Show that,if $T$ represent the unit tangent vector then $\frac {dT}{ds}$is perpendicular to $T$.

Question

In the book written by R.M.Khan it is given that since $T$ is unit modulus $T\frac {dT}{ds}=0$ implies $T$ and $\frac {dT}{ds}$ are perpendicular to each other.

But $\frac {dT}{ds}$ is not a proper vector or in other words a null vector; how can the result be obtained?

Please help to make me understand what actually the statement means.

Answer 1

The derivative of $s \mapsto \Vert T(s) \Vert^2$ is $2 T.\frac{dT}{ds}$.

As $\Vert T \Vert^2=1$ is constant, the derivative is equal to zero which leads to the result.

Answer 2

it means that the original curve, call it $\gamma(s),$ is required to be arc length parametrized, that is $T = \gamma'$ is always of length one, or $T \cdot T = 1.$ The product rule says $$(V \cdot W)' = V' \cdot W + V \cdot W'.$$ So $$ 0 = (T \cdot T)' = T' \cdot T + T \cdot T' = 2 T \cdot T'. $$ So $$ T \cdot T' = 0. $$ Whenever $T' \neq 0,$ one defines a scalar function $\kappa$ and another vector field $N$ with $N \cdot N = 1$ by $$ T' = \kappa N $$ with some care needed when $T'=0.$

What are $T, N, \kappa$ when $$ \gamma(s) = (5 \cos s, 5 \sin s) $$ in the plane?

Answer 3

Suppose vector a is of constant magnitude that is it has a scalar magnitude of varying direction da/dt=0. So we relate this idea with vector T with unit modulus that is dT/ds=0. We can perform a scalar product operation to ensure the vectors are perpendicular. T.dT/ds=|T||dT/ds|cos© Hence |T||dT/ds|cos©=0 cos©=0 /|T||dT/ds| cos©=0 ©=arccos(0) ©=90° This shows that vector T is perpendicular to dT/ds.

Answer 4

$\mathbf{T} = (T_x,T_y,T_z)$ is the tangent vector to the curve by construction. If we're talking about the motion of a particle along a curve, this points in the direction of the velocity.

$\frac{d\mathbf{T}}{ds} = (\frac{dT_x}{ds},\frac{dT_y}{ds},\frac{dT_z}{ds})$ is how the tangent vector changes as we progress along the curve. This is akin to an acceleration vector. It points inwards to the curvature of the motion.

If $T^2 = c$ where $c$ represents any constant length, the derivative with respect to ds will reveal that $T$ and $\frac{d\mathbf{T}}{ds}$ are orthogonal.

$$T^2 = c$$ $$\frac{d}{ds}T^2 = \frac{d}{ds}c$$ $$\frac{d}{ds}(\mathbf{T}\cdot\mathbf{T}) = 0$$ $$\frac{d\mathbf{T}}{ds}\cdot\mathbf{T} + \mathbf{T}\cdot\frac{d\mathbf{T}}{ds}= 0$$

Note that we applied the product rule to the LHS; the derivative is NOT simply $2T$.

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

From the definition of the dot product, $\mathbf{a}\cdot\mathbf{b} = ab\cos\theta$, two vectors are orthogonal when the RHS vanishes. Hence our $\mathbf{T}$ and $\frac{d\mathbf{T}}{ds}$ are orthogonal.

Note that this result is not exclusive to the tangent vector. It's true of any vector $\mathbf{v}$ where $v^2 = c$.