Taking a derivative of a magnitude of a vector












1












$begingroup$


I have found several questions including this as a step in the explanation, but have not been able to find an explicit explanation of how to take a derivative of a magnitude of a vector.



I am trying to take the derivative d/dt ||r'(t)|| but don't know how to address the magnitude signs.






multivariable-calculus vectors







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you have a formula for $r(t)$ and for $r'(t)$? If so get the formula for $||r'(t)||$ which will be a 1-var function of $t,$ then do derivative of that.
    $endgroup$
    – coffeemath
    Jan 4 at 21:34






  • 1




    $begingroup$
    What is your norm? The one derived from the dot product?
    $endgroup$
    – mathcounterexamples.net
    Jan 4 at 21:36










  • $begingroup$
    @mathcounterexamples.net r(t) is a path -- what would the norm derived from the dot product be?
    $endgroup$
    – icesk8er
    Jan 4 at 21:44
















1












$begingroup$


I have found several questions including this as a step in the explanation, but have not been able to find an explicit explanation of how to take a derivative of a magnitude of a vector.



I am trying to take the derivative d/dt ||r'(t)|| but don't know how to address the magnitude signs.






multivariable-calculus vectors







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you have a formula for $r(t)$ and for $r'(t)$? If so get the formula for $||r'(t)||$ which will be a 1-var function of $t,$ then do derivative of that.
    $endgroup$
    – coffeemath
    Jan 4 at 21:34






  • 1




    $begingroup$
    What is your norm? The one derived from the dot product?
    $endgroup$
    – mathcounterexamples.net
    Jan 4 at 21:36










  • $begingroup$
    @mathcounterexamples.net r(t) is a path -- what would the norm derived from the dot product be?
    $endgroup$
    – icesk8er
    Jan 4 at 21:44














1












1








1


0



$begingroup$


I have found several questions including this as a step in the explanation, but have not been able to find an explicit explanation of how to take a derivative of a magnitude of a vector.



I am trying to take the derivative d/dt ||r'(t)|| but don't know how to address the magnitude signs.






multivariable-calculus vectors







share|cite|improve this question









$endgroup$




I have found several questions including this as a step in the explanation, but have not been able to find an explicit explanation of how to take a derivative of a magnitude of a vector.



I am trying to take the derivative d/dt ||r'(t)|| but don't know how to address the magnitude signs.






multivariable-calculus vectors




multivariable-calculus vectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 21:28









icesk8ericesk8er

83




83












  • $begingroup$
    Do you have a formula for $r(t)$ and for $r'(t)$? If so get the formula for $||r'(t)||$ which will be a 1-var function of $t,$ then do derivative of that.
    $endgroup$
    – coffeemath
    Jan 4 at 21:34






  • 1




    $begingroup$
    What is your norm? The one derived from the dot product?
    $endgroup$
    – mathcounterexamples.net
    Jan 4 at 21:36










  • $begingroup$
    @mathcounterexamples.net r(t) is a path -- what would the norm derived from the dot product be?
    $endgroup$
    – icesk8er
    Jan 4 at 21:44


















  • $begingroup$
    Do you have a formula for $r(t)$ and for $r'(t)$? If so get the formula for $||r'(t)||$ which will be a 1-var function of $t,$ then do derivative of that.
    $endgroup$
    – coffeemath
    Jan 4 at 21:34






  • 1




    $begingroup$
    What is your norm? The one derived from the dot product?
    $endgroup$
    – mathcounterexamples.net
    Jan 4 at 21:36










  • $begingroup$
    @mathcounterexamples.net r(t) is a path -- what would the norm derived from the dot product be?
    $endgroup$
    – icesk8er
    Jan 4 at 21:44
















$begingroup$
Do you have a formula for $r(t)$ and for $r'(t)$? If so get the formula for $||r'(t)||$ which will be a 1-var function of $t,$ then do derivative of that.
$endgroup$
– coffeemath
Jan 4 at 21:34




$begingroup$
Do you have a formula for $r(t)$ and for $r'(t)$? If so get the formula for $||r'(t)||$ which will be a 1-var function of $t,$ then do derivative of that.
$endgroup$
– coffeemath
Jan 4 at 21:34




1




1




$begingroup$
What is your norm? The one derived from the dot product?
$endgroup$
– mathcounterexamples.net
Jan 4 at 21:36




