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2

Proof: The first is not hard.

α(t) = α(s)ds

dt

= T(s)ν 

= ν T

The second is again a straightforward calculation

α(t) =dν 

dtT(t) + ν (t)T(t)

= ν T + νκν T

= ν T + κν 2N

Note that this last equation separates acceleration, α

, into two components — a tan-gential component, ν T, and a normal component, κν 2N. These are used in physics andare called the tangential and normal components of acceleration.

Now, back to a computation of the Frenet apparatus for an arbitrary curve. We knowthat

T =α

|α|B = T × N

N = B × T

How is this going to help us?From the previous lemma, we can show

α × α = (ν T) × (ν T + κν 2N)

= νν (T × T) + κν 3(T × N)

= κν 3B.

Thus,

|α × α = κν 3,

since |B| = 1 and we have shown that

B =α × α

|α

×α

|(1)

κ =|α × α|

ν 3=|α × α||α|3 (2)

Now, how do we find the torsion? We will use the third derivative:

α =ν T + κν 2N

= ν T + ν T + κν 2N + 2κνν N + κν 2N

= ν T + ν (κν N) + (κν 2 + 2κνν )N + κν 2(−κν T + τν B)

= (ν  − κ2ν 3)T + (κν 2 + κν 2 + 2κνν )N + κτν 3B

B · α = κτν 3

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0.1. ARBITRARY CURVES  3

So,

(α × α) · α = κν 3B · α

= κν 3(κτν 3)

= κ2ν 6τ 

τ  =(α × α) · α

κ2ν 6=

(α × α) · α

|α × α|2 .

We have just proven the following theorem

Theorem 0.1 For any regular curve α, the following formulas hold.

B =α × α

|α × α| (3)

κ =|α × α||α|3 (4)

τ  = (α

× α

) · α

|α × α|2 (5)

Example 0.1 Consider the curve α(t) = (3t − t3, 3t2, 3t + t3). The necessary derivativesare

α(t) = 3(1 − t2, 2t, 1 + t2)

α(t) = 6(−t, 1, t)

α(t) = 6(−1, 0, 1)

What we need for our computations are:

|α| = 3 

(1 − t2)2 + 2t2 + (1 + t2)2 = 3√

2(1 + t2)

|α| = 6√2α × α = 18(−1 + t2,−2t, 1 + t2)

|α × α| = 18√

2(1 + t2)

(α × α) · α = 216

Now,

κ =|α × α||α|3 =

18√

2(1 + t2)

(3√

2(1 + t2))3

=1

3(1 + t2)2

τ  = (α

× α

) · α

|α × α|2 = 216(18

√2(1 + t2))2

=1

3(1 + t2)2

T =α

|α| =(1 − t2, 2t, 1 + t2)√

2(1 + t2)

B =α× α

|α× α| =(−1 + t2,−2t, 1 + t2)√

2(1 + t2)

N = B × T =(−2t, 1 − t2, 0)

1 + t2