$u_{rr} = f_{xx} cos^{2}(x) + 2f_{xy} cos(x) sin(x) + f_{yy} sin^{2}(x)$












0












$begingroup$


enter image description here



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $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
















0












$begingroup$


enter image description here



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $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














0












0








0





$begingroup$


enter image description here



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







share|cite|improve this question











$endgroup$




enter image description here



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 10:25









A.Γ.

22.9k32656




22.9k32656










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


















  • $begingroup$
    $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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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






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    1 Answer
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    active

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    active

    oldest

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    1












    $begingroup$

    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






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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






        share|cite|improve this answer









        $endgroup$



        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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 10:42









        A.Γ.A.Γ.

        22.9k32656




        22.9k32656






























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