I have some troubles wit the next exercise.
Find the required work for move a particle from the point $(1,0)$ to the point $(-1,0)$ trought the ellipse $x^2+\displaystyle\frac{y^2}{b^2}=1$ against to the force $F(x,y)=(3y^2+2,16x)$. Which value of $b$ minimizes the work?
First, consider $\gamma(t)=(\cos(t),-b\sin(t))$ the parametrization of the ellipse. The sign in the second coordinate is because the field goes counterclockwise, and, then, the parametrization should be goes clockwise.
Then, $$\displaystyle\int_{0}^{\pi}(3b^2\sin^2(t)+2,16\cos(t))\cdot(-\sin(t),-b\cos(t))\,dt$$The integral then is $$\int_{0}^{\pi}-3b^2\sin^3(t)-2\sin(t)-16b\cos^2(t)\,dt$$And the value is $-4b^2-8b\pi-4$ but, this function of $b$ doesn't have minimum, only a maximum. What happened here? What is wrong with my exercise?
Best Answer
You are not suppose to have that negative in the second component. You are computing a work integral over the ellipse. Unless stated otherwise, it is always assumed that curves are given the counter clock-wise orientation. Note that orientation really does matter since the force field may be stronger (or weaker) in the opposing direction.
\begin{align*} &\int_{0}^\pi (3 \sin^2t + 2,16 \cos t) \cdot (- \sin t, b\cos t) \ dt \\ & = \int_{0}^\pi -3b^2 \sin^3t- 2 \sin t + 16 b\cos^2t \ dt \\ & = -4b^2+8 \pi b - 4 \end{align*}