Partial derivatives with a variable held constant

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Consider the cylindrical coordinate system $(rho,theta, z)$. If we define a new variable $zeta = theta - hz$, is there any way of rigorously defining the following partial derivative?
$$
X=fracpartialpartial zbigg|_zeta,rho
$$
Where $|_zeta,rho$ indicates that we are taking the derivative while holding the variables $zeta$ and $rho$ constant. This concept doesn't seem rigorously defined. Thinking of $X$ as a vector field, I imagine that it is the same as the projection of the constant unit vector field $partial/partial z$ onto the tangent spaces of the surfaces $zeta=textconst$ and $rho=textconst$, which would lead me to the conclusion that
$$
fracpartialpartial zbigg|_zeta,rho = fracpartialpartial zbigg|_theta,rho + fracpartialpartial thetabigg|_z,rho
$$
My main question is: How can I define the concept of 'holding a variable constant'?

calculus vector-analysis change-of-basis

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asked Aug 3 at 10:38

Tommy1101

614

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't you need a multiplier of $h$ on one of these partials?
  – ancientmathematician
  Aug 3 at 11:20
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Consider the cylindrical coordinate system $(rho,theta, z)$. If we define a new variable $zeta = theta - hz$, is there any way of rigorously defining the following partial derivative?
$$
X=fracpartialpartial zbigg|_zeta,rho
$$
Where $|_zeta,rho$ indicates that we are taking the derivative while holding the variables $zeta$ and $rho$ constant. This concept doesn't seem rigorously defined. Thinking of $X$ as a vector field, I imagine that it is the same as the projection of the constant unit vector field $partial/partial z$ onto the tangent spaces of the surfaces $zeta=textconst$ and $rho=textconst$, which would lead me to the conclusion that
$$
fracpartialpartial zbigg|_zeta,rho = fracpartialpartial zbigg|_theta,rho + fracpartialpartial thetabigg|_z,rho
$$
My main question is: How can I define the concept of 'holding a variable constant'?

calculus vector-analysis change-of-basis

share|cite|improve this question

asked Aug 3 at 10:38

Tommy1101

614

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't you need a multiplier of $h$ on one of these partials?
  – ancientmathematician
  Aug 3 at 11:20
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Consider the cylindrical coordinate system $(rho,theta, z)$. If we define a new variable $zeta = theta - hz$, is there any way of rigorously defining the following partial derivative?
$$
X=fracpartialpartial zbigg|_zeta,rho
$$
Where $|_zeta,rho$ indicates that we are taking the derivative while holding the variables $zeta$ and $rho$ constant. This concept doesn't seem rigorously defined. Thinking of $X$ as a vector field, I imagine that it is the same as the projection of the constant unit vector field $partial/partial z$ onto the tangent spaces of the surfaces $zeta=textconst$ and $rho=textconst$, which would lead me to the conclusion that
$$
fracpartialpartial zbigg|_zeta,rho = fracpartialpartial zbigg|_theta,rho + fracpartialpartial thetabigg|_z,rho
$$
My main question is: How can I define the concept of 'holding a variable constant'?

calculus vector-analysis change-of-basis

share|cite|improve this question

asked Aug 3 at 10:38

Tommy1101

614

Consider the cylindrical coordinate system $(rho,theta, z)$. If we define a new variable $zeta = theta - hz$, is there any way of rigorously defining the following partial derivative?
$$
X=fracpartialpartial zbigg|_zeta,rho
$$
Where $|_zeta,rho$ indicates that we are taking the derivative while holding the variables $zeta$ and $rho$ constant. This concept doesn't seem rigorously defined. Thinking of $X$ as a vector field, I imagine that it is the same as the projection of the constant unit vector field $partial/partial z$ onto the tangent spaces of the surfaces $zeta=textconst$ and $rho=textconst$, which would lead me to the conclusion that
$$
fracpartialpartial zbigg|_zeta,rho = fracpartialpartial zbigg|_theta,rho + fracpartialpartial thetabigg|_z,rho
$$
My main question is: How can I define the concept of 'holding a variable constant'?

calculus vector-analysis change-of-basis

share|cite|improve this question

asked Aug 3 at 10:38

Tommy1101

614

share|cite|improve this question

share|cite|improve this question

asked Aug 3 at 10:38

Tommy1101

614

asked Aug 3 at 10:38

Tommy1101

614

asked Aug 3 at 10:38

Tommy1101

614

614

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't you need a multiplier of $h$ on one of these partials?
  – ancientmathematician
  Aug 3 at 11:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't you need a multiplier of $h$ on one of these partials?
  – ancientmathematician
  Aug 3 at 11:20
  

  
  
  
  

Don't you need a multiplier of $h$ on one of these partials?
– ancientmathematician
Aug 3 at 11:20

Don't you need a multiplier of $h$ on one of these partials?
– ancientmathematician
Aug 3 at 11:20

add a comment | 

1 Answer
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I suggest that if you want to think clearly about this you oughtn't to introduce "a new variable", you ought to introduce a new triple of variables. The partial derivative of a function wrt any one of a triple of variables is well-defined, there's no need to keep repeating the rather confusing "keeping the other two fixed".

