Let \(f : \mathbb{R}^{3} \to \mathbb{R}\) be a linear function.

1. The kernel of \(f\) is the level set of \(f\) with level zero, that is, \(\operatorname{Ker}(f)=f^{-1}(\{0\}).\) If \(f\) has rank \(1,\) then \(f^{-1}(\{0\})\) has dimension \(2\) by the rank-nullity theorem and hence is a two-dimensional plane through the origin \(0_{\mathbb{R}^3}.\)

2. Let \(c\neq 0\) be different from zero. Then \(f^{-1}(\{c\})\) is an affine subspace whose associated vector space is \(f^{-1}\left(\{0\}\right)\) \[f^{-1}\left(\{c\}\right)=f^{-1}\left(\{0\}\right)+q=\left\{p+q\,|\, p \in \operatorname{Ker}(f)\right\},\] where \(q \in \mathbb{R}^3\) satisfies \(f(q)=c.\)

For \(p=(x,y,z)\) we consider \[f : \mathbb{R}^{3} \to \mathbb{R}, \qquad p \mapsto x^2+y^2+z^2.\] Then for all \(r>0\) the level set \(f^{-1}\left(\{r^2\}\right)\) of \(f\) with level \(r^2\) is the \(2\)-sphere of radius \(r\) centred at \(0_{\mathbb{R}^3}\). We will denote it by \(S^2(r)\) with the convention of writing \(S^2\) when \(r=1.\)

Let \(R>0\) and \(f : \mathbb{R}^3 \to \mathbb{R}\) be the function defined by the rule \[p=(x,y,z) \mapsto \left(R-\sqrt{x^2+y^2}\right)^2+z^2.\] Then for \(r < R\) we consider the level set \(f^{-1}\left(\{r^2\}\right)\) of \(f\) with level \(r^2.\) This level set is called a torus.

Consider the smooth function \[f : \mathbb{R}^3 \to \mathbb{R}, \qquad p=(x,y,z)\mapsto x^2+y^2\] Then for \(r>0\) the level set \(f^{-1}\left(\{r^2\}\right)\) of \(f\) with level \(r^2\) is an (infinite) cylinder of radius \(r\) and central axis \(\{(0,0,z)\,|\, z\in \mathbb{R}\}.\)

For \(a,b \in \mathbb{R}^+\) and \(p=(x,y,z)\) consider \(f : \mathbb{R}^{3}\to \mathbb{R}\) defined by the rule \[p\mapsto \frac{x^2}{a^2}+\frac{y^2}{b^2}-z.\] The level set \(f^{-1}\left(\{0\}\right)\) of \(f\) with level zero is known as an elliptic paraboloid.

For \(c \in \mathbb{R}^+\) and \(p=(x,y,z)\) consider \(f : \mathbb{R}^{3}\to \mathbb{R}\) defined by the rule \[p\mapsto \frac{x^2}{c^2}+\frac{y^2}{c^2}-z^2.\] Let \(\mathcal{X}=\left\{p=(x,y,z) \in \mathbb{R}^3 \,|\, z\geqslant 0\right\}\) and consider the level set \(C=f^{-1}\left(\{0\}\right) \cap \mathcal{X}.\) Then \(C\) is a cone whose vertex (its tip) is \(0_{\mathbb{R}^3}.\) Clearly, we cannot define a tangent plane at the vertex of the cone in any geometrically natural way. Observe that \(f\) is smooth and that the exterior derivative of \(f\) is given by \[\mathrm{d}f=\frac{\partial f }{\partial x}\mathrm{d}x+\frac{\partial f }{\partial y}\mathrm{d}y+\frac{\partial f }{\partial z}\mathrm{d}z=\frac{2x}{c^2}\mathrm{d}x+\frac{2y}{c^2}\mathrm{d}y-2z\mathrm{d}z.\] Recall that \(\mathrm{d}f|_{p}: T_p \mathbb{R}^3 \to \mathbb{R}\) is a linear map satisfying \[\mathrm{d}f|_{p}(\vec{v}_p)=\frac{2x}{c^2}v_1+\frac{2y}{c^2}v_2-2zv_3,\] where we write \[\vec{v}_p=\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}_p\] Therefore \[\operatorname{rank}\mathrm{d}f|_{p}=\left\{\begin{array}{ll} 1, & p \neq 0_{\mathbb{R}^3},\\ 0, & p=0_{\mathbb{R}^3}.\end{array}\right.\] The rank of \(\mathrm{d}f|_{{}{p}}\) fails to be maximal (i.e. \(1\)) precisely at the vertex, where we cannot define the tangent plane.

