(b) The cross product x × y is defined by x × y

Calculus Chapter 13 Summary March 25, 2019

Definitions Let x = (x₁, x₂, x₃) and y = (y₁, y₂, y₃) be vectors in R³. (a) The inner product hx , yi is defined by

hx , yi =

3

X

i=1

x_iy_i.

(b) The cross product x × y is defined by

x × y =

   

x₂ x₃ y₂ y₃    

, −    

x₁ x₃ y₁ y₃     ,

x₁ x₂ y₁ y₂    



= (x₂y₃− x₃y₂, x₃y₁− x₁y₃, x₁y₂− x₂y₁) .

Remarks

(a) Let z = (z₁, z₂, z₃) be a vector in R³. Then hz , x × yi =

 z ,

   

x₂ x₃ y₂ y₃    

, −    

x₁ x₃ y₁ y₃     ,

x₁ x₂ y₁ y₂    



= z1

x2 x3

y₂ y₃    

− z₂    

x1 x3

y₁ y₃    

+ z3

x1 x2

y₁ y₂    

=      

z₁ z₂ z₃ x1 x2 x3

y₁ y₂ y₃       .

(b) By the law of cosines,

x

x − y y

θ

kx − yk² = kxk²+ kyk² − 2kxkkyk cos θ.

This gives

2kxkkyk cos θ = kxk²+ kyk²− kx − yk²

=

3

X

i=1

x²_i +

3

X

i=1

y²_i −

3

X

i=1

x_i− y_i
2

= 2

3

X

i=1

x_iy_i

= 2hx , yi.

Hence we have hx , yi = kxkkyk cos θ, where θ is the angle between x and y.

Calculus Chapter 13 Summary March 25, 2019 (c) If x × y 6= 0 ∈ R³, then x × y is perpendicular to the plane spanned by x and y.

(d) x, y and x × y form a right-handed triple.

(e) kx × yk = kxk kyk sin θ, where θ is the angle between x and y. Hencekx × yk = the area of the parallelogram spanned by x and y.

(f) If x and y are parallel, then x × y = 0 ∈ R³.

Remarks Let x, y : (a, b) → R³ be differentiable functions defined on (a, b) with vector value in R³. Then

(a) d

dthx , yi = hdx

dt , yi + hx , dy dti.

(b) d

dt x × y
= dx

dt × y + x ×dy dt.

(c) If kx(t)k = a positive constant, say r, for all t ∈ (a, b), then d

dthx , xi = dr²

dt = 0. We have hdx

dt , xi = 0 for all t ∈ (a, b) =⇒ dx

dt ⊥ x whenever dx dt 6= 0.

Definitions

(a) Let i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) and

C : r(t) = x(t)i + y(t)j + z(t)k = x(t), y(t), z(t)

for t ∈ [a, b]

be a differentiable curve. The vector r⁰(t), if not 0 ∈ R³, issaid to be tangent to the curve C at the point x(t), y(t), z(t)
.

(b) Suppose that the curve

C : r(t) = x(t)i + y(t)j + z(t)k

is twice differentiable and r⁰(t) is never zero. Then at each point x(t), y(t), z(t)
of the curve, there is a unit tangent vector T (t) defined by

T (t) = r⁰(t)

kr⁰(t)k for t ∈ [a, b].

(c) If T⁰(t) 6= 0,for all t ∈ [a, b], since d

dthT (t) , T (t)i = 0 =⇒ T⁰(t) ⊥ T (t). Define the principal normal vector N (t) by

N (t) = T⁰(t)

kT⁰(t)k for t ∈ [a, b].

(d) Suppose that the curve

C : r(t) = x(t)i + y(t)j + z(t)k

is twice differentiable and r⁰(t) is never zero for all t ∈ [a, b]. Then the arc length function s : [a, b] → [0, l], where l =

Z b a

kr⁰(u)k du, defined by

s(t) = Z t

a

kr⁰(u)k du

is differentiable with the differentiable inverse function t : [0, l] → [a, b] satisfying that ds

dt = kr⁰(t)k and dt

ds = 1

kr⁰(t)k =⇒kdr

dsk = kdr dt

dt

dsk = kr⁰(t)k kr⁰(t)k = 1.

