Taking a derivative of a magnitude of a vector
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
asked Jan 4 at 21:28
icesk8ericesk8er
83
New contributor
icesk8er is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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
asked Jan 4 at 21:28
icesk8ericesk8er
83
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
add a comment |
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
asked Jan 4 at 21:28
icesk8ericesk8er
83
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
asked Jan 4 at 21:28
icesk8ericesk8er
83
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.
asked Jan 4 at 21:28
icesk8ericesk8er
83
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.
asked Jan 4 at 21:28
icesk8ericesk8er
83
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.
asked Jan 4 at 21:28
icesk8ericesk8er
83
asked Jan 4 at 21:28
icesk8ericesk8er
83
83
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.
New contributor
icesk8er is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
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
$$begin{align}
frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
&=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
&=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
&=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
&=frac{v(t) cdot v'(t)}{|v(t)|}
end{align}$$
answered Jan 4 at 21:57
BotondBotond
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
add a comment |
$$
eqalign{
& left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
& 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
& {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
$$
answered Jan 4 at 22:01
G CabG Cab
18.1k31237
add a comment |
If the norm is derived from the dot product $langle cdot , cdot rangle$, then
$$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$
answered Jan 4 at 21:50
mathcounterexamples.netmathcounterexamples.net
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
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
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
$$begin{align}
frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
&=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
&=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
&=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
&=frac{v(t) cdot v'(t)}{|v(t)|}
end{align}$$
answered Jan 4 at 21:57
BotondBotond
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
add a comment |
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
$$begin{align}
frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
&=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
&=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
&=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
&=frac{v(t) cdot v'(t)}{|v(t)|}
end{align}$$
answered Jan 4 at 21:57
BotondBotond
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
add a comment |
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
$$begin{align}
frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
&=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
&=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
&=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
&=frac{v(t) cdot v'(t)}{|v(t)|}
end{align}$$
answered Jan 4 at 21:57
BotondBotond
5,5882732
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
$$begin{align}
frac{mathrm{d}|v(t)|}{mathrm{d}t}&=frac{1}{2|v(t)|}frac{mathrm{d}|v(t)|^2}{mathrm{d}t}\
&=frac{1}{2|v(t)|}frac{mathrm{d}(v(t) cdot v(t))}{mathrm{d}t}\
&=frac{1}{2|v(t)|}left(v(t) cdot v'(t)+v'(t) cdot v(t)right)\
&=frac{1}{2|v(t)|}left(2v(t) cdot v'(t)right)\
&=frac{v(t) cdot v'(t)}{|v(t)|}
end{align}$$
answered Jan 4 at 21:57
BotondBotond
5,5882732
answered Jan 4 at 21:57
BotondBotond
5,5882732
answered Jan 4 at 21:57
BotondBotond
5,5882732
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
add a comment |
(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
add a comment |
$$
eqalign{
& left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
& 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
& {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
$$
answered Jan 4 at 22:01
G CabG Cab
18.1k31237
add a comment |
$$
eqalign{
& left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
& 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
& {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
$$
answered Jan 4 at 22:01
G CabG Cab
18.1k31237
add a comment |
$$
eqalign{
& left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
& 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
& {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
$$
answered Jan 4 at 22:01
G CabG Cab
18.1k31237
$$
eqalign{
& left| {r(t)} right|^2 = r_{,x} (t)^2 + r_{,y} (t)^2 + cdots cr
& 2left| {r(t)} right|{d over {dt}}left| {r(t)} right| = 2left( {r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } right) cr
& {d over {dt}}left| {r(t)} right| = {{r_{,x} (t)r_{,x} '(t) + r_{,y} (t)r_{,y} '(t) + cdots } over {left| {r(t)} right|}} cr}
$$
answered Jan 4 at 22:01
G CabG Cab
18.1k31237
answered Jan 4 at 22:01
G CabG Cab
18.1k31237
answered Jan 4 at 22:01
G CabG Cab
18.1k31237
answered Jan 4 at 22:01
G CabG Cab
18.1k31237
18.1k31237
add a comment |
add a comment |
If the norm is derived from the dot product $langle cdot , cdot rangle$, then
$$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$
answered Jan 4 at 21:50
mathcounterexamples.netmathcounterexamples.net
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
add a comment |
If the norm is derived from the dot product $langle cdot , cdot rangle$, then
$$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$
answered Jan 4 at 21:50
mathcounterexamples.netmathcounterexamples.net
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
add a comment |
If the norm is derived from the dot product $langle cdot , cdot rangle$, then
$$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$
answered Jan 4 at 21:50
mathcounterexamples.netmathcounterexamples.net
25.3k21953
If the norm is derived from the dot product $langle cdot , cdot rangle$, then
$$frac{dVert r^prime(t) Vert}{dt} = frac{langle r^prime(t),r^{prime prime}(t)rangle}{Vert r^prime(t)Vert}.$$
answered Jan 4 at 21:50
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Jan 4 at 21:50
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Jan 4 at 21:50
mathcounterexamples.netmathcounterexamples.net
25.3k21953
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
add a comment |
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
add a comment |
icesk8er is a new contributor. Be nice, and check out our Code of Conduct.
icesk8er is a new contributor. Be nice, and check out our Code of Conduct.
icesk8er is a new contributor. Be nice, and check out our Code of Conduct.
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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