# Revolution and differentiability at a point

Here is a statement thought by myself: (not sure if it's true)

Suppose $z=f(x)$ is a real-valued function defined for $x\ge 0$, and the right side derivative at $x=0$ is zero. If $z=g(x,y)$ is the function obtained by revolving the graph of $z=f(x)$ about the z-axis, then $g$ is differentiable at $(0,0)$ and has a horizontal tangent plane there.

I think $z=g(x,y)$ is just $z=f(r)$, which $r=\sqrt{x^2+y^2}$. In this way I can show that $g_x$ and $g_y$ are zero at $(0,0)$. In fact all directional derivatives would be zero. But I get stuck on the differentiability part, which definition is as follows: (quoted from Thomas Calculus)

A function $z=f(x,y)$ is differentiable at $(x_0,y_0)$ if $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$ exist and $\Delta z$ satisifes an equation of the form $$\Delta z=f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y+\epsilon_1\Delta x+\epsilon_2\Delta y$$ in which each of $\epsilon_1, \epsilon_2 \to 0$ as both $\Delta x,\Delta y \to 0$.

What are $\epsilon_1$ and $\epsilon_2$ supposed to be in this case?

EDIT: I realize I forgot to mention $z=f(x)$ is defined for $x\ge 0$ only. Now it's being added. I am very sorry if this caused confusion.

• (This is delt3. I accidentally lost the account so I set up a new one.) I find that the notion of total differentiability would imply the increment formula. It also applies to my example, since $\lim_{(h,k)\to(0,0)}\frac{f(\sqrt{h^2+k^2})-f(0)}{\sqrt{h^2+k^2}}=\lim_{r\to 0^{+}}\frac{f(r)-f(0)}{r}=0$. Quite a coincidence to me. – delt31 Apr 27 '17 at 7:52

Since you want a patch to have differentiable structure, let $z = f(x)$ be a smooth, injective curve for $x \in (-\epsilon, \epsilon)$. If you allow $\lim_{x \to 0+} f(x) \not = \lim_{x \to 0-} f(x)$ then the curve cannot be connected at the point $(0,f(0))=(x,z)$. If you want to proceed this way, you must know what $f(0)$ is equal to so that when we use that parametrization $\sigma(x,u)$, we can compute $\sigma_x(0,f(0))$ and $\sigma_u(0,f(0)$. You also run into another problem since $\partial z/\partial x$ is an entry in the differential, so you'll need equality of the right and left derivatives. Again With the refined conditions, revolving the profile curve about the $z$-axis gives the parametrization,

$$\sigma(x,u) = (h(x) \cos u, h(x) \sin u, f(x)) = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$

Since you let $\partial z/\partial x (0) = 0$ we have $d_{p} \sigma: T_p \mathbb{R}^2 \to T_{\sigma(p)} S$, where $p = (0,0)$ is defined by,

$$d_{(0,0)}\sigma = \begin{pmatrix} h'(0) & 0 \\ 0 & h(0) \\ 0 & 0 \end{pmatrix}$$

The tangent plane at $\sigma(0,0) = (h(0),0,f(0)):=q$ is the image of $d_{(0,0)} \sigma$. Letting $(x-p,y-p) \in T_p\mathbb{R}^2$ then we have,

$$d_{(0,0)} \sigma \begin{pmatrix} x-p \\ y-p \end{pmatrix} = \begin{pmatrix} h'(0) & 0 \\ 0 & h(0) \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x-p \\ y-p \end{pmatrix} = \begin{pmatrix} h'(0)(x-p)\\ h(0)(y-p) \\0 \end{pmatrix}:=v_p$$

Here $v_p \in T_qS$ and so if $\textbf{N}$is our normal on a small patch containing $q$ then the tangent plane is given by $\left(\textbf{N}\right)^{\perp} = \{v_p: v_p \cdot \textbf{N} = 0\}$.

$$\begin{pmatrix} h'(0)(x-p) \\ h(0)(y-p) \\0 \end{pmatrix} \cdot \begin{pmatrix} n_1 \\ n_2 \\ n_3 \end{pmatrix} = 0 \iff n_1h'(0)x + n_2h(0)y = p(h'(0) + h(0))$$

which is never a horizontal tangent plane in the coordinate system $(x,y,z)$ since we only get vertical or slanted planes. For horizontal tangent planes, you need to looks at points $x_0$ such that,

\begin{align*} &\frac{\partial z}{\partial x}(x_0) = f'(x_0) = 0 \\ & n_1h'(x_0) = 0 \\ & n_2h(x_0) = 0 \end{align*}

Diving into one special case i.e $h'(x_0) = h(x_0) = 0$, we observe that these points have flat neighborhoods since,

$$\sigma_x(x_0,u) = \sigma_u(x_0,u)= (0,0,0)$$

Observe that this is exactly what you want since in the ambient space we have the points $p'=(x_0,y,u)$ with a small-neighborhood $V \ni (x_0,u)$ contained on our curve and in the $z=K$ plane i.e $u = K$. The image to have in mind is like the one below where we see that $p$ must be the endpoints of the curve if we wish for the surface to be smooth. From the description of the points $p'$ on the surface of revolution, it is clear which points satisfy the geometric description given above.

