MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN)
LECTURE NOTES
These lecture notes are intended as an outline for both student and instructor; for much more detailed
exposition on the topics contained herein, the student should consult Chapters 12-14 of our standard
textbook Calculus by Briggs and Cochran (NOT Early Transcendentals!). For the student’s convenience
these notes have the same section titles as Briggs/Cochran, arranged in the same order.
1
Preliminaries
Definition 1.1. The symbol R denotes the set of all real numbers. Geometrically we think of R as the
set of all points on the number line, or as 1-dimensional space.
2
Vectors in the Plane
Definition 2.1. The plane, or Cartesian plane, or xy -plane, is the full set of all ordered pairs of
the form (x, y) where x and y are both real numbers. We denote the plane by R2 . We think of R2 as
2-dimensional space.
The following definition is non-rigorous but well captures our intuitive notion of what a “vector”
should be, and will be helpful when we consider applications of calculus to physical problems. We will
give a more clear, i.e. purely mathematical, definition shortly. (Students who have taken a previous
course in linear algebra may be familiar with a certain very general definition of a vector; the definitions
we will end up working with here in fact a special case of the definition they have already learned.)
Definition 2.2 (Naive Definition of a Vector). A vector is an object which consists of both a (strictly
positive) length (or magnitude) and a direction. We typically denote vectors as lower-case letters
with an arrow hat, e.g. ~u, ~v , w,
~ ~a, ~b, etc. We consider two vectors ~u and ~v to be equal if they have both
the same length and the same direction, and we write ~u = ~v . (For equality, note that we do not require
~u and ~v to have the same location!) We also allow the existence of a unique zero vector, denoted ~0,
which we consider to have 0 length and no direction.
We may represent a vector ~v pictorially by drawing an arrow. We refer to pointy part at the end of the
arrow as the head, and the base of the arrow as the tail. If P and Q are two points in (two-dimensional
or three-dimensional) space, then we denote by P~Q the vector which has its tail at P and its head at
~ .)
Q. (Note that unless P = Q, we always have P~Q 6= QP
A scalar is just a magnitude, with no direction. In other words, a scalar is just a real number (a
member of the set R).
Example 2.3. Using the intuitive definition above, classify the following physical quantities as either a
vector or a scalar.
(a) A wind blowing southwest at 22 miles per hour.
(b) The mass of an apple.
(c) The force exerted by gravity on the apple.
(d) The temperature in Denton, TX at 2pm today.
(e) The velocity of a parked car at rest.
Definition 2.4 (Naive Definition of Scalar Multiplication). Let ~v be a vector and let c be a scalar. The
scalar product of c and ~v , denoted c~v , is defined as follows:
(1) if c > 0 then c~v is the vector which points in the same direction as ~v , and whose length is c times
the length of ~v .
(2) if c < 0 then c~v is the vector which points in the opposite direction from ~v , and whose length is
|c| times the length of ~v .
(3) if c = 0 then c~v = ~0.
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2
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
We call the vector c~v a scalar multiple of ~v .
Example 2.5. Draw any non-zero vector ~v . Then draw the vectors 3~v , 12 ~v , −~v = (−1)~v , − 52 ~v , 0~v , and
−π~v .
Definition 2.6. Two vectors are called parallel if one is a scalar multiple of the other.
Example 2.7. Section 12.1 Example 1.
Definition 2.8 (Naive Definition of Vector Addition and Subtraction). Let ~u and ~v be vectors. If the
tail of ~v is placed at the head of ~u, then the vector sum of ~u and ~v , denoted ~u + ~v , is the vector whose
tail coincides with the tail of ~u and whose head coincides with the head of ~v . (Picture helps here.)
The vector difference of ~u and ~v , denoted ~u − ~v , is defined to be the vector sum ~u + (−1)~v .
Example 2.9. Section 12.1 Example 2.
Next we will give a more rigorous definition of vectors and vector operations by coordinatizing them
in R2 .
Definition 2.10. A vector in the plane is an ordered pair ~v = (v1 , v2 ) in R2 . (We think of the tail of
~v as being the origin (0, 0) and the head as the point (v1 , v2 ).) Two vectors ~u = (u1 , u2 ) and ~v = (v1 , v2 )
are considered equal if both u1 = v1 and u2 = v2 , in which case we write ~v = ~u. The number v1 is
called the x-component of ~v and v2 is called the y-component of ~v .
The magnitude of a vector ~v = (v1 , v2 ), denoted by |~v |, is the number
p
|~v | = v12 + v22 .
Of course by the Pythagorean theorem, the magnitude of ~v is exactly its genuine geometric length.
If c is a scalar, we define the scalar product of c and ~v to be the vector c~v = (cv1 , cv2 ). We define
the vector sum of ~u and ~v to be the vector ~u + ~v = (u1 + v1 , u2 + v2 ), and the vector difference to
be the vector ~u − ~v = (u1 − v1 , u2 − v2 ).
Example 2.11. Let P = (x1 , y1 ) and Q = (x2 , y2 ) be two points in R2 . Compute the magnitude of the
vector P~Q.
Example 2.12. Let ~v = (v1 , v2 ) be any vector and let c be any scalar. Compute the relationship
between the magnitudes |~v | and |c~v |.
Fact 2.13. |c~v | = c|~v | for any vector ~v and any scalar c.
Definition 2.14. Let ~u = (u1 , u2 ) and ~v = (v1 , v2 ) be vectors in R2 and let c be a scalar. Define the
vector sum ~u + ~v , the vector difference ~u − ~v , and the scalar product c~v as follows:
~u + ~v = (u1 + v1 , u2 + v2 );
~u − ~v = (u1 − v1 , u2 − v2 );
c~v = (cv1 , cv2 ).
Example 2.15. Let ~u = (−1, 2) and ~v = (2, 3).
(a) Evaluate |~u + ~v |.
(b) Write 2~u − 3~v as a single vector.
(c) Find two distinct vectors half as long as ~u and parallel to ~u.
Definition 2.16. A unit vector is a vector of length 1. In particular we single out the coordinate
unit vectors, which we permanently denote by ~i = (1, 0) and ~j = (0, 1).
Note that if ~v = (v1 , v2 ) is any vector, then we may write
~v = (v1 , v2 ) = (v1 , 0) + (0, v2 ) = v1 (1, 0) + v2 (0, 1) = v1~i + v2~j.
So in general ~v = (v1 , v2 ) = v1~i + v2~j; this gives us another notation for writing vectors.
Fact 2.17. If ~v is a non-zero vector, then the vector
magnitude 1. In other words
~
v
|~
v|
~
v
|~
v|
=
1
v
|~
v| ~
has the same direction as ~v , and
is the unique unit vector which points in the same direction as ~v .
Example 2.18. Let P = (1, −2) and Q = (6, 10) be two points in the plane.
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
3
(a) Find P~Q and two distinct unit vectors parallel to P~Q.
(b) Find two distinct vectors of length 3 parallel to P~Q.
Example 2.19. Assume the water in a river moves southwest at 4 mi/hr. If a motorboat is traveling
due east at 15 mi/hr relative to the shore, determine the speed of the boat and its heading relative to
the moving water.
Example 2.20. A child pulls a wagon with a force of |F | = 20 lb at an angle of θ = 40◦ to the horizontal.
Find the force vector F .
Example 2.21. A 400-lb engine is suspended from two chains that form 60◦ angles with a horizontal
ceiling. How much weight must each chain withstand?
3
Vectors in Three Dimensions
Definition 3.1. The set of all ordered triples (x, y, z) where x, y, and z are all real numbers is called
xyz-space or three-dimensional space, and denoted permanently by R3 .
Example 3.2.
(1) Plot the points P = (3, 4, 5) and (−2, −3, 4) in xyz-space.
(2) Compute the distance between P and Q.
Definition 3.3. A sphere is the set of all points in three-dimensional space which are exactly a fixed
distance r from a single point (a, b, c) in R3 . (The point (a, b, c) is called the center of the sphere and
the distance r is called the radius.) A ball is the set of all points which are strictly less than a fixed
distance r from a single point (a, b, c). In other words a ball is the interior of a sphere.
