Why is dx/dt = -(∂u/∂t) / (∂u/∂x)? [closed]
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I found that $frac{dx}{dt} = -cfrac{ frac{partial u}{partial t} }{ frac{partial u}{partial x}}$ on the internet. I can´t figure out if it is true and why.
multivariable-calculus partial-derivative implicit-differentiation
edited Dec 12 '18 at 23:51
platty
3,350320
asked Dec 12 '18 at 23:27
LollipopLollipop
378
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closed as off-topic by Saad, Chris Custer, Brahadeesh, KReiser, Leucippus Dec 13 '18 at 7:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Chris Custer, Brahadeesh, KReiser, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I found that $frac{dx}{dt} = -cfrac{ frac{partial u}{partial t} }{ frac{partial u}{partial x}}$ on the internet. I can´t figure out if it is true and why.
multivariable-calculus partial-derivative implicit-differentiation
edited Dec 12 '18 at 23:51
platty
3,350320
asked Dec 12 '18 at 23:27
LollipopLollipop
378
$endgroup$
closed as off-topic by Saad, Chris Custer, Brahadeesh, KReiser, Leucippus Dec 13 '18 at 7:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Chris Custer, Brahadeesh, KReiser, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
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it's the chain rule...
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– Hasan Iqbal
Dec 12 '18 at 23:39
add a comment |
$begingroup$
I found that $frac{dx}{dt} = -cfrac{ frac{partial u}{partial t} }{ frac{partial u}{partial x}}$ on the internet. I can´t figure out if it is true and why.
multivariable-calculus partial-derivative implicit-differentiation
edited Dec 12 '18 at 23:51
platty
3,350320
asked Dec 12 '18 at 23:27
LollipopLollipop
378
$endgroup$
I found that $frac{dx}{dt} = -cfrac{ frac{partial u}{partial t} }{ frac{partial u}{partial x}}$ on the internet. I can´t figure out if it is true and why.
multivariable-calculus partial-derivative implicit-differentiation
multivariable-calculus partial-derivative implicit-differentiation
edited Dec 12 '18 at 23:51
platty
3,350320
asked Dec 12 '18 at 23:27
LollipopLollipop
378
edited Dec 12 '18 at 23:51
platty
3,350320
asked Dec 12 '18 at 23:27
LollipopLollipop
378
edited Dec 12 '18 at 23:51
platty
3,350320
edited Dec 12 '18 at 23:51
platty
3,350320
edited Dec 12 '18 at 23:51
platty
3,350320
3,350320
asked Dec 12 '18 at 23:27
LollipopLollipop
378
asked Dec 12 '18 at 23:27
LollipopLollipop
378
asked Dec 12 '18 at 23:27
LollipopLollipop
378
378
closed as off-topic by Saad, Chris Custer, Brahadeesh, KReiser, Leucippus Dec 13 '18 at 7:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Chris Custer, Brahadeesh, KReiser, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Chris Custer, Brahadeesh, KReiser, Leucippus Dec 13 '18 at 7:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Chris Custer, Brahadeesh, KReiser, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
it's the chain rule...
$endgroup$
– Hasan Iqbal
Dec 12 '18 at 23:39
add a comment |
$begingroup$
it's the chain rule...
$endgroup$
– Hasan Iqbal
Dec 12 '18 at 23:39
$begingroup$
it's the chain rule...
$endgroup$
– Hasan Iqbal
Dec 12 '18 at 23:39
$begingroup$
it's the chain rule...
$endgroup$
– Hasan Iqbal
Dec 12 '18 at 23:39
add a comment |
2 Answers
2
active
oldest
votes
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This is a result of the implicit function theorem. In fact, in some presentations (i.e. Rudin) this is included in the theorem. Look there for definitions/proofs.
answered Dec 13 '18 at 0:56
T. FoT. Fo
496311
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add a comment |
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This expression does not mean anything. So first we must define the variables. I assume that you meant: let $u(alpha,beta)$ be a smooth function of two real variables and let $x$ be a smooth univariate function with real values such that $u(t,x(t))=0$.
Then, $x’(t)=-frac{(partial u)/(partial alpha)(t,x(t))}{(partial u)/(partial beta)(t,x(t))}$.
