Given is the curve $\mathbf{\gamma}(p)$ with components




Then $\frac{d\mathbf{\gamma}(p)}{dp}=\left(\begin{array}{c} \cos(p)\\ -\sin(p)\\ 1\\ \end{array}\right)$

Let $\mathbf{A}(p)=\frac{d\mathbf{\gamma}(p)}{dp}$. Expressing $\mathbf{A}(p)$ via the coordinates we can write

$\mathbf{A}(p)(1)=\left(\begin{array}{c} y(p)\\ -\sqrt{1-y^2(p)}\\ 1\\ \end{array}\right)$, or

$\mathbf{A}(p)(2)=\left(\begin{array}{c} \sqrt{1-x^2(p)}\\ -x(p)\\ 1\\ \end{array}\right)$, or

$\mathbf{A}(p)(3)=\left(\begin{array}{c} y(p)\\ -x(p)\\ 1\\ \end{array}\right)$, or

$\mathbf{A}(p)(4)=\left(\begin{array}{c} \cos(z(p))\\ -\sin(z(p))\\ 1\\ \end{array}\right)$.

They are different ways to write A. Now, write the chain rule

$\frac{d\mathbf{A}(p)}{dp}=\frac{\partial \mathbf{A}}{\partial x} \frac{dx}{dp}+\frac{\partial \mathbf{A}}{\partial y} \frac{dy}{dp}+\frac{\partial \mathbf{A}}{\partial z} \frac{dz}{dp}$

Plug in $\mathbf{A}(p)(1)$ and get

$\frac{d\mathbf{A}(p)}{dp}(1)=\left(\begin{array}{c} -\sin(p)\\ \frac{-y \sin(p)}{\sqrt{1-y^2}}\\ 0\\ \end{array}\right)=\left(\begin{array}{c} -\sin(p)\\ -\cos(p)\\ 0\\ \end{array}\right)$

Plug in $\mathbf{A}(p)(2)$ and get

$\frac{d\mathbf{A}(p)}{dp}(2)= \left(\begin{array}{c} \frac{-x \cos(p)}{\sqrt{1-x^2}}\\ -\cos(p)\\ 0\\ \end{array}\right)=\left(\begin{array}{c} -\sin(p)\\ -\cos(p)\\ 0\\ \end{array}\right)$

Plug in $\mathbf{A}(p)(3)$ and get

$\frac{d\mathbf{A}(p)}{dp}(3)= \left(\begin{array}{c} 0\\ -1\\ 0\\ \end{array}\right)\cos(p)+\left(\begin{array}{c} 1\\ 0\\ 0\\ \end{array}\right)(-\sin(p))+0=\left(\begin{array}{c} -\sin(p)\\ -\cos(p)\\ 0\\ \end{array}\right)$

Plug in $\mathbf{A}(p)(4)$ and get

$\frac{d\mathbf{A}(p)}{dp}(4)=0+0+ \left(\begin{array}{c} -\sin(z)\\ -\cos(z)\\ 0\\ \end{array}\right)=\left(\begin{array}{c} -\sin(p)\\ -\cos(p)\\ 0\\ \end{array}\right)$

In all cases the result ends up the same, but the partial derivatives $\frac{\partial\mathbf{A}}{\partial x_i}$ are ambiguous. Still, adding ambiguous terms together consistently results in a correct answer.

User Med suggested that since $y(p)$ is not independent of $x(p)$ etc., we can apply the following chain rule to remove ambiguity:

$\frac{\partial \mathbf{A}}{\partial x_i}= \frac{d\mathbf{A}}{dp}\frac{dp}{dx_i} = \frac{d\mathbf{A}}{dp}\frac{1}{\frac{dx_i}{dp}}$, which works fine in cases (1), (2) and (4), because $\mathbf{A}$ contains only one space variable.

But is case (3) both partial derivatives w/respect to x and y are nonzero and then this workaround can't be used anymore.

Is there a general approach how to avoid the ambiguity in the partial derivatives? How do I express $\frac{\partial\mathbf{A}}{\partial x_i}$ in case (3)?

  • $\begingroup$ I'm not sure what you mean by saying the partial derivatives $\partial \mathbf{A}/\partial x_i$ are ambiguous. In what sense are they ambiguous? $\endgroup$ – WB-man Apr 11 '17 at 1:33
  • $\begingroup$ They depend on how A was written, and A can be written in a number of equivalent ways $\endgroup$ – user142523 Apr 11 '17 at 13:38

I think you're getting tripped up by the notation, which can easily happen when using the derivative notation $\partial \mathbf A/\partial x$. $\newcommand{\A}{\mathbf A}$

The chain rule says that if $\A(x,y,z)$ is a multivariable function, and then we construct the single-variable function $\A\big(x(p),\ y(p),\ z(p) \big)$ with $x$, $y$, and $z$ functions of another variable $p$, then the derivative of the single-variable function of $p$ is $$ \frac{d}{dp} \A\big(x(p),\ y(p),\ z(p) \big) = \frac{\partial \A}{\partial x} \frac{dx}{dp}\ +\ \frac{\partial \A}{\partial y} \frac{dy}{dp}\ +\ \frac{\partial \A}{\partial z} \frac{dz}{dp} $$ where the notation $\partial \A/\partial x$ in this context denotes the partial derivative of the multivariable function $\A$ with respect to its first component evaluated at the particular vector $\big(x(p), y(p), z(p) \big)$ for every $p$, and similarly for $\partial \A/\partial y$ and $\partial \A/\partial z$.