Example 3.4. (a) Find an equation in three variables whose graph is the sphere centered at the origin
of radius 4.
(b) Find an inequality in three variables whose graph is the ball centered at the origin of radius 4.
(c) Find an equation for the sphere centered at (1, −2, 5) and containing the point (3, 4, −6).
(d) Describe the set of points that satisfy the equation x2 + y 2 + z 2 − 2x + 6y − 8z = −1.
Definition 3.5. A vector in three dimensions is an ordered triple in R3 . The magnitude of a
vector ~v = (v1 , v2 , v3 ) in three dimensions is the quantity
p
|~v = v12 + v22 + v32 .
The real numbers v1 , v2 , and v3 are called the x-component, y-component, and z-component
of ~v respectively. The notions of being parallel and of vector addition, vector subtraction, and
scalar multiplication are analogously to the two-dimensional case.
Definition 3.6. When working in R3 , the unit coordinate vectors are the following three distinguished vectors:
~i = (1, 0, 0);
~j = (0, 1, 0);
~k = (0, 0, 1).
Note that for any vector ~v = (v1 , v2 , v3 ) we have ~v = v1~i + v2~j + v3~k.
Example 3.7. Let ~u = (2, −4, 1) and ~v = (3, 0, −1).
(a)
(b)
(c)
(d)
Find −4~u + 2~v .
Find |~u − ~v |.
Find the unique unit vector with the same direction as ~u − ~v .
Write the unit vector from part (c) as a sum of scalar multiples of ~i, ~j, and ~k.
HOMEWORK (Due 7/9/13): Section 12.1 #18, 20, 26, 28, 30, 32, 34, 38, 40, 45, 51; Section 12.2
#36, 38, 39, 40
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MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
Dot Products
Definition 4.1. Given two vectors ~u = (u1 , u2 , u3 ) and ~v = (v1 , v2 , v3 ), we define the dot product,
denoted ~u · ~v , to be the scalar given by
~u · ~v = u1 v1 + u2 v2 + u3 v3 .
√
√
Example 4.2. (a) Let√~u = ( 3, 1, 0) and √
~v = (1, 3, 0). Compute ~u · ~v .
(b) Let ~u = (2, −1, −2 3) and ~v = (2, −2, 3). Compute ~u · ~v .
The usefulness of the dot product emerges in the upcoming Theorem 4.4; in order to prove Theorem
4.4, we need to recall the following fact about triangles.
Fact 4.3 (Law of Cosines). If a triangle has angle measures A, B, and C and corresponding opposite
side lengths a, b, and c, then the following equalities hold:
c2 = a2 + b2 − 2ab cos C; b2 = a2 + c2 − 2ac cos B; a2 = b2 + c2 − 2bc cos A.
The following theorem says that we can use dot products to compute the angles between given vectors.
Theorem 4.4. Let ~u = (u1 , u2 , u3 ) and ~v = (v1 , v2 , v3 ) be non-zero vectors, and let θ be the angle
between ~u and ~v with 0 ≤ θ ≤ π. Then
~u · ~v = |~u||~v | cos θ.
Proof. Consider the triangle which has ~u and ~v for two of its sides. The third side is equal as a vector
to ~u − ~v , and hence the Law of Cosines (previous fact) implies that
|~u − ~v |2 = |~u|2 + |~v |2 − 2|~u||~v | cos θ.
Now solving for |~u||~v | cos θ in the above, we get:
|~u||~v | cos θ = 21 (|~u − ~v |2 − |~u|2 − |~v |2 ).
p
Now by the definition of magnitude in R3 , we have |~u|2 = ( u21 + u22 + u23 )2 = u21 + u22 + u23 . Similarly
we have |~v |2 = v12 + v22 + v32 and
|~u − ~v |2 = (u1 − v1 )2 + (u2 − v2 )2 + (u3 − v3 )2 .
In that case, expanding out our equality from earlier and simplifying, we get:
1
(|~u − ~v |2 − |~u|2 − |~v |2 )
2
1
= ((u1 − v1 )2 + (u2 − v2 )2 + (u3 − v2 )2 − u21 − u22 − u23 − v12 − v22 − v32 )
2
1
= (2u1 v1 + 2u2 v2 + 2u3 v3 )
2
= u1 v1 + u2 v2 + u3 v3
|~u||~v | cos θ =
= ~u · ~v .
This proves the equality in the theorem.
√
√
Example 4.5. (a) Let√~u = ( 3, 1, 0) and √
~v = (1, 3, 0). Compute the angle θ between ~u and ~v .
(b) Let ~u = (2, −1, −2 3) and ~v = (2, −2, 3). Compute the angle θ between ~u and ~v .
Corollary 4.6. If the angle θ between two vectors ~u and ~v is
π
2
(90 degrees), then ~u · ~v = 0.
Definition 4.7. Two vectors ~u and ~v are called orthogonal if ~u ·~v = 0. (Orthogonal and perpendicular
are synonyms in two dimensions.)
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
5
Definition 4.8. Given two vectors ~u and ~v , define the othogonal projection of ~u onto ~v , denoted
proj~v ~u, to be the vector component of ~u which lies in the direction of ~v . Define the scalar component
of ~u in the direction of ~v , denoted scal~v ~u, to be the length of the vector proj~v ~u.
We have already observed that |~~vv| is the unit vector which points in the same direction as ~v . Elementary geometric considerations show that scal~v ~u = |~u| cos θ, where θ is the angle formed by ~u and ~v . So
we immediately have
proj~v ~u = |~u| cos θ( |~~vv| .
Theorem 4.9. For any two vectors ~u and ~v ,
~u · ~v
proj~v ~u =
~v , and
~v · ~v
~u · ~v
scal~v ~u =
.
|~v |
Example 4.10. Find proj~v ~u and scal~v ~u for the following vectors.
(a) ~u = (4, 1), ~v = (3, 4).
(b) ~u = (−4, −3), ~v = (1, −1).
HOMEWORK (Due 7/10/13): Section 12.2 #17, 18, 19, 23, 24; Section 12.3 #11, 12, 14, 15, 17,
19, 20, 23, 24, 25, 26, 27, 28
Definition 4.11. Let ~u and ~v be vectors in R3 . Define the cross product ~u × ~v to be the vector with
magnitude given by
|~u × ~v | = |~u||~v | sin θ,
where θ is the angle between ~u and ~v (0 ≤ θ ≤ π), and with direction given by the right-hand rule:
when you put the vectors tail to tail and let the fingers of your right hand curl from ~u to ~v , the direction
of ~u × ~v is the direction of your thumb, orthogonal to both ~u and ~v . (Note that the right-hand rule only
makes sense if ~u and ~v are not parallel; if they are parallel and θ = 0 or θ = π, check that the given
magnitude is 0 and hence ~u × ~v = ~0.)
√
Example 4.12. Find the direction and magnitude of ~u × ~v , where ~u = (1, 1, 0) and ~v = (1, 1, 2).
Example 4.13. (a) In general is ~u · ~v = ~v · ~u?
(b) In general is ~u × ~v = ~v × ~u?
Fact 4.14 (Properties of the Cross Product). Let ~u, ~v , and w
~ be any vectors in R3 and let a and b be
any scalars. The following properties all hold.
(a)
(b)
(c)
(d)
~u × ~v = −(~v × ~u)
(a~u) × (b~v ) = ab(~u × ~v )
~u × (~v + w)
~ = (~u × ~v ) + (~u × w)
~
(~u + ~v ) × w
~ = (~u × w)
~ + (~v × w)
~
Example 4.15. Evaluate all possible cross products of the coordinate vectors ~i, ~j, and ~k.
Example 4.16. Let ~u = (u1 , u2 , u3 ) and ~v = (v1 , v2 , v3 ) be any two vectors in R3 . Find a closed formula
for the coordinates of ~u × ~v .
Theorem 4.17 (The Cross Product as a Determinant). Let ~u = (u1 , u2 , u3 ) and ~v = (v1 , v2 , v3 ). Then


~k
~i
~j
u2 u3 ~
u1 u3 ~
u1 u2 ~


~u × ~v = det u1 u2 u3 = det
i − det
j + det
k.
v2 v 3
v1 v3
v1 v2
v1 v2 v3
Example 4.18. Find a vector orthogonal to ~u = −~i + 6~k and ~v = 2~i − 5~j − 3~k.
