Partial Derivative Paradox?

Calculus Level 2

Consider a point parameterized in spherical coordinates. Its coordinates are:

x = r sin ⁡ θ cos ⁡ ϕ x = r \sin{\theta} \cos{\phi} y = r sin ⁡ θ sin ⁡ ϕ y = r \sin{\theta} \sin{\phi} z = r cos ⁡ θ z = r \cos{\theta}

It is known that:

x θ = ∂ x ∂ θ = r cos ⁡ θ cos ⁡ ϕ x_{\theta}=\frac{\partial x}{\partial \theta} = r \cos{\theta} \cos{\phi}

Now consider the spherical coordinate parameters be expressed in terms of Cartesian coordinates. For θ \theta , this turns out to be:

θ = arccos ⁡ ( z x 2 + y 2 + z 2 ) \theta = \arccos\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)

Compute the quantity: θ x = ∂ θ ∂ x \theta_x=\frac{\partial \theta}{\partial x}

After computing, eliminate x x , y y and z z using the relations above to write the resulting expression in terms of r r , θ \theta and ϕ \phi . Is the following true or false? x θ = 1 θ x x_{\theta}=\frac{1}{\theta_x}

In other words, is the following true or false?

∂ x ∂ θ = 1 ∂ θ ∂ x \frac{\partial x}{\partial \theta} = \frac{1}{\frac{\partial \theta}{\partial x}}

Bonus: Explain why? Also, do this little exercise for the other cartesian and spherical parameters to verify your finding.

True False

Karan Chatrath by Karan Chatrath , 27, Netherlands

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3 solutions

Let x = x ( r , θ , Φ ) x=x(r,\theta, Φ) . Then d x = ∂ x ∂ r d r + ∂ x ∂ θ d θ + ∂ x ∂ Φ d Φ = ∂ x ∂ r ( ∂ r ∂ x d x + ∂ r ∂ y d y + ∂ r ∂ z d z ) + ∂ x ∂ θ ( ∂ θ ∂ x d x + ∂ θ ∂ y d y + ∂ θ ∂ z d z ) + ∂ x ∂ Φ ( ∂ Φ ∂ x d x + ∂ Φ ∂ y d y + ∂ Φ ∂ z d z ) = ( ∂ x ∂ r . ∂ r ∂ x + ∂ x ∂ θ . ∂ θ ∂ x + ∂ x ∂ Φ . ∂ Φ ∂ x ) d x + dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta+\frac{\partial x}{\partial Φ}dΦ=\frac{\partial x}{\partial r}(\frac{\partial r}{\partial x}dx+\frac{\partial r}{\partial y}dy+\frac{\partial r}{\partial z}dz)+\frac{\partial x}{\partial \theta}(\frac{\partial \theta}{\partial x}dx+\frac {\partial \theta}{\partial y}dy+\frac{\partial \theta}{\partial z}dz)+\frac{\partial x}{\partial Φ}(\frac{\partial Φ}{\partial x}dx+\frac{\partial Φ}{\partial y}dy+\frac{\partial Φ}{\partial z}dz)=(\frac {\partial x}{\partial r}.\frac{\partial r}{\partial x}+\frac{\partial x}{\partial \theta}.\frac{\partial \theta}{\partial x}+\frac{\partial x}{\partial Φ}.\frac{\partial Φ}{\partial x})dx+ similar terms containing d y dy and d z dz . So

∂ x ∂ r . ∂ r ∂ x + ∂ x ∂ θ . ∂ θ ∂ x + ∂ x ∂ Φ . ∂ Φ ∂ x = 1 \frac{\partial x}{\partial r}.\frac{\partial r}{\partial x}+\frac{\partial x}{\partial \theta }.\frac{\partial \theta}{\partial x}+\frac{\partial x}{\partial Φ}.\frac{\partial Φ}{\partial x}=1 .

Since none of the terms in the L. H. S. is identically zero, therefore none is identically one also.

2 Helpful 2 Interesting 3 Brilliant 0 Confused

Thank you so much for this solution. Really appreciate it.

So basically, the coefficients of d y dy and d z dz will be zero for the above example, am I correct?

Karan Chatrath - 1 year, 1 month ago

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Yes. Same for the other two equations for d y dy and d z dz .

A Former Brilliant Member - 1 year, 1 month ago
Steven Chase
May 3, 2020

Here's what I came up with.

