Parametrizations of a surface of revolution

Consider $S$ the surface of revolution obtained by rotating the planar curve $\alpha(t)=(f(t),0,g(t))$, $t\in I$ which it is contained in the halfplane $y=0, x\geq 0$. We know that $\alpha$ is regular and we have two possible types of curves, i) $\alpha$ is an embedding or ii) $\alpha$ is a simple closed curve. The surface $S$ is $X(I\times{\mathbb R})$, where $$X(t,\theta)=(f(t)\cos\theta,f(t)\sin\theta,g(t)).$$ In order to prove that $S$ is a surface, it appears the problem about how many of parametrizations are needed and how to prove that they are homeomorphisms.

First consider that $\alpha$ is an embedding. We compute the inverse function of $X$ so we will discover what is the right domain of $X$. By the injectivity, it is necessary that $\theta$ moves in an interval of length $2\pi$ at most. Thus a possibility is $X:U_1:=I\times (0,2\pi)\rightarrow X(U_1)$. The set $X(U_1)$ is an open set of $S$ because $$X(U_1)=S-\alpha(I)=S-S\cap(\{y=0,x\geq 0\}).$$ Here we use that $f(t)>0$. Because we need to cover the curve $\alpha(I)$, then the other parametrization is $Y:U_2:=I\times (-\pi,\pi)\rightarrow X(U_2)$. For $Y$ we have to remove `the other side' of $S$, that is $\Phi_\pi(\alpha(I))$, where $\Phi_\theta$ is the rotation of angle $\theta$. In other words, $Y(U_2)=S-(S\cap(\{y=0,x\leq 0\})$.

We compute the inverse of $X$ (or $Y$). We have to write $t$ and $\theta$ in terms of $x,y,z$ from the next equations $$\left\{\begin{array}{l}x=f(t)\cos\theta\\ y=f(t)\sin\theta\\ z=g(t)\end{array}\right.$$ From $x^2+y^2=f(t)^2$, we obtain $f(t)=\sqrt{x^2+y^2}$, and using that $\alpha$ is an embedding, then $t=\alpha^{-1}(\sqrt{x^2+y^2},0,z)$. For $\theta$ we have two possibilities:

1. If we want to write `something' with the arc tangent, then it would be $\theta=\mbox{arc tan}(y/x)$. But in such a case, $\theta$ is defined in $(-\pi/2,\pi/2)$. Because the initial interval is $(0,2\pi)$, we have to change the domain of $X$. We write now $X:U_1:=I\times(-\pi/2,\pi/2)\rightarrow X(U_1)$. The only difference is that we have to prove that $X(U_1)$ is an open set of $S$. But by the picture, $X(U_1)=S-(S\cap \{x\leq 0\})$. Other parametrization is with $I\times (\pi/2,3\pi/2)$ where it is possible to define the inverse of the tangent. And what about the points with $\theta=\pi/2$ or $\theta=3\pi/2$? Here, the $x$-coordinate of the point $(x,y,z)\in S$ vanishes. Now we consider that inverse of the cotangent and taking $\theta={\mbox arc cot}(x/y)$. We need two more domains, namely, $I\times (0,\pi)$ and $I\times (\pi,2\pi)$. Finally, we observe that, using the inverse functions of the tangent of the cotangent function, we need 4 parametrizations: all them write 'very similar', but  the domain goes changing.
2.  If we do not want to use trigonometric functions, we can do the following. If one looks the picture of a surface of revolution, it is clear that the parametrization $X$ can be defined in $I\times (0,2\pi)$, because the angle $\theta$ is well defined. The problem appeared in how to catch the variable $\theta$. The trigonometric functions have added a bit of confusion. Other way is the following.                                                                                                                        Consider $\beta:(0,2\pi)\rightarrow {\mathbb S}^1-\{(0,0\})$ a parametrization of the circle minus one point. The key is the following: the map $\beta$ is a homeomorphism! It is clear that $\beta$ is one-to-one and continuous. One could do the following argument. It is well known that  ${\mathbb S}^1$ minus one point is homeomorphic to the real line ${\mathbb R}$, which it is homeomorphic to the interval $(0,2\pi)$. But the problem is if $\beta$ is, indeed, a homeomorphism. The only trouble is about the inverse. But $\beta$ is an open map because the image of an interval of $(0,2\pi)$   is an open set of ${\mathbb S}^1$.             Once proved that $\beta$ is a homeomorphism, from $x=f(t)\cos\theta$ and $y=f(t)\sin\theta$ we conclude $$\theta=\beta^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right).$$ Then $$X^{-1}(x,y,z)=\left(\alpha^{-1}(\sqrt{x^2+y^2}),\beta^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)\right).$$ Thus we need 2 parametrizations.

