Curve on sphere question

TsAmE

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Aug 28, 2010
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Let a > 0. Does the curve parameterised by r(t)=(asin(t)cos(2t),asin(t)sin(2t),acos(t))\displaystyle r(t) = (asin(t)cos(2t), asin(t)sin(2t), acos(t)), t E R, lie on a sphere in R3\displaystyle R^{3} with its centre at the origin? If so, find the radius of the sphere.

Attempt:

x2+y2+z2=r2\displaystyle x^{2} + y^{2} + z^{2} = r^{2}

a2sin2(t)cos2(2t)+a2sin2(t)sin2(2t)+a2cos2(t)=r2\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
 
Hello, TsAmE!

You're off to a good start . . .


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

Does the curve parameterised by: r(t)=(asintcos2t,  asintsin2t,  acost),  tR\displaystyle \text{Does the curve parameterised by: }\:r(t) \:=\: (a\sin t \cos2t,\; a\sin t\sin2t,\; a\cos t),\;t \in R

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

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

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

. . . . . . . . .=  a2sin2 ⁣tcos2 ⁣2t+a2sin2 ⁣tsin2 ⁣2t+a2cos2 ⁣t\displaystyle =\;a^2\sin^2\!t\cos^2\!2t + a^2\sin^2\!t\sin^2\!2t + a^2\cos^2\!t

. . . . . . . . .=  a2sin2 ⁣t(cos2 ⁣2t+sin2 ⁣2t)This is 1+  a2cos2 ⁣t\displaystyle =\;a^2\sin^2\!t\underbrace{(\cos^2\!2t + \sin^2\!2t)}_{\text{This is 1}} + \;a^2\cos^2\!t

. . . . . . . . .=  a2sin2 ⁣t+a2cos2 ⁣t\displaystyle =\;a^2\sin^2\!t + a^2\cos^2\!t

. . . . . . . . .=  a2(sin2 ⁣t+cos2 ⁣t)This is 1\displaystyle =\;a^2\underbrace{(\sin^2\!t + \cos^2\!t)}_{\text{This is 1}}
. . . . . . . . .=  a2\displaystyle =\;a^2


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

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

 
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