calculus of vector valued functions §13 - …zteitler/teaching/2013c/slides/slides... · calculus...
TRANSCRIPT
Clicker Question:
Which of the following best describes the path of a particledefined by the equations x(t) = cos(t2), y(t) = sin(t2), fort ≥ 0?
A. a circle around which the particle moves faster and faster
B. a parabola on which the particle travels at constant speed
C. a parabola on which the particle travels faster and faster
D. a circle around which the particle moves slower and slower
receiver channel: 41 session ID: bsumath275
Limits.
Limits are componentwise!
Theorem
If r(t) = 〈x(t), y(t), z(t)〉 then
limt→t0
r(t) =⟨
limt→t0
x(t), limt→t0
y(t), limt→t0
z(t)⟩
Theorem
If r(t) = 〈x(t), y(t), z(t)〉 then
limt→t0
r(t) =⟨
limt→t0
x(t), limt→t0
y(t), limt→t0
z(t)⟩
Proof.limt→t0
r(t) = 〈x0, y0, z0〉
⇐⇒ ‖r(t)− 〈x0, y0, z0〉‖ → 0 as t → t0
⇐⇒ ‖〈x(t)− x0, y(t)− y0, z(t)− z0〉‖ → 0
⇐⇒ (x(t)− x0)2 + (y(t)− y0)2 + (z(t)− z0)2 → 0
⇐⇒ x(t)− x0 → 0 and y(t)− y0 → 0 and . . .
⇐⇒ limt→t0
x(t) = x0 and . . .
⇐⇒ 〈x0, y0, z0〉 =⟨
limt→t0
x(t), limt→t0
y(t), limt→t0
z(t)⟩.
Upshot: Derivatives.
Derivatives are defined in terms of limits, so derivatives arecomponentwise too!
r′(t) = limh→0
r(t + h)− r(t)
h
= limh→0
⟨x(t + h)− x(t)
h,y(t + h)− y(t)
h, . . .
⟩=
⟨limh→0
x(t + h)− x(t)
h, limh→0
y(t + h)− y(t)
h, . . .
⟩= 〈x ′(t), y ′(t), z ′(t)〉
Theorem
If r(t) = 〈x(t), y(t), z(t)〉 then r′(t) = 〈x ′(t), y ′(t), z ′(t)〉.
Integrals.
Integrals are also componentwise:∫ b
a
〈x(t), y(t),z(t)〉 dt
=
⟨∫ b
a
x(t) dt ,
∫ b
a
y(t) dt ,
∫ b
a
z(t) dt
⟩=⟨X∣∣∣ba, Y∣∣∣ba, Z∣∣∣ba
⟩where 〈X ,Y ,Z 〉′ = 〈x , y , z〉.
Derivative Properties.
THREE product rules:
Id
dt
(f (t)r(t)
)= f (t)r′(t) + f ′(t)r(t)
(f (t) = scalar function, r(t) = vector valued function)
Id
dt
(r1(t) · r2(t)
)= r1(t) · r′2(t) + r′1(t) · r2(t)
Id
dt
(r1(t)× r2(t)
)= r1(t)× r′2(t) + r′1(t)× r2(t)
Chain Rule:
Id
dt
(r(g(t))
)= r′(g(t)) g ′(t)
Example.
r(t) = 〈t, t2, t3〉, g(t) = cos t,d
dt(r(g(t))) =?
Answer 1: r(g(t)) = 〈cos t, cos2 t, cos3 t〉 so
d
dtr(g(t)) = 〈(cos t)′, (cos2 t)′, (cos3 t)′〉
= 〈− sin t,−2 cos t sin t,−3 cos2 t sin t〉.
Answer: r′(t) = 〈1, 2t, 3t2〉 and g ′(t) = − sin t so
d
dtr(g(t)) = 〈1, 2 cos t, 3 cos2 t〉(− sin t)
Tangent Vector.
r′(t) is a tangent vector to the curve r(t).
Example: For the intersection of x2 − y 2 = z − 1,x2 + y 2 = 4, find points with horizontal tangent line.
Tangent Vector.
r′(t) is a tangent vector to the curve r(t).