If $\alpha$ is a simple closed curve, for covering the curve $\alpha$ by embeddings, we need two parametrizations, namely, $\alpha:(0,T)\rightarrow \alpha(0,T)$ and $\alpha:(T/2,3T/2)\rightarrow \alpha(T/2,3T/2)$. Then for the surface, we need 4 parametrizations.

There is another elegant argument. We observe that when one has proved that $X$ is a parametrization, then we think $Y$ as a `rotation' of $X$. Recall that if $S$ is a parametrization and $\psi$ is a diffeomorphism of ${\mathbb R}^3$, then $\psi(S)$ is a surface. But the parametrizations of $\psi(S)$ are of type  $\psi\circ X$, where $X$ is a parametrization of $S$. With this idea in mind, consider now $\Phi_\theta$ the rotation about the $z$-axis of angle $\theta$. By the definition of $S$, we have $S=\Phi_\theta(S)$. Suppose that we have proved that $X:U_1:=I\times (-\pi/2,\pi/2)\rightarrow {\mathbb R}^3$ is a parametrization of $S$: it is only of a part of $S$, exactly, $V_1=X(U_1)$. Now we take $\theta\in{\mathbb R}$. Because $\Phi_\theta$ is a homeomorphism, it is clear that $\Phi_\theta\circ X$ satisfies the properties of a parametrization, where now the coordinate open is $\Phi_\theta(V_1)$. We point out that $\Phi_\theta(V_1)$ is an open set of $S$. Therefore, if we are going taking many $\theta$, we cover all the surface $S$ by coordinate opens of type $\Phi_\theta(U_1)$. In fact, it is enough only one rotation!, namely, $\Phi_\pi$. With this argument, the effort lies only in the first parametrization $X$: the other ones are obtained `by rotating' $X$.

Finally, we conclude that it is clear that the parametrizations $X$ hold to prove that $S$ is a surface of revolution, for example, taking very small domains, or rotations of $X$. However, the proof is a bit tedious if one wants to write precise arguments.

Surfaces and topology

The surfaces that we are introduced present a variety of possibilities on its topology.

1. There are connected surfaces (sphere) and non-connected surfaces (hyperboloid of two sheets, with two connected components).
2. There are compact surfaces (sphere) and non-compact surfaces (plane).
3. There are closed surfaces (sphere) and non-closed surfaces (a hemisphere).
4. The boundary of the sphere ${\mathbb S}^2$  is the very sphere ${\mathbb S}^2$.
5. Every point of a surface has a neighborhood homeomorphic to ${\mathbb R}^2$. In fact, the coordinate open $V$ of $p\in S$ is homeomorphic to an open set $U\subset{\mathbb R}^2$ via the parametrization $X:U\rightarrow V$. Since $U$ is an open set, there exists a ball $B_r(q)$ around $q=X^{-1}(p)$ with $B_r(q)\subset U$. Then then restriction $$X_{| B_r(q)}:B_r(q)\rightarrow X(B_r(q))$$ is a homeomorphism, being $X(B_r(q))$ an open set of $V$, so, of $S$. This means that $X(B_r(q))$ is an open set around $p$ homeomorphic to ${\mathbb R}^2$. As a consequence, we conclude:
• The interior of a surface is empty.
• The surfaces 'have not boundary point', I mean, for example, the closed hemisphere $T=\{p\in{\mathbb S}^2: z(p)\geq 0\}$ is not a  surface because the above property fails at the points with $z(p)=0$.
• A point is not a surface.
• A surface has a non-countable set of points.
6. If $\phi:{\mathbb R}^3\rightarrow{\mathbb R}^3$ is a diffeomorphism and $S$ is a surface, then $\phi(S)$ is a surface which is a homeomorphic to $S$ thanks to the restriction $\phi_{|S}:S\rightarrow \phi(S)$.
7. An open set of a surface is a surface (proved).
8. Some closed sets of a surface are surfaces; other not. For example, if $S$ is the union of two disjoint spheres, then each sphere is closed and it is a surface. On the other hand, the closed hemisphere is closed in ${\mathbb S}^2$ and it is not a closed set.

Curves-maps; surfaces-sets

I remarked in the classroom the differences between the definition of a curve and a surface: a curve is a differentiable map and a surface is a subset of Euclidean space where there do exist parametrizations. I return again with it.