• Thanks for the answer. I haven't learnt about differentiable structure yet, but I guess you do parametrization of the surface by 2 variables and do some calculus on that? I'm not sure if $(0,0)$ now satisfies the 3 equations. – delt3 Apr 24 '17 at 14:53
• @delt3: $(0,0) = (x,u) = (x,f(x))$ cannot be a point with a horizontal plane since points of this form give slanted or vertical planes which is the first thing I give you. I go one after that to show you which points on a smooth surface of revolution can have horizontal planes. The intuition behind this is the for a curve of finite length, if a point $p$ in not an endpoint then the surface must pinch at this point. – Faraad Armwood Apr 24 '17 at 15:12
• Your intuition about the points which have horizontal tangent planes was almost spot on, its just that we needed to other conditions to actually give us a surface patch about these points so that we can do calculus. – Faraad Armwood Apr 24 '17 at 15:17
• I'm puzzled about the $h(x)$ term... So let's say the function $z=f(x)=x^2$ is parametrized as $(x,x^2)$. Is the surface parametrized by $(x\cos u,x\sin u,x^2)$, which $h(x)=x$, $u$ is the angle as in polar coordinate? – delt3 Apr 24 '17 at 16:29
• Well, I hope it truly helped. your question, although innocent to you, was a good one. There were many ways to hand wave it, but I tried not to. At least now, you've been exposed to other things and hopefully that'll encourage you to read more on the subject. I suggest coming back to this post after you learn more differential geometry and multivariable calculus. More than likely you'll find shorter paths or things I've overlooked. – Faraad Armwood Apr 24 '17 at 17:58

True formula must be $$\Delta z=g_x(x_0,y_0)\Delta x+g_y(x_0,y_0)\Delta y+\epsilon_1\Delta x+\epsilon_2\Delta y$$ where $z=g(x,y)=f(r)$. Then $$g_x=f'(r)\dfrac{\partial r}{\partial x}=f'(r)\dfrac{x}{\sqrt{x^2+y^2}} \text{ and } g_y=f'(r)\dfrac{\partial r}{\partial y}=f'(r)\dfrac{y}{\sqrt{x^2+y^2}}$$ also $\epsilon_1(\Delta x,\Delta y)$ and $\epsilon_2(\Delta x,\Delta y)$ completely determine with $f$. For instance with $z=f(x)=x^2$ then $z=g(x,y)=f(r)=r^2=x^2+y^2$. So $$\Delta z=g(x+\Delta x,y+\Delta y)-g(x,y)=2x\Delta x+2y\Delta y+(\Delta x)^2+(\Delta y)^2$$ therefore $\epsilon_1(\Delta x,\Delta y)=(\Delta x)^2$ and $\epsilon_2(\Delta x,\Delta y)=(\Delta y)^2$, furthermore $$\lim_{\Delta x\to0,\Delta y\to0}\epsilon_1(\Delta x,\Delta y)\to0 \text{ and } \lim_{\Delta x\to0,\Delta y\to0}\epsilon_2(\Delta x,\Delta y)\to0$$

• But what if $f$ is a function-in-general? i.e the expression of f is not given. Is there still a way to express $\epsilon_1, \epsilon_2$? – delt3 Apr 24 '17 at 16:56
• Here $\Delta x, \Delta y$ are just infinitesimal increments. The expression above is just the linearization of $z$. $$\Delta z = g_x(p_0) (x-x_0) + g_y(p_0)(y-y_0) + \epsilon_1 (x-x_0) + \epsilon_2(y-y_0)$$ where $p_0= (x_0,y_0)$. So for a typical problem, you choose $(x,y)$ such that it is really close to $(x_0,y_0)$. You do this since this approximation is limited in its domain about $p_0$. The quantities $(x-x_0)$ and $(y-y_0)$ are therefore very, very small and so we define them to be $\Delta x, \Delta y$ respectively. – Faraad Armwood Apr 24 '17 at 17:04
• Also, typically we just have have $\epsilon(x,y) = \epsilon_1(x-x_0) + \epsilon_2(y-y_0)$. I think it was given this way to show that the linearization picks up on the error for the partials in the $x,y$ directions. Hence we have that, $$\lim_{(x,y) \to (x_0,y_0)} \frac{\epsilon(x,y)}{\sqrt{(x-x_0)^2+(y-y_0)^2}} = 0$$ – Faraad Armwood Apr 24 '17 at 17:06