Math 254 - Vector Calculus Notes

11.1 - 11.2 Vectors

Ordered pairs: (x, y) where x, y are real numbers.

Ordered triples: (x, y, z) where x, y, z are real numbers.

One can refer to an ordered n-tuple as a list of n real numbers:

(x1, x2, ..., xn).

Such an n-tuple can be regarded as an n-dimensional vector with components xi, or as ann-dimensional point with coordinates xi, depending on the context. We use Rn to denotethe set of all ordered n-tuples.

Geometrically, a point is a location, while a non-zero vector (at least one non-zero component)is a direction and magnitude (length) represented by an arrow (with a tail and head).

The zero vector, (0, 0, ..., 0), has no defined direction.

R2 denotes the xy-plane and R3 denotes xyz-space.

-Draw (2,−3) as a point and as a vector.

-Draw (1,−3, 1) as a point and as a vector. Sometimes a angular bracket notation is used toindicate a vector. Sometimes symbols are in bold or have an arrow above them to indicatethat they represent a vector.

An n-dimensional vector u = (u1, u2, ..., un) can be multiplied by a real number c by multi-plying each component by c:

cu = (cu1, cu2, ..., cun).

This is called scalar multiplication, since geometrically, the vector’s length is scaled by thenumber |c| and its direction is reversed if and only if c < 0.

-Draw the vector c(2, 3) for c = −2, .5, 2.

Two vectors are parallel if they are non-zero scalar multiples of each other.

Consider a second n-dimensional vector v = (v1, v2, ..., vn). The vectors u and v can beadded component-wise:

u + v = (u1 + v1, u2 + v2, ..., un + vn).

Geometrically, we can see that (2, 1)+(−1, 3) is equivalent to the arrow derived from puttingthe tail of a “second” vector at the head of the “first” vector with the sum is the vector thatgoes from the tail of the first to the head of the second.

Vector addition is commutative: u + v = v + u. Geometrically, this is the parallelogramlaw. [PICTURE!]

Vector addition is also associative: (u + v) + w = u + (u + w).

Scalar Multiplication distributes: c(u + v) = cu + cv.

Scalar Multiplication is associative: a(cu) = (ac)u

Vector subtraction (points from the head of v to the head of u): u − v = u + (−v).[PICTURE!]

Magnitude of a vector (n-dimensional analog of the Pythagorean Theorem):

|u| =√

u21 + u2

2 + · · ·+ u2n.

Some texts use “double” absolute value notation, but it really isn’t necessary.

If P (x1, x2, ..., xn) and Q(y1, y2, ..., yn) are points then−→PQ (coordinates of P subtract coor-

dinates of Q) is the vector that points from P to Q. Then |−→PQ| is the distance formula!Recall also how to compute the midpoint.

Find the length of the vector (3,−4, 12).

Basic property of length:

|cu| =√

c2u21 + c2u2

2 + · · ·+ c2u2n = |c|

√u2

1 + u22 + · · ·+ u2

n = |c||u|.

For any non-zero vector u, the vector

u =u

|u| ,

has length 1 and is in the same direction as u. A vector of length 1 is called a unit vector.

There are two unit vectors parallel to u, namely ±u.

An application to basic physics: Suppose a 160 lb person is suspended from two identicalchains that form 30-degree angles with the horizontal ceiling. How much tension force ismust each chain withstand?

Well, let T be the tension force in each cable (by symmetry). We know that T sin 30 = 0.5Tis the vertical component of tension force in each chain. So 2(0.5T ) = 160lb. So the tensionforce applied each chain must withstand is 160 lb.

Circles, Disks, Spheres and Balls:

Circle of radius r centered at (h, k): (x− h)2 + (y − k)2 = r2

We really hope this is “old news” by multivariable calculus...

It becomes a closed disk if = is replaced by ≤ and open disk if = is replaced by <.

Sphere of radius r centered at (h, k, `): (x− h)2 + (y − k)2 + (z − `)2 = r2

It becomes a closed ball if = is replaced by ≤ and open ball if = is replaced by <.