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MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
5
Lines and Curves in Space
Definition 5.1. A vector-valued function is a function which takes a real number for input, and
outputs a vector (typically in R2 or R3 ). We will typically denote vector-valued functions with the arrow
hat notation, e.g. ~r(t), f~(t), etc., where t is regarded as the input variable. If ~r(t) outputs vectors in
R3 , we may write
~r(t) = (x(t), y(t), z(t)),
where x, y, and z are all real-valued functions. In this way we regard ~r(t) as an ordered triple of
real-valued parametric functions. The graph of a vector-valued function ~r is the set of all possible
outputs ~r(t). (Note that this definition differs somewhat from our usual definition of a graph.)
Fact 5.2. An equation of the line in three-dimensional space passing through the point (x0 , y0 , z0 ) in the
direction of the vector ~v = (a, b, c) is ~r(t) = r~0 + t~v , or
(x, y, z) = (x0 , y0 , z0 ) + t(a, b, c), for −∞ < t < ∞.
Equivalently, the parametric equations of the line are
x(t) = x0 + at; y(t) = y0 + bt; z(t) = z0 + ct.
Example 5.3. (a) Find an equation of the line passing through (1, 2, 4) in the direction of ~v = (5, −3, 1).
(b) Find an equation of the line passing through (−3, 5, 8) and (4, 2, −1).
Example 5.4 (Helix Curve). Graph the curve described by the equation
~r(t) = 4 cos t~i + sin t~j + t ~k.
2π
Example 5.5 (Roller Coaster Curve). Graph the curve ~r(t) = cos t~i + sin t~j + 0.4 sin 2t~k.
Definition 5.6. Let a be a real number. A vector-valued function ~r(t) approaches the limit L as t
approaches a, written
~
lim = L,
t→a
~ = 0.
provided lim |~r(t) − L|
t→a
The function ~r(t) is continuous at a provided ~r(a) exists, lim ~r(t) exists, and lim ~r(t) = ~r(a). The
t→a
t→a
function ~r(t) is simply called continuous if it is continuous at every point in its domain.
Fact 5.7. Let ~r(t) = (x(t), y(t), z(t)) be a vector-valued function. If lim x(t) = L1 , lim y(t) = L2 , and
t→a
t→a
lim z(t) = L3 , then
t→a
lim ~r(t) = (L1 , L2 , L3 ).
t→a
Also, ~r(t) is continuous at a point a provided x(t), y(t), and z(t) are all continuous at a.
Example 5.8. Consider the function
~r(t) = cos πt~i + sin πt~j + e−t~k for t ≥ 0.
(a) Evaluate lim ~r(t).
t→2
(b) Evaluate lim ~r(t).
t→∞
(c) At what points is ~r continuous?
HOMEWORK (Due 7/11/13): Section 12.4 #10, 11, 15, 17, 23, 24, 26, 30; Section 12.5 #13, 15,
21, 23, 27, 34, 35
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
6
7
Calculus of Vector-Valued Functions
Definition 6.1. Let ~r(t) = (x(t), y(t), z(t)) be a vector-valued function. We define the derivative of
~r(t), denoted ~r0 (t), to be
~r(t + h) − ~r(t)
~r0 (t) = lim
,
h→0
h
if the limit exists. If the derivative exists at a point t then we say ~r is differentiable at t. We also
d
~r(t).
denote the derivative by dt
Note that the derivative ~r0 is a vector-valued function just like ~r. For any given t, the vector ~r0 (t)
points in the same direction as the curve given by ~r(t); for this reason ~r0 (t) is called a tangent vector
at ~r(t). We regard ~r0 (t) as the rate of change of the function ~r(t); for example, if ~r(t) is a position
function in three-dimensional space, then ~r0 (t) is the associated velocity function.
Fact 6.2. Let ~r(t) = x(t)~i + y(t)~j + z(t)~k, where x, y, and z are all differentiable functions of t. Then
~r0 (t) = x0 (t)~i + y 0 (t)~j + z 0 (t)~k.
Example 6.3. Compute the derivative of the following functions.
t3
(a) ~r(t) = (t3 , 3t2 , )
6√
(b) ~r(t) = e−t~i + 10 t~j + 2 cos(3t)~k
Example 6.4. Observe the behavior of ~r = (0, t2 , t3 ) and ~r0 at t = 0 in part (a) of the example above.
Definition 6.5. A vector-valued function ~r(t) is called smooth on an interval if it is differentiable on
that interval, and also ~r0 (t) 6= 0 on that interval.
Let ~r(t) be a smooth vector-valued function on some interval [a, b]. The unit tangent vector for
~r(t) on [a, b] is
~r0 (t)
T~ (t) = 0
.
|~r (t)|
Example 6.6. Find the unit tangent vectors for the following functions.
(a) ~r(t) = (t2 , 4t, ln t) for t > 0.
(b) ~r(t) = (10, 3 cos t, 3 sin t) for 0 ≤ t ≤ 2π.
Fact 6.7 (Derivative Rules). Let ~u(t) and ~v (t) be differentiable vector-valued functions and let f (t) be
a differentiable scalar-valued function. Let ~a be a constant vector.
d
(1) dt
~a = ~0
+ ~v (t)) = ~u0 (t) + ~v 0 (t)
(2)
d
u(t)
dt (~
(3)
d
u(t))
dt (f (t)~
(4)
d
u(f (t))
dt ~
(5)
d
u(t)
dt (~
· ~v (t)) = ~u0 (t) · ~v (t) + ~u(t) · ~v 0 (t)
(6)
d
u(t)
dt (~
× ~v (t)) = ~u0 (t) × ~v (t) + ~u(t) × ~v 0 (t)
= f 0 (t)~u(t) + f (t)~u0 (t)
= f 0 (t)~u0 (f (t))
(Product Rule)
(Chain Rule)
(Dot Product Rule)
(Cross Product Rule)
Example 6.8. Compute the following derivatives, where ~u(t) = t~i+t2~j −t3~k and ~v (t) = sin t~i+2 cos t~j +
cos t~k.
d
(a) dt
~v (t2 )
(b)
d 2
v (t)]
dt [t ~
(c)
d
u(t)
dt [~
· ~v (t)]
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MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
Example 6.9. Compute the first, second, and third derivatives of ~r(t) = (t2 , 8 ln t, 3e−2t ).
~ for which R
~ 0 = ~r. If
Definition 6.10. An antiderivative of a vector-valued function ~r is a function R
~
~r(t) = x(t)~i + y(t)~j + z(t)~k, then R(t)
= X(t)~i + Y (t)~j + Z(t)~k, where X, Y , and Z are antiderivatives
~ of ~r is called the indefinite integral
of x, y, and z respectively. The collection of all antiderivatives R
of ~r, and denoted
R
~r(t)dt.
As is the case with real-valued functions, any two antiderivatives of ~r differ only by some constant
~ So if R
~ is any antiderivative of ~r, we may write
vector C.
R
~ + C,
~
~r(t)dt = R(t)
~ is an arbitrary constant.
where C
R
t ~
√
i + e−3t~j + (sin 4t + 1)~k .
Example 6.11. Evaluate
t2 + 2
Example 6.12. Find ~r(t) such that ~r0 (t) = (e2 , sin t, t) and ~r(0) = ~j.
Definition 6.13. Let ~r(t) = f (t)~i + g(t)~j + h(t)~k be a vector-valued function, where f , g, and h are
integrable on the interval [a, b]. We define the definite integral of ~r(t) across [a, b] to be the vector
Rb
Rb
Rb
Rb
~r(t)dt = [ a f (t)dt]~i + [ a g(t)dt]~j[ a h(t)dt]~k.
a
Rπ
Example 6.14. Compute 0 [~i + 3 cos( 2t )~j − 4t~k]dt.
