If the unit tangent vector T has constant length, under what additional conditions does the derivative of T with respect to time have constant length?

Question

I am taking Calculus 3 and I am learning about the unit tangent vector T. I know that dT/dt does not always have constant length. My question is under what additional conditions does dT/dt have constant length. I have an idea of 4 additional conditions, of which I'll kind of prove below.

Let's assume a 2D example for simplicity. r = x i + y j dr/dt = dx/dt i + dy/dt j ||dr/dt|| = sqrt( (dx/dt)^2 + (dy/dt)^2) = Let's call this speed T = dr/dt / speed = (dx/dt / speed) i + (dy/dt / speed) j Also I will be using x' for dx/dt and x'' for d2x/dt2 dT/dt = (x''*(speed^2) - x'*(x'*x'' + y'*y''))/(speed^3) i + (y''*(speed^2) - y'*(x'*x'' + y'*y''))/(speed^3) j So our first condition will be that speed or ||dr/dt|| is constant.

Let's factor speed out of dT/dt and only focus on the numerators of the x and y components. (x''*(speed^2))^2 + (y''*(speed^2))^2 should be constant if 2 conditions apply:

One speed is constant. We already assumed this with our first condition

Two also (x'')^2 + (y'')^2 is constant.

(x'*(x'*x'' + y'*y''))^2 + (y'*(x'*x'' + y'*y''))^2 should be constant if 2 conditions apply:

One (x')^2 + (y')^2 is constant. We already assumed this when we assumed speed is constant

Two x'*x'' + y'*y'' is constant. This is a new condition.

All together, we get the following conclusion: If we have unit tangent vector T, then dT/dt has constant length if and only if

One speed or (x')^2 + (y')^2 is constant

Two (x'')^2 + (y'')^2 is constant

Three x'*x'' + y'*y'' is constant

Can someone please tell me if this is right? And also provide an intuitive explanation for why these 3 conditions must hold? Thank you for taking the time to read this.

Answer 1

I will use a bit of mechanics here... Acceleration of point is a sum of two components:

$$\vec{a}=\frac{d\vec v}{dt}=\vec{a_t}+\vec{a_n}$$

$$\vec{a_t}=\frac{dv}{dt}\vec{u_t}$$

$$\vec{a_n}=\frac{v^2}{R_k}\vec{u_n}$$

$\vec{u_t},\vec{u_n}$ are unit vectors, first is tangent to trajectory, the second is perpendicular to it. $R_k$ is the radius of the trajectory at the given point.

In your problem $\vec{T}$ is actually velocity $\vec{v}$ of a point along some trajectory in space.

Because $\vec{v}$ has constant length, $dv/dt=0$ which means that $\vec{a_t}=0$. This leaves only the second component so the intensitiy of it is:

$$a=a_n=\frac{v^2}{R_k}$$

You know that $v$ is constant and you want $a$ to be constant. The only possibility is to have constant $R_k$ as well. And the only curve that has constant $R_k$ is a circle or a part of it!

Answer 2

Curvature $\kappa$ is defined as the magnitude of $\dfrac{d\vec T}{ds}$, where $s$ is the arclength of the curve. The chain rule gives us $$\frac{d\vec T}{dt} = \frac{d\vec T}{ds}\frac{ds}{dt} = \kappa\vec N\,\upsilon,$$ where $\upsilon = \dfrac{ds}{dt}$ is the speed and $\vec N$ is called the principal normal.

Your condition that $d\vec T/dt$ have constant magnitude is therefore the condition that $\kappa\upsilon$ be constant. If your speed is constant, this is equivalent to constant $\kappa$. In the plane this is a circle, but in space there are zillions of examples (among them circular helices, but lots more complicated). And if you imagine a particle changing speed inversely proportional to the curvature, then you get zillions more examples.

(At this point you might want to look a little bit at an undergraduate differential geometry text. Immediately accessible is mine, if you're interested. See the beginning of the second section.)