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 an
n-dimensional point with coordinates xi , depending on the context. We use Rn to denote
the 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 to
indicate a vector. Sometimes symbols are in bold or have an arrow above them to indicate
that they represent a vector.
An n-dimensional vector u = (u1 , u2 , ..., un ) can be multiplied by a real number c by multiplying each component by c:
cu = (cu1 , cu2 , ..., cun ).
This is called scalar multiplication, since geometrically, the vector’s length is scaled by the
number |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 be
added 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 putting
the tail of a “second” vector at the head of the “first” vector with the sum is the vector that
goes from the tail of the first to the head of the second.
Vector addition is commutative: u + v = v + u. Geometrically, this is the parallelogram
law. [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):
q
|u| =
u21 + u22 + · · · + 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 P Q (coordinates of P subtract coor−→
dinates of Q) is the vector that points from P to Q. Then |P Q| is the distance formula!
Recall also how to compute the midpoint.
Find the length of the vector (3, −4, 12).
Basic property of length:
q
q
2
2
2
2
2
2
|cu| = c u1 + c u2 + · · · + c un = |c| u21 + u22 + · · · + u2n = |c||u|.
For any non-zero vector u, the vector
û =
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 ±û.
An application to basic physics: Suppose a 160 lb person is suspended from two identical
chains that form 30-degree angles with the horizontal ceiling. How much tension force is
must each chain withstand?
Well, let T be the tension force in each cable (by symmetry). We know that T sin 30 = 0.5T
is the vertical component of tension force in each chain. So 2(0.5T ) = 160lb. So the tension
force 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 + y 2 + 8y + z 2 ≤ 0.
[CTS]
11.3 - Dot Product
The dot product of two vectors u = (u1 , u2 , ..., un ) and v = (v1 , v2 , ..., vn ) is a scalar given
by
u·v =
n
X
ui v i .
i=1
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 scalar
multiple of v that is orthogonal to u − p as seen here:
So
³u · v´
|u||v| cos θ
|u| cos θ
v=
v=
v.
projv (u) = (|u| cos θ)(v̂) =
|v|
|v|2
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 :
î = (1, 0, 0) (or (1, 0))
ĵ = (0, 1, 0) (or (0, 1))
k̂ = (0, 0, 1).
Then (a, b, c) = aî + bĵ + ck̂.
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 perpendicular 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 = (u2 v3 − u3 v2 , u3 v1 − u1 v3 , u1 v2 − u2 v1 ).
There is a method using a matrix determinant:


î ĵ k̂
u × v = det  u1 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 )
−−−−−−−−−−−−−
= (u2 v3 − u3 v2 , u3 v1 − u1 v3 , u1 v2 − u2 v1 ).
Random cross products! Do one where 2-dim. vectors are embedded in 3-space.
|u × v|2 = (u2 v3 − u3 v2 )2 + (u3 v1 − u1 v3 )2 + (u1 v2 − u2 v1 )2 →
|u × v|2 = u22 v32 − 2u2 v2 u3 v3 + u23 v22 + u23 v12 − 2u1 v1 u3 v3 + u21 v32 + u21 v22 − 2u1 v1 u2 v2 + u22 v12 →
|u × v|2 = u21 v12 + u22 v12 + u23 v12 + u21 v22 + u22 v22 + u23 v22 + u21 v32 + u22 v32 + u23 v32 − (u · v)2 →
|u × v|2 = (u21 + u22 + u23 )(v12 + v22 + v32 ) − (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) (c1 u) × (c2 v) = c1 c2 (u × v),
(3) u × (v + w) = u × v + u × w.
Geometrically, the cross product is perpendicular to both inputs. (SHOW!) Its direction is
given 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 degrees
with 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 v
is 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 equating
solutions 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: Draw
a 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 − projv u|.
(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 + tP Q 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). What
period 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| is
arbitrarily 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 scalar
limits have vector-valued analogs.
Evaluate the limit:
µ
lim
t→∞
¶
t2 + 3t − 1
−t
−1
i + e j − tan xk .
5t2 + 1
Determine the domain where the function is continuous:
v(t) =
³√
´
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 the
curve 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 v0 (t) = (v10 (t), v20 (t), ..., vn0 (t)).
