Chapter 13 Vector-Valued Functions and Motion in Space

Chapter 13

Vector-Valued Functions and Motion inSpace

13.1 Vector Functions and Space Curves

Definition 15. A vector valued function (abbr. v.v.f.) is a function whose domain is

and whose range is a set of . The variable is typically t and we can think

of it as a .

Example 32. Try to visualize the curve with given vector equation, then sketch by hand or plot in

Mathematica and indicate/note the direction of increasing t.

(a) ~r(t) = 〈 t2, t2, t 〉

(b) ~r(t) = t ı̂ + cos t ̂ + sin t k̂

Example 33. Find the domain of

(a) ~r(t) =⟨√

t, t2, 3t

⟩

(b) ~r(t) =⟨

sin t, ln t,√

1− t⟩

25

26 CHAPTER 13. VECTOR-VALUED FUNCTIONS AND MOTION IN SPACE

Limits of Vector Functions

If ~r(t) = 〈 f(t), g(t), h(t) 〉 then limt→a

~r(t) =

provided that the limit of each of the component functions .

Note: All of the Calc. I and II limit rules apply!

Example 34. (a) limt→2

⟨t,t2 − 4

t2 − 2t,1

t

⟩=

(b) limt→0

t2 ı̂− 5

t̂ + et k̂ =

(c) limt→∞

⟨5t3 + 7

3t3 − 6t, e−t,

sin t

t

⟩=

Definition 16. A vector function ~r(t) is continuous at a if

Definition 17. The set C of all points (x, y, z) in space where x = f(t), y = g(t), z = h(t) and t ∈ Iand where f, g, and h are continuous real valued functions is called a .

Definition 18. If ~r(t) = 〈 f(t), g(t), h(t) 〉 then the derivative of ~r(t) is

~r ′(t) =

(As long as f, g and h are all differentiable. The usual differentiation rules apply.)

13.1. VECTOR FUNCTIONS AND SPACE CURVES 27

Example 35. Find a vector function that represents the curve of intersection of the two surfaces.

(a) z = x3, y = x3 + z2 + 2

(b) z = x+ y2, z = 2x+ y

(c) y2 + z2 = 9, x = 5y

We will need to find parametric equations that are defined by both surfaces.

Tips:

•

•


Definition 19. ~r(t) is smooth on an interval I if .

Evaluating ~r ′(t) at a specific t value gives the to the curve at the

point as long as ~r ′(t) exists and is not nonzero.

Definition 20. The tangent line to the curve C at a point P is the line through P

the tangent vector ~r ′(t). The unit tangent vector is ~T (t) =

If the position of an object in space is given by the vector function ~r(t), then its velocity is

, its speed is , and its acceleration is .

Note that: ~v(t) = |~v(t)| ~T (t).

Example 36. (a) Find the unit tangent vector for ~r(t) = 〈t, ln t, t2〉 when t = 2.

(b) Find a vector equation for the tangent line to the curve from the previous question at the point(1, 0, 1).

Differentiation Rules

1. ddt

[α~u(t)] = α~u ′(t)

2. ddt

[f(t) ~u(t)] = f ′(t) ~u(t) + f(t) ~u ′(t)

3. ddt

[~u(t) + ~v(t)] = ~u ′(t) + ~v ′(t)

4. ddt

[~u(t)− ~v(t)] = ~u ′(t)− ~v ′(t)

5. ddt

[~u(t) q~v(t)] = ~u ′(t) q~v(t) + ~u(t) q~v ′(t)6. d

dt[~u(t)× ~v(t)] = ~u ′(t)× ~v(t) + ~u(t)× ~v ′(t)

7. ddt

[~u(f(t))] = f ′(t) ~u ′(f(t))

13.1. VECTOR FUNCTIONS AND SPACE CURVES 29

Example 37. For ~u(t) = 〈 t, t2, t3 〉 , ~v(t) = 〈 e2t, t, 6 〉 , f(t) = 1t

find

(a) ddt

[f(t)~u(t)]

(b) ddt

[~u q~v]

Vector Functions of Constant Length

If ~r(t) is a differentiable vvf whose length is constant then we have the following:


13.2 Integrals of Vector Functions and Projectile Motion

Definition 21. For ~r(t) = 〈f(t), g(t), h(t)〉,∫~r(t) dt =

and∫ b

a~r(t) dt =

Example 38. (a)∫〈 t, e2t, cos t 〉 dt =

(b)∫ ⟨

ln t, 2t,√t+ 1

⟩dt =

(c)∫ 2

0〈 t, t2, t3 〉 dt =

Example 39 (Exercise 13.2.18). Find ~r(t) given that

d2~r

dt2= −(̂ı + ̂ + k̂), ~r(0) = 10̂ı + 10̂ + 10k̂,

d~r

dt

∣∣∣∣t=0

= ~0

13.2. INTEGRALS OF VECTOR FUNCTIONS AND PROJECTILE MOTION 31

Ideal Projectile Motion

Definition 22 (Ideal Projectile Motion Equation). Given an object launched with initial velocity v0,launch angle [or “firing angle” or “angle of elevation”] α, and gravitational constant g, the equationdescribing the motion of the object is given by

~r(t) =

(We call this “ideal” because we assume negligible friction, constant gravity, and a motionless Earth.)

Once we have the equation for motion we can also find:

Maximum height:

Flight time:

Range:

We can also adjust for the projectile starting at a general point (x0, y0):

or incorporate outside forces acting on our object.

