$R(t) \cdot R'(t) = 0$, which is what every source I can find tells me. Even though I understand the proof I don't understand the underlying concept. If $R(t)\cdot R'(t) = 0$, then $R'(t)$ is orthogonal to $R(t)$, right?

But you use the same derivative to find the tangent of a curve. Then somehow if you differentiate the tangent itself, you get the normal to the curve.

I really can't wrap my head around this. Could someone help me understand?

  • 3
    $\begingroup$ Obviously this is false for general $R$. For example, say $R(t)=(t,t)$. Then $R'(t)=(1,1)$. Are they orthogonal? $\endgroup$ Feb 25, 2015 at 4:36
  • 3
    $\begingroup$ This isn't true for a generic vector R(t). For instance, the vector $\langle t,0,0\rangle$ is parallel to its first derivative. (On the other hand, it will be true if the magnitude of R doesn't change in time...) $\endgroup$ Feb 25, 2015 at 4:39
  • $\begingroup$ Can you demonstrate the proof that you have understood? I know that the derivative is perpendicular provided that the vector has constant length, in which case the result follows simply by taking the derivative of the length. $\endgroup$
    – Alp Uzman
    Feb 25, 2015 at 4:43
  • $\begingroup$ @Semiclassical: did you miss that I said exactly the same thing? $\endgroup$ Feb 25, 2015 at 4:57
  • $\begingroup$ @Uzman I got the proof from here... quite confused though $\endgroup$ Feb 25, 2015 at 6:41

4 Answers 4


I'm going to assume that you mean to ask why the derivative of a fixed length vector is perpendicular to the vector itself. Here's the idea:

$$\vec{r}(t)\cdot\vec{r}'(t) = \frac{1}{2}\frac{d}{dt}(\vec{r}(t)\cdot\vec{r}(t)) = 0.$$

If you want to think about why it is true, think about an object rotating on a circle around the origin. Draw some diagrams to figure out what the position and velocity vectors are. Position points outward radially if we imagine the center of the circle is the origin of the $xy$ plane. Velocity is the rate of change of position. If you consider the average velocity over some interval of time, it would be a secant vector (if you want to call it that). As you take a limit, the secant vector will become a vector tangent to the circle. See the images below. (Note that there are length considerations for $v_{\text{avg}}$ that I am ignoring, but that's not super important for this heuristic argument.)

enter image description here

enter image description here

enter image description here

  • $\begingroup$ Thanks for helping! I think I get it now :) $\endgroup$ Feb 25, 2015 at 6:50
  • $\begingroup$ Glad to help :) $\endgroup$ Feb 25, 2015 at 11:45
  • $\begingroup$ @CameronWilliams I was looking at your picture, and I have some trouble understanding how it relates. Can you please explain this to me? $\endgroup$
    – user174622
    Jun 29, 2015 at 2:29
  • $\begingroup$ @YagnaPatel If you imagine the two radial vectors getting closer and closer, the difference of the two velocity vectors will be along the length of the curve. $\endgroup$ Jun 29, 2015 at 2:38
  • 1
    $\begingroup$ @AmirrezaRiahi Yes, perhaps for trivial reasons (assuming you're working with the standard dot product): $\vec{r}(t)\cdot\vec{r}(t) = R^2$ is the same as $x_1(t)^2 + x_2(t)^2 + \cdots + x_n(t)^2 = R^2$ which is the equation of a sphere with radius $R$. $\endgroup$ Jun 16, 2020 at 12:07

The derivative of $\vec{v}$ is orthogonal to $\vec{v}$ if and only if $\vec{v}$ has constant magnitude. To see this:

$|\vec{v}|$ is constant $\iff|\vec{v}|^2$ is constant $\iff\sum v_\alpha^2$ is constant $\iff\frac{d}{dt}\sum v_\alpha^2=0$ $\iff\sum \frac{d}{dt}(v_\alpha^2)=0$ $\iff\sum 2v_\alpha\frac{dv_\alpha}{dt}=0$ $\iff\sum v_\alpha\frac{dv_\alpha}{dt}=0$ $\iff \vec{v}\cdot\frac{d\vec{v}}{dt}=0$

You can close your eyes and picture a vector in space with tail at the origin, and head moving... it changes magnitude when and only when the direction it's moving in isn't orthogonal to itself!

  • $\begingroup$ can't dv/dt be zero? $\endgroup$ Aug 5, 2019 at 18:28
  • $\begingroup$ @ashishyadaveee11 If that is the case, then $\vec{v}$ is the constant vector which of course means that $\|\vec{v}\|$ is constant. $\endgroup$ Jun 16, 2020 at 12:10

Perhaps extended comment... I just find the answer here to be short and sweet:

A vector of constant magnitude will be orthogonal to its derivative because:

$$r(t)\cdot r(t) = \Vert r(t)\Vert^2 = C$$

for some $C$ independent of $t.$ Therefore, differentiating with respect to $t$ we obtain

$$2 r(t) \cdot r'(t) = 0$$


Just to elaborate on the accepted answer.

Another perspective is that the dot product gives how much the two vectors are pointing in the same direction, or how much one of the vectors is in the same direction as the other.

Saying that, the tangent vector being the one which points the direction of movement of the radius vector of the curve at a particular point, when the magnitude is constant, the two vectors in question wont point in the same direction at all and thus the dot product $(\overrightarrow v(t), \overrightarrow {v'}(t))=0$. Now imagine how much the magnitude changes if the velocity vector points is in the same direction as the vector itself.

  • $\begingroup$ "Now imagine how much the magnitude changes if the velocity vector points is in the same direction as the vector itself." By exactly the magnitude of the velocity vector? $\endgroup$ Jan 6, 2020 at 13:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .