How do you find the maximum value of $r$ in a polar function?

Question

I have $\, r=\cos\alpha +\sin2\alpha,\quad 0\le\alpha\le\frac{\pi}{2}.$

Do you then find $\dfrac{dr}{d\alpha}$ and let that $=0$ ?

I am after just a few set of instructions.

Answer 1

Taking the first derivative of $r$ vs. $\alpha$ you find: $$\frac{dr}{d\alpha}=-\sin(\alpha)+2\cos(2\alpha)$$ putting this to zero, you find: $$\alpha=.6348668713, 2.506725783, -1.002966954, -2.138625700$$ Take the second derivative of $r$ which is: $$\frac{d^2r}{d\alpha^2}-\cos(\alpha)-4\sin(2\alpha)$$ in these points. If it's negative, this give you a maximum

Answer 2

We have to investigate $$r(\alpha):=\cos\alpha+\sin(2\alpha)\qquad(0\leq\alpha\leq{\pi\over2})\ .$$ We note in passing that $r(0)=1$, $\ r({\pi\over2})=0$ and compute $$r'(\alpha)=-\sin\alpha +2\cos(2\alpha)=-(4u^2+u-2)\ ,$$ where we have put $\sin\alpha=:u$. Solving $r'(\alpha)=0$ gives $$u_\pm={-1\pm\sqrt{33}\over 8}\ ,$$ and since a usable $u$ has to lie in the interval $[0,1]$ we are left with $$\sin\alpha_0=u_+={\sqrt{33}-1\over 8}\doteq0.593\ ,$$ which leads to $$\cos\alpha_0=\sqrt{1-u_+^2}={\sqrt{30+2\sqrt{33}}\over8}\ .$$ In this way we obtain $$r(\alpha_0)=\cos\alpha_0(1+2\sin\alpha_0)={\sqrt{30+2\sqrt{33}}\over8}{\sqrt{33}+3\over 4}\doteq1.760\ .$$ Since $r(\alpha_0)$ is greater than both $r(0)$ and $r({\pi\over2})$ it follows that the obtained value $r(\alpha_0)\doteq1.760$ is the maximal value for $r$ in the given interval.

It is not necessary to compute second derivatives here: The extreme values of $r$ are among $r(0)$, $\>r({\pi\over2})$, and the values of $r$ at the zeros of $r'$ in the interval $\ \bigl]0,{\pi\over2}\bigr[\ $.