Mixing
Taking partial derivatives in new coordinates
$begingroup$
When I was studying multivariable calculus, there would be questions like
Find $frac{partial}{partial r} (r^2 sintheta)$ and evaluate it at $r = 2$.
And basically an acceptable answer (in the multivariable course that I was doing), would be to just "symbolically" differentiate and get $frac{partial}{partial r} (r^2 sintheta) = 2rsintheta$ and then $4sintheta$ when $r = 2$. And usually such functions wouldn't even have a domain specified but usually one would assume that such a function was stated in polar coordinates.
But in learning multivariable analysis and basic differential geometry where I'm forced to use calculus more rigorously, I learnt that there is only one real definition for partial derivatives in $mathbb{R}^n$.
Definition: Let $U subseteq mathbb{R}^n$ be an open set and let $f : U to mathbb{R}$. The $i^{text{th}}$ partial derivative of $f$ at $p in U$ is defined to be $$frac{partial f}{partial x^i} (p) = lim_{t to 0} frac{f(p + te_i)-f(p)}{t}$$ where $e_i$ is the $i$-th basis vector for $mathbb{R}^n$ in other words $e_i = (0, dots, 1, dots, 0)$ with $1$ in the $i$-th position.
Now suppose I have a scenario where I want to calculate the partial derivative of a function defined in polar coordinates.
To be more precise let $P = (0, infty) times mathbb{R}$ (where we want to think of $P$ as being polar coordinates), let $V = (0, infty) times (-pi, pi)$ and let $f : V to mathbb{R}$ be defined by $f(r, theta) = r^2sintheta$. Now the only way I think it would be possible to calculate $frac{partial f}{partial r} (2, 0)$ would be to find a diffeomorphism $F$ from $V$ to an open subset $U$ of $mathbb{R}^2$ and then evaluate the partial derivative of the composite $f circ F^{-1}$.
It turns out that the function $F : (0, infty) times mathbb{R} to mathbb{R}^2$ defined by $F(r, theta) = (rcostheta, rsintheta)$ gives a diffeomorphism from $V$ to $U = {(x, y) in mathbb{R}^2 | x > 0, y in mathbb{R} } subseteq mathbb{R}^2$. I then think that $$frac{partial f}{partial r} (2, 0) = frac{partial (f circ F^{-1})}{partial x} (F(2, 0))$$
Am I correct in saying that? If so why was it so easy to take and evaluate partial derivatives in multivariable calculus courses when the rigorous approach seems a lot more difficult?
multivariable-calculus
asked Jan 9 at 22:39
PerturbativePerturbative
4,32311551
$endgroup$
add a comment |
$begingroup$
When I was studying multivariable calculus, there would be questions like
Find $frac{partial}{partial r} (r^2 sintheta)$ and evaluate it at $r = 2$.
And basically an acceptable answer (in the multivariable course that I was doing), would be to just "symbolically" differentiate and get $frac{partial}{partial r} (r^2 sintheta) = 2rsintheta$ and then $4sintheta$ when $r = 2$. And usually such functions wouldn't even have a domain specified but usually one would assume that such a function was stated in polar coordinates.
But in learning multivariable analysis and basic differential geometry where I'm forced to use calculus more rigorously, I learnt that there is only one real definition for partial derivatives in $mathbb{R}^n$.
Definition: Let $U subseteq mathbb{R}^n$ be an open set and let $f : U to mathbb{R}$. The $i^{text{th}}$ partial derivative of $f$ at $p in U$ is defined to be $$frac{partial f}{partial x^i} (p) = lim_{t to 0} frac{f(p + te_i)-f(p)}{t}$$ where $e_i$ is the $i$-th basis vector for $mathbb{R}^n$ in other words $e_i = (0, dots, 1, dots, 0)$ with $1$ in the $i$-th position.
Now suppose I have a scenario where I want to calculate the partial derivative of a function defined in polar coordinates.
To be more precise let $P = (0, infty) times mathbb{R}$ (where we want to think of $P$ as being polar coordinates), let $V = (0, infty) times (-pi, pi)$ and let $f : V to mathbb{R}$ be defined by $f(r, theta) = r^2sintheta$. Now the only way I think it would be possible to calculate $frac{partial f}{partial r} (2, 0)$ would be to find a diffeomorphism $F$ from $V$ to an open subset $U$ of $mathbb{R}^2$ and then evaluate the partial derivative of the composite $f circ F^{-1}$.
It turns out that the function $F : (0, infty) times mathbb{R} to mathbb{R}^2$ defined by $F(r, theta) = (rcostheta, rsintheta)$ gives a diffeomorphism from $V$ to $U = {(x, y) in mathbb{R}^2 | x > 0, y in mathbb{R} } subseteq mathbb{R}^2$. I then think that $$frac{partial f}{partial r} (2, 0) = frac{partial (f circ F^{-1})}{partial x} (F(2, 0))$$
Am I correct in saying that? If so why was it so easy to take and evaluate partial derivatives in multivariable calculus courses when the rigorous approach seems a lot more difficult?
multivariable-calculus
asked Jan 9 at 22:39
PerturbativePerturbative
4,32311551
$endgroup$
$begingroup$
What are these “polar coordinates” of which you speak? Unless you’ve got information to the contrary, this $f$ is defined on a copy of $mathbb R^2$ with coordinate system in which the coordinates happen to be called $r$ and $theta$. If $f$ were defined in terms of some other coordinate system and you were asked to compute $partial_r f$, that would be another matter.
