What determines the magnitude of a tangent vector?

Main Question or Discussion Point

The unit tangent vector, T(t) = r'(t) / || r'(t) || always has length 1. Alright, so how do we get a sense of the length of the actual tangent vector itself? Its direction is easy to imagine, but I can't understand how its magnitude changes along the curve (does it have something to do with curvature?).

tiny-tim
Homework Helper
welcome to pf!

hi atreide! welcome to pf!
The unit tangent vector, T(t) = r'(t) / || r'(t) || always has length 1. Alright, so how do we get a sense of the length of the actual tangent vector itself? Its direction is easy to imagine, but I can't understand how its magnitude changes along the curve (does it have something to do with curvature?).
no, it has to do with the parameter (the thing you're differentiating wrt, to find the tangent) …

if your parameter is s, the arc-length, so that the curve is (x(s), y(s)), then the magnitude of the derivative wrt s, dr/ds, is always 1

but if your parameter is t, say, then the magnitude (from the chain rule) is ds/dt (or 1/(dt/ds))

lavinia
Gold Member
The unit tangent vector, T(t) = r'(t) / || r'(t) || always has length 1. Alright, so how do we get a sense of the length of the actual tangent vector itself? Its direction is easy to imagine, but I can't understand how its magnitude changes along the curve (does it have something to do with curvature?).
A moving particle has a speed at each point in time. The speed is the length of the tangent vector. T Another particle can move along the same path at different speeds. Its tangent vectors will have different lengths even though the geometric path is the same.

The curve may be curved or straight.

A particle can move at non-constant speed in a straight line.

A moving particle has a speed at each point in time. The speed is the length of the tangent vector. T Another particle can move along the same path at different speeds. Its tangent vectors will have different lengths even though the geometric path is the same.

The curve may be curved or straight.

A particle can move at non-constant speed in a straight line.
So the rate of change of the position vector is related to the parametrization of the curve? I think I get it. So if we have x = t, y = t^2, z = t^3, then the tangent vector would have a larger component in the direction of z, as opposed to x and y?

HallsofIvy
Homework Helper
There is no such thing as "the actual tangent vector." Mathematically, a "tangent vector" to a curve, at a given point, is defined solely by having the same "direction" as the curve at that point. There are an infinite number of "tangent vectors", differing in length (and, in fact, in the opposite direction), at a given point of a curve. Because the length of the derivative (with respect to the parameter) vector depends upon the parameter, the length of the tangent vector contains no information about the curve itself, only about this particular parameterization of the curve. That is one reason why we prefer to use the arclength of the curve as parameter- and, in that case, the length of the derivative vector is 1- we get the unit tangent vector.

But your last sentence implies that you can change the relative size of components of the tangent vector by changing the parameter. That is not true. The ratios of those components depends upon the ratios of the direction angles and, since the direction of the tangent vector is not changed by changing the parameter, they will not change.

If your parameterization of some curve is $x= t$, $y= t^2$, $z= t^3$, then the derivative is the vector $\vec{i}+ 2t\vec{j}+ 3t^2\vec{k}$ but when you say "have a larger component in the direction of z" what other parameterization are you comparing to? And at what point? At (1/3, 1/9, 1/27), the derivative vector is
$$\vec{i}+ \frac{2}{3}\vec{j}+ \frac{1}{9}\vec{k}$$
where it is certainly not true that it has "a larger component in the direction of z" if that was what you meant.

Last edited by a moderator:
Because the length of the derivative (with respect to the parameter) vector depends upon the parameter, the length of the tangent vector contains no information about the curve itself, only about this particular parameterization of the curve. That is one reason why we prefer to use the arclength of the curve as parameter- and, in that case, the length of the derivative vector is 1- we get the unit tangent vector.
Ah, this is exactly what I wanted to know. Thanks for clarifying.

Landau
Because the length of the derivative (with respect to the parameter) vector depends upon the parameter, the length of the tangent vector contains no information about the curve itself, only about this particular parameterization of the curve.
I think there is some ambiguity in the use of the word 'curve'. To me, a 'curve' in a manifold M is a smooth map $$\gamma:I\to M$$ (where I is some interval in R containg 0). You seem to be using 'curve' as the image of such a map, right?

HallsofIvy
Are you saying you would consider $R->R^2$ defined by $x-> (x, x^2)$ to be a different curve than $x-> (x^3,x^6)$? If I were working in differential geometry, I would define a "curve" to be an equivalence class of such maps.