By the definition of curvature (in terms of $t$), $$\kappa(t)=\frac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3},$$ where $r(t)$ represents a linear vertical vector-valued function, such as $r(t)=<0,t>$. Since the derivative of a vertical line is undefined, (e.g. a vertical tangent) it would seem that that $r'(t)$ would similarly be undefined, meaning $\kappa(t)$ would be undefined.
Yet at the same time, the textbook does point out that a linear line's curvature is constant (by nature of it not bending) -- that is, $$\left\|\frac{dT}{ds}\right\|,$$ is constant, or $$\kappa(s)=\left\|\frac{dT}{ds}\right\|=0.$$ Thus, my question is how I can reconcile the two answers, and see where I went wrong in my reasoning that $\kappa(t)$ was undefined.
If $r(t)=\langle0,t\rangle$, then $r'(t)=\langle0,1\rangle$, which is the unit vector pointing up. And $r''(t)=\langle0,0\rangle$, so $$\kappa(t)=\frac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3}=\frac{\|\langle0,1\rangle\times \langle0,0\rangle\|}{\|\langle0,1\rangle\|^3}=\frac{0}{1^3}=0.$$
When you say the derivative of a vertical line is undefined, you're probably confusing the derivative of the function $r(t)$ with the slope of the line. A non-vertical line can be described either by a function $r(t)$ or a function $y(x)$, and the derivative of the $latter$ function, $y'(x)$, is the slope of the line. But a vertical line cannot be described as a function $y(x)$ to begin with.