Some set-up from Spivak's textbook:
Given a differentiable map $f:\mathbb R^n\to \mathbb R^m$, define $f_\ast: \mathbb R_p^n\to \mathbb R_{f(p)}^m$ by $$f_\ast(v_p)=(Df(p)(v))_{f(p)}.$$ Note that the notation $v_p$ stands for the vector $v$ at $p$ (so the RHS is a vector at $f(p)$).
Let $c$ be a differentiable curve in $\mathbb R^n$, i.e., a differentiable function $c:[0,1]\to \mathbb R^n$. Define the tangent vector $v$ of $c$ at $t$ as $c_\ast((e_1)_t)=((c^1)'(t),\dots,(c^n)'(t))_{c(t)}$.
Let $c$ be a curve as above with $|c(t)|=1$ for all $t$. Show that $c(t)_{c(t)}$ and the tangent vector to $c$ at $t$ are perpendicular.
First of all, what does $c(t)_{c(t)}$ stand for? The subscript alludes to the fact that $c(t)$ is considered as a tangent vector at $c(t)$, which makes no sense since $c(t)$ is a curve, not a vector. On the other hand, perpendicularity (I assume w.r.t. the standard dot product) is defined only for vectors. So I have no idea what this problem is asking and how to solve it. Any help is appreciated.
You have $c(t) \in \Bbb R^n$, so $c(t)_{c(t)}$ denotes the vector $c(t)$ starting in the point $c(t)$. Then if $|c(t)|=1$ for all $t$, square that and differentiate to get $c(t) \cdot c'(t) = 0$, and this $\cdot$ is the product in the vector space $\Bbb R^n$. Now, by definition of the inner product $\ast$ in $T_{c(t)}\Bbb R^n = \Bbb R^n_{c(t)}$, we have $$c(t)_{c(t)} \ast c'(t)_{c(t)} = c(t)\cdot c'(t) = 0,$$as wanted.