HOMEWORK (Due 7/15/13): Section 12.6 #7, 8, 11, 13, 14, 19, 20, 21, 22, 26, 28, 29, 30, 31, 32,
36, 42, 43
7
Motion in Space
Example 7.1. Consider the two-dimensional motion given by the position vector
~r(t) = (x(t), y(t)) = (3 cos t, 3 sin t) for 0 ≤ t ≤ 2π.
(a) Sketch the trajectory of the object.
(b) Find the velocity of the object.
(c) Find the speed of the object.
(d) Find the acceleration of the object.
(e) Sketch the position, velocity, and acceleration vectors for t = 0, t = π2 , t = π, and t =
3π
2 .
Example 7.2. Consider the vector functions given by f~(t) = (3t, t+2, t+3) and ~g (t) = 3t2 , t2 +2, t2 +3).
(a) Sketch graphs of both functions. How do the graphs compare?
(b) Compute the velocity functions f~0 and ~g ’. How do they compare?
(c) Graph both speed functions |f~0 | and |~g 0 |.
Definition 7.3. We say that a function ~r(t) models a uniform (constant velocity) straight-line
motion if ~r is of the form ~r(t) = (x0 + at, y0 + bt, z0 + ct) for some constants (x0 , y0 , z0 ) and (a, b, c) in
R3 . In this case the velocity is ~r0 (t) = (a, b, c) and the acceleration is ~r00 (t) = (0, 0, 0).
We say that a function ~r(t) models a circular motion, or moves on a sphere, if |~r(t)| = r for some
fixed radius r ≥ 0, for every t in the domain of ~r.
Example 7.4. An object moves on a trajectory described by ~r(t) = (3 cos t, 5 sin t, 4 cos t), for 0 ≤ t ≤
2π.
(a) Show that the object moves on a sphere, and find the radius of the sphere.
(b) Find the velocity and speed of the object.
Theorem 7.5. Suppose ~r(t) is differentiable and moves on a sphere, i.e. |~r(t)| = r for some fixed r,
for all t in the domain of ~r. Then the position vector ~r(t) and the velocity vector ~r0 (t) are orthogonal at
every t in the domain of ~r.
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
Proof. First notice that since |~r(t)|2 = ~r(t) · ~r(t) = r2 is a constant function, we have
So, applying the dot product rule for derivatives, we have:
d
r(t)
dt ~
9
· ~r(t) = 0.
d
~r(t) · ~r(t)
dt
= ~r0 (t) · ~r(t) + ~r(t) · ~r0 (t)
0=
= 2~r0 (t) · ~r(t).
So we must have ~r0 (t) · ~r(t) = 0, i.e. ~r0 and ~r are orthogonal.
Fact 7.6. Gravity accelerates objects downward at a rate of approximately 32 ft/s2 , or 9.8 m/s2 .
Example 7.7. A baseball is hit from 3 ft above home plate with an initial velocity in ft/s of ~v (0) =
(80, 80).
(a) Find functions ~v (t) and ~r(t) which model the position and velocity of the ball between the time it is
hit and the time when it first hits the ground. (Neglect all forces acting on the ball except gravity.)
(b) Show that the trajectory of the ball is a segment of a parabola.
(c) Assuming a flat playing field, how far does the ball travel horizontally? Plot the trajectory of the
ball. (Answer: t = 5.04s, x = 403ft.)
(d) What is the maximum height of the ball? (Answer: y = 103ft.)
(e) Does the ball clear a 20-ft fence that is 380 ft from home plate (directly under the path of the ball)?
(t = 4.75s, y − 22ft.)
Example 7.8. Suppose an object is launched from the origin (in two dimensions) at some acute angle
α (0 ≤ α ≤ π2 ) with an initial speed of |~v0 | (where ~v0 represents the initial velocity). Further suppose
gravity is the only force acting on the object.
(a) If ~v0 = (u0 , v0 ), use trigonometry to find expressions for u0 and v0 in terms of the direction α and
the speed |v0 |.
(b) Find general expressions for the velocity function ~v (t) and the position function ~r(t).
(c) Find a closed formula for the total flight time of the object.
(d) Find a closed formula for the total horizontal distance traveled by the object. (Which angle α
maximizes the distance? Minimizes the distance?)
(e) Find a closed formula for the maximum height of the object.
Example 7.9. A golf ball is driven down a horizontal fairway with an initial speed of 55m/s at an
initial angle of 25◦ (from a tee with negligible height). Neglect all forces except gravity and assume the
ball’s trajectory lies in a plane.
(a) How far does the ball travel horizontally and when does it land? (236m, 4.7s.)
(b) What is the maximum height of the ball? (27.6m.)
(c) At what angles should the ball be hit to reach a green that is 300 m from the tee? (sin 2α = .972,
α = 38.2◦ or α = 51.8◦ .)
Example 7.10. A small projectile is fired over horizontal ground in an easterly direction with an initial
speed of |~v0 | = 300 m/s at an angle of α = 30◦ above the horizontal. A crosswind blows from south to
north producing an acceleration of the projectile of 0.36 m/s2 to the north.
(a) Where does the projectile land? (t = 30.6 s, (x, y, z) = (7953, 169, 0) m.)
(b) In order to correct for the crosswind and make the projectile land due east of the launch site, at
what angle from due east must the projectile be fired? (−1.21◦ .)
HOMEWORK (Due 7/16/13): Section 12.7 #8, 9, 10, 19, 25, 26, 30, 32, 34, 36, 37, 38, 39
8
Length of Curves
Fact 8.1. Let ~r(t) = (f (t), g(t), h(t)) be a parametrized curve, where f 0 , g 0 , and h0 are continuous, and
the curve is traversed once for a ≤ t ≤ b. The arc length of the curve between (f (a), g(a), h(a)) and
(f (b), g(b), h(b)) is
Rbp
Rb
L = a |~r0 (t)|dt = a f 0 (t)2 + g 0 (t)2 + h0 (t)2 dt.
10
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
We will omit the construction of the above formula here, but please see the intro to Section 12.8 in
Briggs/Cochran for a plausibility argument.
Example 8.2. Compute the circumference of a circle of radius a.
Example 8.3. Find the length of the hypocycloid parametrized by ~r(t) = (cos3 t, sin3 t), where 0 ≤ t ≤
2π.
Example 8.4. An eagle rises at a rate of 100 vertical ft/min on a helical path given by ~r(t) =
(250 cos t, 250 sin t, 100t),
√ where ~r is measured in feet and t is measured in minutes. How far does it
travel in 10 minutes? ( 2502 + 1002 ≈ 269.)
9
Curvature and Normal Vectors
Definition 9.1. Let ~r(t) be a smooth vector-valued function for all t ≥ a and let ~v (t) = ~r0 (t). If |~t| = 1
for all t ≥ a, then we say that ~r is parametrized by arc length.
Keep in mind for the following fact, we usually define ~r in such a way that a = 0.
Fact 9.2. If ~r(t) is parametrized by arc length, and s(t) denotes the length of the curve parametrized by
~r from a to t, then s(t) = t − a for every t ≥ a.
Proof. By our arc length formula,
s(t) =
Rt
a
|~r0 (t)|dt =
Rt
a
|~v (t)|dt.
Since |~v (t)| = 1 for all t ≥ a, we have
Rt
s(t) =
a
1dt = t − a.
Example 9.3. Consider the helix parametrized by ~r(t) = 2 cos t, 2 sin t, 4t). Find the arc length function
s(t). Is ~r parametrized by arc length?
Definition 9.4. Recall from Definition 6.5 that the unit tangent vector associated to a smooth vector~r0 (t)
.
valued function ~r(t) is the function T~ (t) = 0
|~r (t)|
The curvature of a smooth parametrized curve, denoted by κ, is the magnitude of the rate of change
of T~ with respect to arc length s. In other words, if s denotes arc length,
~
κ(s) = | ddsT |.
We wish to have a method of computing curvature which works for any parameter t, and not just for
the arc length parameter s as above. The following theorem gives us such a method.
Theorem 9.5. Let ~r(t) be a smooth parametrized curve, where t is any parameter. If ~v = ~r0 is the
velocity and T~ is the unit tangent vector, then the curvature is
κ(t) =
~
1 dT
|~
v | | dt |
=
~ 0 (t)|
|T
~
r 0 (t)| .
