math 21a midterm i review

. . . . . .

Midterm I Review

Math 21a

March 5, 2008

.

.Image: Flickr user Mr. Theklan

http://flickr.com/photos/theklan/

. . . . . .

Announcements

◮ Midterm covers up to Section 11.4 in text◮ Any topics not covered in class or on HW will not be on test

(i.e., curvature)◮ Complete list of learning objectives on the Midterm I Review

sheet◮ Odd problems, chapter reviews, old exams, reviews

. . . . . .

Outline

Vectors and the Geometry ofSpace

Three-DimensionalCoordinate SystemsVectorsThe Dot ProductThe Cross ProductEquations of Lines and PlanesFunctions and surfacesCylindrical and SphericalCoordinates

Vector FunctionsVector Functions and SpaceCurves

Derivatives and Integrals ofVector FunctionsArc Length (not Curvature)Motion in Space: Velocity andAccelerationParametric Surfaces

Partial DerivativesFunctions of Several VariablesUtility Functions andindifference curvesLimits and ContinuityPartial DerivativesTangent Planes and LinearApproximations

. . . . . .

Three-Dimensional Coordinate SystemsSection 9.1 Learning Objectives

◮ To understand and be able to apply rectangular coordinatesystems in R3 and the right hand rule.

◮ To understand and be able to find the distance d between twopoints (x1, y1, z1) and (x2, y2, z2) in R3 and to be able to use thedistance formula

d =√

(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2.

◮ To understand and be able to apply the equation of a sphere ofradius r and center (x0, y0, z0),

(x − x0)2 + (y − y0)

2 + (z − z0)2 = r2.

. . . . . .

Axes in three-dimensional space

.

.x

.y

.z

.x

.y

.z

.x

.y

.z

.x

.y

.z

. . . . . .

The right-hand rule

.

.

.x

.y

.z

. . . . . .

Distance in space

ExampleFind the distance between the points P1(3, 2, 1) and P2(4, 4, 4).

Solution

.

.x

.y

.z

..P1

. .P2

.1.2

.√

5

.3.d

d =√

5 + 32 =√

1 + 4 + 9

. . . . . .

Distance in space—GeneralTheoremThe distance between (x1, y1, z1) and (x2, y2, z2) is√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

In space, the locus of all points which are a fixed distance from afixed point is a sphere.

. . . . . .

Munging an equation to see its surface

ExampleFind the surface is represented by the equation

x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0

SolutionWe can complete the square:

0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36 − 4 − 16 − 25

= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9

So

(x − 2)2 + (y + 4) + (z − 5)2 = 9

=⇒ |(x, y, z)(2,−4, 5)| = 3

This is a sphere of radius 3, centered at (2,−4, 5).

. . . . . .

VectorsLearning objectives for Section 9.2

◮ To understand and be able to apply the definition of a scalar anda vector.

◮ To be able to represent vectors geometrically as directed linesegments or algebraically as ordered pairs or triples of numbers.

◮ To understand and be able to apply the axioms or vectoraddition and scalar multiplication.

◮ To be understand and be able to determine the length of avector.

◮ To understand the definitions of unit vectors, standard basisvectors, and position vectors and to be able to apply thesedefinitions.

. . . . . .

What is a vector?Definition

◮ A vector is something that has magnitude and direction◮ We denote vectors by boldface (v) or little arrows (⃗v). One is

good for print, one for script◮ Given two points A and B in flatland or spaceland, the vector

which starts at A and ends at B is called the displacementvector

−→AB.

◮ Two vectors are equal if they have the same magnitude anddirection (they need not overlap)

..A

.B

.v

.C

.D

.u

. . . . . .

Vector or scalar?

DefinitionA scalar is another name for a real number.

ExampleWhich of these are vectors or scalars?

(i) Cost of a theater ticket

scalar

(ii) The current in a river

vector

(iii) The initial flight path from Boston to New York

vector

(iv) The population of the world

scalar

. . . . . .

Vector or scalar?

DefinitionA scalar is another name for a real number.

ExampleWhich of these are vectors or scalars?

