
Subtracting this from the total acceleration, what remains is the normal acceleration:
$$ r'' - frac{r' cdot r''}{lVert r'rVert^2} r'.$$
And if we want a unit vector then we need to divide by the magnitude:
$$ frac{r'' - frac{r' cdot r''}{lVert r'rVert^2} r'}
{leftlVert r'' - frac{r' cdot r''}{lVert r'rVert^2} r' rightrVert}.$$
This may help to explain why you might want to use $T$ as an intermediate step.
answered Dec 12 '18 at 17:12
David KDavid K
55.5k345121
$endgroup$
add a comment |
$begingroup$
No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.
answered Dec 12 '18 at 16:45
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
What is the major change? AFAIK the only reason to use T over r' is that it is that it is of unit length, which I think doesn't change the derivative. But this may be a misconception.
$endgroup$
– J Atkin
Dec 12 '18 at 16:48
$begingroup$
When you differentiate a vectorial function $f$ such that each $f(x)$ has norm $1$, then, for each $x$, $f(x)$ and $f'(x)$ are orthogonal. This is not true anymore if you drotp the hypothesis that each $f(x)$ has norm $1$.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 17:08
add a comment |
$begingroup$
No. $mathbf{N}(t) = frac{mathbf{T}'(t)}{|mathbf{T}'(t)|} = frac{frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}}{|mathbf{T}'(t)|}$ (using the quotient rule of differentiation). Since the denominator is just a scalar, analyze the numerator: $frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2} = frac{mathbf{r}''(t)}{|mathbf{r}'(t)|} - mathbf{r}'(t)frac{(|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}$. The expression reduces to a vector in the direction of $mathbf{r}''(t)$ minus a vector in the direction of $mathbf{r}'(t)$. Since these vectors are not in general pointing in the same direction, the resulting vector will not generally be in the direction of $mathbf{r}''(t)$. Therefore, it won't be $frac{mathbf{r}''(t)}{|mathbf{r}''(t)|}$.
Note: by $(|mathbf{r}'(t)|)'(t)$ I mean the function $frac{d|mathbf{r}'(t)|}{dt}$.
answered Dec 12 '18 at 17:11
CyborgOctopusCyborgOctopus
1957
$endgroup$
add a comment |
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3 Answers
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3 Answers
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votes
$begingroup$
The specific reason why this does not work in the simple way you attempted
is that $r''$ includes both the normal acceleration (which is in the direction you want) and the tangential acceleration (which is perpendicular to the direction you want).
You can separate out these two components by projecting $r''$ onto the tangential direction (that is, onto the direction of $r'$). This gives the tangential component of $r''$:
$$ frac{r' cdot r''}{lVert r'rVert^2} r'.$$
Subtracting this from the total acceleration, what remains is the normal acceleration:
$$ r'' - frac{r' cdot r''}{lVert r'rVert^2} r'.$$
And if we want a unit vector then we need to divide by the magnitude:
$$ frac{r'' - frac{r' cdot r''}{lVert r'rVert^2} r'}
{leftlVert r'' - frac{r' cdot r''}{lVert r'rVert^2} r' rightrVert}.$$
This may help to explain why you might want to use $T$ as an intermediate step.
answered Dec 12 '18 at 17:12
David KDavid K
55.5k345121
$endgroup$
add a comment |
$begingroup$
The specific reason why this does not work in the simple way you attempted
is that $r''$ includes both the normal acceleration (which is in the direction you want) and the tangential acceleration (which is perpendicular to the direction you want).
You can separate out these two components by projecting $r''$ onto the tangential direction (that is, onto the direction of $r'$). This gives the tangential component of $r''$:
$$ frac{r' cdot r''}{lVert r'rVert^2} r'.$$
Subtracting this from the total acceleration, what remains is the normal acceleration:
$$ r'' - frac{r' cdot r''}{lVert r'rVert^2} r'.$$
And if we want a unit vector then we need to divide by the magnitude:
$$ frac{r'' - frac{r' cdot r''}{lVert r'rVert^2} r'}
{leftlVert r'' - frac{r' cdot r''}{lVert r'rVert^2} r' rightrVert}.$$
This may help to explain why you might want to use $T$ as an intermediate step.
answered Dec 12 '18 at 17:12
David KDavid K
55.5k345121
$endgroup$
add a comment |
$begingroup$
The specific reason why this does not work in the simple way you attempted
is that $r''$ includes both the normal acceleration (which is in the direction you want) and the tangential acceleration (which is perpendicular to the direction you want).
