Obtaining the norm of the derivative (of $mathbbR to mathbbR^3$ function) from a limit
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Let $mathbfv:mathbbR to mathbbR^n$ be a real vector valued function. I tried to prove that the following equality holds (sometimes I found it helpful when working with physics): $$leftlVert fracdmathbfvdt rightrVert = lim_Delta t to 0fracleftlVert Delta mathbfv rightrVertDelta t$$
Evaluating the limit gives something like:
$$lim_Delta t to 0fracsqrt Delta mathbfv_x^2 + Delta mathbfv_y^2 + Delta mathbfv_z^2 Delta t$$ but from there I've not been able to proced.
multivariable-calculus
asked Aug 7 at 16:12
marco21
10518
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up vote
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Let $mathbfv:mathbbR to mathbbR^n$ be a real vector valued function. I tried to prove that the following equality holds (sometimes I found it helpful when working with physics): $$leftlVert fracdmathbfvdt rightrVert = lim_Delta t to 0fracleftlVert Delta mathbfv rightrVertDelta t$$
Evaluating the limit gives something like:
$$lim_Delta t to 0fracsqrt Delta mathbfv_x^2 + Delta mathbfv_y^2 + Delta mathbfv_z^2 Delta t$$ but from there I've not been able to proced.
multivariable-calculus
asked Aug 7 at 16:12
marco21
10518
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $mathbfv:mathbbR to mathbbR^n$ be a real vector valued function. I tried to prove that the following equality holds (sometimes I found it helpful when working with physics): $$leftlVert fracdmathbfvdt rightrVert = lim_Delta t to 0fracleftlVert Delta mathbfv rightrVertDelta t$$
Evaluating the limit gives something like:
$$lim_Delta t to 0fracsqrt Delta mathbfv_x^2 + Delta mathbfv_y^2 + Delta mathbfv_z^2 Delta t$$ but from there I've not been able to proced.
multivariable-calculus
asked Aug 7 at 16:12
marco21
10518
Let $mathbfv:mathbbR to mathbbR^n$ be a real vector valued function. I tried to prove that the following equality holds (sometimes I found it helpful when working with physics): $$leftlVert fracdmathbfvdt rightrVert = lim_Delta t to 0fracleftlVert Delta mathbfv rightrVertDelta t$$
Evaluating the limit gives something like:
$$lim_Delta t to 0fracsqrt Delta mathbfv_x^2 + Delta mathbfv_y^2 + Delta mathbfv_z^2 Delta t$$ but from there I've not been able to proced.
multivariable-calculus
asked Aug 7 at 16:12
marco21
10518
edited Aug 9 at 21:35
asked Aug 7 at 16:12
marco21
10518
asked Aug 7 at 16:12
marco21
10518
asked Aug 7 at 16:12
marco21
10518
10518
add a comment |Â
add a comment |Â
1 Answer
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This just isn't true. Consider $$f(t) = beginpmatrix cos t \sin t endpmatrix$$
Observe that $|f|=1$ for all $t$, so $fracddt|f| = 0$, but
$$f'(t) = beginpmatrix-sin t \ cos tendpmatrix,$$
which still has norm 1.
For physical intuition, consider a car traveling in a circle around the origin with constant speed. Then the distance of the car from the origin doesn't change, so the derivative of the norm is 0, but the velocity has constant positive magnitude (since the magnitude of the velocity is by definition speed).
answered Aug 7 at 16:32
jgon
8,54611435
But $lim_Delta t to 0fraclVert Delta mathbf v rVertDelta t neq fracddtlVert Delta mathbf v rVert$. The thing that I have tried to prove was if $lim_Delta t to 0fraclVert Delta mathbfv rVertDelta t = leftlVert lim_Delta t to 0fracDelta mathbfvDelta t rightrVert$ holds (maybe assuming $lVert v rVert$ is constant)
– marco21
Aug 8 at 15:59
@marco oh yeah duh, I sorta took the question in the title as the actual question rather than properly reading the whole thing apparently. I'll edit this later or delete it if I don't have an answer
– jgon
Aug 8 at 16:14
1
In that case, this follows from the norm being continuous (I'll edit an explanation later)
– jgon
Aug 8 at 16:16
Thank you! I didn't notice it.
