Curve on sphere question

[TsAmE]

Let a > 0. Does the curve parameterised by \(\displaystyle r(t) = (asin(t)cos(2t), asin(t)sin(2t), acos(t))\), t E R, lie on a sphere in \(\displaystyle R^{3}\) with its centre at the origin? If so, find the radius of the sphere.

Attempt:

\(\displaystyle x^{2} + y^{2} + z^{2} = r^{2}\)

\(\displaystyle a^{2}sin^{2}(t)cos^{2}(2t) + a^{2}sin^{2}(t)sin^{2}(2t) + a^{2}cos^{2}(t) = r^{2}\)

I am not sure what else to do

 

[soroban]

Hello, TsAmE!

You're off to a good start . . .

\(\displaystyle \text{Let }a > 0.\)

\(\displaystyle \text{Does the curve parameterised by: }\:r(t) \:=\: (a\sin t \cos2t,\; a\sin t\sin2t,\; a\cos t),\;t \in R\)

. . \(\displaystyle \text}lie on a sphere in }R^{3}\text{ with its centre at the origin?}\)

\(\displaystyle \text{If so, find the radius of the sphere.}\)

\(\displaystyle x^2 + y^2 + z^2 \:=\a\sin t\cos2t)^2 + (a\sin t\sin2t)^2 + (a\cos t)^2\)

. . . . . . . . .\(\displaystyle =\;a^2\sin^2\!t\cos^2\!2t + a^2\sin^2\!t\sin^2\!2t + a^2\cos^2\!t\)

. . . . . . . . .\(\displaystyle =\;a^2\sin^2\!t\underbrace{(\cos^2\!2t + \sin^2\!2t)}_{\text{This is 1}} + \;a^2\cos^2\!t\)

. . . . . . . . .\(\displaystyle =\;a^2\sin^2\!t + a^2\cos^2\!t\)

. . . . . . . . .\(\displaystyle =\;a^2\underbrace{(\sin^2\!t + \cos^2\!t)}_{\text{This is 1}}\)
. . . . . . . . .\(\displaystyle =\;a^2\)


\(\displaystyle \text{We have: }\:x^2 + y^2 + z^2 \:=\:a^2 \quad\hdots\;\text{ a sphere.}\)

. . \(\displaystyle \text{Its radius is }a.\)