Mixing
Obtain function from its partial derivatives
Clash Royale CLAN TAG#URR8PPP
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I should determine the function $M(H,T)$ by only knowing the derivatives:
$$
left ( fracpartial Mpartial H right )_T=fraca1-T/T_c+3bH^2
$$
where $a$ and $b$ are some real constants
$$
left ( fracpartial Mpartial T right )_H=
frac1T_cfracf(H)(1-T/T_c)^2-
frac12fracM_0T_cfrac1(1-T/T_c)^1/2
$$
with $M_0$, $T_c$, $a$ and $b$ are constants and $f(H)$ is some function
with the property $f(H=0)=0$. The variable $T$ will be bound to the domain $0le Tlt T_c$.
a) determine $f(H)$
b) determine $M(T,H)$
Now I would start by writing down
$$
dM = left ( fracpartial Mpartial H right )_TdH + left ( fracpartial Mpartial T right )_H dT
$$
This tells me how I would obtain the M in terms of the derivatives.
I tried now to do the integration what gives:
$$
M(T,H)=fracaH1-T/T_c+bH^3+fracf(H)1-T/T_c+M_0(1-T/T_c)^1/2
$$
If I compute now again the derivative with respect to H I obtain:
$$
left ( fracpartial Mpartial H right )_T=fraca1-T/T_c+3bH^2+frac11-T/T_cfracpartial f(H)partial H
$$
This is now not the same as at the beginning. From here I don't know how to go on. Is this already wrong or is it somehow possible from this equation to determine $f(H)$.
My first intuition was to say $f(H)$ is just zero, because then the derivative with respect to $H$ would be satisfied and so would be the $f(H=0)=0$ criterion. But this might also be a trivial solution.
calculus pde
edited 1 hour ago
Harry49
5,1812828
asked 3 hours ago
zodiac
224
add a comment |Â
up vote
2
down vote
favorite
I should determine the function $M(H,T)$ by only knowing the derivatives:
$$
left ( fracpartial Mpartial H right )_T=fraca1-T/T_c+3bH^2
$$
where $a$ and $b$ are some real constants
$$
left ( fracpartial Mpartial T right )_H=
frac1T_cfracf(H)(1-T/T_c)^2-
frac12fracM_0T_cfrac1(1-T/T_c)^1/2
$$
with $M_0$, $T_c$, $a$ and $b$ are constants and $f(H)$ is some function
with the property $f(H=0)=0$. The variable $T$ will be bound to the domain $0le Tlt T_c$.
a) determine $f(H)$
b) determine $M(T,H)$
Now I would start by writing down
$$
dM = left ( fracpartial Mpartial H right )_TdH + left ( fracpartial Mpartial T right )_H dT
$$
This tells me how I would obtain the M in terms of the derivatives.
I tried now to do the integration what gives:
$$
M(T,H)=fracaH1-T/T_c+bH^3+fracf(H)1-T/T_c+M_0(1-T/T_c)^1/2
$$
If I compute now again the derivative with respect to H I obtain:
$$
left ( fracpartial Mpartial H right )_T=fraca1-T/T_c+3bH^2+frac11-T/T_cfracpartial f(H)partial H
$$
This is now not the same as at the beginning. From here I don't know how to go on. Is this already wrong or is it somehow possible from this equation to determine $f(H)$.
My first intuition was to say $f(H)$ is just zero, because then the derivative with respect to $H$ would be satisfied and so would be the $f(H=0)=0$ criterion. But this might also be a trivial solution.
calculus pde
edited 1 hour ago
Harry49
5,1812828
asked 3 hours ago
zodiac
224
What's the answer provided?
– Dhamnekar Winod
2 hours ago
@Dhamnekar Winod: I don't have one this is my problem. What I wrote above is everything I got
– zodiac
2 hours ago
I suggest tag your question with "partial differential equations" to have more chances for attention.
