# What is the intuition behind the unit normal vector being the derivative of the unit tangent vector?

I've seen the math, but... It just doesn't make sense to me. How is the slope going to point perpendicular to the vector that is clearly a straight line not going in that direction?

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Take some point $P_1 = C(s_1)$ on a curve $C$. Let $T_1$ be the unit tangent at this point. Now move a tiny distance along the curve, to get to the point $P_2 = C(s_2)$, and let $T_2$ be the unit tangent at this point. Now draw the vector $V = T_2 - T_1$. If $P_1$ and $P_2$ are very close together, $T_1$ and $T_2$ will be almost equal, and $V$ will be roughly perpendicular to both of them (because the triangle with sides $T_1$ and $T_2$ will be isosceles and very skinny. Here's a picture.

But, by the definition of derivative, $$\frac{dT}{ds} = \lim_{s_2 \to s1} \frac{T_2 - T_1}{s_2 - s_1}$$ and this limit will also be roughly perpendicular to $T_1$.

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The unit tangent vector gives the instantaneous velocity. But unless you go in a straight line forever, you will turn. Suppose you turn left. The unit tangent vector still points forward at any given moment, but it is turning left -- its derivative is leftward. The unit normal points left, to indicate the direction that the tangent is changing.

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If you do go in a straight line forever, then what happens? Is our unit normal vector still the derivative? I believe so... So you wouldn't be turning left or right, which means the tangent isn't changing, but what would happen? Is another way to think about this the fact that the unit normal is the double derivative of our vector function, which tends to show the concavity. In this case, the concavity is showing the general trend, which is that the graph is going left or right? I still don't quite see it. If we were to treat our first derivative as a new vector function, that seems to break. –  David Sep 30 '13 at 21:46
@David No. In the case of straight-line motion, the unit tangent vector is a constant vector, so its derivative is the zero vector. –  David H Sep 30 '13 at 22:02
@DavidH, indeed. Then, the unit tangent vector on non-straight-line motion would definitely not be a constant vector. If we were to treat our unit tangent vector as our vector function of interest, then wouldn't the derivative of it be a tangent vector, which is tangent to it? Or, perhaps, I'm thinking of the tangent vector incorrectly. –  David Sep 30 '13 at 22:11
@David See the answer I just submitted. Does that help? –  David H Sep 30 '13 at 22:34

The dot product of the unit tangent vector with itself is of course equal to 1.

$$\mathbf{T} \cdot \mathbf{T} = \|\mathbf{T}\|^2 = 1^2 = 1.$$

Take the derivative of both sides, and remembering the product rule,

$$\frac{d}{ds} ( \mathbf{T} \cdot \mathbf{T} ) = \frac{d}{ds}(1)$$

$$\mathbf{T} \cdot \frac{d \mathbf{T}}{ds} + \frac{d \mathbf{T}}{ds} \cdot \mathbf{T} = 0$$

$$2 \mathbf{T} \cdot \frac{d \mathbf{T}}{ds} = 0$$

or

$$\mathbf{T} \cdot \frac{d \mathbf{T}}{ds} = 0.$$

## That is to say, the derivative of the unit tangent vector is perpendicular to the unit tangent vector, i.e. it's a normal vector.

The essence of a derivative is the approximation of functions by linear equations: $\mathbf{T} (s + \delta s) \approx \mathbf{T} (s) + \delta s \mathbf{T}' (s)$.

Performing this vector addition tip-to-tale, the three vectors $\mathbf{T} (s + \delta s),\space \mathbf{T} (s) \space \text{and} \space \delta s \mathbf{T}' (s)$ form the sides of an isosceles triangle with two legs of length 1 and an arbitrarily small base with length proportional to $\delta s$. As $\delta s$ goes to zero, the angle between the two legs goes to zero, and the base angles go to right angles.

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See, I understand that. I could go through the math, but I want to be able to visually see or verify this, which I'm having a hard time doing (probably because I might be conceptually wrong in what I think a derivative in this case is). –  David Sep 30 '13 at 22:47
@David H: $\Vert T'(s) \Vert \ne 1$ in general; instead, we have the curvature $\kappa = \Vert T'(s) \Vert$. –  Robert Lewis Oct 1 '13 at 0:50
@RobertLewis Wow, of course. I must have the dumb. –  David H Oct 1 '13 at 1:23
@David H: s'all good, we caught that grouder! –  Robert Lewis Oct 1 '13 at 1:32