I also suggest that suppression of the function(s) involved is another source of confusion.

Suppose, then, that you are interested in a function $f(rho,theta,z)$. Suppose that each of these variables can be expressed in terms of a new triple of variables $R,Theta,Z$: explicitly $rho=rho(R,Theta,Z)$, $theta=theta(R,Theta,Z)$ and $z=z(R,Theta,Z)$.

Note: You are interested in the case $rho(R,Theta,Z)=R$, $theta(R,Theta,Z)=Theta+hZ$, $z(R,Theta,Z)=Z$.

Now the change of variables means that we have a new function

$$ F(R,Theta,Z):=f(rho(R,Theta,Z),theta(R,Theta,Z),z(R,Theta,Z)).
$$

Note: You are interested in the partial derivative of $F$ with respect to the third variable, $Z$.

By the Chain rule we have:
$$
fracpartial Fpartial Z
=
fracpartial fpartial rhofracpartial rho partial Z
+
fracpartial fpartial thetafracpartial thetapartial Z
+
fracpartial fpartial zfracpartial zpartial Z
.
$$

Note: In your case
$$
fracpartial rho partial Z=0, fracpartial thetapartial Z=h,
fracpartial zpartial Z=1
$$
and so
$$
fracpartial Fpartial Z
=
hfracpartial fpartial theta
+
fracpartial fpartial z
.
$$

As I say, I think that it is confusing to suppress the functions -- the function of the left is not the function of the right -- and I also think that embroidering the partials with other symbols is confusing. (I know it is done in the old-fashioned books.)

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answered Aug 3 at 11:57

ancientmathematician

4,0681312

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

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votes

up vote
0
down vote

I suggest that if you want to think clearly about this you oughtn't to introduce "a new variable", you ought to introduce a new triple of variables. The partial derivative of a function wrt any one of a triple of variables is well-defined, there's no need to keep repeating the rather confusing "keeping the other two fixed".

I also suggest that suppression of the function(s) involved is another source of confusion.

Suppose, then, that you are interested in a function $f(rho,theta,z)$. Suppose that each of these variables can be expressed in terms of a new triple of variables $R,Theta,Z$: explicitly $rho=rho(R,Theta,Z)$, $theta=theta(R,Theta,Z)$ and $z=z(R,Theta,Z)$.

Note: You are interested in the case $rho(R,Theta,Z)=R$, $theta(R,Theta,Z)=Theta+hZ$, $z(R,Theta,Z)=Z$.

Now the change of variables means that we have a new function

$$ F(R,Theta,Z):=f(rho(R,Theta,Z),theta(R,Theta,Z),z(R,Theta,Z)).
$$

Note: You are interested in the partial derivative of $F$ with respect to the third variable, $Z$.

By the Chain rule we have:
$$
fracpartial Fpartial Z
=
fracpartial fpartial rhofracpartial rho partial Z
+
fracpartial fpartial thetafracpartial thetapartial Z
+
fracpartial fpartial zfracpartial zpartial Z
.
$$

Note: In your case
$$
fracpartial rho partial Z=0, fracpartial thetapartial Z=h,
fracpartial zpartial Z=1
$$
and so
$$
fracpartial Fpartial Z
=
hfracpartial fpartial theta
+
fracpartial fpartial z
.
$$

As I say, I think that it is confusing to suppress the functions -- the function of the left is not the function of the right -- and I also think that embroidering the partials with other symbols is confusing. (I know it is done in the old-fashioned books.)

share|cite|improve this answer

answered Aug 3 at 11:57

ancientmathematician

4,0681312

add a comment | 

up vote
0
down vote

I suggest that if you want to think clearly about this you oughtn't to introduce "a new variable", you ought to introduce a new triple of variables. The partial derivative of a function wrt any one of a triple of variables is well-defined, there's no need to keep repeating the rather confusing "keeping the other two fixed".

I also suggest that suppression of the function(s) involved is another source of confusion.

Suppose, then, that you are interested in a function $f(rho,theta,z)$. Suppose that each of these variables can be expressed in terms of a new triple of variables $R,Theta,Z$: explicitly $rho=rho(R,Theta,Z)$, $theta=theta(R,Theta,Z)$ and $z=z(R,Theta,Z)$.

Note: You are interested in the case $rho(R,Theta,Z)=R$, $theta(R,Theta,Z)=Theta+hZ$, $z(R,Theta,Z)=Z$.

Now the change of variables means that we have a new function

$$ F(R,Theta,Z):=f(rho(R,Theta,Z),theta(R,Theta,Z),z(R,Theta,Z)).
$$

Note: You are interested in the partial derivative of $F$ with respect to the third variable, $Z$.