– dmtri
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I should determine the function $M(H,T)$ by only knowing the derivatives:
$$
left ( fracpartial Mpartial H right )_T=fraca1-T/T_c+3bH^2
$$
where $a$ and $b$ are some real constants
$$
left ( fracpartial Mpartial T right )_H=
frac1T_cfracf(H)(1-T/T_c)^2-
frac12fracM_0T_cfrac1(1-T/T_c)^1/2
$$
with $M_0$, $T_c$, $a$ and $b$ are constants and $f(H)$ is some function
with the property $f(H=0)=0$. The variable $T$ will be bound to the domain $0le Tlt T_c$.
a) determine $f(H)$
b) determine $M(T,H)$
Now I would start by writing down
$$
dM = left ( fracpartial Mpartial H right )_TdH + left ( fracpartial Mpartial T right )_H dT
$$
This tells me how I would obtain the M in terms of the derivatives.
I tried now to do the integration what gives:
$$
M(T,H)=fracaH1-T/T_c+bH^3+fracf(H)1-T/T_c+M_0(1-T/T_c)^1/2
$$
If I compute now again the derivative with respect to H I obtain:
$$
left ( fracpartial Mpartial H right )_T=fraca1-T/T_c+3bH^2+frac11-T/T_cfracpartial f(H)partial H
$$
This is now not the same as at the beginning. From here I don't know how to go on. Is this already wrong or is it somehow possible from this equation to determine $f(H)$.
My first intuition was to say $f(H)$ is just zero, because then the derivative with respect to $H$ would be satisfied and so would be the $f(H=0)=0$ criterion. But this might also be a trivial solution.
calculus pde
edited 1 hour ago
Harry49
5,1812828
asked 3 hours ago
zodiac
224
I should determine the function $M(H,T)$ by only knowing the derivatives:
$$
left ( fracpartial Mpartial H right )_T=fraca1-T/T_c+3bH^2
$$
where $a$ and $b$ are some real constants
$$
left ( fracpartial Mpartial T right )_H=
frac1T_cfracf(H)(1-T/T_c)^2-
frac12fracM_0T_cfrac1(1-T/T_c)^1/2
$$
with $M_0$, $T_c$, $a$ and $b$ are constants and $f(H)$ is some function
with the property $f(H=0)=0$. The variable $T$ will be bound to the domain $0le Tlt T_c$.
a) determine $f(H)$
b) determine $M(T,H)$
Now I would start by writing down
$$
dM = left ( fracpartial Mpartial H right )_TdH + left ( fracpartial Mpartial T right )_H dT
$$
This tells me how I would obtain the M in terms of the derivatives.
I tried now to do the integration what gives:
$$
M(T,H)=fracaH1-T/T_c+bH^3+fracf(H)1-T/T_c+M_0(1-T/T_c)^1/2
$$
If I compute now again the derivative with respect to H I obtain:
$$
left ( fracpartial Mpartial H right )_T=fraca1-T/T_c+3bH^2+frac11-T/T_cfracpartial f(H)partial H
$$
This is now not the same as at the beginning. From here I don't know how to go on. Is this already wrong or is it somehow possible from this equation to determine $f(H)$.
My first intuition was to say $f(H)$ is just zero, because then the derivative with respect to $H$ would be satisfied and so would be the $f(H=0)=0$ criterion. But this might also be a trivial solution.
calculus pde
calculus pde
edited 1 hour ago
Harry49
5,1812828
asked 3 hours ago
zodiac
224
edited 1 hour ago
Harry49
5,1812828
asked 3 hours ago
zodiac
224
edited 1 hour ago
Harry49
5,1812828
edited 1 hour ago
Harry49
5,1812828
edited 1 hour ago
Harry49
5,1812828
5,1812828
asked 3 hours ago
zodiac
224
asked 3 hours ago
zodiac
224
asked 3 hours ago
zodiac
224
224
What's the answer provided?
– Dhamnekar Winod
2 hours ago
@Dhamnekar Winod: I don't have one this is my problem. What I wrote above is everything I got
– zodiac
2 hours ago
I suggest tag your question with "partial differential equations" to have more chances for attention.
– dmtri
1 hour ago
add a comment |Â
What's the answer provided?
