(arrows stand for vectors)

a) Find the T→ N B frame.

b) At the point ((√2)/4, (2)/4π) find the equations of the Osculating, Normal, and Rectifying planes.

c) Also at the point ((√2)/4, (2)/4π) find the equation (as a vector valued function) of the Osculating Circle.

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:

|               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/|rr''|=(3sin(t)cos(t))³/|rr''|.

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>

rr''=

|                     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)>

|rr''|=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...

by Top Rated User (825k points)

I had to abbreviate my solution because of extreme editing difficulties and multiple system failures and crashes.