Unit Tangent Vector: When v 0 (t) exists and is nonzero, it is a tangent vector to v(t) and
we can define the unit tangent vector:
T(t) =
v0 (t)
.
|v0 (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)
(2)
(3)
(4)
(5)
d
(c) = 0.
dt
d
(u(t) + v(t)) = u0 (t) + v0 (t).
dt
d
(f (t)u(t)) = f 0 (t)u(t) + f (t)u0 (t).
dt
d
(u(t) · v(t)) = u0 (t) · v(t) + u(t) · v0 (t).
dt
d
(u(t) × v(t)) = u0 (t) × v(t) + u(t) × v0 (t).
dt
Examples:
(1)
´
√
d ³
cos (1 − t2 ), sin t, te−t , 2t + 1 .
dt
(2) Let v(t) = (cos (2πt), sin (2πt), t). Find v0 (t) and the unit tangent vector. Also find
v00 (t)
(3)
¢
d ¡
(tan−1 t, e2t , 3t2 ) · (1 + t2 , ln t, 1) .
dt
(4) Find an equation to the tangent line to the curve v(t) = (2 cos (2πt), 3 sin (2πt), 6t)
at t = 21 .
Let v(t) be an n-dim. vector-valued function. If V(t) is an n-dim. vector-valued function
such that V0 (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:
Z
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:
Z
b
v(t)dt = V(b) − V(a), where V is an antiderivative of v.
a
Compute:
Z
4
1
¶
µ
1
−t
te i + j + k dt.
t
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 path
or trajectory of the object.
Velocity and acceleration:
v(t) = r0 (t) represents velocity of the object and a(t) = v0 (t) = r00 (t) represents acceleration
of the object.
Speed is the magnitude of velocity: s(t) = |v(t)| = |r0 (t)|.
Consider an object with the path (1 − t, 2 + 3t, 5 − 6t). What is the velocity, speed and
acceleration of this object? What kind of path is this?
Consider an object with the path ( √12 cos t, √12 sin t, √12 t). 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
(r(t) · r(t)) = r0 (t) · r(t) + r(t) · r0 (t) = 2r0 (t) · r(t) →
dt
0 = r0 (t) · r(t) → r0 (t) ⊥ r(t).
We just proved for trajectories stuck on a sphere, the position and velocity vectors are
orthogonal!
Assume r(t) is a trajectory of an object with unit speed: |r0 (t)| = 1. Then
0=
d 0
(r (t) · r0 (t)) = r00 (t) · r0 (t) + r0 (t) · r00 (t) = 2r00 (t) · r0 (t) →
dt
0 = r00 (t) · r0 (t) → r00 (t) ⊥ r0 (t).
We just proved for trajectories with unit speed that the velocity and acceleration vectors are
orthogonal!
Suppose we model 3-dim. motion by having the first two coordinates be east-west and
north-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 no
other forces.
By antidifferentiation,
v(t) = at + v0 = (u0 , v0 , w0 − gt),
represents velocity.
By another antidifferentiation,
a
g
r(t) = t2 + v0 t + r0 = (x0 + u0 t, y0 + v0 t, z0 + w0 t − t2 ),
2
2
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 meters
per second, toward the northeast, making an angle of 30 degrees with the horizontal. A
northerly wind provides an acceleration of 2 m/s2 . Assuming g = 9.8 m/s2 , determine how
far north and east the rocket traveled.
Hint: a = (0, 2, −9.8), v0 = (
√
2
2
·
√
3
2
· 200,
√
2
2
·
√
3
2
· 200, 12 · 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) is
given by the integral:
Z bp
Z
b
(dx1 )2 + (dx2 )2 + · · · + (dxn )2 =
a
a
s
µ

dx1
dt
¶2
µ
+
dx2
dt
¶2
µ
+···+
dxn
dt
¶2

dt.
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 function is differentiable. The the arc-length L of the curve between r(a) and r(b) is given by
the following integral:
Z
L=
b
0
|r (t)|dt =
a
Z bp
a
(x01 (t))2 + (x02 (t))2 + · · · + (x0n (t))2 dt.