You can separate out these two components by projecting $r''$ onto the tangential direction (that is, onto the direction of $r'$). This gives the tangential component of $r''$:
$$ frac{r' cdot r''}{lVert r'rVert^2} r'.$$
Subtracting this from the total acceleration, what remains is the normal acceleration:
$$ r'' - frac{r' cdot r''}{lVert r'rVert^2} r'.$$
And if we want a unit vector then we need to divide by the magnitude:
$$ frac{r'' - frac{r' cdot r''}{lVert r'rVert^2} r'}
{leftlVert r'' - frac{r' cdot r''}{lVert r'rVert^2} r' rightrVert}.$$
This may help to explain why you might want to use $T$ as an intermediate step.
answered Dec 12 '18 at 17:12
David KDavid K
55.5k345121
$endgroup$
The specific reason why this does not work in the simple way you attempted
is that $r''$ includes both the normal acceleration (which is in the direction you want) and the tangential acceleration (which is perpendicular to the direction you want).
You can separate out these two components by projecting $r''$ onto the tangential direction (that is, onto the direction of $r'$). This gives the tangential component of $r''$:
$$ frac{r' cdot r''}{lVert r'rVert^2} r'.$$
Subtracting this from the total acceleration, what remains is the normal acceleration:
$$ r'' - frac{r' cdot r''}{lVert r'rVert^2} r'.$$
And if we want a unit vector then we need to divide by the magnitude:
$$ frac{r'' - frac{r' cdot r''}{lVert r'rVert^2} r'}
{leftlVert r'' - frac{r' cdot r''}{lVert r'rVert^2} r' rightrVert}.$$
This may help to explain why you might want to use $T$ as an intermediate step.
answered Dec 12 '18 at 17:12
David KDavid K
55.5k345121
answered Dec 12 '18 at 17:12
David KDavid K
55.5k345121
answered Dec 12 '18 at 17:12
David KDavid K
55.5k345121
answered Dec 12 '18 at 17:12
David KDavid K
55.5k345121
55.5k345121
add a comment |
add a comment |
$begingroup$
No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.
answered Dec 12 '18 at 16:45
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
What is the major change? AFAIK the only reason to use T over r' is that it is that it is of unit length, which I think doesn't change the derivative. But this may be a misconception.
$endgroup$
– J Atkin
Dec 12 '18 at 16:48
$begingroup$
When you differentiate a vectorial function $f$ such that each $f(x)$ has norm $1$, then, for each $x$, $f(x)$ and $f'(x)$ are orthogonal. This is not true anymore if you drotp the hypothesis that each $f(x)$ has norm $1$.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 17:08
add a comment |
$begingroup$
No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.
answered Dec 12 '18 at 16:45
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
What is the major change? AFAIK the only reason to use T over r' is that it is that it is of unit length, which I think doesn't change the derivative. But this may be a misconception.
$endgroup$
– J Atkin
Dec 12 '18 at 16:48
$begingroup$
When you differentiate a vectorial function $f$ such that each $f(x)$ has norm $1$, then, for each $x$, $f(x)$ and $f'(x)$ are orthogonal. This is not true anymore if you drotp the hypothesis that each $f(x)$ has norm $1$.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 17:08
add a comment |
$begingroup$
No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.
answered Dec 12 '18 at 16:45
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.
answered Dec 12 '18 at 16:45
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 12 '18 at 16:45
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 12 '18 at 16:45
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 12 '18 at 16:45
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
$begingroup$
What is the major change? AFAIK the only reason to use T over r' is that it is that it is of unit length, which I think doesn't change the derivative. But this may be a misconception.
$endgroup$
– J Atkin
Dec 12 '18 at 16:48
$begingroup$
When you differentiate a vectorial function $f$ such that each $f(x)$ has norm $1$, then, for each $x$, $f(x)$ and $f'(x)$ are orthogonal. This is not true anymore if you drotp the hypothesis that each $f(x)$ has norm $1$.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 17:08
add a comment |
$begingroup$
What is the major change? AFAIK the only reason to use T over r' is that it is that it is of unit length, which I think doesn't change the derivative. But this may be a misconception.
$endgroup$
– J Atkin
Dec 12 '18 at 16:48
$begingroup$
When you differentiate a vectorial function $f$ such that each $f(x)$ has norm $1$, then, for each $x$, $f(x)$ and $f'(x)$ are orthogonal. This is not true anymore if you drotp the hypothesis that each $f(x)$ has norm $1$.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 17:08
$begingroup$
What is the major change? AFAIK the only reason to use T over r' is that it is that it is of unit length, which I think doesn't change the derivative. But this may be a misconception.
$endgroup$
– J Atkin
Dec 12 '18 at 16:48
$begingroup$
What is the major change? AFAIK the only reason to use T over r' is that it is that it is of unit length, which I think doesn't change the derivative. But this may be a misconception.