– marco21
Aug 9 at 21:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This just isn't true. Consider $$f(t) = beginpmatrix cos t \sin t endpmatrix$$
Observe that $|f|=1$ for all $t$, so $fracddt|f| = 0$, but
$$f'(t) = beginpmatrix-sin t \ cos tendpmatrix,$$
which still has norm 1.
For physical intuition, consider a car traveling in a circle around the origin with constant speed. Then the distance of the car from the origin doesn't change, so the derivative of the norm is 0, but the velocity has constant positive magnitude (since the magnitude of the velocity is by definition speed).
answered Aug 7 at 16:32
jgon
8,54611435
But $lim_Delta t to 0fraclVert Delta mathbf v rVertDelta t neq fracddtlVert Delta mathbf v rVert$. The thing that I have tried to prove was if $lim_Delta t to 0fraclVert Delta mathbfv rVertDelta t = leftlVert lim_Delta t to 0fracDelta mathbfvDelta t rightrVert$ holds (maybe assuming $lVert v rVert$ is constant)
– marco21
Aug 8 at 15:59
@marco oh yeah duh, I sorta took the question in the title as the actual question rather than properly reading the whole thing apparently. I'll edit this later or delete it if I don't have an answer
– jgon
Aug 8 at 16:14
1
In that case, this follows from the norm being continuous (I'll edit an explanation later)
– jgon
Aug 8 at 16:16
Thank you! I didn't notice it.
– marco21
Aug 9 at 21:31
add a comment |Â
up vote
0
down vote
This just isn't true. Consider $$f(t) = beginpmatrix cos t \sin t endpmatrix$$
Observe that $|f|=1$ for all $t$, so $fracddt|f| = 0$, but
$$f'(t) = beginpmatrix-sin t \ cos tendpmatrix,$$
which still has norm 1.
For physical intuition, consider a car traveling in a circle around the origin with constant speed. Then the distance of the car from the origin doesn't change, so the derivative of the norm is 0, but the velocity has constant positive magnitude (since the magnitude of the velocity is by definition speed).
answered Aug 7 at 16:32
jgon
8,54611435
But $lim_Delta t to 0fraclVert Delta mathbf v rVertDelta t neq fracddtlVert Delta mathbf v rVert$. The thing that I have tried to prove was if $lim_Delta t to 0fraclVert Delta mathbfv rVertDelta t = leftlVert lim_Delta t to 0fracDelta mathbfvDelta t rightrVert$ holds (maybe assuming $lVert v rVert$ is constant)
– marco21
Aug 8 at 15:59
@marco oh yeah duh, I sorta took the question in the title as the actual question rather than properly reading the whole thing apparently. I'll edit this later or delete it if I don't have an answer
– jgon
Aug 8 at 16:14
1
In that case, this follows from the norm being continuous (I'll edit an explanation later)
– jgon
Aug 8 at 16:16
Thank you! I didn't notice it.
– marco21
Aug 9 at 21:31
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This just isn't true. Consider $$f(t) = beginpmatrix cos t \sin t endpmatrix$$
Observe that $|f|=1$ for all $t$, so $fracddt|f| = 0$, but
$$f'(t) = beginpmatrix-sin t \ cos tendpmatrix,$$
which still has norm 1.
For physical intuition, consider a car traveling in a circle around the origin with constant speed. Then the distance of the car from the origin doesn't change, so the derivative of the norm is 0, but the velocity has constant positive magnitude (since the magnitude of the velocity is by definition speed).
answered Aug 7 at 16:32
jgon
8,54611435
This just isn't true. Consider $$f(t) = beginpmatrix cos t \sin t endpmatrix$$
Observe that $|f|=1$ for all $t$, so $fracddt|f| = 0$, but
$$f'(t) = beginpmatrix-sin t \ cos tendpmatrix,$$
which still has norm 1.
