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...