section7. curvature and torsion

� � � � � ∴ B ′ = T ′ × N + T × N ′ ( ∗) � T′ � � = � , so that T ′ × N = 0. T′ � � � B ′ = T × N ′.
and Definition 7 .2 (Torision) 7.2
� � the point r = r ( s ).
� � � ∵ B ( s ) = T ( s) × N ( s)
� � � � � ∴ B ′ = T ′ × N + T × N ′ ( ∗)
Notice that
� � r ′′ N= � r ′′
Thus becomes
� � � � � � � � This indicates B′ ⊥ T , while B′ ⊥ B, so B′ � B × T = N � � There is no harming in assuming B ′ = −τ N � � � � � � B ′ i N = −τ N i N ⇒ τ = − B ′ ⋅ N
� T ( s0 )
� N ( s0 )
principal normal
� � � r ′ ( s0 ) × r ′′ ( s0 ) B ( s0 ) = � r ′′ ( s0 ) � � r ′′( s0 ) � � � 1 N ( s0 ) = � ( r ′( s0 ) × r ′′( s 0 )) × r ′( s 0 ) = � . ′′ ′′ r ( s0 ) r ( s0 ) i ii i � � � � r ( t0 ) × r ( t0 ) � r ( t0 ) � � B ( t ) = T ( t0 ) = i 0 r = r (t) i ii � � � r ( t0 ) × r ( t0 ) r ( t0 )
y′ = 0.8 x , y′′ = 0 .8
1 + 0.64 x 1 R= = κ 0 .8 1 min R ( x ) = = 1. 25 ( as x = 0 ) 0 .8
y
(
2
)
3/2
y = 0. 4 x 2
O
x
Hence, the diameter of the grinding wheel should be chosen slightly smaller than 2.5 cm.
� (1 ) r = ( x ( t ), y ( t ), 0 ) κ =
(2) y =
y( x )
x ̇̇ y ̇− y ̇̇ x ̇ . 2 2 32 [( x ̇) + (y ̇) ] y ′′ κ = . 2 32 [1 + ( y′ ) ]
Example 2 Find the பைடு நூலகம்urvature of the helix � r ( t ) = (a cos t , a sin t , kt )
� B ( s0 )
normal plane
binormal
� r ( s0 )
•
� N ( s0 ) osculating plane
tangent
� T ( s0 )
Frenet fram
� � � T ( s) , N ( s) , B( s)
Example 1 Find the Frenet frame, equations of the oculating plane and the rectifying plane for the helix � r = ( a cos t , a sin t , kt )
7.2 Curvature
∆θ
∆θ
∆s
∆θ =C ∆s
N
P
M
∆θ
circle
Definition and computation of curvature 1. 1.Definition Definition 7.1 (curvature) Suppose Γ : r = r ( s ) ( a ≤ s ≤ b )
� � r ( s0 + ∆ s ) r ( s0 )
π′
•
� � � 1 � � � r ( s0 + ∆ s ) − r ( s0 ) ≈ r ′ ( s0 ) ∆ s + r′′ ( s0 ) ∆ s2 / / k1 r ′ ( s0 ) + k2 r′′ ( s0 ) 2!
(
� T ( s0 )
� B′ = τ
� � τ ( s ) = − B ′ ( s ) ⋅ N ( s ) is called thh torision of the curve at
τ ( s ) = − B′ ⋅ N = − (T × N′ ) ⋅ N
= ( T × N) ⋅ N′
= (T × 1 1 1 T′ ) ⋅ [( )′ T′ + T′′ ] κ κ κ 1 1 1 = (T × T′ ) ⋅ T′′ = 2 [T T′ T′′ ] κ κ κ [r ′( s )r′′( s )r′′′( s )] τ ( s) = 2 r′′( s )
� B ( s0 ) rectifying plane
principal normal
� � � � � N ( s0 ) = B ( s0 ) × T ( s0 ) , T ( s0 ) = r ′ ( s0 ) ,
� � � ρ = r ( s0 ) + λ N ( s0 ).