If a curve $\alpha:I\rightarrow{\mathbb R}^3$ is regular $t_0$, then $\alpha'(t_0)\not=0$. If we write in terms of the differential map of $\alpha$, it means that $(d\alpha)_t:{\mathbb R}\rightarrow {\mathbb R}^3$ is a non-zero linear map. This is equivalent to say that $\mbox{rank}(d\alpha)_t=1$, because $$(d\alpha)_{t_0}(1)=\frac{d}{ds}{\Big |}_{s=0}\alpha(t_0+s)=(x'(t_0),y'(t_0),z'(t_0))\not=(0,0,0).$$ Thus the rank  of $(d\alpha)_t$ is the maximum possible (it would be $0$ or $1$). Furthermore, by using the inverse function theorem, ``the curve is a graph locally around $t_0$''. In fact, it was proved that there exists $\epsilon>0$ such that $$\alpha:J=(t_0-\epsilon,t_0+\epsilon)\rightarrow \alpha(t_0-\epsilon,t_0+\epsilon)$$ coincides with the graph of a function, that is, there exists a differentiable function $f:K\subset {\mathbb R}\rightarrow {\mathbb R}$ such that $\{(x,f(x)):x\in K\}=\alpha(J)$. As a consequence, $\alpha:J\rightarrow\alpha(J)$ is homeomorphic to an interval of ${\mathbb R}$.

Then the map $\alpha$ would play (almost) the same role of parametrizations in a surface. If we want to give the definition of the  analogous $1$-dimensional case of a surface, then a subset $C\subset{\mathbb R}^3$ is a $1$-surface (=curve) if for each point $p\in C$ there exists $I\subset {\mathbb R}$ and a map $X:I\rightarrow V\subset C$ a homeomorphism, where $V$ is an open of $C$ around $p$, $X:I\rightarrow {\mathbb R}^3$ is differentiable and $X'(t)\not=(0,0,0)$.

The question is the definition of curve given in chapter $1$ is now a $1$-surface, more precisely, if the trace $\alpha(I)$ is a such $1$-surface. Then one would think `yes' by taking around each point $p\in C$ the corresponding restriction of $\alpha$ to the suitable interval $J$. However, the only problem is the following: Is $\alpha(J)$ an open set of $C$? because the other properties have been showed. We find the answer in the curve $\alpha(t)=(\cos(t),\sin(2t))$, $t\in {\mathbb R}$.

This curve self-intersects at the origin. Thus it can not be a $1$-surface because this point has not a neighbourhood which is homeomorphic to ${\mathbb R}$. By the inverse function theorem, around $t_0=0$, $\alpha(J)\cong J$ is a graph, but $\alpha(J)$ is not an open set of $\alpha({\mathbb R})$, which it happens exactly in our example, as one can see in the next picture: the red color line is $\alpha(J)$ is not an open set in $C$.

Curvature of a planar curve by reversing its direction

After the last exercise this afternoon, I will revise what is the curvature of a planar curve when we reverse its orientation. In order to simplify the notation, we will assume that the domain $I$ of the planar curve is $I=\mathbb{R}$, so we have $\alpha:\mathbb{R}\rightarrow{\mathbb R}^2$, which we suppose parametrized by the length-arc. In classroom we have said "when we reverse the orientation, then the curvature change the sign at the point". Firstly, we explanation of our intuition.

When we change the orientation on the curve, we are taking $-s$ instead of $s$ as the new parameter the curve, and as we increase $s$, we are going in the initial curve on the contrary direction. Then the result that we want to prove is that when we pass by a point of the curve, its curvature is the opposite if we go along the curve in the other direction. That is, we say that at $s_0$  the curvature of $\alpha$ when we reverse the sense of the parameter is the opposite with respect to the other direction.

The computations are clear and were made in the classroom. Define $\beta(s)=\alpha(-s)$ the curve $\alpha$ parameterized in the opposite direction. Then $T_\beta(s)=-T_\alpha(-s)$, $N_\beta(s)=JT_\beta(s)=-JT_\alpha(-s)=-N_\alpha(-s)$ and $T_\beta'(s)=T'_\alpha(-s)$. Thus $$\kappa_\beta(s)=\langle T'_\beta(s),N_\beta(s)\rangle=-\langle T'_\alpha(-s),N_\alpha(-s)\rangle=-\kappa_\alpha(-s). (*)$$

The problem appears when we utilize the words to describe the above result. The symbol $\kappa_\beta(s)$ is the curvature of $\alpha$ with the reverse direction at the point $s$. The right hand side says that it coincides with the opposite of the curvature of $\alpha$ at $-s$.