Of course, one has the n-dimensional version of a sphere and ball.

Classify the following (draw it too):

x2 − 6x + y2 + 8y + z2 ≤ 0.

[CTS]

11.3 - Dot Product

The dot product of two vectors u = (u1, u2, ..., un) and v = (v1, v2, ..., vn) is a scalar givenby

u · v =n∑

i=1

uivi.

Basic properties of the dot product:

(1) u · v = v · u.

(2) u · (v + w) = u · v + u ·w.

(3) u · u > 0 if and only if u 6= 0. u · u = 0 if and only if u = 0.

(4) u · u = |u|2.

Law of Cosines (draw the triangle!):

|u− v|2 = |u|2 + |v|2 − 2|u||v| cos θ →

(u− v) · (u− v) = u · u + v · v − 2|u||v| cos θ →

−2u · v = −2|u||v| cos θ → u · v = |u||v| cos θ.

Geometric interpretation of the dot product: u · v = |u||v| cos θ.

Moreover, two vectors are orthogonal (perpendicular) if and only if their dot product is zero!

Find two vectors orthogonal to (0, 1, 1) and to each other...

Orthogonal Projections:

The orthogonal projection of u onto v 6= 0, denoted p = projv(u), is defined as the scalarmultiple of v that is orthogonal to u− p as seen here:�� ��� �

����� ��������

So

projv(u) = (|u| cos θ)(v) =|u| cos θ

|v| v =|u||v| cos θ

|v|2 v =(u · vv · v

)v.

Find the projection of (2, 4) onto (5,−1). Find the projection of (1, 2, 3) onto (1,−1, 1).

Let a constant force F be applied to an object, producing a displacement d.

The work done by the force is W = F · d.

A force F = (35,−15) (in lbs) is used to move a sled 50 feet horizontally. Compute the work.

Coordinate unit vectors and engineering notation for R2,R3:

i = (1, 0, 0) (or (1, 0))

j = (0, 1, 0) (or (0, 1))

k = (0, 0, 1).

Then (a, b, c) = ai + bj + ck.

11.4 Cross Product

The dot product, takes two vectors in Rn and produces a scalar.

The cross product takes two vectors in R3 and produces another vector in R3 that is per-pendicular to both input vectors and in a direction determined by the “right-hand rule.”

Let u = (u1, u2, u3) and v = (v1, v2, v3). Then the cross product given by:

u× v = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1).

There is a method using a matrix determinant:

u× v = det

i j ku1 u2 u3

v1 v2 v3

.

Really, the most efficient way to calculate a cross product is to use the following template:

(u1, u2, u3)

x (v1, v2, v3)

−−−−−−−−−−−−−= (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1).

Random cross products! Do one where 2-dim. vectors are embedded in 3-space.

|u× v|2 = (u2v3 − u3v2)2 + (u3v1 − u1v3)

2 + (u1v2 − u2v1)2 →

|u× v|2 = u22v

23 − 2u2v2u3v3 + u2

3v22 + u2

3v21 − 2u1v1u3v3 + u2

1v23 + u2

1v22 − 2u1v1u2v2 + u2

2v21 →

|u× v|2 = u21v

21 + u2

2v21 + u2

3v21 + u2

1v22 + u2

2v22 + u2

3v22 + u2

1v23 + u2

2v23 + u2

3v23 − (u · v)2 →

|u× v|2 = (u21 + u2

2 + u23)(v

21 + v2

2 + v23)− (u · v)2 →

|u× v|2 = |u|2|v|2 − |u|2|v|2 cos2 θ →

|u× v|2 = |u|2|v|2(1− cos2 θ) →

|u× v|2 = |u|2|v|2 sin2 θ → |u× v| = |u||v| sin θ.

Properties of the cross product:

(1) u× v = −v × u,

(2) (c1u)× (c2v) = c1c2(u× v),

(3) u× (v + w) = u× v + u×w.