θ = a c o s ( z x 2 + y 2 + z 2 ) \theta = acos\Big( \frac{z}{\sqrt{x^2 + y^2 + z^2}} \Big)

∂ θ ∂ x = − 1 1 − z 2 x 2 + y 2 + z 2 ( − z 2 ) ( x 2 + y 2 + z 2 ) − 3 / 2 ( 2 x ) \frac{\partial{\theta}}{\partial{x}} = \frac{-1}{\sqrt{1 - \frac{z^2}{x^2 + y^2 + z^2}}} (-\frac{z}{2}) (x^2 + y^2 + z^2)^{-3/2} (2 x)

Making some substitutions:

∂ θ ∂ x = x z 1 − c o s 2 θ r − 3 = x z sin ⁡ θ r − 3 = r 2 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ sin ⁡ θ r − 3 = cos ⁡ θ cos ⁡ ϕ r ≠ 1 r cos ⁡ θ cos ⁡ ϕ \frac{\partial{\theta}}{\partial{x}} = \frac{x z}{\sqrt{1 - cos^2 \theta}} r^{-3} = \frac{x z}{\sin \theta} r^{-3} = \frac{r^2 \sin \theta \, \cos \theta \, \cos \phi}{\sin \theta} r^{-3} = \frac{\cos \theta \, \cos \phi}{r} \neq \frac{1}{ r \cos \theta \, \cos \phi}

Intuitively, I expected this reciprocal property to be true. So that's a surprise

Suppose we try a simpler example:

f = x 2 + y x = f − y ∂ f ∂ x = 2 x ∂ x ∂ f = 1 2 ( f − y ) − 1 / 2 = 1 2 x f = x^2 + y \\ x = \sqrt{f - y} \\ \frac{\partial{f}}{\partial{x}} = 2x \\ \frac{\partial{x}}{\partial{f}} = \frac{1}{2} (f - y)^{-1/2} = \frac{1}{2 x}

In the simpler case, the reciprocal rule works. I am reading some articles about the requirements for the reciprocal property to be true, but I haven't come to any sound conclusions yet.

2 Helpful 2 Interesting 2 Brilliant 0 Confused

@Steven Chase I guess the answer , fluke. And it got correct(I am first solver also) Well, I think we can also use some logical ways to answer this (True or False) questions. If it was true then why should author write the problem. It means there is something fault. BTW nice problem.

A Former Brilliant Member - 1 year, 1 month ago

Thanks for the solution. I have been dwelling on this too and am referring to articles. Nothing helpful so far. If you arrive at a reason, please do share it with me.

Karan Chatrath - 1 year, 1 month ago

Perhaps I'm missing something (it's late here, and social isolation with two young children is taking its toll) but isn't this a simpler example still? x = t + u y = t − u \begin{aligned} x&=t+u \\ y&=t-u \end{aligned}

We have t = 1 2 ( x + y ) t=\frac12 (x+y) , so t x = 1 2 t_x=\frac12 , but of course x t = 1 x_t=1 . At least, I think this is an example of the same thing!

Chris Lewis - 1 year, 1 month ago

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Yes, it is an example of the same thing. Thanks for sharing.

Karan Chatrath - 1 year, 1 month ago
Uros Stojkovic
May 11, 2020

Shortcut: use chain rule . cos ⁡ θ = z r − sin ⁡ θ ∂ θ ∂ x = − z r 2 ∂ r ∂ x x 2 + y 2 r ∂ θ ∂ x = z r 2 x r ∂ θ ∂ x = z r 2 x x 2 + y 2 ∂ θ ∂ x = cos ⁡ θ cos ⁡ φ r \begin{aligned} \cos{\theta} &= \frac{z}{r} \\ -\sin{\theta}\frac{\partial \theta }{\partial x} &= -\frac{z}{r^{2}}\frac{\partial r}{\partial x} \\ \frac{\sqrt{x^{2}+y^{2}}}{r}\frac{\partial \theta }{\partial x} &= \frac{z}{r^{2}} \frac{x}{r} \\ \frac{\partial \theta }{\partial x} &=\frac{z}{r^{2}}\frac{x}{\sqrt{x^{2}+y^{2}}} \\ \frac{\partial \theta }{\partial x} &= \frac{\cos{\theta}\cos{\varphi}}{r}\end{aligned} The reason why the reciprocal rule isn't valid here is because we use two completely different set of coordinates to describe position. So, we have x ( r , θ , φ ) x(r, \theta, \varphi) , but not x ( r , y , z ) x(r, y, z) . Changing only r r coordinate yields two different scenarios in these two cases - in the former it corresponds to moving in radial direction, but in the latter it actually corresponds to moving in x-direction. Hence, the other coordinates also matter! For example, the counter-example which @Steven Chase gave is of this latter kind while the example by @Chris Lewis is of the former kind.

1 Helpful 1 Interesting 1 Brilliant 0 Confused

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