For physical intuition, consider a car traveling in a circle around the origin with constant speed. Then the distance of the car from the origin doesn't change, so the derivative of the norm is 0, but the velocity has constant positive magnitude (since the magnitude of the velocity is by definition speed).
answered Aug 7 at 16:32
jgon
8,54611435
answered Aug 7 at 16:32
jgon
8,54611435
answered Aug 7 at 16:32
jgon
8,54611435
answered Aug 7 at 16:32
jgon
8,54611435
8,54611435
But $lim_Delta t to 0fraclVert Delta mathbf v rVertDelta t neq fracddtlVert Delta mathbf v rVert$. The thing that I have tried to prove was if $lim_Delta t to 0fraclVert Delta mathbfv rVertDelta t = leftlVert lim_Delta t to 0fracDelta mathbfvDelta t rightrVert$ holds (maybe assuming $lVert v rVert$ is constant)
– marco21
Aug 8 at 15:59
@marco oh yeah duh, I sorta took the question in the title as the actual question rather than properly reading the whole thing apparently. I'll edit this later or delete it if I don't have an answer
– jgon
Aug 8 at 16:14
1
In that case, this follows from the norm being continuous (I'll edit an explanation later)
– jgon
Aug 8 at 16:16
Thank you! I didn't notice it.
– marco21
Aug 9 at 21:31
add a comment |Â
But $lim_Delta t to 0fraclVert Delta mathbf v rVertDelta t neq fracddtlVert Delta mathbf v rVert$. The thing that I have tried to prove was if $lim_Delta t to 0fraclVert Delta mathbfv rVertDelta t = leftlVert lim_Delta t to 0fracDelta mathbfvDelta t rightrVert$ holds (maybe assuming $lVert v rVert$ is constant)
– marco21
Aug 8 at 15:59
@marco oh yeah duh, I sorta took the question in the title as the actual question rather than properly reading the whole thing apparently. I'll edit this later or delete it if I don't have an answer
– jgon
Aug 8 at 16:14
1
In that case, this follows from the norm being continuous (I'll edit an explanation later)
– jgon
Aug 8 at 16:16
Thank you! I didn't notice it.
– marco21
Aug 9 at 21:31
But $lim_Delta t to 0fraclVert Delta mathbf v rVertDelta t neq fracddtlVert Delta mathbf v rVert$. The thing that I have tried to prove was if $lim_Delta t to 0fraclVert Delta mathbfv rVertDelta t = leftlVert lim_Delta t to 0fracDelta mathbfvDelta t rightrVert$ holds (maybe assuming $lVert v rVert$ is constant)
– marco21
Aug 8 at 15:59
But $lim_Delta t to 0fraclVert Delta mathbf v rVertDelta t neq fracddtlVert Delta mathbf v rVert$. The thing that I have tried to prove was if $lim_Delta t to 0fraclVert Delta mathbfv rVertDelta t = leftlVert lim_Delta t to 0fracDelta mathbfvDelta t rightrVert$ holds (maybe assuming $lVert v rVert$ is constant)
– marco21
Aug 8 at 15:59
@marco oh yeah duh, I sorta took the question in the title as the actual question rather than properly reading the whole thing apparently. I'll edit this later or delete it if I don't have an answer
– jgon
Aug 8 at 16:14
@marco oh yeah duh, I sorta took the question in the title as the actual question rather than properly reading the whole thing apparently. I'll edit this later or delete it if I don't have an answer
– jgon
Aug 8 at 16:14
1
1
In that case, this follows from the norm being continuous (I'll edit an explanation later)
– jgon
Aug 8 at 16:16
In that case, this follows from the norm being continuous (I'll edit an explanation later)
– jgon
Aug 8 at 16:16
Thank you! I didn't notice it.
– marco21
Aug 9 at 21:31
Thank you! I didn't notice it.
– marco21
Aug 9 at 21:31
add a comment |Â
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