Geometrically, the cross product is perpendicular to both inputs. (SHOW!) Its direction isgiven by the “right-hand” rule.

Find both unit vectors normal (perpendicular) to both (1, 0, 2) and (−1, 1, 3).

The torque about point O produced by a force F applied to point P at the head of r =−→OP

is given by

τ = r× F.

To tighten a bolt, a force of 30 N is applied to a 20 cm wrench at an angle of 45 degreeswith the wrench. Calculate the magnitude of the torque.

Also, |u× v| = u||v| sin θ gives the area of the parallelogram spanned by u,v.

[PICTURE]

Furthermore, the volume of the parallelepiped spanned by u,v,w is |u·(v×w)| = |u||v||w|(sin θ)| cos α|where θ is the angle between v,w and α is the angle between u and v ×w.

[PICTURE]

Find the area of the parallelogram with vertices (1, 2), (5, 4), (3, 9), (7, 11).

Find the volume of the parallelepiped spanned by the vectors (0, 1, 2), (3,−1, 5), (1, 1, 1).

11.5 Lines and Curves

A function from one variable to many is a vector-valued function.

Such a function has a form v(t) = (x1(t), x2(t), ..., xn(t)).

The graph of the outputs gives a 1-dimensional curve in n-dimensional space.

For example, graph (cos t, sin t).

What about the graph of (t, t3)?

Lines:

If each component function is linear then the curve is a line in n-dimensional space. Moreover,the line can be described by

`(t) = r0 + tv,

where r0 is a vector that from the origin goes to the line (so also a point on the line) and vis any vector parallel to the line.

The form of the line given on the previous page is the “vector form.” The “parametric form”is list of equations, one for each component. The “synthetic form” is given by equatingsolutions for t in each of the component equations.

Examples:

(1) Find an equation of the line in R4 through (1, 0, 1, 2) and (0,−8, 5,−13).

(2) Find an equation of the line through the origin and perpendicular to both (1, 2, 3)and (−4, 5,−1).

(3) Calculate the distance between the point (1, 1, 1) and the line (−t, 7, 5t). Hint: Drawa picture! Find any vector u that points from a point on the line to the point (1, 1, 1).Find its projection onto a vector v that is parallel to the line. Then the distance is|u− projvu|.

(4) Find a “synthetic form” for the equation of the line `(t) = (1,−2, 5) + t(−1, 2, 4).

The equation of a line segment between point P (x0, y0, z0) and Q(x1, y1, z1) can be given by

s(t) = tQ + (1− t)P = P + t−→PQ for 0 ≤ t ≤ 1.

Find an equation for the line segment joining (1, 7,−4) to (−2,−1, 3).

Curves:

In general, v(t) = (x1(t), x2(t), ..., xn(t)) is a curve in n-dim. space.

A helix : h(t) = (cos (2πt), sin (2πt), t).

What does this graph look like? Consider its projection onto the xy-plane (z = 0). Whatperiod does that shadow have?

Limits and Continuity of vector-valued functions:

A vector-valued function v(t) approaches a vector limit L as t approaches a, if |v(t)− L| isarbitrarily close to 0 for t sufficiently close to a, but not equal to a.

If v(a) equals the limit of v(t) as t → a, then v(t) is continuous at t = a.

Suppose v(t) = (v1(t), v2(t), ..., vn(t)) and each vi has a limit Li as t → a. Set L =(L1, L2, ..., Ln). Then as t → a,

|v(t)− L|2 = (v1(t)− L1)2 + (v2(t)− L2)

2 + · · ·+ (vn(t)− Ln)2 → 0+,

making the limit of v(t) the vector L.

So just calculate the limits of the component functions! Also, all the standard laws for scalarlimits have vector-valued analogs.

Evaluate the limit:

limt→∞

(t2 + 3t− 1

5t2 + 1i + e−tj− tan−1 xk

).

Determine the domain where the function is continuous:

v(t) =(√

9− t2, ln (t− 1), et)

.