Proof. Let s denote arc length of ~r. Note that the relationship between the derivatives of T~ with respect
to s and to t is given by the chain rule:
~
dT
dt
=
~
dT
ds
·
ds
dt .
Now by the Fundamental Theorem of Calculus, we have
~
dT
ds
=
1
|~
v|
So by our original definition of curvature, we have
·
~
dT
dt .
ds
dt
= |~v |, and hence, solving for
~
dT
ds ,
we get
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
~
κ(t) = | ddsT | =
~
1 dT
|~
v | | dt |
=
11
~ 0 (t)|
|T
|~
r0 |
as claimed.
Example 9.6. Which has greater curvature: A circle with a large radius or a circle with a small radius?
Theorem 9.7. Let ~r be the position of an object moving on a smooth curve. Let ~v = ~r0 be the velocity
and ~a = ~v 0 the acceleration of the object. Then the curvature at a point on the curve is
|~a × ~v |
.
κ=
|~v |3
Proof. Since T~ =
~
v
|~
v| ,
write
~v = |~v |T~ .
Differentiate both sides above using the product rule.
~
d
~a = ( dt
|~v |)T~ + |~v | ddtT .
Next compute the cross product ~a × ~v :
dT~
d
|~v |)T~ + |~v | ] × [|~v |T~ ]
dt
dt
d ~
dT~
= ( |~v |T × |~v |T~ + |~v |
× |~v |T~
dt
dt
dT~
= ~0 + |~v |2 (
× T~ )
dt
dT~
= |~v |2 (
× T~ ).
dt
~a × ~v = [(
~
Now notice that T~ and ddtT are orthogonal since T~ moves in a sphere! (See Theorem 7.5. So the angle
~
~
between T~ and ddtT is π2 , and hence the cross product ddtT × T~ has magnitude
~
~
~
~
| ddtT × T~ | = | ddtT ||T~ | sin π2 = | ddtT | · 1 · 1 = | ddtT |.
Putting everything together, we have
dT~
× T~ |
dt
dT~
= |~v |2 | |
dt
1 dT~
= |~v |3 ·
| |
|~v | dt
|~a × ~v | = |~v |2 |
= |~v |3 κ.
So κ =
|~a × ~v |
as claimed.
|~v |3
Example 9.8. Find the curvature of the parabola ~r(t) = (t, at2 ).
Example 9.9. Find the curvature of the helix ~r(t) = a cos t, a sin t, bt), where a, b > 0 are real numbers.
12
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
Definition 9.10. Let ~r be a smooth parametrized curve. The principal unit normal vector associated to ~r is
~
~ = dT /ds = 1 · dT~ .
N
κ
ds
|dT~ /ds|
We think of the principal unit normal vector as the (unit vector) direction in which the graph is
curving. In practice it suffices to use the equivalent formula:
~
~ = dT /dt .
N
~
|dT /dt|
Example 9.11. Find the principal unit normal vector for the helix ~r(t) = (a cos t, a sin t, bt), where
a, b > 0 are real numbers.
Fact 9.12. Let ~r be a smooth parametrized curve with unit tangent vector T~ and principal unit normal
~ . Then T~ and N
~ are orthogonal.
vector N
~
~
Proof. Since T~ moves in a sphere, it is orthogonal to its own derivative ddtT by Theorem 7.5. Since N
~
~ are orthogonal.
points in the same direction as ddtT by definition, T~ and N
Theorem 9.13. Let ~r be a smooth parametrized curve describing the motion of an object through space.
Then the acceleration of the function ~a = ~r00 has a unique representation as the sum of its tangential
component aT~ and its normal component aN~ :
~ + a ~ T~ ,
~a = a ~ N
N
where aN~ = κ|~v |2 =
|~a × ~v |
and aT~ =
|~v |
T
d2 s
dt2 .
Proof. We begin with the fact that T~ = |~~vv| and hence ~v = T~ |~v | = T~ ds
dt . Differentiating both sides and
using the product rule and chain rule respectively, we get
d~v
dt
ds
d
= (T~ )
dt dt
dT~ ds ~ d2 s
=
+T 2
dt dt
dt
dT~ ds ds ~ d2 s
=
+T 2.
ds dt dt
dt
~a =
Since
~
dT
ds
~ and
= κN
ds
dt
= |~v |, we get
~ +
~a = κ|~v |2 N
d2 s ~
dt2 T
and we are done.
Example 9.14. The driver of a car follows the parabolic trajectory ~r(t) = (t, t2 ) for −2 ≤ t ≤ 2 through
a sharp bend in the road. Find the tangential and normal components of the acceleration of the car.
HOMEWORK (Due 7/17/13): Section 12.8 #7, 10, 13, 14; Section 12.9 #9, 10, 13, 15, 17, 18, 24, 26
Test 1 7/18/13. No HW due!
HOMEWORK: (Due 7/22/13): Section 12.9 #24, 26, 28, 29, 30, 32, 37, 38, 40
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
10
13
Planes and Surfaces
In this section we begin to study functions which depend on multiple real inputs (or vector inputs),
most typically of the form f (x, y), and their graphs. We also study equations in three variables and
their graphs. We start with perhaps the simplest possible graphs in R3 :
Definition 10.1. Given a fixed point P0 = (x0 , y0 , z0 ) in R3 and a non-zero vector ~n = (a, b, c) in R2 ,
the set of all points P = (x, y, z) in R3 for which P − P − 0 is orthogonal to ~n is called a plane. (The
vector ~n is called the normal vector to the plane.)
Example 10.2. Given a point P0 and a normal vector ~n as in the above definition, find an equation in
three variables whose graph is the plane determined by P0 and ~n.
Fact 10.3. The plane passing through P0 = (x0 , y0 , z0 ) with normal vector ~n = (a, b, c) is described by
the equation
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.
Example 10.4. Find an equation of the plane passing through (2, −3, 4) and with normal vector ~n =
(−1, 2, 3).
Example 10.5. Find an equation of the plane that passes through the points (2, −1, 3), (1, 4, 0), and
(0, −1, 5).
Definition 10.6. Two distinct planes are called parallel if their respective normal vectors are parallel.
Two planes are orthogonal if their respective normal vectors are orthogonal.
Example 10.7. Which of the following distinct planes are parallel and which are orthogonal?
Q: 2x − 3y + 6z = 12
R: −x + 23 y − 3z = 14
S: 6x + 8y + 2z = 1
T : −9x − 12y − 3z = 7
Example 10.8. Find a parametrized vector-valued function whose graph is the line of intersection of
the planes Q: x + 2y + z = 5 and R: 2x + y − z = 7.
Definition 10.9. Given a curve C in a plane P and a line ` not in P , a cylinder is the surface consisting
of all lines parallel to ` that pass through C. (Note that this is a rare occasion where our definition
probably does not coincide with the student’s intuition from a previous math course.)
Definition 10.10. A trace of a surface is the set of points at which the surface intersects a plane which
is parallel to one of the coordinate planes. (Informally, a trace is a cross-section of a surface.) The traces
in the coordinate planes are called the xy-trace, the xz-trace, and the yz-trace.
Example 10.11. Use traces to sketch graphs of the following cylinders in R3 .
(a) x2 + 4y 2 = 16
(b) z − sin x = 0
See Section 13.1, Table 13.1 (p. 774) in Briggs/Cochran for a full summary of the next few examples,
which we will probably have limited in-class time for.
Example 10.12 (Ellipsoid). An ellipsoid is the graph of an equation of the form
Graph the ellipsoid with a = 3, b = 4, and c = 5.
x2
a2
+
y2
b2
+
z2
c2
= 1.
Example 10.13 (Elliptic Paraboloid). An elliptic paraboloid is the graph of an equation of the form
2
2
z = xa2 + yb2 . Graph the elliptic paraboloid with a = 4 and b = 2.
Example 10.14 (Hyperboloid of One Sheet). Graph the equation
x2
4
+
Example 10.15 (Hyperbolic Paraboloid). Graph the equation z = x2 −
Example 10.16 (Elliptic Cone). Graph
y2
4
y2
9
− z 2 = 1.
y2
4 .
