Find the biggest potential P (Particle) can have along this the curve C?

Question

The curve $C$ is given by $x=t×cos(t)$, and $y=sin(t)$ $C$ $R^2$, where $t$ $∈$ $R≥0$.

1. Find the parametrization of the curve.
2. 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.

Answer 1

1. The answer is in the question: \begin{cases} x=t\cos t\\ y=\sin t \end{cases}
2. $P$ 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.