11.6 Calculus of Curves

When it exists, the derivative of a vector-valued function v(t) yields a vector tangent to thecurve with magnitude equal to the speed: the rate of change in distance traveled by v(t)with respect to t.

An easy calculation shows that if v(t) = (v1(t), v2(t), ..., vn(t)) then v′(t) = (v′1(t), v′2(t), ..., v

′n(t)).

Unit Tangent Vector: When v′(t) exists and is nonzero, it is a tangent vector to v(t) andwe can define the unit tangent vector:

T(t) =v′(t)|v′(t)| .

Rules for derivatives (proofs are left to the reader): Let u,v be vector-valued functions of t,let f be a scalar-valued differentiable function of t, and let c be a constant vector.

(1)d

dt(c) = 0.

(2)d

dt(u(t) + v(t)) = u′(t) + v′(t).

(3)d

dt(f(t)u(t)) = f ′(t)u(t) + f(t)u′(t).

(4)d

dt(u(t) · v(t)) = u′(t) · v(t) + u(t) · v′(t).

(5)d

dt(u(t)× v(t)) = u′(t)× v(t) + u(t)× v′(t).

Examples:

(1)d

dt

(cos (1− t2), sin t, te−t,

√2t + 1

).

(2) Let v(t) = (cos (2πt), sin (2πt), t). Find v′(t) and the unit tangent vector. Also findv′′(t)

(3)d

dt

((tan−1 t, e2t, 3t2) · (1 + t2, ln t, 1)

).

(4) Find an equation to the tangent line to the curve v(t) = (2 cos (2πt), 3 sin (2πt), 6t)at t = 1

2.

Let v(t) be an n-dim. vector-valued function. If V(t) is an n-dim. vector-valued functionsuch that V′(t) = v(t) then, we call V and antiderivative of v. To compute an antiderivative,just compute an antiderivative for each of the component functions!

The indefinite integral is the set of all antiderivatives denoted in the following way:

∫v(t)dt = V(t) + C, where V is an antiderivative of v and C is a constant.

The definite integral is the calculated in the following way:

∫ b

a

v(t)dt = V(b)−V(a), where V is an antiderivative of v.

Compute:

∫ 4

1

(te−ti + j +

1

tk

)dt.

If v represents velocity, then the definite integral is the net displacement.

11.7 Motion

Let r(t) represent position of an object against time t. The whole curve represents the pathor trajectory of the object.

Velocity and acceleration:

v(t) = r′(t) represents velocity of the object and a(t) = v′(t) = r′′(t) represents accelerationof the object.

Speed is the magnitude of velocity: s(t) = |v(t)| = |r′(t)|.

Consider an object with the path (1 − t, 2 + 3t, 5 − 6t). What is the velocity, speed andacceleration of this object? What kind of path is this?

Consider an object with the path ( 1√2cos t, 1√

2sin t, 1√

2t). What is the velocity, speed and

acceleration of this object? What kind of path is this?

Assume r(t) is a trajectory of an object with constant magnitude |r(t)|. Then

0 =d

dt(r(t) · r(t)) = r′(t) · r(t) + r(t) · r′(t) = 2r′(t) · r(t) →

0 = r′(t) · r(t) → r′(t) ⊥ r(t).

We just proved for trajectories stuck on a sphere, the position and velocity vectors areorthogonal!

Assume r(t) is a trajectory of an object with unit speed: |r′(t)| = 1. Then

0 =d

dt(r′(t) · r′(t)) = r′′(t) · r′(t) + r′(t) · r′′(t) = 2r′′(t) · r′(t) →

0 = r′′(t) · r′(t) → r′′(t) ⊥ r′(t).

We just proved for trajectories with unit speed that the velocity and acceleration vectors areorthogonal!

Suppose we model 3-dim. motion by having the first two coordinates be east-west andnorth-south respectively, while the third coordinate represents up-down.

