My definition of resultant velocity:

If a certain object, at some instant of time, moves with speed $v_x$ in the x-direction, and with speed $v_y$ in the y-direction, then it has a resultant velocity which is the hypotenuse of the triangle formed by the two vectors: one in pure x-direction with magnitude $v_x$ and the other in purely y-direction with magnitude $v_y$.

Thus, $v_x$ and $v_y$ are components of the resultant velocity vector.

One way to represent how the three vectors relate in magnitude is by the classic Pythagorean theorem: (1) $$v_{res}^2 = v_x^2 + v_y^2$$

However, the object's position also follows the Pythagorean theorem (for ease-of-calculation let's say at $t = 0$, the object is at the origin, yielding: $$r(t)^2 = x(t)^2 + y(t)^2$$

Differentiating with respect to $t$ on both sides, and re-arranging yields: $$ r\dot r = x\dot x + y\dot y $$(2) $$ \dot r = v_{res} = \frac{x\dot x + y\dot y}r$$

Of course (1) and (2) are not equivalent - but if they are both derivations for the resultant velocity of an object - why are they not the same? I suspect that the two setups are representing different scenarios (like the first one is a simple relative speed problem and the latter is a related rates problem involving, perhaps, 2 objects).

  • $\begingroup$ What do you mean by "the object's position also follows the Pythagorean theorem"? The position of the object is a point in plane, don't you need to write $r(t)=(x(t),y(t))$? Then taking derivative you get $r'(t)=(x'(t),y'(t))=(v_x,v_y)$ and then $v_{res}=\Vert r'(t)\Vert = \sqrt{v_x^2+v_y^2}$. $\endgroup$ – Levent Aug 3 '20 at 21:27

They are not the same in general because derivative of the norm is not the same as norm of the derivative.

Speed $v$ is the norm of the velocity vector $\vec{v}$, i.e. $$v = \|\vec{v}\| = \left\|\frac{d}{dt}\vec{r}\right\|= \|(\dot{x},\dot{y})\| = \sqrt{\dot{x}^2+\dot{y}^2}$$ Your second concept is the derivative of the norm of the position vector $\vec{r}$, i.e. $$\dot{r} = \frac{dr}{dt} = \frac{d}{dt} \|\vec{r}\| = \frac{d}{dt}\|(x,y)\|=\frac{d}{dt}\sqrt{x^2+y^2}.$$

For a simple example, consider a circular motion given by $\vec{r}(t) = (\cos t, \sin t)$. The velocity is $$\vec{v}(t) = (-\sin t,\cos t) \implies v = \|v\| = 1.$$ Your other concept is $$r = \|\vec{r}\| = 1 \implies \dot{r} = 0$$ so clearly $v \ne \dot{r}$.

It is interesting to see that it always holds $\dot{r} \le v$. Namely, we have $$2r\dot{r}=\frac{d}{dt}(r^2) = \frac{d}{dt}\|\vec{r}\|^2 = \frac{d}{dt}(\vec{r}\cdot\vec{r}) = 2\dot{\vec{r}}\cdot \vec{r} = 2\vec{v}\cdot\vec{r}$$ and hence Cauchy-Schwartz inequality implies $$r\dot{r} = \vec{v}\cdot\vec{r} \le \|\vec{v}\|\|\vec{r}\| = vr \implies \dot{r} \le v.$$

  • $\begingroup$ Does the "derivative of the norm" have any significance? As in, what tangible way is it (possibly) applied? $\endgroup$ – Fragondruit Aug 4 '20 at 7:21
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    $\begingroup$ @Fragondruit $r$ denotes the distance from the origin so $\dot{r}$ describes how fast is a particle moving from the origin. For example, in the polar coordinate system the position vector is given by $\vec{r} = r\hat{r}$ and the velocity is $\vec{v} = \dot{r}\hat{r} + r\dot{\phi}\hat{\phi}$. The term $\dot{r}\hat{r}$ is the radial component of the velocity (how fast the particle approaches/recedes from the origin), and $r\dot{\phi}\hat{\phi}$ describes how fast the particle circles around the origin. $\endgroup$ – mechanodroid Aug 4 '20 at 9:27

The position of a moving object in the plane can be represented in cartesian coordinates as

$$ p = (x(t), y(t)) $$

It is a vector quantity. The velocity is obtained as

$$ \dot p = (\dot x(t), \dot y(t)) = (v_x(t), v_y(t)) $$

which is a vector while the speed is $||\dot p|| = \sqrt{v_x^2+v_y^2}$

now if you derive the scalar quantity $||p||^2$ the result is

$$ \frac{d}{dt}||p||^2 = 2p\cdot \dot p = 2 x v_x+2y v_y \ne ||\dot p||^2 = v_x^2+v_y^2 $$

which is a scalar


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