Alternative way to define the Normal Vector












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 16:44







J Atkin

















asked Dec 12 '18 at 16:35









J AtkinJ Atkin

1205




1205












  • $begingroup$
    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
















$begingroup$
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










3 Answers
3






active

oldest

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1












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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



















    1












    $begingroup$

    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 $frac{d|mathbf{r}'(t)|}{dt}$.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
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      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 17:12









          David KDavid K

          55.5k345121




          55.5k345121























              1












              $begingroup$

              No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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
















              1












              $begingroup$

              No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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














              1












              1








              1





              $begingroup$

              No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.






              share|cite|improve this answer









              $endgroup$



              No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $overrightarrow{T}(t)$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              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


















              • $begingroup$
                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











              1












              $begingroup$

              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 $frac{d|mathbf{r}'(t)|}{dt}$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                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 $frac{d|mathbf{r}'(t)|}{dt}$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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 $frac{d|mathbf{r}'(t)|}{dt}$.






                  share|cite|improve this answer









                  $endgroup$



                  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 $frac{d|mathbf{r}'(t)|}{dt}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 17:11









                  CyborgOctopusCyborgOctopus

                  1957




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