An example. Consider the curve $\alpha(t)=(t,t^2+t^3)$. This curve is not parameterized by the length-arc, but the above argument (which it involves only the curvature) holds for any curve. The curvature of $\alpha$ is $$\alpha'(t)=(1,2t+3t^2), \alpha''(t)=(0,2+6t)\Rightarrow\kappa_\alpha(t)=\frac{2+6t}{(1+(2t+3t^2)^2)^{3/2}}.$$ We fix the point $t=1$, that is, $\alpha(1)=(1,2)$. Then the curvature at $t=1$ is $\kappa_\alpha(1)=8/26^{3/2}$.  We reverse the orientation of $\alpha$ by doing $\beta(t)=(-t,t^2-t^3)$. Now the point $t=1$ for $\alpha$ corresponds with $t=-1$. The curvature of $\beta$ is $$\kappa_\beta(t)=\frac{-2+6t}{(1+(2t-3t^2)^2)^{3/2}}.$$ Then $\kappa_\beta(-1)=-8/26^{3/2}$, which corresponds with the formula  (*).

To finish, I propose a question. Suppose now that the curve is the graph of a function $y=f(x)$, where we know that the curvature of the curve is $f''(x)/(1+f'(x)^2)^{3/2}$. If we move in the opposite direction, then the curve is the graph of the function $y=f(-x)$. Using the above formula,  the curvature would be $f''(-x)/(1+f'(-x)^2)^{3/2}$ because the first derivative gives $-f'(-x)$, but the second one yields $f''(-x)$. But according to (*), this should give $-f''(-x)/(1+f'(-x)^2)^{3/2}$, the oppossite sign.

Question: what is it happening?

Curvature and symmetry about a point

Motivated by one of the proposed exercises today, we observe that the curve $y=x^3$, which is symmetric about the origin of the plane, satisfies that the curvature at the point $-x$ is the opposite in $x$: look in which side lies the graph of $y=x^3$ around $x$. Thus we prove the next result:

Result. If $\alpha:I\rightarrow{\mathbb R}^2$ is a planar curve which is symmetric about  a point $p_0\in\alpha(I)$, then $\kappa(-s)=-\kappa(s)$. Here we are assuming that $\alpha(0)=p$.

After a translation and without loss of generality, we suppose that $p_0=(0,0)$. Say that $\alpha$ is symmetric about the origin means $\alpha(-s)= M\alpha(s)$, where $M$ is the symmetry, that is, $Mx=-x$. Thus $\alpha(-s)=-\alpha(s)$.  Let $\beta(s)=\alpha(-s)$. Then $\kappa_\beta(s)=-\kappa_\alpha(-s)$. On the other hand, the curve $\gamma(s)=-\alpha(s)$ is a direct rigid motion of $\alpha$, so $\kappa_\gamma(s)=\kappa_\alpha(s)$. Since $\kappa_\gamma(s)=\kappa_\beta(s)$, we conclude $$-\kappa_\alpha(-s)=\kappa_\alpha(s).$$
This can be checked for curves that are graphs of $y=f(x)$ where we know that $$\kappa_\alpha(x)=\frac{f''(x)}{(1+f'(x)^2)^{3/2}}.$$ Denote $\alpha$ the curve $y=f(x)$. As $f$ is symmetric about the origin, then $f(-x)=-f(x)$. The curvature of $y=g(x)$, where $g(x)=f(-x)$ is
$$g'(x)=-f'(-x), g''(x)=f''(-x)\Rightarrow \kappa_g(x)=\frac{f''(-x)}{(1+f'(-x)^2)^{3/2}}.$$
The curvature of $y=-h(x)$ is $$h'(x)=-f'(x), h''(x)=-f''(x)\Rightarrow \kappa_h(x)=\frac{-f''(x)}{(1+f'(x)^2)^{3/2}}.$$ Since the curvatures of both graphs coincide, we have $$\kappa_f(-x)=\frac{f''(-x)}{(1+f'(-x)^2)^{3/2}}=\frac{-f''(x)}{(1+f'(x)^2)^{3/2}}=-\kappa_f(x).$$
The converse is also true:

Theorem. If $\alpha:I=(-a,a)\rightarrow{\mathbb R}^2$ is a planar curve  such that $\kappa(-s)=-\kappa(s)$, then the graphic of $\alpha$ is symmetric about  the point $p=\alpha(0)$.

