I apologize in advance for the length of this question....
Now, consider the sphere $$ S:=\{ (x,y,z):x^2+y^2+z^2=1 \}$$ in $\mathbb{R}^3$. Now in Chicone's book "Ordinary differential equations" on pp. 53 proving that $S$ is an invariant manifold of some ODE boils down to proving that $S$ is everywhere tangent to the vector field $$ (x,y,z)\mapsto (x^2-Ax,y^2-Ay,z^2-Az),$$where $A:=x^3+y^3+z^3$.
Keep in mind that at this point only the concept of manifold but not the concept of tangent space is introduced. His argument why the sentence written in italic holds is the following:
My questions are:
1) Can someone please give me reference (or recall for me) this definition of tangency of vectors to spheres from euclidean geometry ? (Later on the concept of tangent space at a manifold is introduced, but I don't want to make use of later concepts in his book to understand earlier ones; as he phrased it, there should be somewhere in basic euclidean geometry a definition of this type of tangency).
2) I don't understand his argument at all, because in my basic geometry class things like normal lines weren't introduced - only normal vectors, which I know how to calculate. Also how can I derive how the outer normal unit field looks like ?
3) A minor question (if someone has the time - if not ignore this one): What does generally mean that a vector field is tangent to a manifold, how can I visualize this ? (A reference to a (well discussed) definition would suffice here)
Regarding (1): Tangency at spheres. We work in $\mathbb R^n$ with Euclidean scalar product $\langle\cdot,\cdot\rangle$ in an orthonormal basis. Then, we have the sphere of radius $r$ centered at the origin: $S_r = \{x\in \mathbb R^n|\langle x,x\rangle = r^2\}.$ Pick a point $y$ in $S_r$ and a vector $v \in \mathbb R^n\setminus\{0\}$ and consider the line $z = y+\lambda v$ with parameter $\lambda \in \mathbb R.$ This line is tangent to $S_r$ iff it intersects $S_r$ only in $y.$ In order to calculate the intersections of $z$ and $S_r,$ we solve $$ \langle y+\lambda v,y+\lambda v\rangle = r^2 $$ for $\lambda.$ We calculate $$ r^2 = \langle y+\lambda v,y+\lambda v\rangle = \langle y,y\rangle + 2\lambda\langle y,v\rangle + \lambda^2\langle v,v\rangle, $$ which, by using $\langle y,y\rangle = r^2,$ is equivalent to $$ 0 = 2\lambda\langle y,v\rangle + \lambda^2\langle v,v\rangle. $$ In order for $z$ to intersect $S_r$ only in $y,$ this equation must have the unique solution $\lambda = 0.$ This is the case iff $\langle y,v\rangle = 0.$ Also note that $\langle v,v\rangle \neq 0$ since $v\neq 0.$ So we have shown that the vector $v$ is tangent to $S_r$ in the point $y \in S_r$ iff $\langle y,v\rangle = 0.$ If you think about all this for a very short time, you will realize that it makes sense to drop the requirement $v\neq 0.$
Regarding (2): Vector field and unit sphere. We consider the unit sphere $S_1 \subseteq \mathbb R^3$ and the vector field $V(x,y,z) = (x^2-Ax,y^2-Ay,z^2-Az)$ for $(x,y,z) \in \mathbb R^3.$ with $A = x^3+y^3+z^3.$ We want that for any $(x,y,z) \in S_1,$ $V(x,y,z)$ is tangent to $S_1$ in the point $(x,y,z).$ According to (1), we just have to calculate $$ \begin{align} \langle (x,y,z),V(x,y,z) \rangle & = x^3-Ax^2+y^3-Ay^2+z^3-Az^2 \\ & = A-A(x^2+y^2+z^2) \\ & = A - A\cdot 1 \\ & = 0, \end{align} $$ as desired. Note that we have used $(x,y,z) \in S_1.$
Regarding (3): Manifolds and tangent vectors. Say, you have a smooth surface element M (or some higher-dimensional analogue) embedded in some $\mathbb R^n.$ Think of a sphere or a torus. Consider a smooth curve $c: I\rightarrow M, t \mapsto c(t)$ where $I \subseteq \mathbb R$ is some open interval. Then we can differentiate $c$ with respect to $t$ as in multivariate calculus and get a vector $c'(t)$ at the point $c(t) \in M.$ Then $c'(t)$ is a tangent vector at $M$ in $c(t).$ In fact, you can get all tangent vectors in this way. Just to convince you, let's apply this construction to the unit sphere. So, suppose that $c$ runs entirely in the unit sphere, i.e. $\langle c(t),c(t)\rangle \equiv 1$ is constant. By differenting, we get $$ 0 = \frac{d}{dt}1 = \frac{d}{dt}\langle c(t),c(t)\rangle = 2\langle c'(t),c(t)\rangle, $$ which is exactly our condition from (1) for tangent vectors at the unit sphere. Nice, isn't it? :-) In fact, this calculation is one of the slick tricks from differential geometry. :-) :-) You can find all this in much more detail in any good book on that subject.