Note that the expression $\frac {\d }{\d t} |\vec {r}(t)|$ is to be understood as $\frac {\d }{\d t} |\vec {r}(t)|$, which measures how the distance the particle is from the origin changes in time. The righthand side is the speed at which the particle is moving; this generally does not tell us how the particle is moving away from the origin. For instance, if a particle moves in a circle of radius $1$, the distance from the origin is always $1$, but the particle could be moving around the circle in many different ways.

Algebraically, note what must be computed on each side.

• $\frac {\d }{\d t} |\vec {r}(t)| = \frac {\d }{\d t} \sqrt {x(t)^2+y(t)^2+z(t)^2}$
• $|\vec {r}'(t)| =\sqrt {\left (\frac {\d x}{\d t}\right )^2+\left (\frac {\d y}{\d t}\right )^2+\left (\frac {\d z}{\d t}\right )^2}$

The chain rule is needed to compute the derivative of the first expression, and there is no reason to believe that the result will be the expression on the righthand side. To see that this is the case, consider $\vec {r}(t) = \vector {\cos (\omega t),\sin (\omega t)}$, which traces out a circle of radius $1$ in a counterclockwise fashion, and compute both expressions.

Intuitively, $| \vec {r}(t) | = 2t^2+4$ tells us that the distance from the origin a point $(x(t),y(t),z(t))$ is at time $t$. Knowing this distance should have no relationship with exactly how fast the point on the curve is moving.

For a specific counterexample, take $\vec {r}(t)=\vector {(2t^2+4)\cos (t),(2t^2+4)\sin (t)}$. Then, $|\vec {r}(t)| = 2t^2+4$, but $\vec {r}'(t) = \vector {4t\cos (t)-(2t^2+4)\sin (t), 4t\sin (t)+(2t^2+4)\cos (t)}$. Now, if $|\vec {r}'(t)|=4t$ for all $t$, it certainly should be true that $|\vec {r}'(\pi )|=4\pi$ (where the choice $t=\pi$ was made since we can check whether $|\vec {r}'(t)| = 4t$ easily there). However, $\vec {r}(\pi ) = \vector {-4\pi ,-2\pi ^2+4}$, so $|\vec {r}'(\pi )| = \sqrt {16\pi ^2+(2\pi ^2+4)^2} > 4 \pi$.

Intuitively, if $\vec {r}(t)$ is a unit vector for all time $t$, this means that the point on the curve at any time $t$ is $1$ unit away from the origin. In three dimensions, this means that the point could be moving along a sphere of radius $1$. No matter how quickly it is moving, as long as it stays on the sphere, it will be $1$ unit away from the origin. Thus, knowing the magnitude of $\vec {r}(t)$ should not give an information about the magnitude of $\vec {r}'(t)$ (i.e. the speed).

For a specific counterexample, take $\vec {r}(t) = \vector {\cos (\omega t),\sin (\omega t)}$ (this describes a particle moving counterclockwise in a circle with angular speed $\omega$). You can check the following.

• $|\vec {r}(t)| = 1$.
• $\vec {r}'(t) = \vector {-\omega \sin (\omega t), \omega \cos (\omega t)}$, so $|\vec {r}'(t)| = \omega$.

Make sure to pay attention to what we are given and being asked to find. We are given that the distance the point on the curve is from the origin is always $2$. The derivative $\dfrac {\d }{\d t} |\vec {r}(t) |$ will give us how this magnitude changes in time, so it must be $0$. Note in particular that $\dfrac {\d }{\d t} |\vec {r}(t) |$ is not the speed. Said more explicitly,

$$\frac {\d }{\d t} |\vec {r}(t)| \neq |\vec {r}'(t)|.$$