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Calculus Chapter 13 Summary March 25, 2019 (e) Suppose that the curve

C : r(t) = x(t)i + y(t)j + z(t)k is twice differentiable and r⁰(t) is never zero for all t ∈ [a, b].

(i) The function R(s) defined by

R(s) = r(t(s)), for s ∈ [0, l] parametrizes C by arc length.

(ii) In term of the arc length parameter s, the unit tangent vector T (s) is defined by T (s) = T (t(s)) = dr

ds for s ∈ [0, l], where l =

Z b a

kr⁰(u)k du = the arc lebgth of C.

(iii) Since kT (s)k² = hT (s) , T (s)i = 1 for all s ∈ [0, l], we have hdT

ds , T (s)i = 0, i.e. along the curve, T only changes in direction and does not change in length. The number

κ(s) =

dT ds

=

d²r ds²

= kr⁰⁰(s)k

is calledthe curvature of C at r(t(s)).

(iv) If T⁰(s) 6= 0, for all s ∈ [0, l], the principal unit normal vector n(s) is defined by n(s) = T⁰(s)

kT⁰(s)k = r⁰⁰(s)

kr⁰⁰(s)k = r⁰⁰(s)

κ(s)k for s ∈ [0, l].

Note that T (s) ⊥ n(s) for all s ∈ [0, l].

(v) If T⁰(s) 6= 0, for all s ∈ [0, l], the binormal vector b(s) is defined by b(s) = T (s) × n(s) for s ∈ [0, l].

Thus {T (s), n(s), b(s)} form an orthonormal basis of R³ at r(t(s)).

(vi) Since kb(s)k = 1 for all s ∈ [0, l], we have

hb⁰(s) , b(s)i = 0, i.e. b⁰(s) ⊥ b(s).

Also since

b⁰(s) = T⁰(s) × n(s) + T (s) × n⁰(s) = κn(s) × n(s) + T (s) × n⁰(s) = T (s) × n⁰(s), we have

b⁰(s) ⊥ T (s).

Hence there is a function τ : [0, l] → R, called the torsion of C at r(t(s)) defined by b⁰(s) = τ (s)n(s) for s ∈ [0, l].

(f) Suppose that the curve

C : r(t) = x(t)i + y(t)j + z(t)k

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Calculus Chapter 13 Summary March 25, 2019 is twice differentiable and r⁰(t) is never zero for all t ∈ [a, b]. Since

r⁰(t) = dr ds

ds dt = dr

dskr⁰(t)k = kr⁰(t)k T (s), r⁰⁰(t) = d

dt

 dr ds

ds dt



= d²r ds²

 ds dt

2

+dr ds

d²s

dt² = κ(s) ds dt

2

n(s) + d²s dt² T (s), we have

|r⁰(t) × r⁰⁰(t)| = κ(s) ds dt

3

|T (s) × n(s)| =⇒κ(s(t)) = |r⁰(t) × r⁰⁰(t)|

|r⁰(t)|³ . (g) Let C be parametrized by arc length s. Since

kT (s)k² = hT (s) , T (s)i = 1 for all s ∈ [0, l] =⇒ T (s) ∈ {(x, y) ∈ R² | x²+ y² = 1}, We may define an angular function

θ = θ(s) : [0, l] → R such that the function

T : [0, l] → S¹ = {(x, y) ∈ R² | x²+ y² = 1} = the unit circle is given by

T (s) = cos θ(s) , sin θ(s)
, where θ(s) is the angle measured from positive x−axis to T (s).

Hence we have κ(s)=

dT ds

=

− sin θ(s) , cos θ(s)

|θ⁰(s)|= |θ⁰(s)| for s ∈ [0, l]

and

Z l

0

κ(s) ds = Z l

0

|θ⁰(s)| ds = total variation of angles.

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