$endgroup$
– J Atkin
Dec 12 '18 at 16:48
$begingroup$
When you differentiate a vectorial function $f$ such that each $f(x)$ has norm $1$, then, for each $x$, $f(x)$ and $f'(x)$ are orthogonal. This is not true anymore if you drotp the hypothesis that each $f(x)$ has norm $1$.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 17:08
$begingroup$
When you differentiate a vectorial function $f$ such that each $f(x)$ has norm $1$, then, for each $x$, $f(x)$ and $f'(x)$ are orthogonal. This is not true anymore if you drotp the hypothesis that each $f(x)$ has norm $1$.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 17:08
add a comment |
$begingroup$
No. $mathbf{N}(t) = frac{mathbf{T}'(t)}{|mathbf{T}'(t)|} = frac{frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}}{|mathbf{T}'(t)|}$ (using the quotient rule of differentiation). Since the denominator is just a scalar, analyze the numerator: $frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2} = frac{mathbf{r}''(t)}{|mathbf{r}'(t)|} - mathbf{r}'(t)frac{(|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}$. The expression reduces to a vector in the direction of $mathbf{r}''(t)$ minus a vector in the direction of $mathbf{r}'(t)$. Since these vectors are not in general pointing in the same direction, the resulting vector will not generally be in the direction of $mathbf{r}''(t)$. Therefore, it won't be $frac{mathbf{r}''(t)}{|mathbf{r}''(t)|}$.
Note: by $(|mathbf{r}'(t)|)'(t)$ I mean the function $frac{d|mathbf{r}'(t)|}{dt}$.
answered Dec 12 '18 at 17:11
CyborgOctopusCyborgOctopus
1957
$endgroup$
add a comment |
$begingroup$
No. $mathbf{N}(t) = frac{mathbf{T}'(t)}{|mathbf{T}'(t)|} = frac{frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}}{|mathbf{T}'(t)|}$ (using the quotient rule of differentiation). Since the denominator is just a scalar, analyze the numerator: $frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2} = frac{mathbf{r}''(t)}{|mathbf{r}'(t)|} - mathbf{r}'(t)frac{(|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}$. The expression reduces to a vector in the direction of $mathbf{r}''(t)$ minus a vector in the direction of $mathbf{r}'(t)$. Since these vectors are not in general pointing in the same direction, the resulting vector will not generally be in the direction of $mathbf{r}''(t)$. Therefore, it won't be $frac{mathbf{r}''(t)}{|mathbf{r}''(t)|}$.
Note: by $(|mathbf{r}'(t)|)'(t)$ I mean the function $frac{d|mathbf{r}'(t)|}{dt}$.
answered Dec 12 '18 at 17:11
CyborgOctopusCyborgOctopus
1957
$endgroup$
add a comment |
$begingroup$
No. $mathbf{N}(t) = frac{mathbf{T}'(t)}{|mathbf{T}'(t)|} = frac{frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}}{|mathbf{T}'(t)|}$ (using the quotient rule of differentiation). Since the denominator is just a scalar, analyze the numerator: $frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2} = frac{mathbf{r}''(t)}{|mathbf{r}'(t)|} - mathbf{r}'(t)frac{(|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}$. The expression reduces to a vector in the direction of $mathbf{r}''(t)$ minus a vector in the direction of $mathbf{r}'(t)$. Since these vectors are not in general pointing in the same direction, the resulting vector will not generally be in the direction of $mathbf{r}''(t)$. Therefore, it won't be $frac{mathbf{r}''(t)}{|mathbf{r}''(t)|}$.
Note: by $(|mathbf{r}'(t)|)'(t)$ I mean the function $frac{d|mathbf{r}'(t)|}{dt}$.
answered Dec 12 '18 at 17:11
CyborgOctopusCyborgOctopus
1957
$endgroup$
No. $mathbf{N}(t) = frac{mathbf{T}'(t)}{|mathbf{T}'(t)|} = frac{frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}}{|mathbf{T}'(t)|}$ (using the quotient rule of differentiation). Since the denominator is just a scalar, analyze the numerator: $frac{mathbf{r}''(t)|mathbf{r}'(t)| - mathbf{r}'(t) (|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2} = frac{mathbf{r}''(t)}{|mathbf{r}'(t)|} - mathbf{r}'(t)frac{(|mathbf{r}'|)'(t)}{|mathbf{r}'(t)|^2}$. The expression reduces to a vector in the direction of $mathbf{r}''(t)$ minus a vector in the direction of $mathbf{r}'(t)$. Since these vectors are not in general pointing in the same direction, the resulting vector will not generally be in the direction of $mathbf{r}''(t)$. Therefore, it won't be $frac{mathbf{r}''(t)}{|mathbf{r}''(t)|}$.
Note: by $(|mathbf{r}'(t)|)'(t)$ I mean the function $frac{d|mathbf{r}'(t)|}{dt}$.
answered Dec 12 '18 at 17:11
CyborgOctopusCyborgOctopus
1957
answered Dec 12 '18 at 17:11
CyborgOctopusCyborgOctopus
1957
answered Dec 12 '18 at 17:11
CyborgOctopusCyborgOctopus
1957
answered Dec 12 '18 at 17:11
CyborgOctopusCyborgOctopus
1957
1957
add a comment |
add a comment |
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$begingroup$
Is this because the quotient could be a function of
t? So, in the specific case where the first derivative ofris a constant, I would be correct?$endgroup$
– J Atkin
Dec 12 '18 at 16:55