$endgroup$
– amd
Jan 9 at 23:53
1
$begingroup$
"Symbolically" differentiating $r^2 sin(theta)$ with respect to r gives $2r sin(theta)$, not $r sin(theta)$.
$endgroup$
– user247327
Jan 10 at 0:01
$begingroup$
@user247327 Whoops thanks for spotting that
$endgroup$
– Perturbative
Jan 10 at 1:03
add a comment |
$begingroup$
When I was studying multivariable calculus, there would be questions like
Find $frac{partial}{partial r} (r^2 sintheta)$ and evaluate it at $r = 2$.
And basically an acceptable answer (in the multivariable course that I was doing), would be to just "symbolically" differentiate and get $frac{partial}{partial r} (r^2 sintheta) = 2rsintheta$ and then $4sintheta$ when $r = 2$. And usually such functions wouldn't even have a domain specified but usually one would assume that such a function was stated in polar coordinates.
But in learning multivariable analysis and basic differential geometry where I'm forced to use calculus more rigorously, I learnt that there is only one real definition for partial derivatives in $mathbb{R}^n$.
Definition: Let $U subseteq mathbb{R}^n$ be an open set and let $f : U to mathbb{R}$. The $i^{text{th}}$ partial derivative of $f$ at $p in U$ is defined to be $$frac{partial f}{partial x^i} (p) = lim_{t to 0} frac{f(p + te_i)-f(p)}{t}$$ where $e_i$ is the $i$-th basis vector for $mathbb{R}^n$ in other words $e_i = (0, dots, 1, dots, 0)$ with $1$ in the $i$-th position.
Now suppose I have a scenario where I want to calculate the partial derivative of a function defined in polar coordinates.
To be more precise let $P = (0, infty) times mathbb{R}$ (where we want to think of $P$ as being polar coordinates), let $V = (0, infty) times (-pi, pi)$ and let $f : V to mathbb{R}$ be defined by $f(r, theta) = r^2sintheta$. Now the only way I think it would be possible to calculate $frac{partial f}{partial r} (2, 0)$ would be to find a diffeomorphism $F$ from $V$ to an open subset $U$ of $mathbb{R}^2$ and then evaluate the partial derivative of the composite $f circ F^{-1}$.
It turns out that the function $F : (0, infty) times mathbb{R} to mathbb{R}^2$ defined by $F(r, theta) = (rcostheta, rsintheta)$ gives a diffeomorphism from $V$ to $U = {(x, y) in mathbb{R}^2 | x > 0, y in mathbb{R} } subseteq mathbb{R}^2$. I then think that $$frac{partial f}{partial r} (2, 0) = frac{partial (f circ F^{-1})}{partial x} (F(2, 0))$$
Am I correct in saying that? If so why was it so easy to take and evaluate partial derivatives in multivariable calculus courses when the rigorous approach seems a lot more difficult?
multivariable-calculus
asked Jan 9 at 22:39
PerturbativePerturbative
4,32311551
$endgroup$
When I was studying multivariable calculus, there would be questions like
Find $frac{partial}{partial r} (r^2 sintheta)$ and evaluate it at $r = 2$.
And basically an acceptable answer (in the multivariable course that I was doing), would be to just "symbolically" differentiate and get $frac{partial}{partial r} (r^2 sintheta) = 2rsintheta$ and then $4sintheta$ when $r = 2$. And usually such functions wouldn't even have a domain specified but usually one would assume that such a function was stated in polar coordinates.
But in learning multivariable analysis and basic differential geometry where I'm forced to use calculus more rigorously, I learnt that there is only one real definition for partial derivatives in $mathbb{R}^n$.
Definition: Let $U subseteq mathbb{R}^n$ be an open set and let $f : U to mathbb{R}$. The $i^{text{th}}$ partial derivative of $f$ at $p in U$ is defined to be $$frac{partial f}{partial x^i} (p) = lim_{t to 0} frac{f(p + te_i)-f(p)}{t}$$ where $e_i$ is the $i$-th basis vector for $mathbb{R}^n$ in other words $e_i = (0, dots, 1, dots, 0)$ with $1$ in the $i$-th position.
Now suppose I have a scenario where I want to calculate the partial derivative of a function defined in polar coordinates.
To be more precise let $P = (0, infty) times mathbb{R}$ (where we want to think of $P$ as being polar coordinates), let $V = (0, infty) times (-pi, pi)$ and let $f : V to mathbb{R}$ be defined by $f(r, theta) = r^2sintheta$. Now the only way I think it would be possible to calculate $frac{partial f}{partial r} (2, 0)$ would be to find a diffeomorphism $F$ from $V$ to an open subset $U$ of $mathbb{R}^2$ and then evaluate the partial derivative of the composite $f circ F^{-1}$.