Let \(U\subset \mathbb{R}^2\) be open and \(h : U \to \mathbb{R}\) a smooth function. Then the graph of \(h\) \[\mathcal{G}_h:=\left\{(q,h(q))\,|\, q \in U\right\}\] is an embedded surface in \(\mathbb{R}^{3}.\) Indeed, consider \(\mathcal{X}=U\times \mathbb{R}\subset \mathbb{R}^{3}\) and \[f : \mathcal{X} \to \mathbb{R}, \qquad p=(q,t) \mapsto h(q)-t.\] for \(q \in U\) and \(t \in \mathbb{R}.\) Then \[f^{-1}\left(\{0\}\right)=\left\{(q,h(q))\,|\, q \in U\right\}=\mathcal{G}_h\] and writing \(q=(u,v),\) we have \[\mathrm{d}f=\frac{\partial h}{\partial u} \mathrm{d}u+\frac{\partial h}{\partial v} \mathrm{d}v-\mathrm{d}t.\] Therefore, \(\mathrm{d}f|_{p=(q,t)}\) has rank \(1\) for all \(p=(q,t) \in U \times \mathbb{R}\) and \(M=f^{-1}\left(\{0\}\right)\) is an embedded surface. The tangent space at \((q,h(q))\) for \(q \in U\) is given by \[T_{(q,h(q))}M=\left\{\vec{v}_{(q,h(q))} \in T_{(q,h(q))}\mathbb{R}^{3} \ \Big|\, v_{3}=\frac{\partial h}{\partial u}(q)v_1+\frac{\partial h}{\partial v}(q)v_2\right\}.\]

Let \(M=f^{-1}\left(\{c\}\right)\) be an embedded surface. The map \[N : M \to TM^{\perp}, \qquad p \mapsto =\frac{\operatorname{grad}f(p)}{\Vert \operatorname{grad}f(p)\Vert}\] is a smooth unit normal field on \(M.\)

Consider the cylinder of radius \(r\) and central axis \(\{(0,0,z)\,|\,z\in \mathbb{R}\}\) \[M=\left\{(x,y,z) \in \mathbb{R}^3\,|\, x^2+y^2=r^2\right\}=f^{-1}\left(\{r^2\}\right),\] where \(f : \mathbb{R}^3\to \mathbb{R}\) is given by \(f(p)=x^2+y^2.\) For \(b \in \mathbb{R}\) consider the helix \[\gamma : \mathbb{R}\to M \subset \mathbb{R}^3, \qquad t \mapsto (r\cos(t),r\sin(t),bt).\] Then, writing \(p=(x,y,z)\) we have \[\operatorname{grad}f(p)=\begin{pmatrix} 2x \\ 2y \\ 0 \end{pmatrix}_{p}\] as well as \[\dot{\gamma}(t)=\begin{pmatrix} -r \sin(t) \\r\cos(t) \\ b \end{pmatrix}_{\gamma(t)}\quad \text{and}\quad \ddot{\gamma}(t)=-r\begin{pmatrix}\cos(t) \\ \sin(t) \\ 0 \end{pmatrix}_{\gamma(t)}=-\frac{1}{2} \operatorname{grad}f(\gamma(t)).\] Since \(\operatorname{grad}f(p)\) is a basis of \(T_pM^{\perp}\) for all \(p \in M,\) it follows that \(\ddot{\gamma}(t) \in T_{\gamma(t)}M^{\perp}\) for all \(t \in \mathbb{R},\) hence \(\gamma\) is a geodesic.

