On the differentiability of $f(r,\theta,h)=(r\cos\theta,r\sin\theta,h)$

Question

My question is:

Show that $f(r,\theta,h)=(r\cos\theta,r\sin\theta,h)$ is differentiable in $\mathbb{R}^3$?

My answer is: Since $r\cos\theta$, $r\sin\theta$ and $h$ are differentiable in $\mathbb{R}^3$ then $f$ is differentiable.

Does my answer is correct or no? Thank you.

To illustrate why the existence of partials is not enough, let $f$ be defined on the plane and have value 0 on the axes and 1 everywhere else. Then the partials exists at $(0,0)$ but the function is not even continuous at $(0,0)$. — copper.hat, May 16 '17 at 15:21
You are correct. In our case all derivatives are continuous. — Theory Nombre, May 16 '17 at 15:31

score 1 · Accepted Answer · answered May 16 '17 at 16:34

$f(r,\theta,h)=(rcos(\theta),rsin(\theta),h)$ actually defines a smooth, vector-valued function function on $\mathbb{R}^3$, (i.e. $f\in C^{\infty}(\mathbb{R}^3, \mathbb{R}^3))$.

Note that you can differentiate the component functions, and each one of them are continuously differentiable. You can find derivatives of all orders. As it turns out, the component functions are just the familiar (hopefully) cylindrical coordinates. You can convert them to Cartesian coordinates and get $f(x,y,z)=(x,y,h)$ where $x=rcos(\theta)$, $y=rsin(\theta)$, $z=h$. You can apply the same argument: show that each component function is differentiable and continuous, and conclude we have a smooth (infinitely differentiable with all derivatives continuous) vector field.

Thank you for details. – Theory Nombre May 16 '17 at 16:52 — Theory Nombre, May 16 '17 at 16:52

On the differentiability of $f(r,\theta,h)=(r\cos\theta,r\sin\theta,h)$

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