After a translation, we suppose $\alpha(0)=(0,0)$. Define two curves: $\beta(s)=\alpha(-s)$ and $\gamma(s)=-\alpha(s)$. We compute their curvatures. For $\beta$, $\kappa_\beta(s)=-\kappa(-s)$ and for $\gamma$, $\kappa_\gamma(s)=\kappa(s)$ because the symmetry is a direct rigid motion.  By hypothesis, $\kappa_\gamma(s)=\kappa_\beta(s)$ so there is a direct rigid motion $M$ such that $\gamma(s)=M\beta(s)$. Because $\gamma(0)=\alpha(0)=\beta(0)$ and $\gamma'(0)=\beta'(0)$ (check!), then their normal vector also coincide at $s=0$. Then $M$ must be the identity, proving the result.

The intermediate value theorem

I consider in this entry the exercise 2 of the homework page and some people asked. The interest of this exercise is to show what we gain in this course comparing we know from the high-school or Calculus I. We learnt the following intermediate value theorem: if $f$f is a smooth function in $[a,b]$, then there exists  $\xi\in (a,b)$ such that $f(b)-f(a)=f'(\xi)(b-a)$. If we write this equality as  $$f'(\xi)=\frac{f(b)-f(a)}{b-a}$$ the results asserts that there exists an intermediate point between $a$ and $b$ where the tangent line at $x=\xi$ is parallel to the straight-line joining the points $(a,f(a))$ and $(b,f(b))$: the about identity says that the slopes of both lines coincide.

Exercise 1 says that this result holds for any planar curve without to be a graph of a function $y=f(x)$, as it shows the next picture.

A possible starting point of the proof would be applying the intermediate value theorem to the functions  $x=x(t)$ and $y=y(t)$, where $\alpha(t)=(x(t),y(t))$. Suppose $t\in [a,b]$. Then there exists $\xi_1,\xi_2\in (a,b)$ such that $$x(b)-x(a)=x'(\xi_1)(b-a), \ y(b)-y(a)=y'(\xi)(b-a).$$ Hence $$x'(\xi_1)=\frac{x(b)-x(a)}{b-a},\ y'(\xi_2)=\frac{y(b)-y(a)}{b-a}.$$
A vector determining the line $\alpha(a)$ with $\alpha(b)$ is $$\alpha(b)-\alpha(a)=(x(b)-x(a),y(b)-y(a))$$ and a vector of the tangent line at $t$ is $(x'(t),y'(t))$. And we want to prove that the first one is proportional to the second one at some point $t_0\in (a,b)$. But this point is not  $\xi_1$ or $\xi_2$ because they do not need to coincide!

Thus the proof follows other ideas. In the picture we observe that the point that we are looking for is the farthest point of the trace  $\alpha(a,b)$ with respect to the line $R$ joining  $\alpha(a)$ with $\alpha(b)$. Then define $$f:[a,b]\rightarrow\mathbb{R},\ f(s)=\langle \alpha(s)-\alpha(a),N(a)\rangle$$ the distance between $\alpha(s)$ and $R$ and take $s_0\in (a,b)$ the maximum of $f$ (if $\alpha(s_0)=0$, then $\alpha([a,b])\subset R$ and the result is immediate). Let us point out that the maximum is attained because $[a,b]$ is a compact set. In particular, $f'(s_0)=0$ and this yields $$0=f'(s_0)=\langle \alpha'(s_0),N(a)\rangle,$$ which proves the result.

In fact the reasoning says that it holds for any critical point of $f$ in $(a,b)$, as it appears in the next figure.

Evolute of the parabola

I propose an easy example of evolutes using Mathematica to draw the curves. For this, I take the known parabola $y=x^2$ and its parametrization is $\alpha(t)=(t,t^2)$. This curve  is not parametrized by the length-arc so I use the formula for arbitrary curves, obtaining
$$\kappa(t)=\frac{2}{\left(4 t^2+1\right)^{3/2}}.$$
For the normal vector of $\alpha$, we first calculate the tangent vector
$$T_\alpha(t)=\frac{\alpha'(t)}{|\alpha'(t)|}\Rightarrow N_\alpha(t)=J T_\alpha(t)=\left(-\frac{2 t}{\sqrt{4 t^2+1}},\frac{1}{\sqrt{4 t^2+1}}\right).$$ Finally we have $$\beta(t)=\alpha(t)+N(t)/\kappa(t)=\left(-4 t^3,3 t^2+\frac{1}{2}\right).$$ The picture of both curves is:

where the blue color corresponds with the parabola and the green one is its evolute. We check the property proved in classroom that the tangent line of  $\alpha$ is the normal line of $\beta$ at the corresponding point. We do it at $t=1$, with $\alpha(1)=(1,1)$ and $\beta(1)=(-4,7/2)$.

Question: Let do with the curve $y=\sin(x)$ at the point $(0,0)$.