r(t) = [cos(t^3)sin(t), sin(t)sin(t^3), cos(t)], where t is an element of (0,2pi).
Show that the graph of this function is on the surface of a sphere. Then find it's radius.
T(t) = r'(t)/norm[r'(t)]
equation for a sphere in 3-space: r^2=x^2+y^2+z^2
The Attempt at a Solution
My thought is, a tangent vector must lie on the surface of this function (which is a sphere), and therefore the tangent vector at some point (t) would be sufficient proof. But that seems too easy.
Secondly, to find the radius of a sphere, set up the parametric equations in the form of x= cos(t^3)sin(t), y = sin(t^3)sin(t), z = cos(t), in the form of the equation for a sphere:
(cos(t^3)sin(t))^2 + (sin(t^3)sin(t))^2 + (cos(t))^2 = r^2
then substitute one of the elements to find the radius squared?