# A Problem with velocity vector

I am having a conceptual problem. I understand why the definition of the velocity of a body moving in one dimension is the derivate of its position coordinate. But I don't get why the velocity vector is defined as the derivative of the position vector. Is it that the derivative of a position vector was just found to be equal to the velocity vector or was the derivative of a vector defined so that the velocity vector equals the derivative of position vector?

• Hello, and welcome to Physics SE! I assume you're talking about vectors with different directions. Correct? – Mauro Giliberti Jun 27 at 6:43
• Yh. In two dimensions or more – Ahmod Ahmed Jun 27 at 6:57
• Question, have you had any teaching in vector calculus? Just curious because that's usually my go to explanation. – Triatticus Jun 27 at 7:50
• @AhmodAhmed: I do not get your question. Do you have trouble to understand the vector nature of the velocity, or do you have a problem with the fact that the velocity is the time derivative of the position? – Semoi Jun 27 at 8:25
• Indeed I did mean that, and yes as the page indicates I personally use it as a stand in for multivariate calculus also (at my university they are one and the same). But you can take what I said as multivariate calculus instead if that seems more appropriate to you. – Triatticus Jun 27 at 14:49

## 5 Answers

The logic goes something like this:

• A position vector consists of 2 or 3 components expressed in a Cartesian coordinate system.
• The time derivative of each component results in the speed along each of the components
• Combined all the speeds define the velocity vector.

$$\boldsymbol{v} = \pmatrix{ \dot{x} \\ \dot{y} \\ \dot{z} } = \frac{\rm d}{{\rm d}t} \pmatrix{x \\ y \\ z} = \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{r}$$

The deep reason is that differentiation is a linear operator, meaning the derivative of a vector equals the vector of all derivatives.

To start understanding vectors intuitively, just think of them as mathematical objects that combine dimensions together. If you do understand why the velocity is the change of position with respect to time in one dimension, think of two separate velocities, one in the x-axis, and one in the y-axis. In time t, the given object moves a distance $$v_x \times t$$ from the origin, in the x direction and $$v_y \times t$$ in the y direction. The combination of these two displacements is the total displacement of the object, shown conveniently by a single vector. When we changed from velocity to displacement here by multiplying with time t, we didn't change anything about what the velocity is. So the derivative of this position vector we defined is the combination of the two x and y velocities we initially started with.

The object being a vector doesn't affect its change with time in any way. It only has a direction now, that moves through both the x and y axes (or xyz if it is a 3d vector), instead of being confined to one axis. Hope that helps!

Velocity tells you how fast the position changes. Intuitively the velocity vector should have the following properties

1. The direction of the vector should tell you the direction of movement
2. The magnitude should tell you how fast the position is changing. Or to be more precise, the magnitude should tell you the speed.

When you have two position vectors $$\vec x_1,\vec x_2$$ the difference $$\vec x_2-\vec x_1$$ gives a vector that points from $$\vec x_1$$ to $$\vec x_2$$. Now assume you have some position vector $$\vec r(t)$$. So to find a candidate for our velocity vector I propose $$\vec r(t+\Delta t)-\vec r(t)$$. This is a vector that points from the current position to the future position so it satisfies our first requirement. The magnitude is not quite right: it scales with $$\Delta t$$. But the solution to this should be familiar from calculus. Divide by $$\Delta t$$ and take the limit $$\Delta t\rightarrow 0$$. $$\vec v(t)=\lim_{\Delta t\rightarrow 0}\frac 1 {\Delta t}\big(\vec r(t+\Delta t)-\vec r(t)\big)=\frac{d \vec r}{dt}$$ So I guess that "the derivative of a position vector was just found to be equal to be equal to the velocity vector" would be the most accurate.

So far we haven't shown the second requirement yet explicitly, only hand-wavy. To do this a little better first consider a particle that is restricted in 1 dimension. In that case we only have $$v_x=dx/dt$$ and this definitely fulfulls the second requirement because it is the definition of speed. Now switch to a case that is not constrained in 1 dimension by rotating the vector. Rotating a vector doesn't change its magnitude so the second requirement is still fulfilled.

• Thanks. I think you ve answered me – Ahmod Ahmed Jun 27 at 19:00

It would be possible to answer this question in different ways. Let me try to use an argument which unifies the 1D and multidimensional cases.

Among the possible ways to define the velocity at time $$t$$ in 1D is

that number that, when multiplied by a small time $$\Delta t$$, will provide the best (linear) approximation of the position at time $$t + \Delta t$$.

Here, best means that any approximation $$x(t+\Delta t) \simeq x(t) + w(t) \Delta t$$ will correspond to an error going to zero like $$\Delta t$$, but in the case where $$w(t) = \dot x(t)$$, which corresponds to an error of order at least $$\Delta t^2$$. This is the formal statement of the conceptual idea that velocity provides the information about the instantaneous change of position.

Exactly the same argument can be applied to motions in more than two dimensions. Positions are now represented by the values of a vector function $${\bf r}(t)$$. Changes of positions in a multidimensional space are also represented by vectors $${\bf w}$$. For example, in 2D a local change of coordinates can be written, at the first order in $$\Delta t$$ as: \begin{align} x(t+\Delta t) &= x(t) + w_x(t) \Delta t \\ y(t+\Delta t) &= y(t) + w_y(t) \Delta t. \end{align} Here again, the best linear approximation corresponds to $$w_x(t) = \dot x(t)$$ and $$w_y(t) = \dot y(t)$$. In vector notation $${\bf w}(t) = {\bf \dot r}(t)$$.

It's pretty much straight-forward conclusion of vector and calculus algebra. Position vector is defined as : $$\mathbf r = x\, \mathbf{\hat i} + y\,\mathbf{\hat j} + z\,\mathbf{\hat k}$$

If you value picture, then this is position vector in 3D space :

Speed is position vector derivative against time by definition, so : \begin{align} \mathbf v &= \frac {d\mathbf r}{dt} \\&= \frac {d}{dt} \left(x\, \mathbf{\hat i} + y\,\mathbf{\hat j} + z\,\mathbf{\hat k}\right) \\&=\frac {d}{dt}x\, \mathbf{\hat i} + \frac {d}{dt}y\, \mathbf{\hat j} + \frac {d}{dt}z\, \mathbf{\hat k} \\&=v_x\,\mathbf{\hat i} + v_y\,\mathbf{\hat j} + v_z\,\mathbf{\hat k} \end{align}

So as you see mathematically both $$\mathbf r,\mathbf v$$ structures are identical,- both are vectors composed of summation of three scaled unit vectors in 3D space. Only scaling scalar values has different Physical meaning- in one case it's a body position magnitudes $$(x,y,z)$$ in Euclidean space along different axis. In another - speed projection magnitudes $$(v_x,v_y,v_z)$$ in same Euclidean axis.