# Uniform Circular motion and rate of change of Vectors related to it

When I was studying uniform circular motion and thinking of ways to derive the expressions of related vectors like centripetal and centrifugal acceleration I noticed a certain oddity that occurred in my derivation. I want to know if it is simply a coincidence or whether there is some way to explain it.

Let us consider a circle of uniform radius $r$ with the $r$ making an angle of $\alpha$. Now let us consider two unit vectors $\vec e_r$ along the radial direction with the tail connected to the tip of the $r$ vector. Also the other is $\vec e_t$ where $\vec e_t$ is along the tangent of the circle.

Now we get on resolving them into the components;
$$\vec e_t = cos\alpha \hat j -sin\alpha \hat i$$ $$\vec e_r = cos \alpha \hat i + sin \alpha \hat j$$ $$\vec r = \vert r \vert \cdot \vec e_r$$ Now differentiating w.r.t time, we can derive all off the expressions required however I noticed that the $\vec e_r$ when differentiated w.r.t angle $\alpha$ we get,
$$\frac {d e_r}{d \alpha} = \vec e_t$$ Is this simply a coincidence or is there some sort of concept that can be unearthed here?

Also, can someone tell me someway I can create an image to make the figure clear?

The tangent vector $\vec{T}(\alpha)$ for a curve parameterized by some parameter $\alpha$, say $\vec{r}(\alpha)$, is given by the derivative with respect to $\alpha$, $$\vec{T}(\alpha)\equiv\frac{d}{d\alpha}\vec{r}(\alpha).$$ Your radial unit vector $\hat{e}_r(\alpha)$ is just the unit length vector pointing to the point on the circle parameterized by $\alpha$, so it's natural that taking the $\alpha$ derivative of $\hat{e}_r$ gives you $\hat{e}_t$, the unit vector along the tangent to the circle.
• Why is the tangent to the curve parameterized by some parameter $\alpha$ given by $\frac {d}{d \alpha}$ $\vec r (\alpha)$? Or is that a question for a mathematics forum? – Prakhar Nagpal Jul 1 '18 at 15:48
It seems that this formula just reflects, in the vector form, a relationship between the length of an arc, the radius and the angle subtended by the arc: $S=r\theta$.
The diagram below, shows two unit radial vectors, $\vec e_r$ and $\vec e_r+d\vec e_r$, separated by a small angle $d\alpha$. It also shows a difference vector, $d\vec e_r$, pointing roughly in the direction of a unit tangent vector $\vec e_t$.
With respect to this diagram, your formula says that the magnitude of the difference vector, $d\vec e_r$, is equal to the product of the unit radius, $1$, and the angle $d\alpha$ ($S=r\theta$), while its direction coincides with the direction of the tangent vector $\vec e_t$, which is a good approximation for small angles.