Hy, I have to ask someone for help with this problem.
I have a curve with this implicit equation:
$$\left ( x^2 + y^2 \right )^2 = x^3 + y^3$$
I have to find the most distant coordinates from the center of coordinate system, and the most distant coordinates from y-axis.
The graph looks like this Wolfram graph plot
Do I have to use Lagrange method to solve this and how to set the equation up if so ?
Thank you very much !
On
Convert to polar coordinates:
$$\begin{align}(r^2)^2&=r^3(\cos^3\theta + \sin^3\theta)\\ r&=\cos^3\theta + \sin^3\theta\qquad \because r\neq0\\ \frac {dr}{d\theta}&=-3\cos^2\theta \sin\theta+3\sin^3\theta\cos\theta=0\qquad \quad \text{at max/min}\\ \sin\theta\cos\theta(\sin\theta-\cos\theta)&=0\\ &\Rightarrow \begin{cases} \sin\theta&=0\qquad \qquad \qquad \quad \Rightarrow \theta=n\pi\\ \cos\theta&=0\qquad \qquad \qquad \quad \Rightarrow \theta=(n+\frac 12)\pi\\ \sin\theta&=\cos\theta\Rightarrow \tan\theta=1\Rightarrow \theta=(n+\frac 14)\pi \end{cases} \end{align}$$
Checking each case with using $\frac {d^2r}{d\theta^2}$ will confirm that the maximum value for $r$ occurs at $$\theta=2n\pi, (2n+\frac12)\pi$$ where $$r=1$$ i.e. the most distant coordinates from the origin are, in $(x,y)$ terms, $$(1,0), (0,1)\qquad \blacksquare$$
Yes, the Lagrange method should work quite well with this. You need to maximize distance; but I'll give you a hint that it is easier to maximize the square of distance. So, you must maximize $$f(x,y)=x^2+y^2 $$ subject to the constraint $$g(x,y) = (x^2+y^2)^2-x^3-y^3=0 $$