This question is an exact duplicate of:

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 $u_T$.


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

Can I conclude that


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?


marked as duplicate by ACuriousMind, ja72, user36790, Qmechanic May 6 '16 at 19:22

This question was marked as an exact duplicate of an existing question.


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.


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