The critical thing to note is that $\A$ denotes a multivariable function that takes any triple $(x,y,z)$ and maps it to another triple. But $\A\big(x(p),\ y(p),\ z(p) \big)$ denotes the single-variable function that takes a number $p$ and returns a triple. In other words, it is a composition of a multivariable function with a single-variable function, thus resulting in a single-variable function. That means the notation $\frac{\partial}{\partial x} \A\big(x(p),\ y(p),\ z(p) \big)$ is meaningless: how can we take the partial derivative of a single-variable function?

The difference for your case is that you defined (before you ever mentioned the chain rule) that $\A(p)$ equals something. That makes $\A$ a single-variable function. Thus (strictly speaking) the expression $\partial \A/\partial x$ is meaningless. Although you note that $\A$ can be rewritten so that it looks like a multivariable function. Here's one rewriting (the one you're chiefly interested in): $$ \mathbf{A}(p)=\left(\begin{array}{c} y(p)\\ -x(p)\\ 1\\ \end{array}\right) $$ Even written like this, $\A$ is still merely a single-variable function. But it fits the template of a multivariable function. Namely this one: $$(x,y,z) \mapsto \begin{pmatrix} y \\ -x \\ 1 \end{pmatrix} $$ $\newcommand{\T}{\mathbf T}$ I'll call this function $\T$ for template. In that case, we may write $$ \A(p) = \T\big(x(p),\ y(p),\ z(p) \big) $$ and write the chain rule as \begin{align} \frac{d\A}{dp} &= \frac{\partial \T}{\partial x} \frac{dx}{dp} + \frac{\partial \T}{\partial y} \frac{dy}{dp} + \frac{\partial \T}{\partial z} \frac{dz}{dp} \\ &= \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} \cos p + \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \cdot -\sin p + \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \cdot 1 \\ &= \begin{pmatrix} -\sin p \\ -\cos p \\ 0 \end{pmatrix} \end{align} as expected.

Of course, it is a lot of trouble to separately define $\T$, and it looks more intuitive to just write $\partial \A/\partial x$. That's fine as long as it's clear from context what your template function is, and you know what's really going on under the hood.

  • $\begingroup$ Thank you for your answer. I also wrote an attempted answer, could you comment on it? I agree with your argument about the definitions of A and T. I am only wondering if the partial derivatives of T can become independent of how A was written. $\endgroup$ – user142523 Apr 12 '17 at 15:31
  • $\begingroup$ The question here is, does a vector field defined along a curve have spatial derivatives? Since the curve can be a spatial curve, one feels the answer should be a 'yes', but how do we express them then $\endgroup$ – user142523 Apr 12 '17 at 16:00

This is mainly in response to the answer you provided (and the questions you ask therein). I'd put it as a comment, but it's too long. Hopefully it will help. $\newcommand{\A}{\mathbf A} \newcommand{\T}{\mathbf T}$

As I mentioned before, you defined $\A$ as a function of $p$. Therefore, strictly speaking, the notation $\partial \A/\partial x$ has no meaning.

So we need to define a meaning for this notation. I take the meaning you're after is "the instantaneous rate of change of the vector $\A$ when $x$ changes by a very small amount". In other words: $$ \frac{\partial \A}{\partial x} := \lim_{\Delta x \to 0} \frac{\Delta \A}{\Delta x} = \lim_{\Delta p \to 0} \frac{\Delta \A / \Delta p}{\Delta x / \Delta p} = \frac{d\A/dp}{dx/dp} $$ which is the formula you wrote. It should look similar to the formula used to compute the derivative $dy/dx$ when $x$ and $y$ are defined parametrically in terms of a variable $t$.