Example 40 (Exercise 12.3.26). A baseball is thrown from the stands 32 ft above the field at an angleof 30◦ up from the horizontal. When and how far away will the ball strike the ground if its initial speedis 32 ft/sec?


13.3 Arc Length And Curvature

Definition 23. The arc length of a smooth curve ~r(t) = 〈 f(t), g(t), h(t) 〉 traversed exactly once betweent = a and t = b is

L =

∫ b

a

√(f ′(t))2 + (g′(t))2 + (h′(t))2 dt

or=

∫ b

a

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt

or=

∫ b

a

|~r ′(t)| dt

or=

∫ b

a

|~v| dt

Example 41. Find the arc length of ~r(t) = 〈t, t2, 23t3〉 between (0, 0, 0) and

(2, 4, 16

3

).

A curve can be represented by more than one vector function depending on

and we can switch from one to another. For example if ~r(t) = 〈t, 1t, 4t2〉

2 ≤ t ≤ 5 is the vector function for the curve C, we can let t = to get another vector function

for C: ~r(t) = .

Note: Arc length is , in other words no matter

which parametrization we use, we’ll get the same for the same section of

curve.

13.4. CURVATURE AND NORMAL VECTORS OF A CURVE 33

Definition 24. The arc length parameter s is

s(t) =

∫ t

t0

|~r ′(τ)| dτ

where s(t) is the length of the curve given by ~r(t) between ~r(t0) and ~r(t).

If we take the derivative of both sides of the arc length function we get

which can be interpreted as the speed along the curve.

It is useful to [re]parameterize a curve with respect to because arc length

is independent of parameter so it won’t affect the outcome if we change to a different coordinate system.

To do this we’ll

.

Example 42. Reparameterize the curve ~r(t) = (5 − t) ı̂ + t ̂ + (3t + 7) k̂ with respect to arc lengthmeasured from the point t = 0 in the direction of increasing t.

13.4 Curvature and Normal Vectors of a Curve

Definition 25. Given a smooth vector function ~r(t) with unit tangent vector ~T and parametrized with

respect to arc length the curvature of ~r(t) is κ =

∣∣∣∣∣d~Tds∣∣∣∣∣.


Since reparametrizing in terms of arc length is hard, we’d like a formula where the parameter is t. . .

Thus we have κ(t) =

Another more convenient formula is (see proof in the book on pg. 880):

Computational Curvature Formula: κ(t) =|~r ′(t)× ~r ′′(t)||~r ′(t)|3

For a plane curve y = f(x) we have

Plane Curve Curvature Formula: κ(x) =|f ′′(x)|

[1 + (f ′(x))2]3/2

Example 43. (a) Find κ(t) and κ(1) for ~r(t) =⟨t2, 2t,−1

3t3⟩

(b) Find κ(x) and κ(0) for f(x) = x− 14x2.

13.4. CURVATURE AND NORMAL VECTORS OF A CURVE 35

The unit tangent vector points .

The principal unit normal vector points .

Unit Normal Vector: ~N(t) =~T ′(t)

|~T ′(t)|

Binormal Vector: ~B(t) = ~T (t)× ~N(t)

Example 44. Find ~T (t), ~N(t) and ~B(t) for ~r(t) = 〈3 sin t, 4t− 7, 3 cos t〉.


Definition 26 (Circle of Curvature). The circle of curvature or osculating circle at a point P on aplane curve where κ 6= 0 is the circle in the plane of the curve that

1. is tangent to the curve at P , having the same tangent line as the curve

2. has the same curvature as the curve at P

3. has center that lies toward the concave or inner side of the curve (in the direction of ~N(t)).

The radius of curvature is

ρ =1

κ

The plane containing ~T (t) and ~N(t) is called the osculating plane.

Example 45.

13.5 Tangential and Normal Components of Acceleration

If the position of an object in space is given by the vector function ~r(t), then its velocity is ,

its speed is , and its acceleration is .

13.5. TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION 37

We are often interested in resolving the acceleration vector into two components, one in the

direction of the tangent vector and the other in the direction of the normal vector. Recall that:

~v(t) = |~v(t)| ~T (t). Using our derivative rules we get:

Then, substituting in using our equations for curvature and the normal vector we find:

~a(t) = |~v|′ ~T + κ |~v|2 ~N

For computational purposes it may be easier to use: aT =~r ′(t) � ~r ′′(t)|~r ′(t)|

and aN =|~r ′(t)× ~r ′′(t)||~r ′(t)|

Example 46. If the position function of an object is given by ~r(t) = 〈3 sin t, 4t−7, 3 cos t〉, then find itsvelocity and acceleration vectors. Also find the tangential and normal components of the accelerationvector.


Definition 27 (Torsion). The torsion function of a smooth curve is

τ =

Informally, torsion measures the ‘twistiness’ of the curve. More formally, it measures the rate atwhich the osculating plane turns about ~T .

Theorem 5 (Torsion formula in R3). Given ~r(t) = 〈x, y, z 〉 (where x, y, z are functions of t) wecan compute torsion as

τ =

∣∣∣∣∣∣x′ y′ z′

x′′ y′′ z′′

x′′′ y′′′ z′′′

∣∣∣∣∣∣|~r ′ × ~r ′′|2

as long as ~r ′ × ~r ′′ 6= ~0.

Example 47.

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Chapter 13 Vector-Valued Functions and Motion in Space