This curve resembles a square with vertices at (±1,±1) with concave sides sitting on the plane z=π. The osculating circle at (√2/4,√2/4,π) is centred at (√2,√2,π) with radius 3/2 in the same plane of course. Below is the formal derivation of various quantities, expressions and equations.
a) TNB FRAME
Vector r(t)=<cos³(t),sin³(t),π>.
Tangent unit vector T(t)=r'(t)/|r'(t)|.
r'(t)=<-3cos²(t)sin(t),3sin²(t)cos(t),0>
|r'(t)|=√(9cos⁴(t)sin²(t)+9sin⁴(t)cos²(t)),
|r'(t)|=3sin(t)cos(t)√(cos²(t)+sin²(t)),
|r'(t)|=3sin(t)cos(t).
T(t)=<-3cos²(t)sin(t),3sin²(t)cos(t),0>/(3sin(t)cos(t)),
T(t)=<-cos(t),sin(t),0>.
T'(t)=<sin(t),cos(t),0>
|T'(t)|=√(sin²(t)+cos²(t))=1.
Normal unit vector N(t)=T'(t)/|T'(t)|.
N(t)=<sin(t),cos(t),0>
Binormal unit vector B(t)=T(t)×N(t) (vector cross product).
Use determinant to compute cross product:
| i j k |
| -cos(t) sin(t) 0 |=
| sin(t) cos(t) 0 |
B(t)=<0,0,-cos²(t)-sin²(t)>=<0,0,-1>
TNB frame is:
T(t)=<-cos(t),sin(t),0>
N(t)=<sin(t),cos(t),0>
B(t)=<0,0,-1>
b) We need to find t such that cos³(t)=sin³(t)=√2/4, which confirms:
r(t)=<√2/4,√2/4,π> as the given point.
When t=π/4, sin(t)=cos(t)=√2/2 and (√2/2)³=2√2/8=√2/4.
With this value of t we can compute T(π/4)=<-√2/2,√2/2,0>,
N(π/4)=<√2/2,√2/2,0>. B(π/4)=<0,0,-1> as before.
OSCULATING PLANE
This plane contains T and N. We use the normal to this plane, which is B, to find the equation of the plane and the fact that the dot-product is zero because of orthogonality:
<0,0,-1>᛫<x-√2/4,y-√2/4,z-π>=0,
π-z=0, or z=π.
NORMAL PLANE
This plane contains N and B, so its normal is T.
<-√2/2,√2/2,0>᛫<x-√2/4,y-√2/4,z-π>=0,
-x√2/2+¼+y√2/2-¼=0,
-x+y=0 or y=x.
RECTIFYING PLANE
This plane contains T and B with normal N.
<√2/2,√2/2,0>᛫<x-√2/4,y-√2/4,z-π>=0,
x√2/2-¼+y√2/2-¼=0,
(x+y)/√2-½=0,
x+y-√2/2=0 or x+y=√2/2.
c) The radius ρ of the osculating circle is the inverse of the curvature, so:
ρ=|r'|³/|r'×r''|=(3sin(t)cos(t))³/|r'×r''|.
r'(t)=<-3cos²(t)sin(t),3sin²(t)cos(t),0>
r''(t)=<-3cos³(t)+6cos(t)sin²(t),-3sin³(t)+6sin(t)cos²(t),0>
r'×r''=
| i j k|
| -3cos²(t)sin(t) 3sin²(t)cos(t) 0 | =
| -3cos³(t)+6cos(t)sin²(t) -3sin³(t)+6sin(t)cos²(t) 0 |
<0,0,-3cos²(t)sin(t)(-3sin³(t)+6sin(t)cos²(t))-3sin²(t)cos(t)(-3cos³(t)+6cos(t)sin²(t))>=
<0,0,9cos²(t)sin⁴(t)-18sin²(t)cos⁴(t)+9sin²(t)cos⁴(t)-18sin⁴(t)cos²(t)>=
<0,0,-9cos²(t)sin⁴(t)-9sin²(t)cos⁴(t)>=<0,0,-9sin²(t)cos²(t)>
|r'×r''|=9sin²(t)cos²(t).
ρ=27sin³(t)cos³(t)/(9sin²(t)cos²(t))=3sin(t)cos(t).
At t=π/4, ρ=3/2. From this we can see that the centre of the osculating circle is 3/2 units from the point given by r(π/4).
But the circle lies in the osculating plane z=π (parallel to the x-y plane) and passes through the point r(t)=<√2/4,√2/4,π>.
The vector equation of the osculating circle is:
r=<3cos(t)/2+√2,3sin(t)/2+√2,π>
More to follow in due course...