It turns out that the function $F : (0, infty) times mathbb{R} to mathbb{R}^2$ defined by $F(r, theta) = (rcostheta, rsintheta)$ gives a diffeomorphism from $V$ to $U = {(x, y) in mathbb{R}^2 | x > 0, y in mathbb{R} } subseteq mathbb{R}^2$. I then think that $$frac{partial f}{partial r} (2, 0) = frac{partial (f circ F^{-1})}{partial x} (F(2, 0))$$
Am I correct in saying that? If so why was it so easy to take and evaluate partial derivatives in multivariable calculus courses when the rigorous approach seems a lot more difficult?
multivariable-calculus
multivariable-calculus
asked Jan 9 at 22:39
PerturbativePerturbative
4,32311551
asked Jan 9 at 22:39
PerturbativePerturbative
4,32311551
edited Jan 10 at 1:02
Perturbative
asked Jan 9 at 22:39
PerturbativePerturbative
4,32311551
asked Jan 9 at 22:39
PerturbativePerturbative
4,32311551
asked Jan 9 at 22:39
PerturbativePerturbative
4,32311551
4,32311551
$begingroup$
What are these “polar coordinates” of which you speak? Unless you’ve got information to the contrary, this $f$ is defined on a copy of $mathbb R^2$ with coordinate system in which the coordinates happen to be called $r$ and $theta$. If $f$ were defined in terms of some other coordinate system and you were asked to compute $partial_r f$, that would be another matter.
$endgroup$
– amd
Jan 9 at 23:53
1
$begingroup$
"Symbolically" differentiating $r^2 sin(theta)$ with respect to r gives $2r sin(theta)$, not $r sin(theta)$.
$endgroup$
– user247327
Jan 10 at 0:01
$begingroup$
@user247327 Whoops thanks for spotting that
$endgroup$
– Perturbative
Jan 10 at 1:03
add a comment |
$begingroup$
What are these “polar coordinates” of which you speak? Unless you’ve got information to the contrary, this $f$ is defined on a copy of $mathbb R^2$ with coordinate system in which the coordinates happen to be called $r$ and $theta$. If $f$ were defined in terms of some other coordinate system and you were asked to compute $partial_r f$, that would be another matter.
$endgroup$
– amd
Jan 9 at 23:53
1
$begingroup$
"Symbolically" differentiating $r^2 sin(theta)$ with respect to r gives $2r sin(theta)$, not $r sin(theta)$.
$endgroup$
– user247327
Jan 10 at 0:01
$begingroup$
@user247327 Whoops thanks for spotting that
$endgroup$
– Perturbative
Jan 10 at 1:03
$begingroup$
What are these “polar coordinates” of which you speak? Unless you’ve got information to the contrary, this $f$ is defined on a copy of $mathbb R^2$ with coordinate system in which the coordinates happen to be called $r$ and $theta$. If $f$ were defined in terms of some other coordinate system and you were asked to compute $partial_r f$, that would be another matter.
$endgroup$
– amd
Jan 9 at 23:53
$begingroup$
What are these “polar coordinates” of which you speak? Unless you’ve got information to the contrary, this $f$ is defined on a copy of $mathbb R^2$ with coordinate system in which the coordinates happen to be called $r$ and $theta$. If $f$ were defined in terms of some other coordinate system and you were asked to compute $partial_r f$, that would be another matter.
$endgroup$
– amd
Jan 9 at 23:53
1
1
$begingroup$
"Symbolically" differentiating $r^2 sin(theta)$ with respect to r gives $2r sin(theta)$, not $r sin(theta)$.
$endgroup$
– user247327
Jan 10 at 0:01
$begingroup$
"Symbolically" differentiating $r^2 sin(theta)$ with respect to r gives $2r sin(theta)$, not $r sin(theta)$.
$endgroup$
– user247327
Jan 10 at 0:01
$begingroup$
@user247327 Whoops thanks for spotting that
$endgroup$
– Perturbative
Jan 10 at 1:03
$begingroup$
@user247327 Whoops thanks for spotting that
$endgroup$
– Perturbative
Jan 10 at 1:03
add a comment |
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$begingroup$
What are these “polar coordinates” of which you speak? Unless you’ve got information to the contrary, this $f$ is defined on a copy of $mathbb R^2$ with coordinate system in which the coordinates happen to be called $r$ and $theta$. If $f$ were defined in terms of some other coordinate system and you were asked to compute $partial_r f$, that would be another matter.
$endgroup$
– amd
Jan 9 at 23:53
1
$begingroup$
"Symbolically" differentiating $r^2 sin(theta)$ with respect to r gives $2r sin(theta)$, not $r sin(theta)$.
$endgroup$
– user247327
Jan 10 at 0:01
$begingroup$
@user247327 Whoops thanks for spotting that
$endgroup$
– Perturbative
Jan 10 at 1:03