Now, you wrote $d\A/dp$ using the multivariable chain rule as $$ \frac{d\A}{dp} = \frac{\partial \A}{\partial x} \frac{dx}{dp} + \frac{\partial \A}{\partial y} \frac{dy}{dp} + \frac{\partial \A}{\partial z} \frac{dz}{dp} $$ But the weird thing is, if we solve the equation $$ \frac{\partial \A}{\partial x} = \frac{d\A/dp}{dx/dp} $$ for $d\A/dp$, we get $$ \frac{d\A}{dp} = \frac{\partial \A}{\partial x} \cdot \frac{dx}{dp} $$ without the additional terms! What happened to them?? I think this is where the notational confusion comes in. As I mentioned in my other answer, the multivariable chain rule doesn't really apply to $\A$ since $\A$ is single-variable. However, $\A$ can be made to "look" multivariable if you can match it to a template function $\T$. With the template function (which is multivariable), the chain rule can be properly written as $$ \frac{d\A}{dp} = \frac{\partial \T}{\partial x} \frac{dx}{dp} + \frac{\partial \T}{\partial y} \frac{dy}{dp} + \frac{\partial \T}{\partial z} \frac{dz}{dp} $$ The thing is, in this answer here, we've given a different definition to $\partial \A/\partial x$, which is incompatible with the template function definition. That is, $$ \frac{\partial \A}{\partial x} \neq \frac{\partial \T}{\partial x} $$ You can see why if we consider your third case, with the template function $$ \T(x,y,z) = (y, -x,\ 1) $$ whose partial with respect to $x$ is $$ \frac{\partial \T}{\partial x} = (0, -1,\ 0) $$ But as for $\partial \A / \partial x$: \begin{align} \frac{\partial \A}{\partial x} &= \frac{d\A/dp}{dx/dp} = \begin{pmatrix} -\sin p \\ -\cos p \\ 0 \end{pmatrix} \cdot \frac{1}{\cos p} \\ &= \begin{pmatrix} -\tan p \\ -1 \\ 0 \end{pmatrix} \end{align} which are clearly different for almost all $p$ values.

Interestingly, I think $\partial \A/\partial x$ does equal $\partial \T/\partial x$ in the other cases, but only by a fluke since the template function, though technically dependent on $x$, $y$, and $z$, is really only meaningfully dependent on one of those variables and is constant in terms of the other variables. Thus, for those cases, $\T$ is formally single-variable, and you can get away with confusing the definitions of $\A$ and $\T$ (because the partials with respect to the other variables will be zero, and hence the additional terms of the multivariable chain rule will vanish, but again, I think this is just a fluke of the triviality of the template function).


Attempted answer:

This is what I attempted to avoid the ambiguity... Consider unit-speed curves.

The curve




is unit-speed and $\mathbf{A}(p)$ is its tangent vector with norm 1.

Above, $\mathbf{A}(p)$ was written in four equivalent ways. But (3) is incorrect, because its components don't contain the relationship they have through $p$ (notice that $\mathbf{A}_x(p),\mathbf{A}_y(p),\mathbf{A}_z(p)$ are interdependent via $p$). The vector (3)

$\left(\begin{array}{c} y\\ -x\\ 1\\ \end{array}\right)$

can be considered incorrect, because its norm is not equal to 1, unless we substitute in $x(p)$ and $y(p)$ their definitions. All other cases (1), (2) and (4) are fine, because their components show how they are inter-related via $p$ and their norm is 1, thus showing that the tangent vector they represent belongs to a unit-speed curve.

Then, let $\mathbf{A}(p)$ be written for now to depend on only one of the spatial variables $x$, $y$ or $z$. This way we arrive at the expression (as per user Med's suggestion)

$\frac{\partial \mathbf{A}}{\partial x_i}= \frac{d \mathbf{A}}{dx_i}=\frac{d \mathbf{A}}{dp} \frac{dp}{dx_i}$

which yields the non-ambiguous formula for the spatial derivatives of $\mathbf{A}(p)$:

$\frac{\partial \mathbf{A}}{\partial x_i}=\frac{\mathbf{A}^{\prime}}{x_i^{\prime}}$

(prime means d/dp). This formula always produces the same result no matter which spatial variable we chose to express $\mathbf{A}$ into!!! One needs to remember, that when we take $\frac{\partial}{\partial x}$ of say, (1), $y$ depends on $x$ via $p$ and thus the partial derivative is nonzero.

Notice that $\frac{\partial (1)}{\partial x}=\frac{\partial (2)}{\partial x}=\frac{\partial (3)}{\partial x}=\frac{\partial (4)}{\partial x}$

It even works on (3) too.

Notice that only one derivative $\frac{\partial \mathbf{A}}{\partial x_i}$ should be treated as nonzero, all others $\frac{\partial \mathbf{A}}{\partial x_j}$ (where $j\ne i$) should be treated as equal to zero. That is, the notation should be consistent. If we chose to express $\mathbf{A}$ via $x$, we should stay with this notation. Because in this way we ensure that the chain rule is meaningful:

$\frac{d\mathbf{A}}{dp}=\frac{\partial \mathbf{A}}{\partial x}\frac{dx}{dp}+\frac{\partial \mathbf{A}}{\partial y}\frac{dy}{dp}+\frac{\partial \mathbf{A}}{\partial z}\frac{dz}{dp} = \frac{\mathbf{A}^{\prime}}{x^{\prime}}x^{\prime}+0 +0= \mathbf{A}^{\prime}=\frac{d\mathbf{A}}{dp}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.