I'm working on some calc III problems, and found that the unit tangent vector of
$$\langle t + \dfrac{1}{t}, 2\ln(t)\rangle $$
is
$$\left\langle \dfrac{t^2 - 1}{t^2 + 1}, \dfrac{2t}{t^2 + 1} \right\rangle.$$
This is weird to me! I recognize this expression as the same equation as projecting the line onto the circle, i.e. the two-dimensional stereographic projection. Is this a coincidence, or is there something special happening here? I'm very suspicious.
On
Let me begin with a bit nitpicking: The function $\rho(t) = \left(\dfrac{t^2 - 1}{t^2 + 1}, \dfrac{2t}{t^2 + 1}\right)$ does not exactly occur in the context of two-dimensional stereographic projection. The latter is usually understood as the stereographic projection from the north pole of the unit circle $S^1 = \{(x,y) \mid x^2 + y^2 = 1 \}$ to the $x$-axis. Its inverse is given by the homeomorphism $\sigma : \mathbb R \to S^1 \setminus \{(0,1)\}, \sigma(t) = \left(\dfrac{2t}{t^2 + 1}, \dfrac{t^2 - 1}{t^2 + 1}\right)$.
You consider the curve $\phi: (0,\infty) \to \mathbb R^2, \phi(t) = \left(t + \dfrac{1}{t}, 2\ln(t)\right)$ and observe that its "unit tangent vector curve" is $\rho : (0,\infty) \to \mathbb R^2, \sigma(t) = \left(\dfrac{t^2 - 1}{t^2 + 1},\dfrac{2t}{t^2 + 1} \right)$. This function differs from $\sigma$ by its smaller domain and the interchanged coordinates. Geometrically $\rho$ is a restriction of the inverse of the stereographic projection from the "east point" $(1,0)$ of $S^1$ to the $y$-axis. It parameterizes the open upper half circle in clockwise direction. Anyway, it does not really matter that $\rho$ is formally something else than $\sigma$.
Ravi Fernandos's comment gives the appropriate answer to your question. Although it may at first glance look surprising that the the unit tangent vectors of the curve $\phi$ give us the inverse $\rho$ of the "eastern stereographic projection", it can't be really surprising that the function $\rho$ occurs for some curve $(0,\infty) \to \mathbb R^2$. Thus we may regard it as a mere coincidence that in your calculus III problems you have to consider the particular curve $\phi$.
But perhaps it is more interesting to understand how we can find a curve producing $\rho$ as its unit tangent vector curve. We shall see that the solution is not unique.
Approach 1.
Both coordinate functions are rational functions. It is well-known how to determine their antiderivatives and we see that $\rho$ is the derivative of the curve
$$\alpha(t) = \left(t - 2\arctan t, \ln(t^2 +1) \right) .$$
Approach 2.
It is well-known that the Weierstrass substitution $t = \tau(x) = \tan(x/2)$ gives us the function $$f = \rho \circ \tau: (0,\pi) \to \mathbb R^2, f(x) = (-\cos x, \sin x) .$$ This is just a reparameterization of $\rho$. Obviously we have $f(x) = (-\sin'x, -\cos' x) = g'(x)$ with $g(x) = (-\sin x, -\cos x)$. Thus $f$ is the unit tangent vector curve of $g : (0,\pi) \to \mathbb R^2$ which is not at all surprising. Note that $g$ parameterizes the open left half circle in clockwise direction.
Let us now substitute $ x = \tau^{-1}(t)$. We get the function $$F = g \circ \tau^{-1} : (0,\infty) \to \mathbb R^2. $$ Then $$F'(t) = g'(\tau^{-1}(t))(\tau^{-1})'(t) = f\tau^{-1}(t) \frac{1}{\tau'(\tau^{-1}(t))} = \frac{\rho(t)}{\tau'(\tau^{-1}(t))} .$$ This shows that the unit tangent vector curve of $F$ is nothing else than $\rho$. Explicitly we get $$F(t) = \left(\dfrac{-2t}{t^2 + 1} , \dfrac{t^2 - 1}{t^2 + 1} \right) , \\ F'(t) = \frac{2}{t^2+1}\rho(t) .$$
Perhaps we could also have guessed that the above $F$ will do, without reparameterizing $\rho$ into the trigonometric representation. But I think it is useful to understand the connection of the functions $ \dfrac{t^2 - 1}{t^2 + 1}$ and $\dfrac{2t}{t^2 + 1}$ with $\sin$ and $\cos$.
Approach 3.
Write $$\rho(t) = \frac{t^2}{t^2 +1}\left(\frac{t^2-1}{t^2}, \frac{2}{t}\right) = \frac{t^2}{t^2 +1}\left(1 - \frac{1}{t^2}, \frac{2}{t}\right)$$ or equivalently $$\left(1 + \frac{1}{t^2}\right)\rho(t) = \left(1 - \frac{1}{t^2}, \frac{2}{t}\right) .$$
The function on the RHS is obviously the derivative of $\phi(t) = \left(t + \frac{1}{t}, 2 \ln t \right)$ which immediately implies that $\rho$ is the unit tangent vector curve of $\phi$, the factor $\left(1 + \frac{1}{t^2}\right)$ being the norm of $\phi'(t)$.
Approach 4.
Write $$\rho(t) = \frac{1}{t^2 +1}\left(t^2-1, 2t \right)$$ or equivalently $$(t^2+1)\rho(t) = (t^2-1,2t) .$$
The function on the RHS is obviously the derivative of $\psi(t) = \left(t^3/3 -t , t^2 \right)$ which immediately implies that $\rho$ is the unit tangent vector curve of $\psi$, the factor $t^2 +1$ being the norm of $\psi'(t)$.
Your second equation parametrizes $\sin$ and $\cos$.
As the integral of these functions is the $\pm$ the other one (with a zero constant), the mapping is a transform of the original.