$u_{rr} = f_{xx} cos^{2}(x) + 2f_{xy} cos(x) sin(x) + f_{yy} sin^{2}(x)$
$begingroup$
For this Question is if were to do $U_r$ after the first partial derivative. It will result in a function $U_r(x,y,θ)$ why is it that upon the second time I do a partial derivative, I need not consider it as the following $U_{rr}=frac{∂Ur}{∂x}·frac{∂x}{∂r}+frac{∂Ur}{∂y}·frac{∂y}{∂r}+frac{∂Ur}{∂θ}·frac{∂θ}{∂r}$? Instead what was considered to be the answer was $U_{rr}=frac{∂Ur}{∂x}·frac{∂x}{∂r}+frac{∂Ur}{∂y}·frac{∂y}{∂r}.$ Do we not need to consider the last term? and if so how would I set up a function to do the partial derivative of $frac{∂θ}{∂r}$?
Thanks
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:25
A.Γ.
22.9k32656
asked Dec 29 '18 at 8:27
Radical particalRadical partical
31
$endgroup$
add a comment |
$begingroup$
For this Question is if were to do $U_r$ after the first partial derivative. It will result in a function $U_r(x,y,θ)$ why is it that upon the second time I do a partial derivative, I need not consider it as the following $U_{rr}=frac{∂Ur}{∂x}·frac{∂x}{∂r}+frac{∂Ur}{∂y}·frac{∂y}{∂r}+frac{∂Ur}{∂θ}·frac{∂θ}{∂r}$? Instead what was considered to be the answer was $U_{rr}=frac{∂Ur}{∂x}·frac{∂x}{∂r}+frac{∂Ur}{∂y}·frac{∂y}{∂r}.$ Do we not need to consider the last term? and if so how would I set up a function to do the partial derivative of $frac{∂θ}{∂r}$?
Thanks
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:25
A.Γ.
22.9k32656
asked Dec 29 '18 at 8:27
Radical particalRadical partical
31
$endgroup$
$begingroup$
$theta $ is not a function of $r $.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 8:32
$begingroup$
Just because of that we do not need to consider the last part and it is omitted?
$endgroup$
– Radical partical
Dec 29 '18 at 9:24
$begingroup$
Since $theta $ is not a function of $r $, we have $frac{∂θ}{∂r}=0$.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 9:48
add a comment |
$begingroup$
For this Question is if were to do $U_r$ after the first partial derivative. It will result in a function $U_r(x,y,θ)$ why is it that upon the second time I do a partial derivative, I need not consider it as the following $U_{rr}=frac{∂Ur}{∂x}·frac{∂x}{∂r}+frac{∂Ur}{∂y}·frac{∂y}{∂r}+frac{∂Ur}{∂θ}·frac{∂θ}{∂r}$? Instead what was considered to be the answer was $U_{rr}=frac{∂Ur}{∂x}·frac{∂x}{∂r}+frac{∂Ur}{∂y}·frac{∂y}{∂r}.$ Do we not need to consider the last term? and if so how would I set up a function to do the partial derivative of $frac{∂θ}{∂r}$?
Thanks
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:25
A.Γ.
22.9k32656
asked Dec 29 '18 at 8:27
Radical particalRadical partical
31
$endgroup$
For this Question is if were to do $U_r$ after the first partial derivative. It will result in a function $U_r(x,y,θ)$ why is it that upon the second time I do a partial derivative, I need not consider it as the following $U_{rr}=frac{∂Ur}{∂x}·frac{∂x}{∂r}+frac{∂Ur}{∂y}·frac{∂y}{∂r}+frac{∂Ur}{∂θ}·frac{∂θ}{∂r}$? Instead what was considered to be the answer was $U_{rr}=frac{∂Ur}{∂x}·frac{∂x}{∂r}+frac{∂Ur}{∂y}·frac{∂y}{∂r}.$ Do we not need to consider the last term? and if so how would I set up a function to do the partial derivative of $frac{∂θ}{∂r}$?
Thanks
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:25
A.Γ.
22.9k32656
asked Dec 29 '18 at 8:27
Radical particalRadical partical
31
edited Dec 29 '18 at 10:25
A.Γ.
22.9k32656
asked Dec 29 '18 at 8:27
Radical particalRadical partical
31
edited Dec 29 '18 at 10:25
A.Γ.
22.9k32656
edited Dec 29 '18 at 10:25
A.Γ.
22.9k32656
edited Dec 29 '18 at 10:25
A.Γ.