Let an object have the initial position r0 = (x0, y0, z0), the initial velocity v0 = (u0, v0, w0).Suppose the object is subject to constant acceleration due to gravity a = (0, 0,−g) with noother forces.

By antidifferentiation,v(t) = at + v0 = (u0, v0, w0 − gt),

represents velocity.

By another antidifferentiation,

r(t) =a

2t2 + v0t + r0 = (x0 + u0t, y0 + v0t, z0 + w0t− g

2t2),

represents position.

By solving for the positive root of the vertical coordinate, one can solve for the time in flight:

By plugging this time back into r(t) one can find the point at which the object lands!

A rocket is launched from 10 meters above the ground, with an initial velocity of 200 metersper second, toward the northeast, making an angle of 30 degrees with the horizontal. Anortherly wind provides an acceleration of 2 m/s2. Assuming g = 9.8 m/s2, determine howfar north and east the rocket traveled.

Hint: a = (0, 2,−9.8), v0 = (√

22·√

32· 200,

√2

2·√

32· 200, 1

2· 200), and r0 = (0, 0, 10).

11.8 Arc Length

Consider the curve r(t) = (x1(t), x2(t), ..., xn(t)) with differentiable component functions.By the generalized Pythagorean theorem, the arc length of this curve from r(a) to r(b) isgiven by the integral:

∫ b

a

√(dx1)2 + (dx2)2 + · · ·+ (dxn)2 =

∫ b

a

√(dx1

dt

)2

+

(dx2

dt

)2

+ · · ·+(

dxn

dt

)2dt.

The arc-length is the integral of the speed along the path.

Suppose that r(t) = (x1(t), x2(t), ..., xn(t)) represents a trajectory and each component func-tion is differentiable. The the arc-length L of the curve between r(a) and r(b) is given bythe following integral:

L =

∫ b

a

|r′(t)|dt =

∫ b

a

√(x′1(t))2 + (x′2(t))2 + · · ·+ (x′n(t))2dt.

Calculate the arc length of r(t) = (cos (2πt), sin (2πt), t) from (1, 0, 0) to (1, 0, 1).

Polar coordinates: One can use coordinates (r, θ) to describe points in the xy-plane, where ris the distance to the origin and θ is an angle measured counter-clockwise from the positivex-axis. In terms of the standard, rectangular coordinates (x, y),

x = r cos θ and y = r sin θ.

and

r =√

x2 + y2 and θ = tan−1 (y

x) adjusted to the right quadrant, or ± π

2.

Suppose r = f(θ). Then x = f(θ) cos θ and y = f(θ) sin θ. So

L =

∫ β

α

√(dx

dθ

)2

+

(dy

dθ

)2

dθ →

L =

∫ β

α

√(f ′(θ) cos θ − f(θ) sin θ)2 + (f ′(θ) sin θ + f(θ) cos θ)2dθ →

L =

∫ β

α

√(f ′(θ) cos θ)2 + (f(θ) sin θ)2 + (f ′(θ) sin θ)2 + (f(θ) cos θ)2dθ →

L =

∫ β

α

√(f ′(θ))2 + (f(θ))2dθ.

Consider the cardioid r = 1 + cos θ. Find the arc length of the upper half.

Recall: cos2 x = 12(1 + cos 2x).

L =

∫ π

0

√(− sin θ)2 + (1 + cos θ)2dθ =

∫ π

0

√2 + 2 cos θdθ =

∫ π

0

√2(1 + cos θ)dθ →

L =

∫ π

0

√2

(2 cos2

θ

2

)dθ = 2

∫ π

0

cos

(θ

2

)dθ = 4 · sin

(θ

2

)|π0= 4.

11.9 Curvature and Normal Vectors

What is the difference between the trajectories r1 = (cos t, sin t) and r2 = (cos 2πt, sin 2πt)?Both parameterize the unit circle.

As t goes from 0 to 2π, r1 goes through the unit circle once. So t is equal to the arc lengthfrom r1(0) to r1(t). We call such a parametrization an arc-length parametrization.

