I'm asked to prove that the trace of the curve $\alpha(t)=(r\sin^2(t), r \sin(t),r \cos(t))$ lies on a sphere.
What I tried so far:
Letting $\alpha(t)=(x(t),y(t),z(t))$, we have that $x^2(t)+y^2(t)+z^2(t)=r^2(\sin^4(t)+1)$ which does not satisfy the equation of a sphere. Also, I plotted the curve and I'm not sure whether the trace does actually lie on a sphere.
Any hints? Thanks!
On
There is no such sphere. If there was, let $(a,b,c)$ be its center. Then the distances from $(a,b,c)$ to $p(0)(=(0,0,1))$, to $p\left(\frac\pi2\right)(=(1,1,0))$, to $p\left(-\frac\pi2\right)(=(1,-1,0))$, and to $p(\pi)(=(0,0,-1))$ would all be equal. This means that $(a,b,c)$ is a solution of the system$$\left\{\begin{array}{l}a^2+b^2+(c-1)^2=(a-1)^2+(b-1)^2+c^2\\a^2+b^2+(c-1)^2=(a-1)^2+(b+1)^2+c^2\\a^2+b^2+(c-1)^2=a^2+b^2+(c+1)^2\end{array}\right.$$This sytem has one ond only one solution, which is $\left(\frac12,0,0\right)$. Then the radius of the sphere would be the distance from $\left(\frac12,0,0\right)$ to $(0,0,1)$, which is $\frac{\sqrt5}2$. But the distance from $p\left(\frac\pi6\right)\left(=\left(\frac34,\frac{\sqrt3}2,\frac12\right)\right)$ to $\left(\frac12,0,0\right)$ is $\frac{\sqrt{17}}4\ne\frac{\sqrt5}2$.
Suppose that $C=(a,b,c)$ is the center of the sphere on which $\alpha$ is supposed to lie. The square of the distance $d(C,\alpha)$ has to be constant. Therefore by differentiation
$$\langle \alpha^\prime(t), \alpha(t) - C \rangle$$ has to be equal to zero for all $t \in \mathbb R$. I.e.
$$\begin{aligned} 2\sin t \cos t(a - r\sin^2 t) + \cos t(b -r \sin t) - \sin t(c - r \cos t)&=\\ 2\sin t \cos t(a - r\sin^2 t) + b \cos t - c\sin t=0 \end{aligned}$$
Using $t = 0$ we get $b=0$ and $c =0$ follows from plugging in $t= \pi/2$. We're left with the equation $$\sin t \cos t(a - r\sin^2 t) = 0$$ which should be valid for all $t \in \mathbb R$. And that can't be as it would imply $a - r\sin^2 t=0$ for all $t\in (0,\pi/2)$.
Therefore, the curve doesn't lie on a sphere.