Find $d$ for which $f$ gets its minimal value

Question

Let $\beta>\alpha>0$ and let $A,B,C,D$ be the vertices of the rectangle $[\alpha,\beta]\times[\alpha,\beta]$, such that $A$ is the closest one to the origin, and the path $A\to B\to C\to D\to A$ is oriented clockwise. Let $\vec{F}(x,y)=(y,x)$.

A particle is located at the vertex $B$ and is moving along a curve $\gamma$ which is given by the following parametrization:

$$\left\{\begin{aligned}\gamma(t)=\bigg(\beta-(\beta-\alpha)\cos(t)+e^{-t^2}\sin^3 (2t)& ,\beta-(\beta-\alpha)\sin(t)+(dt^2-\pi t)^3\bigg)\\ t\in[0,\frac \pi 2]\end{aligned}\right.$$

When $d\in\mathbb{R}$.

Let $f(d)$ be the function:

$$f(d)=\left|\int_\gamma \vec{F}\cdot d\vec{r}\right|$$

Find $d$ for which $f$ gets its minimal value. Where does the particle stop, given the value of $d$ you found?

Hint: What does the given rectangle has to do with the problem?

Note: I usually expand on what I tried to do and how I approached the problem. However, for this particular problem, I really could not think of a way to approach it. Calculating the integral and getting an expression that is dependent on $d$ seemed the straightforward solution, but it seems impossible given the above parametrization.

Thank you very much!

Answer 1

Notice that your force $\vec{F}$ is conservative since $\vec{F} = -\nabla U$ where $U(x,y) = -xy$. Therefore the work is independent of the path taken, it is equal to the difference of potential energies at endpoints.

We have

$$\gamma(0) = (\alpha, \beta)= B$$ $$\gamma\left(\frac\pi2\right) = \left(\beta, \alpha + \left(\frac{\pi}2\right)^6(d-2)^3\right)$$

so $$\left|W\right| = \left|\int_\gamma \vec{F}\cdot d\vec{r}\right| = \left|U\left(\gamma(0)\right) - U\left(\gamma\left(\frac\pi2\right)\right)\right| =\left|-\alpha\beta +\alpha\beta - \beta\left(\frac{\pi}2\right)^6(d-2)^3\right| = \beta\left(\frac{\pi}2\right)^6|d-2|^3$$

We see that $\left|W\right|$ is minimal and equal to $0$ when $d = 2$, and then the particle stops at $(\beta, \alpha) = D$.