# [SOLVED]Tangential and normal acceleration of a particle moving in a plane curve in the cartesian coordinates

#### Dhamnekar Winod

##### Active member
How did the author compute the highlighted term 2 from the highlighted term 1 in the following answer to the given question?

If $\rho =\frac{d\psi}{ds}$, then the term 2 should be $\upsilon^2 \frac{d\hat{T}}{d\psi}\rho$, but instead, it was written $\frac{\upsilon^2}{\rho}\frac{d\hat{T}}{d\psi}$

How is that computed? How to compute radius of curvature($\kappa$) if $\frac{d\hat{T}}{ds}= \kappa\hat{N}$

Last edited:

#### topsquark

##### Well-known member
MHB Math Helper
It's a typo. Look at the units real quick. s has units of length and $$\displaystyle \psi$$ in is radians (or so I suppose, to make the derivatives right), which means $$\displaystyle \psi$$ is essentially unitless. So $$\displaystyle \dfrac{d \psi }{ds}$$ has units of 1 / length. Thus it's reasonable to suppose that $$\displaystyle \dfrac{d \psi }{ds} = \dfrac{1}{ \rho }$$. You can look up the formula online... I checked it.

-Dan

#### Klaas van Aarsen

##### MHB Seeker
Staff member
If we traverse an angle $d\psi$ on a circle with radius $\rho$, we traverse an arc length of $ds=\rho\,d\psi$.
In other words, by definition we have $\rho = \frac{ds}{d\psi}$, which we can also write as $\frac{d\psi}{ds}=\frac 1\rho$ by the inverse function theorem.