Find the biggest potential P (Particle) can have along this the curve C? The curve $C$ is given by $x=t×cos(t)$, and $y=sin(t)$ $C$ $R^2$, 
where $t$  $∈$ $R≥0$.


*

*Find the parametrization of the curve. 

*Find the biggest potential P can have along this curve. 


This may be a part of the exercise as well:


*

*Vector field: $(x+2xy)i+(y+x^2-y^2)j$


I am not sure where to even begin. I am not sure how to apply what I know of Gradients, Line-integrals, Lagrange multipliers, etc to this problem. Any hints are very appreciated. 

*

*The answer is in the question:
\begin{cases}
x=t\cos t\\
y=\sin t 
\end{cases}

*$P$ is under the influence of $\vec{F}=(x+2xy,y+x^2-y^2)$. A potential of $\vec{F}$ is a mapping $f(x,y):\mathbb{R}\rightarrow \mathbb{R}$ such that
$$
\vec{F}=\nabla f
$$
Solving for $f$ yields
$$
f(x,y)=\frac{x^2}{2}+x^2y+\frac{y^2}{2}-\frac{y^3}{3}+K,
$$
where $K$ is an arbitrary constant in $\mathbb{R}$. Now, you want to maximize this potential on the curve, i.e. you want to maximize $f(x,y)$ subject to
\begin{cases}
x=t\cos t\\
y=\sin t 
\end{cases}
Substituting $x$ and $y$ by these expressions yields
$$
f(t)=\frac{t^2}{2}\cos^2(t)+t^2\cos^2(t)^2\sin(t)+\frac{\sin^2(t)}{2}-\frac{\sin(t)^3}{3}+K
$$


This function, however, is unbounded:

So either I misinterpreted the question, either something is wrong.
