I am given that $\hat{t}=\dfrac{\hat{x}+y'\hat{y}}{\sqrt{1+y'^2}}$ and $\hat{n}=\dfrac{y'\hat{x}-\hat{y}}{\sqrt{1+y'^2}}$ are the tangential and normal vector in frenet frame. We are considering only 2 dimensional case, where a particle moves along a curve.
Then the author state a equation (second law of motion) ${\vec{F}}=mv\left(\dfrac{dv}{ds}\hat{t}+v\dfrac{d\hat{t}}{ds}\right)=mv\dfrac{dv}{ds}\hat{t}+mv^2\kappa \hat{n}$, which i am able to understand how it comes.
However afterwards, the author defines (Kinetic energy) $K=\frac12mv^2(x,y)$ and claims that $\vec{F}=\nabla K$ I am not able to arrive at the normal $\hat{n}$ component of the result. Can anyone help me and also is there a gradient operator in terms of frenet frame?
Just simplify,
\begin{align*} dW &= \mathbf{F} \cdot d\mathbf{r} \\ &= \mathbf{F} \cdot \mathbf{T} \, ds \\\ &= mv \, dv \end{align*}