– Dhamnekar Winod
2 hours ago
@Dhamnekar Winod: I don't have one this is my problem. What I wrote above is everything I got
– zodiac
2 hours ago
I suggest tag your question with "partial differential equations" to have more chances for attention.
– dmtri
1 hour ago
What's the answer provided?
– Dhamnekar Winod
2 hours ago
What's the answer provided?
– Dhamnekar Winod
2 hours ago
@Dhamnekar Winod: I don't have one this is my problem. What I wrote above is everything I got
– zodiac
2 hours ago
@Dhamnekar Winod: I don't have one this is my problem. What I wrote above is everything I got
– zodiac
2 hours ago
I suggest tag your question with "partial differential equations" to have more chances for attention.
– dmtri
1 hour ago
I suggest tag your question with "partial differential equations" to have more chances for attention.
– dmtri
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Let us integrate the first equation with respect to $H$ while $T$ is seen as a constant:
$$
M(T,H) = fraca H1-T/T_c + bH^3 + phi(T)
$$
where $phi(T)$ is an "integration constant" depending on $T$. Now, the partial derivative w.r.t. $T$ is
$$
fracpartial Mpartial T = fraca HT_cfrac1(1-T/T_c)^2 + phi'(T)
$$
where $phi'$ is the derivative of $phi$. Therefore, $f(H)=aH$ and
$phi'(T) = -frac12fracM_0T_c(1-T/T_c)^-1/2$. The
integration of this last equation with respect to $T$ gives $phi(T)$ and introduces an integration constant $C$. Finally, $M$ is defined up to a constant $C$ by
$$
M(T,H) = fraca H1-T/T_c + bH^3 + M_0, (1-T/T_c)^1/2 + C, .
$$
answered 1 hour ago
Harry49
5,1812828
add a comment |Â
up vote
2
down vote
Hint:
Perform the integrations
$$M(T,H)=int D_h(T,H),dT+C_h(H)=int D_t(T,H),dH+C_t(T))
\=I_h(T,H)+C_h(H)=I_t(T,H)+C_t(T),$$
and determine the functions $C_h,C_t$ that fit.
answered 38 mins ago
Yves Daoust
116k667211
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let us integrate the first equation with respect to $H$ while $T$ is seen as a constant:
$$
M(T,H) = fraca H1-T/T_c + bH^3 + phi(T)
$$
where $phi(T)$ is an "integration constant" depending on $T$. Now, the partial derivative w.r.t. $T$ is
$$
fracpartial Mpartial T = fraca HT_cfrac1(1-T/T_c)^2 + phi'(T)
$$
where $phi'$ is the derivative of $phi$. Therefore, $f(H)=aH$ and
$phi'(T) = -frac12fracM_0T_c(1-T/T_c)^-1/2$. The
integration of this last equation with respect to $T$ gives $phi(T)$ and introduces an integration constant $C$. Finally, $M$ is defined up to a constant $C$ by
$$
M(T,H) = fraca H1-T/T_c + bH^3 + M_0, (1-T/T_c)^1/2 + C, .
$$
answered 1 hour ago
Harry49
5,1812828
add a comment |Â
up vote
4
down vote
accepted
Let us integrate the first equation with respect to $H$ while $T$ is seen as a constant:
$$
M(T,H) = fraca H1-T/T_c + bH^3 + phi(T)
$$
where $phi(T)$ is an "integration constant" depending on $T$. Now, the partial derivative w.r.t. $T$ is
$$
fracpartial Mpartial T = fraca HT_cfrac1(1-T/T_c)^2 + phi'(T)
$$
where $phi'$ is the derivative of $phi$. Therefore, $f(H)=aH$ and
$phi'(T) = -frac12fracM_0T_c(1-T/T_c)^-1/2$. The
integration of this last equation with respect to $T$ gives $phi(T)$ and introduces an integration constant $C$. Finally, $M$ is defined up to a constant $C$ by
$$
M(T,H) = fraca H1-T/T_c + bH^3 + M_0, (1-T/T_c)^1/2 + C, .