Example: For the intersection of x2 − y 2 = z − 1,x2 + y 2 = 4, find points with horizontal tangent line.Answer: We found the parametrization
r(t) = 〈2 cos t, 2 sin t, 4 cos 2t + 1〉
The tangent vector is horizontal ⇐⇒ z ′(t) = 0
⇐⇒−8 sin 2t = 0 ⇐⇒ sin 2t = 0 ⇐⇒ t = 0, π/2, π, 3π/2.So the points are
r(0) = 〈2, 0, 5〉r(π/2) = 〈0, 2,−3〉r(π) = 〈−2, 0, 5〉
r(3π/2) = 〈0,−2,−3〉
Tangent Vector.
r′(t) is a tangent vector to the curve r(t).
Example: For the intersection of x2 − y 2 = z − 1,x2 + y 2 = 4, find points with horizontal tangent line.Answer: We found the parametrization
r(t) = 〈2 cos t, 2 sin t, 4 cos 2t + 1〉
The tangent vector is horizontal ⇐⇒ z ′(t) = 0 ⇐⇒−8 sin 2t = 0 ⇐⇒ sin 2t = 0
⇐⇒ t = 0, π/2, π, 3π/2.So the points are
r(0) = 〈2, 0, 5〉r(π/2) = 〈0, 2,−3〉r(π) = 〈−2, 0, 5〉
r(3π/2) = 〈0,−2,−3〉
Tangent Vector.
r′(t) is a tangent vector to the curve r(t).
Example: For the intersection of x2 − y 2 = z − 1,x2 + y 2 = 4, find points with horizontal tangent line.Answer: We found the parametrization
r(t) = 〈2 cos t, 2 sin t, 4 cos 2t + 1〉
The tangent vector is horizontal ⇐⇒ z ′(t) = 0 ⇐⇒−8 sin 2t = 0 ⇐⇒ sin 2t = 0 ⇐⇒ t = 0, π/2, π, 3π/2.
So the points are
r(0) = 〈2, 0, 5〉r(π/2) = 〈0, 2,−3〉r(π) = 〈−2, 0, 5〉
r(3π/2) = 〈0,−2,−3〉
Tangent Vector.
r′(t) is a tangent vector to the curve r(t).
Example: For the intersection of x2 − y 2 = z − 1,x2 + y 2 = 4, find points with horizontal tangent line.Answer: We found the parametrization
r(t) = 〈2 cos t, 2 sin t, 4 cos 2t + 1〉
The tangent vector is horizontal ⇐⇒ z ′(t) = 0 ⇐⇒−8 sin 2t = 0 ⇐⇒ sin 2t = 0 ⇐⇒ t = 0, π/2, π, 3π/2.So the points are
r(0) = 〈2, 0, 5〉r(π/2) = 〈0, 2,−3〉r(π) = 〈−2, 0, 5〉
r(3π/2) = 〈0,−2,−3〉
Warning.
A horizontal tangent does not necessarily mean the derivativeis zero.
It just means one component of the derivative is zero.
Tangent Line.
Example: For r(t) = 〈t, t2, t3〉, parametrize the tangent lineat t = 1.
Answer: The tangent line passes through r(1) = 〈1, 1, 1〉 andhas direction vector r′(1) = 〈1, 2, 3〉. So the tangent line is
L(t) = (1, 1, 1) + t〈1, 2, 3〉 = 〈1 + t, 1 + 2t, 1 + 3t〉.
Clicker Question: Derivative of Cross Product.
The derivative of the cross product is the cross product of thederivatives.
A. True, and I am very confident.
B. True, but I am not confident.
C. False, but I am not confident.
D. False, and I am very confident.
receiver channel: 41 session ID: bsumath275
Clicker Question: Derivatives — Scalar or Vector?
r1(t) and r2(t) are vector-valued functions.State whether the following derivatives are scalars or vectors:
d
dtr1(t),
d
dt
(r1(t) · r2(t)
),
d
dt
(r1(t)× r2(t)
)A. scalar, vector, vector
B. vector, scalar, vector
C. vector, vector, scalar
D. vector, vector, vector
receiver channel: 41 session ID: bsumath275