Direction of velocity vector in 3D space

Question

According to a well-known textbook (Halliday & Resnick), the direction of a velocity vector, $\vec v$, at any instant is the direction of the tangent to a particle's path at that instant, as is illustrated below in 2D.

According to the same textbook, the same holds for 3D. However, the tangent to a curve in 3D is not a line, but a plane! A vector could be in a plane and still take on any direction betwen $0^{\circ}$ and $360^{\circ}$ within that plane.

How do we define and determine the direction of $\vec v$ given a particle's path in 3D? (If possible, include illustrations in your answers).

Answer 1

Example with Figures on @JEB 's answer.

Let the regular (smooth) curve with parametric equation \begin{equation} \mathbf{x}\left(t\right)=\bigl[x_{1}\left(t\right),x_{2}\left(t\right),x_{3}\left(t\right)\bigr]=\left(5\cos t,5\sin t,2t\right) \tag{01} \end{equation} The parameter $\:t\:$ would represent time in case that this curve is the trajectory of a particle.

Now, the vector \begin{equation} \dfrac{\mathrm d\mathbf{x}}{\mathrm dt}=\Biggl(\dfrac{\mathrm dx_{1}}{\mathrm dt},\dfrac{\mathrm dx_{2}}{\mathrm dt},\dfrac{\mathrm dx_{3}}{\mathrm dt}\Biggr)=\left(-5\sin t,5\cos t,2\right) \tag{02} \end{equation} is tangent to the curve at the point $\:\mathbf{x}\left(t\right)\:$ and well-defined without any indeterminacy. In case of particle motion this is the velocity vector of the particle.

In order to normalize this vector we have \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert=\sqrt{29} \tag{03} \end{equation} This norm, the speed of the particle, is a function of $\:t\:$ in general. Here accidentally is constant. From (02) and (03) we produce the unit vector \begin{equation} \mathbf{t}=\dfrac{\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert}=\sqrt{\frac{1}{29}}\left(-5\sin t,5\cos t,2\right) \tag{04} \end{equation} The vector $\:\mathbf{t}\left(t\right)\:$ is the unit tangent vector to the curve at point $\:\mathbf{x}\left(t\right)$. \begin{equation} \boxed{\:\mathbf{t}=\sqrt{\frac{1}{29}}\left(-5\sin t,5\cos t,2\right)\:} \tag{05} \end{equation} Differentiating again we have \begin{equation} \dfrac{\mathrm d\mathbf{t}}{\mathrm dt}=\sqrt{\frac{1}{29}}\left(-5\cos t,-5\sin t,0\right) \tag{06} \end{equation} a vector normal to $\:\mathbf{t}\:$ with norm \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}\right\Vert=5\sqrt{\frac{1}{29}} \tag{07} \end{equation} Again this norm is a function of $\:t\:$ in general. From (06) and (07) we produce the unit vector \begin{equation} \mathbf{n}=\dfrac{\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}}{\left\Vert\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}\right\Vert}=\left(-\cos t,-\sin t,0\right) \tag{08} \end{equation} The vector $\:\mathbf{n}\left(t\right)\:$ is the principal normal unit vector to the curve at point $\:\mathbf{x}\left(t\right)$. \begin{equation} \boxed{\:\mathbf{n}=\left(-\cos t,-\sin t,0\right)\vphantom{\sqrt{\frac{1}{29}}}\:} \tag{09} \end{equation}

Finally we construct the unit vector \begin{equation} \mathbf{b}=\mathbf{t}\boldsymbol{\times}\mathbf{n}=\sqrt{\frac{1}{29}} \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ -5\sin t & 5\cos t & 2\vphantom{\dfrac{\dfrac{}{}}{}}\\ -\cos t & -\sin t & 0 \vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} =\sqrt{\frac{1}{29}}\left(2\sin t,-2\cos t,5\right) \tag{10} \end{equation} so \begin{equation} \boxed{\:\mathbf{b}=\sqrt{\frac{1}{29}}\left(2\sin t,-2\cos t,5\right)\vphantom{\sqrt{\frac{1}{29}}}\:} \tag{11} \end{equation} The vector $\:\mathbf{b}\left(t\right)\:$ is the unit binormal vector to the curve at point $\:\mathbf{x}\left(t\right)$.

The triad of vectors $\:\left(\mathbf{t},\mathbf{n},\mathbf{b}\right)\:$ forms a right-handed orthonormal triplet, as shown in Figures, called the moving trihedron. For the three planes, sides of the trihedron, we have the following terminology
\begin{align} \text{plane }\left(\mathbf{t},\mathbf{n}\right) & =\textit{Osculating plane} \tag{12a}\\ \text{plane }\! \left(\mathbf{n},\mathbf{b}\right) & =\textit{Normal plane} \tag{12b}\\ \text{plane }\left(\mathbf{b},\mathbf{t}\right) & =\textit{Rectifying plane} \tag{12c} \end{align}

----- 3D image 1 ----- 3D image 2 ----- 2D video ----- 3D video ----- The moving trihedron 3D video

Answer 2

Check out the Frenet-Serret formalism: a space-curve ${\bf x}(t)$ has an orthonormal basis at all points called the "TNB" (Tangent, Normal, Binormal), which are unit vectors along:

${\bf T} \propto \frac{d{\bf x}(t)}{dt} = {\bf v}(t)$

${\bf N} \propto \frac{d{\bf T}(t)}{dt}$

${\bf B} = {\bf T \times N}$.

Answer 3

I don't know what your book is telling you a tangent is. But what is actually meant, and what will tell you how to draw the velocity vector, is this. Consider a short part of the path the object takes near the point you want the tangent. If you have a short enough bit of the path, it is a straight line. Extend that line in the direction the particle was moving and stick an arrow on the end of it. You have the velocity vector.

In the case that taking a short path near the point is not a straight line, for example the path of the object makes a sharp right turn exactly at that point, then you will not be able to draw a velocity vector this way. But that is OK, because at that point in the motion, the velocity of the object is not well defined so your failure to be able to draw it reflects that fact.

Answer 4

An example should elucidate things. Imagine a particle with a position vector,

$$\vec x = \begin{pmatrix} t\\ t\\ ct \end{pmatrix}$$

parametrised by time $t \in [0,\infty)$, with $c\in\mathbb R$. This corresponds to a particle travelling radially outward; its projection onto the $x-y$ plane makes a $45^\circ$ angle from the origin, but it has some slope $c$ in the $z$-component so it is not equidistant from all three axes.

You know the velocity vector's direction immediately, it's intuitively simply in the direction of the path, and it must be,

$$\vec v =\begin{pmatrix} 1\\ 1\\ c \end{pmatrix} = \frac{\mathrm d}{\mathrm dt }\begin{pmatrix} t\\ t\\ ct \end{pmatrix}.$$

Now, if we have a curved path, the same rule applies: simply take the derivative. You will find the direction is along a portion of the infinitesimal path which is straight.

Answer 5

Since asked in Physics.SE I'll project the Physics point of view. This can be answered intuitively. Imagine yourself flying aimlessly in air. At certain instant you are heading to somewhere. Where will your velocity vector point to?

Find a plane containing smallest part of path curve at that instant. The velocity vector must lie in the same plane.

Note: For exactly straight paths it'll be limiting case. Therefore make those impractical straight paths slightly curved ;), then do the above procedure.