Vector Calculus Proof Preamble:
dr/ds≡T
where T is a unit vector tangent to a curve, C,  with arc length
κ
is known as the curvature. It is the proportionality constant defined by T′=κN,
where the prime denotes the derivative with respect to arc length.
J. (statement)
If C is not a straight line, as one moves along C one changes direction, which may be estimated as
ΔT∼κNΔs.
Also
Δθ∼Δs/ρ,
∣ΔT/T∣=|ΔT|∼Δθ
must also be true. It follows that
ρ=1/κ,
I am having trouble figuring out if the above statement J. is true or false. Specifically I know the first two lines of the statement do hold true, however, I am a little unsure of the third line |ΔT / T| = |ΔT|~Δθ. How does |ΔT / T| just reduce down to |ΔT| and why is that approximately equal to Δθ? Thank you so much to whoever can help me with this in advance.
 A: $\mathbf T$ and $\Delta \mathbf T$ are vectors, so the expression $\frac {\Delta \mathbf T}{\mathbf T}$ doesn't make sense, as you cannot divide by vectors. What you really mean here is $$\frac{\|\Delta \mathbf T\|}{\|\mathbf T\|}$$
But recall that $\mathbf T$ is a unit vector. I.e., $\|\mathbf T \| = 1$.
As for your questions concerning $\Delta \theta$ and $\rho$, you have not defined either one, so their truth or falsity is also not defined. We can simply define $\Delta \theta = \|\Delta \mathbf T\|$, and then define $\rho = \frac {\Delta s}{\Delta \theta}$ and be done with it.

But presumably, you'd like to tie these variables to geometric meanings. So we need to trace these down. Fix an arbitrary $s$ and let $P = \mathbf r(s)$ be the corresponding point on $C$, with tangent and normal vectors $\bf T$ and $\bf N$, and curvature $\kappa$. Now move a small distance $\Delta s$ along the curve and define $P_1 = \mathbf r(s+ \Delta s)$, with tangent and normal vectors $\mathbf T_1$ and $\mathbf N_1$. Define $\Delta \mathbf T = \mathbf T_1 -\mathbf T$. We are assuming that the tangents (and therefore the normals) are not parallel.
If we draw the lines passing through $P$ parallel to $\mathbf N$ and passing through $P_1$ parallel to $\mathbf N_1$, since they are not parallel, there is some point $Q$ on the first line that is the closest approach to the second (if the curve is not planar, we can't assume the lines intersect). Draw line segments from $Q$ to $P$ and $P_1$. This gives us something like:

Let $\Delta \theta$ be the angle $\angle PQP_1$ and let $\rho$ be the distance from $P$ to $Q$. Now before you can discuss the difference between $\mathbf T$ and $\mathbf T_1$, they have to be vectors eminating from the same point. So translate them both to $Q$, then take $\Delta \mathbf T = \mathbf T_1 - \mathbf T$. Because $\overline{PQ}$ was chosen parallel to $\mathbf N$, it is perpendicular to $\mathbf T$. If $\Delta s$ is small enough, then $\overline {P_1Q}$ will be very close to perpendicular to $\mathbf N_1$ as well. Hence the angle between $\mathbf T_1$ and $\mathbf T$ will be very close to $\Delta \theta$.
Again, if $\Delta s$ is small enough, $P_1Q \approx PQ = \rho$ and $\Delta \theta \approx \frac {\Delta s}{\rho}$ (by the definition of radian measure of angles). Applying the same calculation to the triangle formed by the tangent vectors, we also have that $\|\Delta \mathbf T\|$ approximates the arclength of the circle of radius $1 = \|\mathbf T\| = \|\mathbf T_1\|$ about $Q$ between the two vectors. And therefore $\Delta \theta \approx \|\Delta \mathbf T\|$