First of all, the unit normal vector $N$ is in general not the derivative of the unit tangent vector $T$. What is true is that the derivative of $T$ with respect to the arc length along the curve $s$ is a vector $T'(s)$ normal to $T(s)$ in the sense that $T'(s) \cdot T(s) = 0$, and if $T'(s) \ne 0$, so that $\Vert T'(s) \Vert \ne 0$, we can define the unit normal $N(s)$ by the equation

$N(s) = \kappa^{-1}T'(s), \tag{1}$

where

$\kappa = \Vert T'(s) \Vert \tag{2}$

is the magnitude of $T'(s)$. (1) is then equivalent to

$N(s) = \kappa T'(s), \tag{3}$

the way (1) is usually written; it is one of the Frenet-Serret formulas which are the standard differential geometric means of addressing the properties of curves in $\Bbb R^3$. It is very important, in utilizing these formulas, to take derivatives with respect to $s$, the arc length of the curve $\gamma(s)$ under consideration. Otherwise, though we will still have $T'(s) \cdot T(s) = 0$, since $T(s) \cdot T(s)$ is constant, implying $T'(s) \cdot T(s) = \frac{1}{2}(T(s) \cdot T(s))' = 0$, $\kappa$ will not necessarily reflect the true rate of bending of the curve $\gamma(s)$. This is perhaps most easily seen via the observation that, if $r$ is another parametrization of $\gamma$, then along $\gamma$, $\frac{d}{dr} = \frac{ds}{dr} \frac{d}{ds}$. This means, for example, that

$\frac{dT}{dr} = \frac{ds}{dr} \frac{dT}{ds} = \frac{ds}{dr} \kappa N; \tag{4}$

if we accept $\Vert \frac{dT}{dr} \Vert$ as the definition of curvature, it's value can vary extensively with the choice of $r$, and its geometrical meaning is obscured. Taking the curve parameter as arc length $s$ standardizes the definition of $\kappa$ by isolating it from the somewhat arbitrary choices of possible parametrizations of $\gamma$.

In the above, the unit vector $N$ gives the direction of bending of the curve $\gamma$, and the curvature $\kappa$ gives its magnitude. Intuition can be developed for this, in terms of elementary physics, by imagining one driving in a car at a fairly high but constant rate of speed. If the vehicle is traveling at say, $100 \,\text{km/hr}$, we can "unitize" this rate by defining a new distance measure, the umpla: $1 \, \text{umpla} = 100 \, \text{km}$; thus "unit speed" in umpla-units is $100 \, \text{km/hr}$; quite rapid, actually. We imagine the car as turning while maintaining this speed; then the car is undergoing an acceleration in a direction perpendicular to its motion; the magnitude of this acceleration is $\kappa$ and its direction is $N$. Now imagine, if you will, for any $s_0$ a circle of radius $\kappa^{-1}(s_0)$ centered at $\gamma(s_0) + \kappa^{-1}(s_0)N(s_0)$, where we hold $s_0$ fixed. This circle touches the curve $\gamma(s)$ at $\gamma(s_0)$. A particle traveling at unit speed on such a circle has velocity vector $T(s_0)$ at $\gamma(s_0)$ and will experience a centripetal acceleration, in the direction $N(s_0)$, given by the classic formula $\frac{v^2}{r}$ with $v = 1$ and $r = \kappa^{-1}(s_0)$, exactly the same as the acceleration $\kappa N$ of $\gamma$ at $s = s_0$; the greater $\kappa(s_0)$ is, smaller $r$ will be, and vice versa; tight turns produce large such accelerations; you can almost feel yourself thrown against the door on a tight right.

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An example from physics: gravity pulls Earth towards the sun, and the earth travels in a (vaguely) circular orbit. So, the attractive force acting on the earth is always a radial vector (pointing towards the center of the orbit), yet the velocity vector that Earth travels with is always tangent to the circle traced out by the orbit.

Force is related to acceleration, and acceleration is the derivative of velocity.

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Ah, and we could generalize this for all curves such that if it isn't perfectly circular, then we are still going towards any one direction at one time (it just so happens that the unit normal vector won't always go towards the origin in a non-perfectly circular orbit... perhaps). Why is this different from derivatives in calc 1 though, where we're used to derivatives explicitly being slope? I feel as though it's odd that we'd come across this. –  David Sep 30 '13 at 23:04
Actually, is it the fact that we're dealing with vectors, and vectors could be anywhere we want them to be, so the derivative is still the slope, but that doesn't necessarily have to look anything like what the tangent looks like. They're different functions, related to one another through the slope. In vectors, it just so happens that the slope of our tangent vector will always be orthogonal to the tangent vector. Yes? If not, then our slope of the tangent vector is orthogonal to the original function we are worried about. –  David Sep 30 '13 at 23:06