Calculus of Vector Valued Functions

§13.2

16 September 2013

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)〉.

Example.

r(t) = 〈t275, ln t, cos(t2013)〉

r′(t) =

⟨275t274 ,

1

t, − sin(t2013) · 2013t2012

⟩

Example.

r(t) = 〈t275, ln t, cos(t2013)〉

r′(t) =⟨

275t274 ,

1

t, − sin(t2013) · 2013t2012

⟩

Example.

r(t) = 〈t275, ln t, cos(t2013)〉

r′(t) =⟨

275t274 ,1

t,

− sin(t2013) · 2013t2012⟩

Example.

r(t) = 〈t275, ln t, cos(t2013)〉

r′(t) =⟨

275t274 ,1

t, − sin(t2013)

· 2013t2012⟩

Example.

r(t) = 〈t275, ln t, cos(t2013)〉

r′(t) =⟨

275t274 ,1

t, − sin(t2013) · 2013t2012

⟩

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〉.

Worksheet.

Worksheet #1–3

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〉.

Worksheet.

Worksheet #4–6

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