# How to find tangential and normal unit vector to a trajectory from the equation? [duplicate]

Considering a 2D motion in a plane and the equation of the trajectory of a point $y=f(x)$, I don't understand how exactly $\frac{\mathrm{d}y}{\mathrm{d}x}$ can be used.

In particular if I'm looking for the tangential acceleration, I need the tangential unit vector $u_T$.

Since

$$\lvert\vec{v}_y \rvert=\frac{\mathrm{d}f(x(t))}{\mathrm{d}t}=\frac{\mathrm{d}f(x)}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}t}=\frac{\mathrm{d}y}{\mathrm{d}x} \lvert \vec{v}_x \rvert$$

Can I conclude that

$$u_T=\frac{\vec{v}}{\lvert\vec{v}\rvert}=\frac{u_x+\frac{\mathrm{d}y}{\mathrm{d}x}u_y}{\sqrt{1+\bigl(\frac{\mathrm{d}y}{\mathrm{d}x}\bigr)^2}}$$

If so, can I do something similar for trajectories (like circles) in the form $f(x,y)=0$?

Furthermore how can I get the normal unit vector $u_N$ (for istance to find the normal acceleration) from the $u_T$ previously found?

I know that $u_N=\frac{\mathrm{d}u_T/\mathrm{d}t}{\lvert\mathrm{d}u_T/\mathrm{d}t\rvert}$ but the derivative is with respect to time here.

Am I missing something?

• Given a planar vector $(e_x,e_y)$ the normal direction is $(-e_y,e_x)$ May 5 '16 at 20:15
• Possible duplicate of Tangential acceleration in circular motion? May 5 '16 at 20:17

At every point the tangential direction is the unit vector of the velocity vector. If you have the velocity components $\boldsymbol{v} = (\dot{x}, \dot{y})$ at every instant, the you decompose this into a magnitude (speed $v$) and direction $\hat{\boldsymbol{e}}$

\begin{align} v & = \sqrt{\dot{x}^2+\dot{y}^2} \\ \hat{\boldsymbol{e}} & = \begin{pmatrix} \frac{\dot{x}}{v} & \frac{\dot{y}}{v} \end{pmatrix} \\ \boldsymbol{v} & = v\; \hat{\boldsymbol{e}} \end{align}

The normal direction is simply

$$\hat{\boldsymbol{n}} = \begin{pmatrix} -\frac{\dot{y}}{v} & \frac{\dot{x}}{v} \end{pmatrix}$$

The easy way of doing this is to parametrice the trajectory. We have the cartesian definition, so let $\textbf{r} : \mathbb{R} \rightarrow {\mathbb{R}}^{2}$ be: $$\textbf{r}(t)=(t,f(t)) \quad\quad t\in (-\infty,\infty)$$

So we got the vector position as function of "time". Then, the velocity and acceleration vectors are defined by: $$\textbf{v}(t)=\frac{d}{dt}\textbf{r}(t) \quad \quad \textbf{a}(t)=\frac{d}{dt}\textbf{v}(t)$$

Now, you can get the tangential acceleration with the projection over the velocity, that is always tangential to the trajectory: $${a}_{t}=\textbf{a} \cdot \frac{\textbf{v}}{v}$$ Where $v$ is the length of the velocity vector.