+ z 2 = 4x2 .
Example 10.17 (Hyperboloid of Two Sheets). Graph −16x2 − 4y 2 + z 2 + 64x − 80 = 0.
HOMEWORK (Due 7/23/13): Section 13.1 #12, 16, 23, 27, 29, 33, 35, 38, 42, 45, 49, 57
14
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
11
Graphs and Level Curves
Definition 11.1. A function in two variables f (x, y) is a function which assigns a unique (real)
output to each input pair (x, y) from a particular set D in R2 . The set D is called the domain of f and
is often implicit rather than explicitly described. The range of f is the set of all real numbers z = (x, y)
which are assumed as (x, y) ranges over the domain D.
The graph of a function in two variables is the set of all triples (x, y, z) in R3 for which z = f (x, y).
Definition 11.2 (Set-Builder Notation). To facilitate the next example and the homework problems,
we recall for the student the use of set-builder notation. For convenience we will describe informally
using examples, rather than give a formal definition. For an example, suppose we wish to formally
describe the set D of all ordered pairs (x, y) for which x is twice y. Then we may write
A = {(x, y) : x = 2y}.
The above notation should be read as The set of all (x, y) in R2 such that x = 2y, which clearly and
precisely defines our set A. For another example, we could write
Y = {x ∈ R : 5 ≤ x < 10},
which reads The set of all x in R such that 5 is less than or equal to x and x is strictly less than 10.
The student should easily verify that, using the interval notation, Y = [5, 10). In general, given a set A
and a precise mathematical sentence P (x) about a variable x, the set-builder notation should be read
as follows.
{
x∈A
:
P (x)}
“The set of all elements x in A such that sentence P (x) is true for the element x.
p
Example 11.3. Let g(x, y) = 4 − x2 − y 2 .
(a) Find the domain and range of g.
(b) Sketch the graph of g.
Definition 11.4. Let f (x, y) be a function. Given any fixed real number z0 , the level curve of f at
z0 is the set of all (x, y) in R2 for which f (x, y) = z0 .
Example 11.5. Sketch some of the level curves of the following functions.
(a) f (x, y) = y − x2 − 1
2
2
(b) f (x, y) = e−x −y
12
Limits and Continuity
Definition 12.1. Let f (x, y) be a function in two variables. The function f has the limit L as (x, y)
approaches (a, b), denoted
lim
f (x, y) = L,
(x,y)→(a,b)
if given any > 0, there exists a δ > 0 such that
|f (x, y) − L| < whenever |(x, y) − (a, b)| < δ.
A function f (x, y) is continuous at a point (a, b) provided f (a, b) exists,
lim
(x,y)→(a,b)
lim
f (x, y) = f (a, b).
(x,y)→(a,b)
Example 12.2. Evaluate
lim
(3x2 y +
√
xy, if it exists.
(x,y)→(2,8)
Example 12.3. Evaluate
lim
(x,y)→(0,0)
3x−2 y 2 , if it exists.
f (x, y) exists, and
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
15
Definition 12.4. A disk in R2 is the set {(x, y) : |(x, y) − (a, b)| < r} for some fixed point (a, b) and
some fixed radius r.
Let R be a region of R2 . A point P in R is called an interior point if there is some disk about P
which is contained entirely in R.
A point Q is called a boundary point if every disk containing Q contains both a point in R and a
point not in R.
A region R is open if it consists entirely of interior points. A region is closed if it contains all its
boundary points.
Example 12.5. Evaluate
xy − 4y 2
√
√ , if it exists.
x−2 y
(x,y)→(4,1)
Example 12.6. Evaluate
(x + y)2
, if it exists.
(x,y)→(0,0) x2 + y 2
lim
lim
HOMEWORK (Due 7/24/13): Section 13.2 #12, 14, 15, 22, 27, 28, 29, 32, 33, 34; Section 13.3
#11, 12, 16, 25, 26
13
Partial Derivatives
Definition 13.1. Let f (x, y) be a function in two variables. The partial derivative of f with respect
to x is
f (x + h, y) − f (x, y)
δ
f (x, y) = lim
,
fx (x, y) = δx
h→0
h
provided the limit exists. The partial derivative of f with respect to y is
f (x, y + h) − f (x, y)
δ
fy (x, y) = δy
,
f (x, y) = lim
h→0
h
provided the limit exists.
Example 13.2. Let f (x, y) = x2 − y 2 + 4.
δf
(a) Compute δf
δx and δy .
(b) Evaluate each derivative at (2, −4).
Example 13.3. Compute the partial derivatives of the following functions.
(a) f (x, y) = sin xy
(b) g(x, y) = x2 exy
Definition 13.4. Let f (x, y) be a function in two variables. The second-order partial derivatives
of f are the following four functions (written with both available notations):
δf
δ2 f
δ
δx δx = δx2 of (fx )x = fxx ;
δf
δ2 f
δ
δy δy = δy 2 of (fy )y = fyy ;
δf
δ2 f
δ
δx δy = δxδy of (fy )x = fyx ;
δf
δ2 f
δ
δx δy = δyδx of (fx )y = fxy .
The latter two derivatives above are called mixed partial derivatives.
Example 13.5. Compute the second-order partial derivatives of f (x, y) = 3x4 y − 2xy + 5xy 3 .
Example 13.6 (Special Example). Let

2
2
 xy(x − y ) , if (x, y) 6= (0, 0)
2
2
f (x, y) =
x +y

0,
if (x, y) = (0, 0).
16
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
(a) Is f continuous?
(b) Compute fxy (0, 0) and fyx (0, 0). Are the mixed partial derivatives equal to one another?
The above example shows that in general fxy need not equal fyx . However, the following theorem
says that in most cases we care about, the equality fxy = fyx really does hold. The proof is beyond the
scope of this course and is omitted.
Theorem 13.7. Assume that f is defined on an open set D of R2 , and fxy and fyx are continuous at
every point of D. Then fxy = fyx at every point of D.
HOMEWORK (Due 7/25/13): Section 13.3 #19, 20, 22; Section 13.4 #8, 10, 11, 12, 15, 18, 21, 22,
27, 28, 29, 32, 36, 48, 49
14
The Chain Rule
Theorem 14.1 (Chain Rule (One Independent Variable)). Let z be a differentiable function of x and
y, where x and y are differentiable functions of t. Then
dz
dt
=
δz dx
δx dt
+
δz dy
δy dt .
Example 14.2. Let z = x2 − 3y 2 + 20, where x = 2 cos t and y = 2 sin t. Find
t = π4 .
dz
dt
and evaluate it at
Theorem 14.3 (Chain Rule (Two Independent Variables)). Let z be a differentiable function of x and
y, where x and y are differentiable functions of s and t. Then
δz
δs
=
δz δx
δx δs
+
δz δy
δy δs
and
δz
δt
=
δz δx
δx δt
+
δz δy
δy δt .
Example 14.4. Let z = sin 2x cos 3y, where x = s + t and y = s − t. Evaluate
δz
δs
and
δz
δt .
Example 14.5 (Special Exercise - Implicit Differentiation). Use a function in two variables and the
dy
chain rule to find dx
, where sin(xy) + πy 2 = x.
Fact 14.6. Let F be a differentiable function in two variables, and suppose a relationship between y and
x is defined implicitly by the rule F (x, y) = 0. If Fy 6= 0, then
dy
dx
Example 14.7. Find
15
dy
dx ,
x
= −F
Fy .
where x4 + 3x2 y 2 − y = 10.
Directional Derivatives and the Gradient
Definition 15.1. Let f be differentiable at (a, b) and let ~u = (cos θ, sin θ) be a unit vector in R2 . The
directional derivative of f at (a, b) in the direction of ~u is
f (a + h cos θ, b + h sin θ) − f (a, b)
,
D~u f (a, b) = lim
h→0
h
provided the limit exists.
Theorem 15.2. Let f be differentiable at (a, b) and let ~u = (u1 , u2 ) be a unit vector in R2 . Then
D~u f (a, b) = (fx (a, b), fy (a, b)) · (u1 , u2 ).
Proof. Define a new function g(s) (real inputs and real outputs) by the rule
g(s) = f (a + su1 , b + su2 ).