22.9k32656
22.9k32656
asked Dec 29 '18 at 8:27
Radical particalRadical partical
31
asked Dec 29 '18 at 8:27
Radical particalRadical partical
31
asked Dec 29 '18 at 8:27
Radical particalRadical partical
31
31
$begingroup$
$theta $ is not a function of $r $.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 8:32
$begingroup$
Just because of that we do not need to consider the last part and it is omitted?
$endgroup$
– Radical partical
Dec 29 '18 at 9:24
$begingroup$
Since $theta $ is not a function of $r $, we have $frac{∂θ}{∂r}=0$.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 9:48
add a comment |
$begingroup$
$theta $ is not a function of $r $.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 8:32
$begingroup$
Just because of that we do not need to consider the last part and it is omitted?
$endgroup$
– Radical partical
Dec 29 '18 at 9:24
$begingroup$
Since $theta $ is not a function of $r $, we have $frac{∂θ}{∂r}=0$.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 9:48
$begingroup$
$theta $ is not a function of $r $.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 8:32
$begingroup$
$theta $ is not a function of $r $.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 8:32
$begingroup$
Just because of that we do not need to consider the last part and it is omitted?
$endgroup$
– Radical partical
Dec 29 '18 at 9:24
$begingroup$
Just because of that we do not need to consider the last part and it is omitted?
$endgroup$
– Radical partical
Dec 29 '18 at 9:24
$begingroup$
Since $theta $ is not a function of $r $, we have $frac{∂θ}{∂r}=0$.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 9:48
$begingroup$
Since $theta $ is not a function of $r $, we have $frac{∂θ}{∂r}=0$.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 9:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The function $u$ is constructed as a composition
$$
u=f(x,y),qquad x=rcostheta, y=rsintheta.
$$
If we write all the variable dependencies carefully we will get
$$
u(r,theta)=f(x(r,theta),y(r,theta)).
$$
Now using the chain rule
$$
frac{partial u}{partial r}=frac{partial f}{partial x}frac{partial x}{partial r}+frac{partial f}{partial y}frac{partial y}{partial r}=frac{partial f}{partial x}costheta+frac{partial f}{partial y}sintheta.
$$
Here the partial derivatives of $f$ are again compositions that depend on $x(r,theta)$ and $y(r,theta)$, which makes the second differentiation slightly more messy, but it is the same chain rule
$$
frac{partial^2 u}{partial r^2}=frac{partial}{partial r}left(frac{partial f}{partial x}right)costheta+frac{partial}{partial r}left(frac{partial f}{partial y}right)sintheta
$$
where e.g.
$$
frac{partial}{partial r}left(frac{partial f}{partial x}right)=frac{partial}{partial x}left(frac{partial f}{partial x}right)frac{partial x}{partial r}+frac{partial}{partial y}left(frac{partial f}{partial x}right)frac{partial y}{partial r}
$$
etc
answered Dec 29 '18 at 10:42
A.Γ.A.Γ.
22.9k32656
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The function $u$ is constructed as a composition
$$
u=f(x,y),qquad x=rcostheta, y=rsintheta.
$$
If we write all the variable dependencies carefully we will get
$$
u(r,theta)=f(x(r,theta),y(r,theta)).
$$
Now using the chain rule
$$
frac{partial u}{partial r}=frac{partial f}{partial x}frac{partial x}{partial r}+frac{partial f}{partial y}frac{partial y}{partial r}=frac{partial f}{partial x}costheta+frac{partial f}{partial y}sintheta.
$$
Here the partial derivatives of $f$ are again compositions that depend on $x(r,theta)$ and $y(r,theta)$, which makes the second differentiation slightly more messy, but it is the same chain rule
$$
frac{partial^2 u}{partial r^2}=frac{partial}{partial r}left(frac{partial f}{partial x}right)costheta+frac{partial}{partial r}left(frac{partial f}{partial y}right)sintheta
$$
where e.g.
$$
frac{partial}{partial r}left(frac{partial f}{partial x}right)=frac{partial}{partial x}left(frac{partial f}{partial x}right)frac{partial x}{partial r}+frac{partial}{partial y}left(frac{partial f}{partial x}right)frac{partial y}{partial r}
$$
etc
answered Dec 29 '18 at 10:42
A.Γ.A.Γ.
