Show $C(t)=(2\cos^3(t),\sin(2t),2\sin(t))$ is on the intersection of $x^2+y^2+z^2=4$ and $x^2+y^2=2x$

Question

Given a curve defined as: $$C(t)=(2\cos^2(t),\sin(2t),2\sin(t))$$ for $0\le t\le2\pi$

Show that the curve is on the intersection of the sphere $x^2+y^2+z^2=4$ and $x^2+y^2=2x$.

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The intersection of the two equation satisfies:

$$x=2-\frac{z^2}{2}$$

I think the curve is on the the intersection of the two curves if : $$2\cos^3(t)=2-\frac{4\sin^2(t)}{2}$$

$$\cos^3(t)=\cos^2(t)$$

For all $0\le t\le2\pi$.

But solving the equation implies this is not true in general.

So what is the main strategy and how should I continue?

Answer 1

If the curve is $$C(t) = (2 \cos^2 t, \sin (2t), 2 \sin t), $$ then, as @Logos said, it is just a matter of substituting it in the equations and checking that it indeed lies on the surfaces.

Answer 2

Check if it belongs to the first surface

Replacing the curve in the first equation we get $$x^2 + y^2 + z^2 = 4\cos^4(t )+ \sin^2(2t) + 4\sin^2(t) =4\cos^4(t )+ 4\sin^2(t)\cos^2(t) + 4\sin^2(t)$$ which is $$4\cos^4(t )+ \sin^2(2t) + 4\sin^2(t) =4\cos^2(t)[\cos^2(t) + \sin^2(t)] + 4\sin^2(t) = 4\cos^2(t) + 4\sin^2(t) = 4$$ So $C(t)$ is on the first surface.

Check if it belongs to the second surface

Replacing the curve in the second equation we get $$x^2 + y^2 = 4\cos^4(t )+ \sin^2(2t) =4\cos^4(t )+ 4\sin^2(t)\cos^2(t) =4\cos^2(t)[\cos^2(t) + \sin^2(t)]=4\cos^2(t)=2x$$ So $C(t)$ is on the second surface.