The discussion focuses on finding the value of parameter t in the parametric equations x = 6 cos t, y = 6 sin t, and z = 6 cos 2t at the point (3√3, 3, 3). The derivative of the position vector r(t) is given as r'(t) = (-6 sin t, 6 cos t, -12 sin 2t). To determine the value of t, the equation (6 cos t, 6 sin t, 6 cos 2t) must be solved for the specified point, leading to the equations 6 cos t = 3√3, 6 sin t = 3, and 6 cos 2t = 3.
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Solve (6cost t, 6sint, 6cos2t) = (3√3, 3, 3) for t .Jimerd said:Homework Statement
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
x = 6 cos t, y = 6 sin t, z = 6 cos 2t; (3√3, 3, 3)
Homework Equations
The Attempt at a Solution
So I understand that r(t)= 6cost t, 6sint, 6cos2t
and r'(t)= -6sint, 6cost, -12sin2t
So here is the questions, how do I find what the value of t is at points (3√3, 3, 3)