22.9k32656
$endgroup$
add a comment |
$begingroup$
The function $u$ is constructed as a composition
$$
u=f(x,y),qquad x=rcostheta, y=rsintheta.
$$
If we write all the variable dependencies carefully we will get
$$
u(r,theta)=f(x(r,theta),y(r,theta)).
$$
Now using the chain rule
$$
frac{partial u}{partial r}=frac{partial f}{partial x}frac{partial x}{partial r}+frac{partial f}{partial y}frac{partial y}{partial r}=frac{partial f}{partial x}costheta+frac{partial f}{partial y}sintheta.
$$
Here the partial derivatives of $f$ are again compositions that depend on $x(r,theta)$ and $y(r,theta)$, which makes the second differentiation slightly more messy, but it is the same chain rule
$$
frac{partial^2 u}{partial r^2}=frac{partial}{partial r}left(frac{partial f}{partial x}right)costheta+frac{partial}{partial r}left(frac{partial f}{partial y}right)sintheta
$$
where e.g.
$$
frac{partial}{partial r}left(frac{partial f}{partial x}right)=frac{partial}{partial x}left(frac{partial f}{partial x}right)frac{partial x}{partial r}+frac{partial}{partial y}left(frac{partial f}{partial x}right)frac{partial y}{partial r}
$$
etc
answered Dec 29 '18 at 10:42
A.Γ.A.Γ.
22.9k32656
$endgroup$
add a comment |
$begingroup$
The function $u$ is constructed as a composition
$$
u=f(x,y),qquad x=rcostheta, y=rsintheta.
$$
If we write all the variable dependencies carefully we will get
$$
u(r,theta)=f(x(r,theta),y(r,theta)).
$$
Now using the chain rule
$$
frac{partial u}{partial r}=frac{partial f}{partial x}frac{partial x}{partial r}+frac{partial f}{partial y}frac{partial y}{partial r}=frac{partial f}{partial x}costheta+frac{partial f}{partial y}sintheta.
$$
Here the partial derivatives of $f$ are again compositions that depend on $x(r,theta)$ and $y(r,theta)$, which makes the second differentiation slightly more messy, but it is the same chain rule
$$
frac{partial^2 u}{partial r^2}=frac{partial}{partial r}left(frac{partial f}{partial x}right)costheta+frac{partial}{partial r}left(frac{partial f}{partial y}right)sintheta
$$
where e.g.
$$
frac{partial}{partial r}left(frac{partial f}{partial x}right)=frac{partial}{partial x}left(frac{partial f}{partial x}right)frac{partial x}{partial r}+frac{partial}{partial y}left(frac{partial f}{partial x}right)frac{partial y}{partial r}
$$
etc
answered Dec 29 '18 at 10:42
A.Γ.A.Γ.
22.9k32656
$endgroup$
The function $u$ is constructed as a composition
$$
u=f(x,y),qquad x=rcostheta, y=rsintheta.
$$
If we write all the variable dependencies carefully we will get
$$
u(r,theta)=f(x(r,theta),y(r,theta)).
$$
Now using the chain rule
$$
frac{partial u}{partial r}=frac{partial f}{partial x}frac{partial x}{partial r}+frac{partial f}{partial y}frac{partial y}{partial r}=frac{partial f}{partial x}costheta+frac{partial f}{partial y}sintheta.
$$
Here the partial derivatives of $f$ are again compositions that depend on $x(r,theta)$ and $y(r,theta)$, which makes the second differentiation slightly more messy, but it is the same chain rule
$$
frac{partial^2 u}{partial r^2}=frac{partial}{partial r}left(frac{partial f}{partial x}right)costheta+frac{partial}{partial r}left(frac{partial f}{partial y}right)sintheta
$$
where e.g.
$$
frac{partial}{partial r}left(frac{partial f}{partial x}right)=frac{partial}{partial x}left(frac{partial f}{partial x}right)frac{partial x}{partial r}+frac{partial}{partial y}left(frac{partial f}{partial x}right)frac{partial y}{partial r}
$$
etc
answered Dec 29 '18 at 10:42
A.Γ.A.Γ.
22.9k32656
answered Dec 29 '18 at 10:42
A.Γ.A.Γ.
22.9k32656
answered Dec 29 '18 at 10:42
A.Γ.A.Γ.
22.9k32656
answered Dec 29 '18 at 10:42
A.Γ.A.Γ.
22.9k32656
22.9k32656
add a comment |
add a comment |
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$begingroup$
$theta $ is not a function of $r $.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 8:32
$begingroup$
Just because of that we do not need to consider the last part and it is omitted?
$endgroup$
– Radical partical
Dec 29 '18 at 9:24
$begingroup$
Since $theta $ is not a function of $r $, we have $frac{∂θ}{∂r}=0$.
$endgroup$
– Thomas Shelby
Dec 29 '18 at 9:48