Geometrically, the graph of g is a “cross-section” of the graph of f passing through the point (a, b)
parallel to vector ~u. It is clear from the definition of the directional derivative that
D~u f (a, b) = g 0 (0).
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
17
Now setting x(s) = a + su1 and y(s) = b + su2 (so g(s) = (x(s), y(s)) and applying the chain rule, we
get
D~u f (a, b) = g 0 (0)
= fx (x(0), y(0))x0 (0) + fy (x(0), y(0))y 0 (0)
= fx (a, b)u1 + fy (a, b)u2 )
= (fx (a, b), fy (a, b)) · (u1 , u2 ).
Example 15.3. Consider the paraboloid z = f (x, y) = 41 (x2 +2y 2 )+2 and the unit vectors ~u = ( √12 , √12 )
√
and ~v = ( 12 , − 23 ).
(a) Find the directional derivative of f at (3, 2) in the directions of ~u and ~v .
(b) Graph the surface and interpret the directional derivatives.
Definition 15.4. Let f (x, y) be differentiable. The gradient of f at (x, y) is the function
∇f (x, y) = (fx (x, y), fy (x, y)).
Example 15.5. Find ∇f and ∇f (3, 2) for f (x, y) = x2 + 2xy − y 3 .
2
2
Example 15.6. Let f (x, y) = 3 − x10 + xy
10 .
(a) Compute ∇f (3, −1).
(b) Compute D~u f (3, −1) where ~u = ( √12 , √12 ).
(c) Compute the directional derivative of f at (3, −1) in the direction of the vector (3, 4).
Theorem 15.7. Let f be differentiable at (a, b).
(1) f has its maximum rate of increase at (a, b) in the direction of the gradient ∇f (a, b). The rate
of increase in this direction is |∇f (a, b)|.
(2) f has its maximum rate of decrease at (a, b) in the direction of −∇f (a, b). The rate of decrease
in this direction is −|∇f (a, b)|.
(3) If ~u is orthogonal to ∇f (a, b), then D~u f (a, b) = 0.
Proof. For any unit vector ~u, we have
D~u = ∇f (a, b) · ~u
= |∇f (a, b)||~u| cos θ
= |∇f (a, b)| cos θ,
where θ is the angle between ∇f (a, b) and ~u. Then cos θ is maximized when θ = 0 and minimized when
θ = π, which proves statements (1) and (2) above. If ∇f (a, b) and ~u are orthogonal then θ = π2 and
hence cos θ = 0; this shows statement (3).
Example 15.8. Consider the bowl-shaped paraboloid z = f (x, y) = 4 + x2 + 3y 2 .
(a) If you are located at the point (2, − 21 , 35
4 ) on the paraboloid, in which direction should you move in
order to ascend the surface at the maximum rate? How quickly will you ascend?
(b) If you are at the point (3, 1, 16), in which directions may you walk in order to neither gain nor lose
height?
HOMEWORK (Due 7/29/13): Section 13.5 #8, 10, 12, 18, 20, 22, 28, 30, 32; Section 13.6 #10, 12,
14, 16, 18, 21, 22
18
16
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
Tangent Planes
Definition 16.1. Let f (x, y) be differentiable at (a, b). The plane tangent to the graph of f at
(a, b) is the graph of the equation
z − f (a, b) = fx (a, b)(x − a) + fy (a, b)(y − b).
More generally, if F (x, y, z) is a function of three variables and the equation F (x, y, z) = 0 defines a
surface, then the tangent plane at a point (a, b, c) on the surface is the graph of the equation
Fx (a, b, c)(x − a) + Fy (a, b, c)(y − b) + Fz (a, b, c)(z − c) = 0.
Example 16.2. Find an equation for the plane tangent to the paraboloid f (x, y) = 32 − 3x2 − 4y 2 at
the point (2, 1, 16).
2
2
Example 16.3. Consider the ellipsoid defined by x9 + y25 + z 2 = 1.
(1) Find the equation of the plane tangent to the ellipsoid at (0, 4, 53 ).
(2) At what points on the ellipsoid is the tangent plane horizontal?
17
Maximum/Minimum Problems
Definition 17.1. Let f be a function in two variables. We say f has a local maximum at (a, b) is
there is some disk D containing (a, b) such that f (a, b) ≥ f (x, y) for all (x, y) in D. We say that f has
a local minimum at (a, b) if there is some disk D containing (a, b) such that f (a, b) ≤ f (x, y) for all
(x, y) in D. In either case, we say that f has a local extremum at (a, b).
A point (a, b) is a critical point of f if either
(1) fx (a, b) = fy (a, b) = 0 or
(2) one (or both) of fx and fy does not exist at (a, b).
Fact 17.2. If f has a local maximum or minimum at (a, b), then (a, b) is a critical point of f .
Example 17.3. Find the critical points of f (x, y) = xy(x − 2)(y + 3).
Definition 17.4. Let f be a function in two variables. We say that f has a saddle point at (a, b)
if (a, b) is a critical point, but for every disk D containing (a, b) there are points (x, y) in D for which
f (x, y) > f (a, b) and points (x, y) in D for which f (x, y) < f (a, b) (in other words f has neither a min
nor a max at (a, b)).
Theorem 17.5 (Second Derivative Test). Suppose that the second partial derivatives of f (x, y) are
continuous in a disk containing (a, b), where fx (a, b) = fy (a, b) = 0. Set
2
D(x, y) = [fxx fyy − fxy
](x, y).
(1)
(2)
(3)
(4)
If
If
If
If
D(a, b) > 0 and fxx (a, b) < 0, then f has a local maximum value at (a, b).
D(a, b) > 0 and fxx (a, b) > 0, then f has a local minimum value at (a, b).
D(a, b) < 0, then f has a saddle point at (a, b).
D(a, b) = 0, the test is inconclusive.
Definition 17.6. The quantity D(x, y) in the above
Theorem 17.5 is called the discriminant of f .
fxx fxy
D(x, y) is the determinant of the Hessian matrix
.
fyx fyy
Example 17.7. Classify all the critical points of f (x, y) = x2 + 2y 2 − 4x + 4y + 6.
Example 17.8. Classify all the critical points of f (x, y) = xy(x − 2)(y + 3).
Example 17.9. A shipping company handles rectangular boxes provided the sum of the length, width,
and height of the box does not exceed 96 in. Find the dimensions of the box that meets the condition
at has the largest volume.
Definition 17.10. If f (x, y) ≤ f (a, b) for all (x, y) in the domain of f , then f has an absolute
maximum at (a, b). If f (x, y) ≥ f (a, b) for all (x, y) in the domain of f , then f has an absolute
minimum at (a, b).
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
19
Example 17.11. Find the absolute maximum and minimum values, if they exist, of f (x, y) = 4−x2 −y 2
on the open disk R = {(x, y) : x2 + y 2 < 1}.
Fact 17.12 (Extreme Value Theorem). If f is continuous on a closed bounded region R in R2 , then f
obtains an absolute maximum and absolute minimum value on R.
Example 17.13. Find the absolute maximum and minimum values of f (x, y) = x2 + y 2 − 2x + 2y + 5
on the set R = {(x, y) : x2 + y 2 ≤ 4}.
Example 17.14. Find the absolute maximum and minimum values of f (x, y) = 6 − x2 − 4y 2 on the set
R = {(x, y) : −2 ≤ x ≤ 2, −1 ≤ y ≤ 1}.
HOMEWORK (Due 7/30/13): Section 13.7 #11, 12, 17, 18; Section 13.8 #15, 16, 19, 20, 22, 23,
29, 31
18
Double Integrals over Rectangular Regions
Definition 18.1. Let f be a function in two variables, and let R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d} be a
d−c
rectangular region in R2 . Given any positive integer n, set ∆x = b−a
n and ∆y = n . For each integer
k with 0 ≤ k ≤ n, set
xk = a + k∆y and yk = c + k∆y.
For each pair of integers j, k with 1 ≤ j ≤ n and 1 ≤ k ≤ n, let (xj , y k ) be a point chosen arbitrarily
from the rectangular region {(x, y) : xj−1 ≤ x ≤ xj , yk−1 ≤ y ≤ yk }.