� r ( s0 ) •
� � � ρ = r ( s0 ) + λ r ′( s0 )
� r ( s0 )
normal plane
� � � normal plane ( ρ − r ( s 0 )) ⋅ r ′ ( s 0 ) = 0
•
tangent
� T ( s0 )
(2)Osculating plane and binormal
∆θ s is a natural parameter, lim is called the curvature ∆ s →0 ∆ s ∆θ κ = lim . of Γ at the point M, denote κ . i.e. ∆s →0 ∆ s
�
�
Theorem 7.1 ( formula for curvature)
� � Γ : r = r (s)
(1) Normal plane and tangent line � � � Suppose r ′ ( s ) ≠ 0, T ( s 0 ) = r ′ ( s 0 ), is a unit tangent vector in the same direction with the positive direction of Γ, then tangent line
� � Suppose Γ : r = r ( s ) , s is the natural parameter. If � � � ( 2 ) and r ′′ ( s ) ≠ 0, then κ ( s ) = r ′′ ( s ) . r ( s) ∈ C
Hint
� � ∆θ ∆ T ∆ θ = lim ∆ T κ ( s ) = lim = lim ∆ s →0 ∆s ∆s →0 ∆ s ∆s →0 ∆ θ ∆s
2．Radius of curvature and circle of curvature
Γ
Q
��� � 1 PQ = κ
T
� rQ � rP
N
P
o
circle of curvature and oculating plane center of curvature and radius of curvature � � � 1 rQ = rP ( s ) + R ( s ) N ( s ) R= κ
y ′′ κ = [1 + ( y ′ ) 2 ]3 2
A
B
D
1 ( 1 + 9a x R= = κ 6a x
2
4
)
3 /2
→ ∞ as x → 0
C
A
B
Example 4 Suppose the cross section of the inner surface of a workepiece is a parabola y = 0.4 x 2cm . We wish to grid its inner surface with a grinding wheel. How large should the diameter of the drinding wheel be? Solution Indeed, the problerm is how to make the radius of the grinding wheel is not large than the mimimum of the radius of the curvature of the workpiece at any point.
= � � r ′ ( s 0 ) × r ′′ ( s 0 ) � r ′′ ( s 0 )
� � r ′ ( s 0 ) × r ′′ ( s 0 ) � � r ′ ( s 0 ) × r ′′ ( s 0 )
osculating plane binormal line
� � � ( ρ − r ( s 0 )) • B ( s 0 ) = 0 � � � ρ = r ( s0 ) + λ B (s0 )
7.3 Torsion（自学）
Torsion is to measure the degree of bend of a curve.
� Because κ ( s ) = T ′ ( s )
� B′ ( s )
� B( s)
binormal
reflects the turing speed of the tangent vector. Similaly, reflect the turing speed of the binormal vector.
Chapter 5 The Mutivariable Differential Calculus and its Application
5.7 Curvature and turison of space curve
Homework: P136 Ex5.7(A) 8(2)(4)；9;11
7.1 Frenet frame
� � � π ′ : T ( s0 ) ,[ r ( s0 + ∆ s ) − r ( s0 ) ]
� B ( s0 )
∆ s → 0, π ′ → π
� � � � ′ nπ ′ = r ( s0 ) × [r ( s0 + ∆ s ) − r ( s0 )]
(closest to tangent line)
osculating plane
)
� � � ∆ s → 0 , nπ = r ′ ( s0 ) × r ′′ ( s0 ) � � r ′ ( s0 ) ⊥ r ′′ ( s0 ) P99 Ex4.
� B ( s0 ) =
binormal vector
� unit binormal vector B ( s0 )
(3) Rectifying plane and princilpal normal principal normal vector
� N ( s0 ) = � � B ( s0 ) × T ( s0 )
� � � rectifying plane ( ρ − r ( s 0 )) ⋅ N ( s 0 ) = 0
� r ( s0 )
•
� N ( s)
osculating plane � T ( s)
So that it depicts the degree of deviation of Γ from its osculsating plane.
� B( s)
is an unit vector, and
� � B ′ ( s ) ⊥ B ( s ).
� � = T ′ ( s ) = r ′′ ( s )
Corollary
� � � r r ̇ (t ) × ̇ ̇( t ) Γ : r = r ( t ) ∈ C , r ′ ( t ) ≠ 0, then κ ( t ) = 3 r ̇ (t )
special case:plane curve
Example 3 Why does a train always sway when it enters a curve from a straight route? How may one reduce this sway?
D
C
mv 2 is not continuous. F= R
y = ax 3 ( a > 0 )