# Direction of velocity vector in two dimensions

I'm having some problems understanding the velocity vector in two dimensions. Any given trajectory it's expressed by a curve. Here we have two metrics to calculate how much the point has traveled:

• Displacement: a vector that only takes into account the initial position and the final position.
• Distance: a scalar that tells us about the amount of space the point has traveled.

In one dimension the velocity can be expressed simply as $$v = \frac{dx}{dt}$$ because we're moving along one axis and the two are the same. However, in two or more dimensions we have a problem with the definition of velocity because, since it's a scalar we have no clue about its direction. The displacement vector is like this:

if we define velocity as $$\vec{v} = \frac{\Delta\vec{r}}{dt}$$ we make an error but taking the limit we get that error close to $$0$$. My books explains how essentially $$d\vec{r} = ds\ \hat{u_T}$$, where $$\hat{u_T}$$ is a versor tangent to the trajectory. This makes sense because if the displacement is infinitesimal then it must be tangent to the trajectory so it's like if we move an infinitesimal step along the tangent direction. Here confusion kicks in, because it also says that now the velocity vector can be expressed as $$\vec{v} = \frac{ds}{dt}\hat{u_T}$$ so it means that the velocity vector can be expressed as a vector tangent to the trajectory that has the magnitude of the instantaneous velocity. I don't get why this is the case, if we consider pointwise the vector $$d\vec{r}$$

we can see that $$\hat{u_T}$$ changes in accordance with position, thus depends on time, so I thought the correct expression was $$\vec{v} = \frac{d(s\ \hat{u_T})}{dt} = \frac{ds}{dt}\hat{u_T} + s \frac{d\hat{u_T}}{dt}$$

Why is this not the case? What am I missing? Was everything right or I misunderstood something?

• In your equation given in the end, you are multiplying a differential quantity (ds) with a finite quantity ($d\hat u_T/dt$), to get another finite quantity (v), thus this term is of no consequence in the limit $ds\rightarrow 0$. Nov 1, 2022 at 17:41
• This probably comes to the same thing as the previous comment, but $\frac {ds \hat u_r}{dt}$ means $\frac {ds }{dt} \hat u_r$ and not, as you seem to be assuming, $\frac{d(s \hat u_r)}{dt}$. Nov 1, 2022 at 17:45
• @PhilipWood Can you expand a little on your comment because I too think that the expression $d(s\hat u_t)/dt$ is the appropriate one before taking the limit of course. For example, this is the same thing we do in deriving kinematical quantities in polar coordinates. Nov 1, 2022 at 17:52
• @JoseManuel Oh God I just realized that the expression you wrote in your post is wrong, the correct one is $v=(ds/dt)\hat u_T+s (d\hat u_T/dt)$ and not $v=(ds/dt)\hat u_T+*ds* (d\hat u_T/dt)$. And yes indeed this is the most general expression one gets for a velocity. (1/2) Nov 1, 2022 at 18:58
• @JoseManuel Your book most probably assumes that the particle is constrained to move on a given path in which case the tangential component does not cause the motion of the particle, it's still there but there are constraint forces in addition. Eli essentially assumes this above. Kleppner and Kolenkow explain this really well. Apologies for any confusion caused. (2/2) Nov 1, 2022 at 18:59

First, a factual issue: velocity is not a scalar. It is a vector. Speed is a scalar (and this is just the magnitude of velocity). In 3D Cartesian coordinates, we can write the velocity vector as $$\vec{v}=\frac{d{\vec{r}(t)}}{dt}=\frac{dr_x}{dt}\hat{x}+\frac{dr_y}{dt}\hat{y}+\frac{d r_z}{dt}\hat{z}$$ where $$r_x$$ is the magnitude of $$r$$ in the $$\hat{x}$$ direction, and so on. (If $$\vec{r}=\vec{x}+\vec{y}+\vec{z}$$, then $$r_x$$ is just $$|\vec{x}|=x$$, and so forth.) This expression tells us how the magnitude of each vector component of $$\vec{r}$$ is changing with time.
We can write this in terms of a unit vector tangent to the trajectory $$\vec{r}(t)$$. Usually we might denote this unit vector by $$\hat{v}$$, but your book uses $$\hat{u_T}$$ to illustrate the point that this vector is tangent to the position curve. We can write the tangent unit vector $$\hat{u_T}$$ like this: $$\hat{v}=\hat{u_T}=\frac{\vec{v}}{|\vec{v}|}=\frac{1}{\sqrt{\dot{r_x}^2+\dot{r_y}^2+\dot{r_z}^2}}(\dot{r_x}\hat{x}+\dot{r_y}\hat{y}+\dot{r_z}\hat{z})$$ where I have used the common physics notation $$\dot{r}=dr/dt$$ for each component. This expression just says that we are dividing each component of the vector by the total magnitude of the vector to rescale our vector such that it has unit length. The proportion of the vector that points in each direction is unchanged; we are just rescaling the vector such that the magnitude is 1.
Now, your question is: Why is this unit vector for the velocity tangent to the position curve $$\vec{r}(t)$$? When we plot the trajectory of an object over time (like you did in your question), the velocity is the direction the object is instantaneously moving and the speed with which it is moving. This is what is meant by the expression $$d\vec{r}/dt$$ (change in position vector over an infinitesimal time). The object can't instantaneously be moving in a direction that is not tangent to its current trajectory. That would be equivalent to an object moving from point A to point B instantaneously without first traveling the distance between them. We sometimes make this point by saying the trajectory of the object must be continuous.
Your confusion about the expression $$\vec{v}=ds/dt\cdot\hat{u_T}$$ is resolved by noting that $$ds/dt$$ is the scalar distance traveled in an infinitesimal amount of time, and $$\hat{u_T}$$ is the vector pointing in the direction of this motion. Since $$\hat{u_T}$$ is just the direction of the change in position, we can understand that $$\hat{u_T}$$ is not changing in the same way that $$ds/dt$$ is not changing. We are dealing with an infinitesimal, which means we are evaluating the instantaneous motion, so neither the distance nor direction of this motion is changing. The scalar distance $$s$$ is just the length of the position vector, so $$ds=d|\vec{r}|=dr_x+dr_y+dr_z$$. So we can write $$\vec{v}=\frac{ds}{dt}\hat{u_T}=\frac{d|\vec{r}|}{dt}\hat{u_T}=(\frac{dr_x}{dt}+\frac{dr_y}{dx}+\frac{dr_z}{dz})\hat{u_T}$$ where you can compute $$\hat{u_T}$$ in the way I showed above.