We define the double integral, or double definite integral, of f over R to be
n
n X
X
RR
f (xj , y k )∆x∆y,
f (x, y)d(x, y) = lim
R
n→∞
j=1 k=1
provided the limit exists and is independent of the choices of (xk , y k ). (Note: Our notation differs slightly
from Briggs/Cochran here.) If the limit exists we say that f is integrable over R. If f is non-negative
on R, then the double definite integral corresponds to the volume of the solid bounded by the graph of
f over R.
Theorem 18.2 (Fubini’s Theorem). Let f be continuous on the rectangular region R = {(x, y) : a ≤
x ≤ b, c ≤ y ≤ d}. Then
RR
RdRb
RbRd
f (x, y)d(x, y) = c a f (x, y)dxdy = a c f (x, y)dydx.
R
Example 18.3. Find the volume of the solid bounded by the surface z = 4 + 9x2 y 2 over the region
R = {(x, y) : −1 ≤ x ≤ 1, 0 ≤ y ≤ 2}. Use both possible orders of integration.
RR
Example 18.4. Evaluate
xexy d(x, y), where R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ ln 2}.
R
HOMEWORK (Due 7/31/13): Section 13.8 #37, 38, 39; Section 14.1 #6, 7, 8, 10, 11, 13, 15, 16,
19, 20, 21, 22
19
Double Integrals over General Regions
Fact 19.1. Let g and h be continuous functions in one variable. Suppose R is a region in R2 bounded
below and above by the graphs of y = g(x) and y = h(x) respectively, and the lines x = a and x = b. If
f is continuous on R, then
RR
R b R h(x)
f (x, y)d(x, y) = a g(x) f (x, y)dydx.
R
Alternatively, if R is bounded on the left and right by the graphs of x = g(y) and x = h(y) respectively,
and the lines y = c and y = d, and f is continuous on R, then
RR
R d R h(y)
f (x, y)d(x, y) = c g(y) f (x, y)dxdy.
R
20
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
Example 19.2. Compute the integral
y = 3x2 and y = 16 − x2 .
RR
R
2x2 yd(x, y), where R is the region bounded by the parabolas
Example 19.3. Compute the volume of the solid below the surface f (x, y) = 2+ y1 and above the region
R in the xy-plane bounded by the lines y = x, y = 8 − x, and y = 1.
R √π R √π
Example 19.4. Evaluate 0
sin x2 dxdy.
y
HOMEWORK (Due 8/6/13): Section 14.2 #14, 15, 16, 18, 19, 20, 29, 30, 33, 43, 49, 50, 51, 55
20
Double Integrals in Polar Coordinates
Definition 20.1. To each pair (r, θ) of real numbers with r ≥ 0, we (implicitly) associate the pair (x, y)
in R2 where x = r cos θ and y = r sin θ. We refer to (r, θ) as the polar coordinates of (x, y). Notice
that any pair (x, y) has infinitely many polar coordinatizations, as (r, θ + 2πn) gives the same (x, y) for
any choice of integer n.
A polar rectangle is a set of the form {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β}, where a, b, α, β are real
numbers with β − α ≤ 2π.
Theorem 20.2. Let f be continuous on the region R = {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ} (where (rθ)
corresponds to a point (x, y) = (r cos θ, r sin θ) as in the above definition). Then
RR
RβRb
f (r, θ)d(x, y) = α a f (r, θ)rdrdθ.
R
Example 20.3. Find the volume of the solid bounded by the paraboloid z = 9 − x2 − y 2 and the
xy-plane.
Example 20.4. Find the volume of the region bounded beneath the surface z = xy + 10 and above the
annular region R = {(r, θ) : 2 ≤ r ≤ 4, 0 ≤ θ ≤ 2π}.
R 3 R √9−x2 p
x2 + y 2 dydx.
Example 20.5. Compute 0 0
Fact 20.6. Let f be continuous on the region R = {(r, θ) : 0 ≤ g(θ) ≤ r ≤ h(θ), α ≤ θ ≤ β}, where g
and h are continuous functions of θ. Then
RR
R β R h(θ)
f (r, θ)d(x, y) = α g(θ) f (r, θ)rdrdθ.
R
21
Triple Integrals
Definition 21.1. Let D = {(x, y, z) : a ≤ x ≤ b, g(x) ≤ y ≤ h(x), G(x, y) ≤ z ≤ H(x, y)}, where g, h,
G, H are continuous functions. Then
RRR
R b R h(x) R H(x,y)
f (x, y, z)d(x, y, z) = a g(x) G(x,y) f (x, y, z)dzdydx.
D
(Other orders of integration are handled similarly.)
R2 R2Re 2
Example 21.2. Evaluate −2 1 1 xyz dzdxdy.
R 1 R √1−x2 R √1−x2 −y2
Example 21.3. Evaluate 0 0
2xzdzdydx.
0
Example 21.4. A solid box D is bounded by the planes x = 0, x = 3, y = 0, y = 2, z = 0, and z = 1.
The density of the box decreases linearly in the z-direction and is given by f (x, y, z) = 2 − z. Find the
mass of the box. (Hint: Mass is the integral of density over the box.)
Example 21.5. Compute the volume of the region D bounded by the paraboloids y = x2 + z 2 and
y = 16 − 3x2 − z 2 .
HOMEWORK (Due 8/7/13): Section 14.3 #11, 14, 16, 19, 21, 22, 24, 25; Section 14.4 #8, 10, 21,
25, 28, 31
MULTIVARIABLE CALCULUS MATH 2730.001 SUMMER 2013 (COHEN) LECTURE NOTES
22
21
Change of Variables in Multiple Integrals
Definition 22.1. A transformation of two variables is a function T which takes pairs (u, v) in R2 for
input and returns pairs (x, y) = T (u, v) for output. We also call T a map or mapping. If S is the
domain of T (a region of R2 ) then we denote the range of T by T (S). T (S) is another region of R2 ,
called the image of S under T .
Example 22.2. Consider the transformation from polar coordinates to rectangular coordinates given
by
T (r, θ) = (r cos θ, r sin θ).
Find the image under this transformation of the rectangle S = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤
π
2 }.
Definition 22.3. A transformation of two variables T on a domain S is called one-to-one, or injective,
if T (u1 , v1 ) = T (u2 , v2 ) only when (u1 , v1 ) = (u2 , v2 ).
Example 22.4. Let T be as in the previous example.
(1) Let S = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ π2 }. Is T one-to-one on S?
(2) Let S = {(r, θ : 1 ≤ r ≤ 2, 0 ≤ θ ≤ π} . Is T one-to-one on S?
Definition 22.5. Given a transformation T (u, v) = (g(u, v), h(u, v)), where g and h are differentiable,
the Jacobian determinant (or just Jacobian) of T is
δx δx δv
J(u, v) = det δu
.
δy
δy
δu
δv
Example 22.6. Compute the Jacobian of the transformation T (r, θ) = (r cos θ, r sin θ).
Theorem 22.7. Let T (u, v) = (g(u, v), h(u, v)) be a transformation of two variables with a closed
bounded domain S. Assume T is one-to-one on the interior of S and that g and h have continuous first
partial derivatives there. Set R = T (S). If f is continuous on R, then
RR
RR
f (x, y)d(x, y) =
f (T (u, v))|J(u, v)|d(u, v).
R
S
RR p
Example 22.8. Evaluate the integral
2x(y − 2x)d(x, y), where R is the parallelogram in the
R
xy-plane with vertices (0, 0), (0, 1), (2, 4), and (2, 5). Use the transformation T (u, v) = (2u, 4u + v).
Z Z r
x−y
d(x, y), where R is the square with vertices (0, 0), (1, −1),
Example 22.9. Compute
x
+
y+1
R
(2, 0), and (1, 1).
RR 2
Example 22.10. Compute
y d(x, y), where R is the region bounded by the parabolas x = y 2 ,
R
x = y 2 − 4, x = 9 − y 2 , and x = 16 − y 2 .
HOMEWORK (Due 8/8/13): Section 14.7 #13, 14, 15, 27, 28, 29, 31, 32, 33