$$
answered 1 hour ago
Harry49
5,1812828
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let us integrate the first equation with respect to $H$ while $T$ is seen as a constant:
$$
M(T,H) = fraca H1-T/T_c + bH^3 + phi(T)
$$
where $phi(T)$ is an "integration constant" depending on $T$. Now, the partial derivative w.r.t. $T$ is
$$
fracpartial Mpartial T = fraca HT_cfrac1(1-T/T_c)^2 + phi'(T)
$$
where $phi'$ is the derivative of $phi$. Therefore, $f(H)=aH$ and
$phi'(T) = -frac12fracM_0T_c(1-T/T_c)^-1/2$. The
integration of this last equation with respect to $T$ gives $phi(T)$ and introduces an integration constant $C$. Finally, $M$ is defined up to a constant $C$ by
$$
M(T,H) = fraca H1-T/T_c + bH^3 + M_0, (1-T/T_c)^1/2 + C, .
$$
answered 1 hour ago
Harry49
5,1812828
Let us integrate the first equation with respect to $H$ while $T$ is seen as a constant:
$$
M(T,H) = fraca H1-T/T_c + bH^3 + phi(T)
$$
where $phi(T)$ is an "integration constant" depending on $T$. Now, the partial derivative w.r.t. $T$ is
$$
fracpartial Mpartial T = fraca HT_cfrac1(1-T/T_c)^2 + phi'(T)
$$
where $phi'$ is the derivative of $phi$. Therefore, $f(H)=aH$ and
$phi'(T) = -frac12fracM_0T_c(1-T/T_c)^-1/2$. The
integration of this last equation with respect to $T$ gives $phi(T)$ and introduces an integration constant $C$. Finally, $M$ is defined up to a constant $C$ by
$$
M(T,H) = fraca H1-T/T_c + bH^3 + M_0, (1-T/T_c)^1/2 + C, .
$$
answered 1 hour ago
Harry49
5,1812828
edited 56 mins ago
answered 1 hour ago
Harry49
5,1812828
answered 1 hour ago
Harry49
5,1812828
answered 1 hour ago
Harry49
5,1812828
5,1812828
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint:
Perform the integrations
$$M(T,H)=int D_h(T,H),dT+C_h(H)=int D_t(T,H),dH+C_t(T))
\=I_h(T,H)+C_h(H)=I_t(T,H)+C_t(T),$$
and determine the functions $C_h,C_t$ that fit.
answered 38 mins ago
Yves Daoust
116k667211
add a comment |Â
up vote
2
down vote
Hint:
Perform the integrations
$$M(T,H)=int D_h(T,H),dT+C_h(H)=int D_t(T,H),dH+C_t(T))
\=I_h(T,H)+C_h(H)=I_t(T,H)+C_t(T),$$
and determine the functions $C_h,C_t$ that fit.
answered 38 mins ago
Yves Daoust
116k667211
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
Perform the integrations
$$M(T,H)=int D_h(T,H),dT+C_h(H)=int D_t(T,H),dH+C_t(T))
\=I_h(T,H)+C_h(H)=I_t(T,H)+C_t(T),$$
and determine the functions $C_h,C_t$ that fit.
answered 38 mins ago
Yves Daoust
116k667211
Hint:
Perform the integrations
$$M(T,H)=int D_h(T,H),dT+C_h(H)=int D_t(T,H),dH+C_t(T))
\=I_h(T,H)+C_h(H)=I_t(T,H)+C_t(T),$$
and determine the functions $C_h,C_t$ that fit.
answered 38 mins ago
Yves Daoust
116k667211
answered 38 mins ago
Yves Daoust
116k667211
answered 38 mins ago
Yves Daoust
116k667211
answered 38 mins ago
Yves Daoust
116k667211
116k667211
add a comment |Â
add a comment |Â
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What's the answer provided?
– Dhamnekar Winod
2 hours ago
@Dhamnekar Winod: I don't have one this is my problem. What I wrote above is everything I got
– zodiac
2 hours ago
I suggest tag your question with "partial differential equations" to have more chances for attention.
– dmtri
1 hour ago