Singular points of $\alpha(t)=(t^2\cos(t),t^2\sin(t))$

Question

I have to find the singular points of $$\alpha(t)=(t^2\cos(t),t^2\sin(t))$$ with $-2\pi\leq t\leq2\pi$. Its first derivative is $$\alpha'(t)=(2t\cos(t)-t^2\sin(t),2t\sin(t)+t^2\cos(t))$$ and so $t$ is singular if $\alpha'(t)=0$: $$\left\{\begin{array}{l}2t\cos(t)-t^2\sin(t)=0 \\ 2t\cos(t)+t^2\sin(t)=0 \end{array}\right.$$ which clearly show $$t=0$$ is a singular point. But, my question is: in both equations, when I factor them, I got $$\left\{\begin{array}{l}2\cos(t)-t\sin(t)=0 \\ 2\cos(t)+t\sin(t)=0 \end{array}\right.$$ and so $$\cos(t)=\frac{t\sin(t)}{2}$$ and then, putting it in the second equation: $$2\sin(t)+\frac{t^{2}\sin(t)}{2}=0\Rightarrow\sin(t)(4+t^2)=0$$ therefore $$\sin(t)=0\Rightarrow t=n\pi$$ But I know that the only singular point is $t=0$. What is wrong here?

Answer 1

You have noted that your derivatives are incorrect. They should be the ones below, which is what I will work with. $$ \begin{cases} 2t\cos(t) - t^2\sin(t)=0 \\ 2t\sin(t) + t^2\cos(t)=0 \end{cases} $$ The first equation is $t(2\cos t - t \sin t)=0$, which implies that either $t=0$ or $2\cos t = t \sin t$. Notice $t=0$ works in the second equation so that $t=0$ is a singular point. Now let's assume that $2 \cos t= t \sin t$. From the second equation, we have $t(2 \sin t + t \cos t)=0$, so either $t=0$ (which we already know is a solution) or $2 \sin t= -t \cos t$. But we know that $2 \cos t = t \sin t$, so that $$ 4 \cos t= 2 \cdot 2\cos t= 2 \cdot t \sin t= t \cdot 2\sin t= -t^2 \cos t $$ Then we know $t^2 \cos t + 4 \cos t= 0$. This is the same as $\cos t (t^2+4)=0$. Clearly, $t^2+4= 0$ produces no solution so that it must be that $\cos t= 0$. Since you are on $-2\pi \leq t \leq 2\pi$, this means $t= \pm \dfrac{\pi}{2}$. Checking, however, we see that these are extraneous solutions as they satisfy neither equation. Therefore, we are only left with the solution $t=0$.