This is straightforward when you differentiate wrt $t$ the equation $u(t,x(t))=0$.
answered Dec 12 '18 at 23:34
MindlackMindlack
4,900211
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a result of the implicit function theorem. In fact, in some presentations (i.e. Rudin) this is included in the theorem. Look there for definitions/proofs.
answered Dec 13 '18 at 0:56
T. FoT. Fo
496311
$endgroup$
add a comment |
$begingroup$
This is a result of the implicit function theorem. In fact, in some presentations (i.e. Rudin) this is included in the theorem. Look there for definitions/proofs.
answered Dec 13 '18 at 0:56
T. FoT. Fo
496311
$endgroup$
add a comment |
$begingroup$
This is a result of the implicit function theorem. In fact, in some presentations (i.e. Rudin) this is included in the theorem. Look there for definitions/proofs.
answered Dec 13 '18 at 0:56
T. FoT. Fo
496311
$endgroup$
This is a result of the implicit function theorem. In fact, in some presentations (i.e. Rudin) this is included in the theorem. Look there for definitions/proofs.
answered Dec 13 '18 at 0:56
T. FoT. Fo
496311
answered Dec 13 '18 at 0:56
T. FoT. Fo
496311
answered Dec 13 '18 at 0:56
T. FoT. Fo
496311
answered Dec 13 '18 at 0:56
T. FoT. Fo
496311
496311
add a comment |
add a comment |
$begingroup$
This expression does not mean anything. So first we must define the variables. I assume that you meant: let $u(alpha,beta)$ be a smooth function of two real variables and let $x$ be a smooth univariate function with real values such that $u(t,x(t))=0$.
Then, $x’(t)=-frac{(partial u)/(partial alpha)(t,x(t))}{(partial u)/(partial beta)(t,x(t))}$.
This is straightforward when you differentiate wrt $t$ the equation $u(t,x(t))=0$.
answered Dec 12 '18 at 23:34
MindlackMindlack
4,900211
$endgroup$
add a comment |
$begingroup$
This expression does not mean anything. So first we must define the variables. I assume that you meant: let $u(alpha,beta)$ be a smooth function of two real variables and let $x$ be a smooth univariate function with real values such that $u(t,x(t))=0$.
Then, $x’(t)=-frac{(partial u)/(partial alpha)(t,x(t))}{(partial u)/(partial beta)(t,x(t))}$.
This is straightforward when you differentiate wrt $t$ the equation $u(t,x(t))=0$.
answered Dec 12 '18 at 23:34
MindlackMindlack
4,900211
$endgroup$
add a comment |
$begingroup$
This expression does not mean anything. So first we must define the variables. I assume that you meant: let $u(alpha,beta)$ be a smooth function of two real variables and let $x$ be a smooth univariate function with real values such that $u(t,x(t))=0$.
Then, $x’(t)=-frac{(partial u)/(partial alpha)(t,x(t))}{(partial u)/(partial beta)(t,x(t))}$.
This is straightforward when you differentiate wrt $t$ the equation $u(t,x(t))=0$.
answered Dec 12 '18 at 23:34
MindlackMindlack
4,900211
$endgroup$
This expression does not mean anything. So first we must define the variables. I assume that you meant: let $u(alpha,beta)$ be a smooth function of two real variables and let $x$ be a smooth univariate function with real values such that $u(t,x(t))=0$.
Then, $x’(t)=-frac{(partial u)/(partial alpha)(t,x(t))}{(partial u)/(partial beta)(t,x(t))}$.
This is straightforward when you differentiate wrt $t$ the equation $u(t,x(t))=0$.
answered Dec 12 '18 at 23:34
MindlackMindlack
4,900211
answered Dec 12 '18 at 23:34
MindlackMindlack
4,900211
answered Dec 12 '18 at 23:34
MindlackMindlack
4,900211
answered Dec 12 '18 at 23:34
MindlackMindlack
4,900211
4,900211
add a comment |
add a comment |
$begingroup$
it's the chain rule...
$endgroup$
– Hasan Iqbal
Dec 12 '18 at 23:39