Taking a derivative of a magnitude of a vector












1














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







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  • 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.
    – coffeemath
    Jan 4 at 21:34






  • 1




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










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
















1














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







New contributor




icesk8er is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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.
    – coffeemath
    Jan 4 at 21:34






  • 1




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










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














1












1








1


0





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







New contributor




icesk8er is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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







New contributor




icesk8er is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




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Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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asked Jan 4 at 21:28









icesk8ericesk8er

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  • 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.
    – coffeemath
    Jan 4 at 21:34






  • 1




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










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


















  • 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.
    – coffeemath
    Jan 4 at 21:34






  • 1




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










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
















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.
– coffeemath
Jan 4 at 21:34




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.
– coffeemath
Jan 4 at 21:34




1




1




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




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












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




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










3 Answers
3






active

oldest

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2














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





















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



















1














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






share|cite|improve this answer





























    0














    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





















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










    • 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 $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
      – 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














    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





















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
















    2














    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





















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














    2












    2








    2






    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












    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

    5,5882732




    5,5882732












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


















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
















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




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











    1














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






    share|cite|improve this answer


























      1














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






      share|cite|improve this answer
























        1












        1








        1






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






        share|cite|improve this answer












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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 22:01









        G CabG Cab

        18.1k31237




        18.1k31237























            0














            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





















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










            • 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 $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
              – mathcounterexamples.net
              Jan 4 at 21:56


















            0














            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





















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










            • 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 $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
              – mathcounterexamples.net
              Jan 4 at 21:56
















            0












            0








            0






            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












            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

            25.3k21953




            25.3k21953












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










            • 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 $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
              – mathcounterexamples.net
              Jan 4 at 21:56




















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










            • 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 $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
              – mathcounterexamples.net
              Jan 4 at 21:56


















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




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












            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 $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
            – mathcounterexamples.net
            Jan 4 at 21:56






            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 $Vert x Vert = sqrt{x_1^2 + dots +x_n^2}$
            – mathcounterexamples.net
            Jan 4 at 21:56












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