The intersection of \(S^2(r)\) with a \(2\)-dimensional vector subspace \(U\subset \mathbb{R}^3\) is called a great circle. Let \(\{w_1,w_2\}\) be an orthonormal basis of \(U.\) Then \(U\cap S^2(r)\) is the image of the curve \[\gamma : \mathbb{R}\to S^2(r)\subset \mathbb{R}^3, \qquad t \mapsto r\cos(t)w_1+r\sin(t)w_2.\] Then \[\ddot{\gamma}(t)=-r\left(\cos(t)\vec{w}_1+\sin(t)\vec{w}_2\right)_{\gamma(t)}\] where here \(\vec{w}_i\) denotes the vector obtained by thinking of \(w_i\) as a column vector. Since \(S^2(r)=f^{-1}\left(\{r^2\}\right)\) for the function \(f : p=(x,y,z) \mapsto f(p)=x^2+y^2+z^2,\) we have \[\operatorname{grad}f(p)=\begin{pmatrix} 2x \\ 2y \\ 2z\end{pmatrix}_p=2\vec{p}_p\] where again \(\vec{p}\) denotes \(p,\) but thought of as a column vector. Consequently, we have \[\operatorname{grad}f(\gamma(t))=2r\left(\cos(t)\vec{w}_1+\sin(t)\vec{w}_2\right)_{\gamma(t)}=-2\ddot{\gamma}(t),\] which shows that \(\gamma\) is a geodesic in \(S^2(r).\)

Consider \(S^2\) and \(\gamma\) to be the “equator” \[\gamma : [0,2\pi] \to S^2,\qquad t \mapsto (\cos(t),\sin(t),0).\] Observe that the vector fields along \(\gamma\) defined by the rule \[E_1(t)=\begin{pmatrix} -\sin(t) \\ \cos(t) \\ 0 \end{pmatrix}_{\gamma(t)} \quad \text{and} \quad E_2(t)=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}_{\gamma(t)}\] span \(T_{\gamma(t)}S^2\) for all \(t \in [0,2\pi].\) Furthermore \[N(t)=\begin{pmatrix} \cos(t) \\ \sin(t) \\ 0 \end{pmatrix}_{\gamma(t)}\] spans \((T_{\gamma(t)}S^2)^{\perp}\) for all \(t \in [0,2\pi]\) and \(\{E_1(t),E_2(t),N(t)\}\) is an orthonormal basis of \(T_{\gamma(t)}\mathbb{R}^3\) for all \(t \in [0,2\pi].\) Any smooth vector field \(X : [0,2\pi] \to TS^2\) along \(\gamma\) is of the form \[X=s_1E_1+s_2E_2\] for smooth functions \(s_1,s_2 : \mathbb{R}\to \mathbb{R}\) which are periodic with period \(2\pi.\) From this we compute \[\dot{X}(t)=s_1^{\prime}(t)E_1(t)+s_2^{\prime}(t)E_2(t)-s_1(t)N(t)\] Hence we have \[\begin{gathered} \frac{\mathrm{D}X}{\mathrm{d}t}(t)=s_1^{\prime}(t)E_1(t)+s_2^{\prime}(t)E_2(t)-s_1(t)N(t)\\-\langle s_1^{\prime}(t)E_1(t)+s_2^{\prime}(t)E_2(t)-s_1(t)N(t),N(t)\rangle N(t) \end{gathered}\] which simplifies to become \[\frac{\mathrm{D}X}{\mathrm{d}t}(t)=s_1^{\prime}(t)E_1(t)+s_2^{\prime}(t)E_2(t).\]