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Serret – Frenet formula

 Let x = x(s) be a space curve and p be any point on this curve. If t, b, n are unit tangent, unit principle normal and unit binormal vectors respectively of the curve at the point P, then the Serret – Frenet formula are given by,

(a) t′ = Kn   (K= kappa)

(b) n′ = -kt +?b

(c) b′ = -?n

Proof: We have t, n, b are mutually orthogonal unit vectors at a point p on the space curve x = x(s), so

t.t = n.n = b.b =1

t.n = n.b=b.t =0

t˄n = b, n˄b = t, b˄t = n

If x = x(u), then we have

ser2

ser3

Therefore t′ lies on the osculating plane at the point P on the space curve x = x(u)

Again, t . t = 1

t′ . t + t . t′ = 0

⇒ 2t . t′ = 0

t . t′ = 0

That is t′ is perpendicular to t. Also, t′ is along n.

Hence, we can write, t′ = K n —————(i)

Where K is a scalar function and K is called the curvature of x = x(s) at S.

Consider, t. b = 0

t′ . b+b′ . t =0

⇒Kn.b+t . b′=0 [by (i)]

⇒ 0 + t . b′ =0

t . b′ =0

b′ is perpendicular to t.

Again consider, b . b =1

b′ . b + b . b′ =0

⇒ 2b . b′ = 0

b . b′ =0

b′ is perpendicular to b.

Thus we see that b′ is perpendicular to b at t, so b′ is along n. Hence along we can write,

b′ = λn

⇒|b′| =| λ n |

⇒|?| = |λ| [∵?=|b| and |n| =1]

⇒ λ = ± ?

⇒ λ = -?

b′ = -?n ————————-(ii)

Where ? is a scalar and ? is called torsion of x = x(s) at S.

Now, n = b ˄ t

n′ = b′ ˄ t + b ˄ t

= -? n ˄ t + b ˄ K n

= -? (-b) + K (-t)

= -Kt + ?b

Matrix form of Serret – Frenet formula: Serret – Frenet Formula can be written as

t′ = 0 . t + K n + 0 . b

n′ = -Kt + 0. n + ? b

b′ = 0 . t – ? n + 0 . b

This can be written in the matrix form,

ser1

 

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