Let $K$ be the level curve $f(x,y)=0$ of $f(x,y)=(x^2+y^2)^2-6x^2y$
$K$ is given in polar coordinates: $r=3\sin(2\theta)\cos(\theta)$, $\theta \in[0,\pi]$
Determine the cartesian coordinates of the points on $K$, furthest from the origin.
I've been trying for quite some time and have no clue what direction to look in. I tried some things with the gradient of $f(x,y)$, the derivative of $r$. I found the critical points and plugged them into the polar equation, but when graphing, it was clear those weren't the points. I'm truly lost here.
Note: This was an ex-exam question so I do not have a solution for this question.
$$r=3\sin(2t)\cos(t)$$ The maximum of $r$ is when $r'=0$ and $r''<0$
$r'=6 \cos (t) \cos (2 t)-3 \sin (t) \sin (2 t)=3 \cos (t) (3 \cos (2 t)-1)$
$r'=0\to 3 \cos (t) (3 \cos (2 t)-1)=0$
$\cos t=0\to t =\frac{\pi}{2}$
$3 \cos (2 t)-1=0\to \cos 2t=\frac{1}{3}\to t=\frac{1}{2}\arccos\left(\frac{1}{3}\right)\approx 35° $
To show it is an actual maximum compute second derivative
$r''=-12 \sin (t) \cos (2 t)-15 \sin (2 t) \cos (t)=-\frac{3}{2} (\sin (t)+9 \sin (3 t))$
At $t=\frac{\pi}{2}$ $r''=12$, $r=0$ minimum
At $t=\frac{1}{2}\arccos\left(\frac{1}{3}\right)$, $r''\approx -13.9$ thus it is a maximum.
$r=2 \sqrt{2} \cos \left(\frac{1}{2} \arccos\left(\frac{1}{3}\right)\right)$
$t=\frac{1}{2}\arccos\left(\frac{1}{3}\right)$
To compute cartesian coordinates we plug back the values we have found in the equations
$x=r\cos t;\;y=r\sin t$
$x=2 \sqrt{2} \cos ^2\left(\frac{1}{2} \arccos\left(\frac{1}{3}\right)\right),y=2 \sqrt{2} \cos \left(\frac{1}{2} \arccos\left(\frac{1}{3}\right)\right) \sin \left(\frac{1}{2} \arccos\left(\frac{1}{3}\right)\right)$
Using trig bisection formula we arrive to
$$x=\frac{4 \sqrt{2}}{3},y=\frac{4}{3}$$