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 r
is the distance to the origin and θ is an angle measured counter-clockwise from the positive
x-axis. In terms of the standard, rectangular coordinates (x, y),
x = r cos θ and y = r sin θ.
and
r=
p
π
y
x2 + y 2 and θ = tan−1 ( ) adjusted to the right quadrant, or ± .
x
2
Suppose r = f (θ). Then x = f (θ) cos θ and y = f (θ) sin θ. So
Z
sµ
β
L=
α
Z
dx
dθ
¶2
µ
+
dy
dθ
¶2
dθ →
q
β
(f 0 (θ) cos θ − f (θ) sin θ)2 + (f 0 (θ) sin θ + f (θ) cos θ)2 dθ →
L=
α
Z
β
q
(f 0 (θ) cos θ)2 + (f (θ) sin θ)2 + (f 0 (θ) sin θ)2 + (f (θ) cos θ)2 dθ →
L=
α
Z
β
L=
q
(f 0 (θ))2 + (f (θ))2 dθ.
α
Consider the cardioid r = 1 + cos θ. Find the arc length of the upper half.
Recall: cos2 x = 12 (1 + cos 2x).
Z
π
Z
q
(− sin θ)2
L=
0
π
2
+ (1 + cos θ) dθ =
0
Z
π
L=
0
√
Z
π
2 + 2 cos θdθ =
p
2(1 + cos θ)dθ →
0
s µ
¶
µ ¶
µ ¶
Z π
θ
θ π
θ
2
2 2 cos
dθ = 2
cos
dθ = 4 · sin
| = 4.
2
2
2 0
0
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 length
from 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):
Z
t
s(t) =
|r0 (z)|dz.
0
If s(t) can be solved for the inverse function t(s), then by substitution, we get r(t(s)) and
thus 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) =
r0 (t)
.
|r0 (t)|
|T(t)| = 1, so T(t) only changes direction, never magnitude. So |T0 (t)| measures how quickly
the curve is changing direction, but it depends on the parameter t. So curvature is defined
to 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:
¯ ¯
¯ dT ¯
κ(s) = ¯¯ ¯¯ ,
ds
where s denotes arc-length and T denotes the unit tangent vector.
We derive some more practical curvature formulas. Using the chain rule,
dT
dt
=
¯ ¯ ¯¯ dT ¯¯
¯ dT ¯ ¯ dt ¯ |T0 (t)|
|T0 (t)|
κ(s) = ¯¯ ¯¯ = ¯ ds
= 0
,
¯= 0
¯ dt ¯
ds
s (t)
|r (t)|
noting that s0 (t) = |r0 (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) = r0 (t) and a(t) = v0 (t) = r00 (t).
d
d
a = (v) =
(|v(t)|T(t)) =
dt
dt
µµ
a×v =
µ
¶
d
dT
|v(t)| T + |v|
→
dt
dt
¶
¶
d
dT
dT
|v(t)| T + |v|
× |v|T = |v|
× |v|T →
dt
dt
dt
¯ ¯
¯ dT ¯
|a × v|
|a × v| = |v| ¯¯ ¯¯ → |T0 (t)| =
.
dt
|v|2
2
by the cross product identity, since T ⊥ T0 as |T| = 1.
So, for paths r(t) in R3 , it follows that
κ(t) =
|T0 (t)|
|r00 × r0 |
=
.
|r0 (t)|
|r0 |3
Calculate the curvature of the helix r(t) = (cos t, sin t, 2t).
dT ds
.
ds dt
Hence,
Principal Unit Normal Vector:
N(t) =
T 0 (t)
.
|T 0 (t)|
Clearly |N| = 1 and T ⊥ N. Furthermore, N points in the direction of the change in T, so
it 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) where
the function f (x) is twice differentiable.
Components of acceleration:
v = |v|T → v = s0 (t)T → a = s00 (t)T + s0 (t)T0 (t) → a = s00 (t)T + |v||T0 (t)|N.
In 3-dimensions:
a = s00 (t)T +
|a × v|
N.
|v|
So the tangential component of acceleration is
a · T = s00 (t).
The normal component of acceleration is
a · N = |v||T0 (t)|.
By playing with the normal component we get that the curvature is also equal to the normal
component of acceleration divided by the square of the speed:
a·N
κ=
.
|v|2
Suppose a racecar driver navigates a turn at a constant speed of 120 mph. Suppose the
turn can be approximated as part of a circle with a radius of 100 ft. What is the normal
component of the drivers acceleration?
ELFY: The circle (100 cos t, 100 sin t) has constant curvature
1
.
100
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 of
acceleration is zero, so the acceleration is perpendicular to the velocity, and thus |a| = a · N.