Consider the following curve.
x = sin(2t), y = -cos(2t), z = 4t
Using the given parametric equations, give the corresponding vector equation r(t).
r(t) = (sin(2t), -cos(2t), 4t)
Find r'(t) and ||r'(t)||.
r'(t) = (2cos(2t), 2sin(2t), 4)
||r'(t)|| = \sqrt{20}
Find the equation of the normal plane of the given curve at the point (0, 1, 2?).
x - 2z = -4?
Now consider the osculating plane of the given curve at the point (0, 1, 2?). Determine each of the following.
T(t) = (\frac{1}{\sqrt{5}}cos(2t), -\frac{1}{\sqrt{5}}sin(2t), \frac{2}{\sqrt{5}})
T'(t) = \frac{1}{\sqrt{5}}(-2sin(2t), 2cos(2t), 0)
||T'(t)|| = \frac{2}{\sqrt{5}}
N(t) = (1, 0, 2)
Find the equation of the osculating plane of the given curve at the point (0, 1, 2?).