This is not the case for r2 as it goes through 2π units of arc length as t goes from 0 to 1.

Let r(t) be a path. Consider the arc-length function for the arc length from r(0) to r(t):

s(t) =

∫ t

0

|r′(z)|dz.

If s(t) can be solved for the inverse function t(s), then by substitution, we get r(t(s)) andthus it can be parameterized by its own arc-length s.

Find an arc-length parametrization for r(t) = (3 cos t, 3 sin t, t). Find one for r(t) = (4t, 3t, 2t3/2)

Curvature:

Recall the unit tangent vector:

T(t) =r′(t)|r′(t)| .

|T(t)| = 1, so T(t) only changes direction, never magnitude. So |T′(t)| measures how quicklythe curve is changing direction, but it depends on the parameter t. So curvature is definedto be the rate at which the direction of T changes with respect to arc-length.

So, if r(t) is a path, the curvature is given by:

κ(s) =

∣∣∣∣dT

ds

∣∣∣∣ ,

where s denotes arc-length and T denotes the unit tangent vector.

We derive some more practical curvature formulas. Using the chain rule, dTdt

= dTds

dsdt

. Hence,

κ(s) =

∣∣∣∣dT

ds

∣∣∣∣ =

∣∣∣∣∣dTdtdsdt

∣∣∣∣∣ =|T′(t)|s′(t)

=|T′(t)||r′(t)| ,

noting that s′(t) = |r′(t)| by the fundamental theorem of calculus.

Calculate the curvature of (cos3 t, sin3 t) for t in [0, π2].

Calculate the curvature of (1, 3t, 1− 4t, 5 ln(sec t)) for t in (0, π2).

Let r(t) be a path in R3. Let v(t) = r′(t) and a(t) = v′(t) = r′′(t).

a =d

dt(v) =

d

dt(|v(t)|T(t)) =

(d

dt|v(t)|

)T + |v|dT

dt→

a× v =

((d

dt|v(t)|

)T + |v|dT

dt

)× |v|T = |v|dT

dt× |v|T →

|a× v| = |v|2∣∣∣∣dT

dt

∣∣∣∣ → |T′(t)| = |a× v||v|2 .

by the cross product identity, since T ⊥ T′ as |T| = 1.

So, for paths r(t) in R3, it follows that

κ(t) =|T′(t)||r′(t)| =

|r′′ × r′||r′|3 .

Calculate the curvature of the helix r(t) = (cos t, sin t, 2t).

Principal Unit Normal Vector:

N(t) =T ′(t)|T ′(t)| .

Clearly |N| = 1 and T ⊥ N. Furthermore, N points in the direction of the change in T, soit points to the “inside” of the curve.

For both T and N for r(t) = (2t, t2).

(ELFR) Come up with a formula to determine the curvature in the curve y = f(x) wherethe function f(x) is twice differentiable.

Components of acceleration:

v = |v|T → v = s′(t)T → a = s′′(t)T + s′(t)T′(t) → a = s′′(t)T + |v||T′(t)|N.

In 3-dimensions:

a = s′′(t)T +|a× v||v| N.

So the tangential component of acceleration is

a ·T = s′′(t).

The normal component of acceleration is

a ·N = |v||T′(t)|.

By playing with the normal component we get that the curvature is also equal to the normalcomponent of acceleration divided by the square of the speed:

κ =a ·N|v|2 .

Suppose a racecar driver navigates a turn at a constant speed of 120 mph. Suppose theturn can be approximated as part of a circle with a radius of 100 ft. What is the normalcomponent of the drivers acceleration?

ELFY: The circle (100 cos t, 100 sin t) has constant curvature 1100

.

Therefore: a ·N = κ|v|2 = (100)(120(5200/3600))2 ≈ 300 ft/s2... (a little over 9-g’s).

We also know, since the driver maintains constant speed, that the tangential component ofacceleration is zero, so the acceleration is perpendicular to the velocity, and thus |a| = a ·N.