$begingroup$
What is your norm? The one derived from the dot product?
$endgroup$
– mathcounterexamples.net
Jan 4 at 21:36












$begingroup$
@mathcounterexamples.net r(t) is a path -- what would the norm derived from the dot product be?
$endgroup$
– icesk8er
Jan 4 at 21:44




$begingroup$
@mathcounterexamples.net r(t) is a path -- what would the norm derived from the dot product be?
$endgroup$
– icesk8er
Jan 4 at 21:44










3 Answers
3






active

oldest

votes


















2












$begingroup$

I will assume that you are working with the euclidian norm and the dot product. Let $v(t)$ be a vector:
$$v(t) cdot v(t)=|v(t)|^2$$
And
$$frac{mathrm{d}|v(t)|^2}{mathrm{d}t}=2|v(t)|frac{mathrm{d}|v(t)|}{mathrm{d}t}$$
Which implies that
$$begin{align}
frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
&=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
&=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
&=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
&=frac{v(t) cdot v'(t)}{|v(t)|}
end{align}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (If anyone is unfamiliar with the notation, $v^prime(t) = dot{v}(t) = frac{d v(t)}{d t}$.)
    $endgroup$
    – Nominal Animal
    Jan 5 at 16:12



















1












$begingroup$

$$
eqalign{
& left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
& 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
& {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If the norm is derived from the dot product $langle cdot , cdot rangle$, then



    $$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you explain how you got that?
      $endgroup$
      – icesk8er
      Jan 4 at 21:52










    • $begingroup$
      Using the chain rule. A more simple way is to use the coordinates and the fact that for a vector $x=(x_1, dots, x_n)$ (in dimension $n$), you have $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
      $endgroup$
      – mathcounterexamples.net
      Jan 4 at 21:56













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    I will assume that you are working with the euclidian norm and the dot product. Let $v(t)$ be a vector:
    $$v(t) cdot v(t)=|v(t)|^2$$
    And
    $$frac{mathrm{d}|v(t)|^2}{mathrm{d}t}=2|v(t)|frac{mathrm{d}|v(t)|}{mathrm{d}t}$$
    Which implies that
    $$begin{align}
    frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
    &=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
    &=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
    &=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
    &=frac{v(t) cdot v'(t)}{|v(t)|}
    end{align}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      (If anyone is unfamiliar with the notation, $v^prime(t) = dot{v}(t) = frac{d v(t)}{d t}$.)
      $endgroup$
      – Nominal Animal
      Jan 5 at 16:12
















    2












    $begingroup$

    I will assume that you are working with the euclidian norm and the dot product. Let $v(t)$ be a vector:
    $$v(t) cdot v(t)=|v(t)|^2$$
    And
    $$frac{mathrm{d}|v(t)|^2}{mathrm{d}t}=2|v(t)|frac{mathrm{d}|v(t)|}{mathrm{d}t}$$
    Which implies that
    $$begin{align}
    frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
    &=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
    &=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
    &=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
    &=frac{v(t) cdot v'(t)}{|v(t)|}
    end{align}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      (If anyone is unfamiliar with the notation, $v^prime(t) = dot{v}(t) = frac{d v(t)}{d t}$.)
      $endgroup$
      – Nominal Animal
      Jan 5 at 16:12














    2












    2








    2





    $begingroup$

    I will assume that you are working with the euclidian norm and the dot product. Let $v(t)$ be a vector:
    $$v(t) cdot v(t)=|v(t)|^2$$
    And
    $$frac{mathrm{d}|v(t)|^2}{mathrm{d}t}=2|v(t)|frac{mathrm{d}|v(t)|}{mathrm{d}t}$$
    Which implies that
    $$begin{align}
    frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
    &=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
    &=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
    &=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
    &=frac{v(t) cdot v'(t)}{|v(t)|}
    end{align}$$






    share|cite|improve this answer









    $endgroup$



    I will assume that you are working with the euclidian norm and the dot product. Let $v(t)$ be a vector:
    $$v(t) cdot v(t)=|v(t)|^2$$
    And
    $$frac{mathrm{d}|v(t)|^2}{mathrm{d}t}=2|v(t)|frac{mathrm{d}|v(t)|}{mathrm{d}t}$$
    Which implies that
    $$begin{align}
    frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
    &=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
    &=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
    &=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
    &=frac{v(t) cdot v'(t)}{|v(t)|}
    end{align}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 at 21:57