By the Chain rule we have:
$$
fracpartial Fpartial Z
=
fracpartial fpartial rhofracpartial rho partial Z
+
fracpartial fpartial thetafracpartial thetapartial Z
+
fracpartial fpartial zfracpartial zpartial Z
.
$$

Note: In your case
$$
fracpartial rho partial Z=0, fracpartial thetapartial Z=h,
fracpartial zpartial Z=1
$$
and so
$$
fracpartial Fpartial Z
=
hfracpartial fpartial theta
+
fracpartial fpartial z
.
$$

As I say, I think that it is confusing to suppress the functions -- the function of the left is not the function of the right -- and I also think that embroidering the partials with other symbols is confusing. (I know it is done in the old-fashioned books.)

share|cite|improve this answer

answered Aug 3 at 11:57

ancientmathematician

4,0681312

add a comment | 

up vote
0
down vote

up vote
0
down vote

I suggest that if you want to think clearly about this you oughtn't to introduce "a new variable", you ought to introduce a new triple of variables. The partial derivative of a function wrt any one of a triple of variables is well-defined, there's no need to keep repeating the rather confusing "keeping the other two fixed".

I also suggest that suppression of the function(s) involved is another source of confusion.

Suppose, then, that you are interested in a function $f(rho,theta,z)$. Suppose that each of these variables can be expressed in terms of a new triple of variables $R,Theta,Z$: explicitly $rho=rho(R,Theta,Z)$, $theta=theta(R,Theta,Z)$ and $z=z(R,Theta,Z)$.

Note: You are interested in the case $rho(R,Theta,Z)=R$, $theta(R,Theta,Z)=Theta+hZ$, $z(R,Theta,Z)=Z$.

Now the change of variables means that we have a new function

$$ F(R,Theta,Z):=f(rho(R,Theta,Z),theta(R,Theta,Z),z(R,Theta,Z)).
$$

Note: You are interested in the partial derivative of $F$ with respect to the third variable, $Z$.

By the Chain rule we have:
$$
fracpartial Fpartial Z
=
fracpartial fpartial rhofracpartial rho partial Z
+
fracpartial fpartial thetafracpartial thetapartial Z
+
fracpartial fpartial zfracpartial zpartial Z
.
$$

Note: In your case
$$
fracpartial rho partial Z=0, fracpartial thetapartial Z=h,
fracpartial zpartial Z=1
$$
and so
$$
fracpartial Fpartial Z
=
hfracpartial fpartial theta
+
fracpartial fpartial z
.
$$

As I say, I think that it is confusing to suppress the functions -- the function of the left is not the function of the right -- and I also think that embroidering the partials with other symbols is confusing. (I know it is done in the old-fashioned books.)

share|cite|improve this answer

answered Aug 3 at 11:57

ancientmathematician

4,0681312

I suggest that if you want to think clearly about this you oughtn't to introduce "a new variable", you ought to introduce a new triple of variables. The partial derivative of a function wrt any one of a triple of variables is well-defined, there's no need to keep repeating the rather confusing "keeping the other two fixed".

I also suggest that suppression of the function(s) involved is another source of confusion.

Suppose, then, that you are interested in a function $f(rho,theta,z)$. Suppose that each of these variables can be expressed in terms of a new triple of variables $R,Theta,Z$: explicitly $rho=rho(R,Theta,Z)$, $theta=theta(R,Theta,Z)$ and $z=z(R,Theta,Z)$.

Note: You are interested in the case $rho(R,Theta,Z)=R$, $theta(R,Theta,Z)=Theta+hZ$, $z(R,Theta,Z)=Z$.

Now the change of variables means that we have a new function

$$ F(R,Theta,Z):=f(rho(R,Theta,Z),theta(R,Theta,Z),z(R,Theta,Z)).
$$

Note: You are interested in the partial derivative of $F$ with respect to the third variable, $Z$.

By the Chain rule we have:
$$
fracpartial Fpartial Z
=
fracpartial fpartial rhofracpartial rho partial Z
+
fracpartial fpartial thetafracpartial thetapartial Z
+
fracpartial fpartial zfracpartial zpartial Z
.
$$

Note: In your case
$$
fracpartial rho partial Z=0, fracpartial thetapartial Z=h,
fracpartial zpartial Z=1
$$
and so
$$
fracpartial Fpartial Z
=
hfracpartial fpartial theta
+
fracpartial fpartial z
.
$$

As I say, I think that it is confusing to suppress the functions -- the function of the left is not the function of the right -- and I also think that embroidering the partials with other symbols is confusing. (I know it is done in the old-fashioned books.)

share|cite|improve this answer

answered Aug 3 at 11:57

ancientmathematician

4,0681312

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 11:57

ancientmathematician

4,0681312

answered Aug 3 at 11:57

ancientmathematician

4,0681312

answered Aug 3 at 11:57

ancientmathematician

4,0681312

4,0681312

add a comment | 

add a comment | 

 

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