How to differentiate $int_{B(t)} f(x,t) dx$ with respect to $t$?
$begingroup$
$int_{B(t)} f(x,t) dx$ is given and assume $x$ is an $n$ dimensional vector variable, $t$ is positive real variable and $f(x,t)$ is a sufficiently smooth function of both $x$ and $t$. Also $B(t)$ is a time varying region in the $d$ dimensional Euclidean space with sufficient smoothness conditions. Then what is the formula for differentiating the integral $int_{0}^{t} f(x,t) dx$ with respect to $t$? Could anyone please explain?
calculus multivariable-calculus
asked Dec 6 '18 at 16:11
KeithKeith
1,425922
$endgroup$
add a comment |
$begingroup$
$int_{B(t)} f(x,t) dx$ is given and assume $x$ is an $n$ dimensional vector variable, $t$ is positive real variable and $f(x,t)$ is a sufficiently smooth function of both $x$ and $t$. Also $B(t)$ is a time varying region in the $d$ dimensional Euclidean space with sufficient smoothness conditions. Then what is the formula for differentiating the integral $int_{0}^{t} f(x,t) dx$ with respect to $t$? Could anyone please explain?
calculus multivariable-calculus
asked Dec 6 '18 at 16:11
KeithKeith
1,425922
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– Cameron Williams
Dec 6 '18 at 16:15
1
$begingroup$
Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:21
$begingroup$
Oops... I made a mistake. I will correct it now.
$endgroup$
– Keith
Dec 6 '18 at 16:50
add a comment |
$begingroup$
$int_{B(t)} f(x,t) dx$ is given and assume $x$ is an $n$ dimensional vector variable, $t$ is positive real variable and $f(x,t)$ is a sufficiently smooth function of both $x$ and $t$. Also $B(t)$ is a time varying region in the $d$ dimensional Euclidean space with sufficient smoothness conditions. Then what is the formula for differentiating the integral $int_{0}^{t} f(x,t) dx$ with respect to $t$? Could anyone please explain?
calculus multivariable-calculus
asked Dec 6 '18 at 16:11
KeithKeith
1,425922
$endgroup$
$int_{B(t)} f(x,t) dx$ is given and assume $x$ is an $n$ dimensional vector variable, $t$ is positive real variable and $f(x,t)$ is a sufficiently smooth function of both $x$ and $t$. Also $B(t)$ is a time varying region in the $d$ dimensional Euclidean space with sufficient smoothness conditions. Then what is the formula for differentiating the integral $int_{0}^{t} f(x,t) dx$ with respect to $t$? Could anyone please explain?
calculus multivariable-calculus
calculus multivariable-calculus
asked Dec 6 '18 at 16:11
KeithKeith
1,425922
asked Dec 6 '18 at 16:11
KeithKeith
1,425922
edited Dec 6 '18 at 16:53
Keith
asked Dec 6 '18 at 16:11
KeithKeith
1,425922
asked Dec 6 '18 at 16:11
KeithKeith
1,425922
asked Dec 6 '18 at 16:11
KeithKeith
1,425922
1,425922
1
$begingroup$
en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– Cameron Williams
Dec 6 '18 at 16:15
1
$begingroup$
Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:21
$begingroup$
Oops... I made a mistake. I will correct it now.
$endgroup$
– Keith
Dec 6 '18 at 16:50
add a comment |
1
$begingroup$
en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– Cameron Williams
Dec 6 '18 at 16:15
1
$begingroup$
Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:21
$begingroup$
Oops... I made a mistake. I will correct it now.
$endgroup$
– Keith
Dec 6 '18 at 16:50
1
1
$begingroup$
en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– Cameron Williams
Dec 6 '18 at 16:15
$begingroup$
en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– Cameron Williams
Dec 6 '18 at 16:15
1
1
$begingroup$
Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:21
$begingroup$
Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:21
$begingroup$
Oops... I made a mistake. I will correct it now.
$endgroup$
– Keith
Dec 6 '18 at 16:50
$begingroup$
Oops... I made a mistake. I will correct it now.
$endgroup$
– Keith
Dec 6 '18 at 16:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Some clever guy has already thought about it: look here.
ADDENDUM
In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.
answered Dec 6 '18 at 16:16
FedericoFederico
5,029514
$endgroup$
$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18
$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20
$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25
$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26
$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29
|
show 4 more comments
Your Answer
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$begingroup$
Some clever guy has already thought about it: look here.
ADDENDUM
In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.
answered Dec 6 '18 at 16:16
FedericoFederico
5,029514
$endgroup$
$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18
$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20
$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25
$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26
$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29
|
show 4 more comments
$begingroup$
Some clever guy has already thought about it: look here.
ADDENDUM
In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.
answered Dec 6 '18 at 16:16
FedericoFederico
5,029514
$endgroup$
$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18
$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20
$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25
$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26
$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29
|
show 4 more comments
$begingroup$
Some clever guy has already thought about it: look here.
ADDENDUM
In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.
answered Dec 6 '18 at 16:16
FedericoFederico
5,029514
$endgroup$
Some clever guy has already thought about it: look here.
ADDENDUM
In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.
answered Dec 6 '18 at 16:16
FedericoFederico
5,029514
edited Dec 6 '18 at 16:54
answered Dec 6 '18 at 16:16
FedericoFederico
5,029514
answered Dec 6 '18 at 16:16
FedericoFederico
5,029514
answered Dec 6 '18 at 16:16
FedericoFederico
5,029514
5,029514
$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18
$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20
$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25
$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26
$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29
|
show 4 more comments
$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18
$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20
$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25
$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26
$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29
$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18
$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18
$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20
$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20
$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25
$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25
$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26
$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26
$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29
$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29
|
show 4 more comments
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1
$begingroup$
en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– Cameron Williams
Dec 6 '18 at 16:15
1
$begingroup$
Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:21
$begingroup$
Oops... I made a mistake. I will correct it now.
$endgroup$
– Keith
Dec 6 '18 at 16:50