finding equation of the osculating plane

Question

find the equation of the osculating plane at t=pi/4 on the curve: r(t)=cos(2t)i+sin(2t)j+tk

Answer 1

x(t) = cos(2t); x(π/4) = cos(π/2) = 0; y(t) = sin(2t); y(π/4) = sin(π/2) = 1; z(t) = t; z(π/4) = π/4.

x'(t) = -2sin(2t); x'(π/4) = -2; y'(t) = 2cos(2t), y'(π/4) = 0; z'(t) = 1, z'(π/4) = 1

x''(t) = -4cos(2t), x''(π/4) = 0; y''(t) = -4sin(2t), y''(π/4) = -4; z''(t) = 0, z''(π/4) = 0

Now equation of osculating plane:

Ix - 0 y - 1 z - π/4I

I -2 0 1 I = 0; or 4x - (y - 1)·0 + (z - π/4)·8 = 0; 4x + 8z - 2π = 0; 2x + 4z - π = 0.

I 0 -4 0 I