    BotondBotond

    6,0632832




    6,0632832












    • $begingroup$
      (If anyone is unfamiliar with the notation, $v^prime(t) = dot{v}(t) = frac{d v(t)}{d t}$.)
      $endgroup$
      – Nominal Animal
      Jan 5 at 16:12


















    • $begingroup$
      (If anyone is unfamiliar with the notation, $v^prime(t) = dot{v}(t) = frac{d v(t)}{d t}$.)
      $endgroup$
      – Nominal Animal
      Jan 5 at 16:12
















    $begingroup$
    (If anyone is unfamiliar with the notation, $v^prime(t) = dot{v}(t) = frac{d v(t)}{d t}$.)
    $endgroup$
    – Nominal Animal
    Jan 5 at 16:12




    $begingroup$
    (If anyone is unfamiliar with the notation, $v^prime(t) = dot{v}(t) = frac{d v(t)}{d t}$.)
    $endgroup$
    – Nominal Animal
    Jan 5 at 16:12











    1












    $begingroup$

    $$
    eqalign{
    & left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
    & 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
    & {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$
      eqalign{
      & left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
      & 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
      & {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$
        eqalign{
        & left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
        & 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
        & {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
        $$






        share|cite|improve this answer









        $endgroup$



        $$
        eqalign{
        & left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
        & 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
        & {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 22:01









        G CabG Cab

        20.1k31340




        20.1k31340























            0












            $begingroup$

            If the norm is derived from the dot product $langle cdot , cdot rangle$, then



            $$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Could you explain how you got that?
              $endgroup$
              – icesk8er
              Jan 4 at 21:52










            • $begingroup$
              Using the chain rule. A more simple way is to use the coordinates and the fact that for a vector $x=(x_1, dots, x_n)$ (in dimension $n$), you have $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
              $endgroup$
              – mathcounterexamples.net
              Jan 4 at 21:56


















            0












            $begingroup$

            If the norm is derived from the dot product $langle cdot , cdot rangle$, then



            $$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Could you explain how you got that?
              $endgroup$
              – icesk8er
              Jan 4 at 21:52










            • $begingroup$
              Using the chain rule. A more simple way is to use the coordinates and the fact that for a vector $x=(x_1, dots, x_n)$ (in dimension $n$), you have $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
              $endgroup$
              – mathcounterexamples.net
              Jan 4 at 21:56
















            0












            0








            0





            $begingroup$

            If the norm is derived from the dot product $langle cdot , cdot rangle$, then



            $$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$






            share|cite|improve this answer









            $endgroup$



            If the norm is derived from the dot product $langle cdot , cdot rangle$, then



            $$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 21:50









            mathcounterexamples.netmathcounterexamples.net

            27k22158




            27k22158












            • $begingroup$
              Could you explain how you got that?
              $endgroup$
              – icesk8er
              Jan 4 at 21:52










            • $begingroup$
              Using the chain rule. A more simple way is to use the coordinates and the fact that for a vector $x=(x_1, dots, x_n)$ (in dimension $n$), you have $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
              $endgroup$
              – mathcounterexamples.net
              Jan 4 at 21:56




















            • $begingroup$
              Could you explain how you got that?
              $endgroup$
              – icesk8er
              Jan 4 at 21:52










            • $begingroup$
              Using the chain rule. A more simple way is to use the coordinates and the fact that for a vector $x=(x_1, dots, x_n)$ (in dimension $n$), you have $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
              $endgroup$
              – mathcounterexamples.net
              Jan 4 at 21:56


















            $begingroup$
            Could you explain how you got that?
            $endgroup$
            – icesk8er
            Jan 4 at 21:52




            $begingroup$
            Could you explain how you got that?
            $endgroup$
            – icesk8er
            Jan 4 at 21:52












            $begingroup$
            Using the chain rule. A more simple way is to use the coordinates and the fact that for a vector $x=(x_1, dots, x_n)$ (in dimension $n$), you have $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
            $endgroup$
            – mathcounterexamples.net
            Jan 4 at 21:56






            $begingroup$
            Using the chain rule. A more simple way is to use the coordinates and the fact that for a vector $x=(x_1, dots, x_n)$ (in dimension $n$), you have $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
            $endgroup$
            – mathcounterexamples.net
            Jan 4 at 21:56




















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            Måne

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            För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
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            Storängen

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            För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
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            VLT Carioca

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            VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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