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, 2017 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, 2017 at 13:38

3 Answers 3


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, 2017 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, 2017 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:

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


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