(i) Cost of a theater ticket scalar

(ii) The current in a river vector

(iii) The initial flight path from Boston to New York vector

(iv) The population of the world scalar

. . . . . .

Vector addition

DefinitionIf u and v are vectors positioned so the initial point of v is theterminal point of u, the sum u + v is the vector whose initial point isthe initial point of u and whose terminal point is the terminal pointof v.

.

.u

.v.u + v

The triangle law.

.u

.v.u

.v.u + v

The parallelogram law

. . . . . .

Opposite and difference

DefinitionGiven vectors u and v,

◮ the opposite of v is the vector −v that has the same length asv but points in the opposite direction

◮ the difference u − v is the sum u + (−v)

.

.u

.v

.−v.u − v

. . . . . .

Scaling vectors

DefinitionIf c is a nonzero scalar and v is a vector, the scalar multiple cv isthe vector whose

◮ length is |c| times the length of v◮ direction is the same as v if c > 0 and opposite v if c < 0

If c = 0, cv = 0.

..

.v.2v

.− 12v

. . . . . .

Components defined

Definition

◮ Given a vector a, it’s often useful to move the tail to O andmeasure the coordinates of the head. These are called thecomponents of a, and we write them like this:

a = ⟨a1, a2, a3⟩

or just two components if the vector is the plane. Note theangle brackets!

◮ Given a point P in the plane or space, the position vector ofP is the vector

−→OP.

FactGiven points A(x1, y1, z1) and B(x2, y2, z2) in space, the vector

−→AB has

components−→AB = ⟨x2 − x1, y2 − y1, z2 − z1⟩

. . . . . .

Vector algebra in components

TheoremIf a = ⟨a1, a2, a3⟩ and b = ⟨b1, b2, b3⟩, and c is a scalar, then

◮ a + b = ⟨a1 + b1, a2 + b2, a3 + b3⟩◮ a − b = ⟨a1 − b1, a2 − b2, a3 − b3⟩◮ ca = ⟨ca1, ca2, ca3⟩

. . . . . .

Properties

TheoremGiven vectors a, b, and c and scalars c and d, we have

1. a + b = b + a2. a + (b + c) = (a + b) + c3. a + 0 = a4. a + (−a) = 05. c(a + b) = ca + cb

6. (c + d)a = ca + da

7. (cd)a = c(da)

8. 1a = a

These can be verified geometrically.

. . . . . .

Length

DefinitionGiven a vector v, its length is the distance between its initial andterminal points.

FactThe length of a vector is the square root of the sum of the squares of itscomponents:

|⟨a1, a2, a3⟩| =√

a21 + a2

2 + a23

. . . . . .

Useful vectors

Definition

◮ A unit vector is a vector whose length is one◮ We define the standard basis vectors i = ⟨1, 0, 0⟩,

j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩. In script, they’re often written as ı̂, ȷ̂,k̂.

FactAny vector a can be written as a linear combination of the standard basisvectors

⟨a1, a2, a3⟩ = a1i + a2j + a3k.

. . . . . .

The Dot Product ILearning objectives for Section 9.3

◮ To understand and be able to use the definitions of the dotproduct to measure the length of a vector and the angle θbetween two vectors. Given two vectors a = ⟨a1, a2, a3⟩ andb = ⟨b1, b2, b3⟩,

a · b = |a||b| cos θ = a1b1 + a2b2 + a3b3.

◮ To understand and be able to apply properties of the dotproduct.

. . . . . .

The Dot Product IILearning objectives for Section 9.3

◮ To understand that two vectors are orthogonal if their dotproduct is zero.

◮ To understand and be able to determine the projection of avector a onto a vector b,

proja b =a · b|a|2

a

◮ To be able to use vectors and the dot product in applications.

. . . . . .

DefinitionIf a and b are any two vectors in the plane or in space, the dotproduct (or scalar product) between them is the quantity

a · b = |a| |b| cos θ,

where θ is the angle between them.Another way to say this is that a · b is |b| times the length of theprojection of a onto b.

.

.a

.b.a · b is |b| times this length

In components, if a = ⟨a1, a2, a3⟩ and b = ⟨b1, b2, b3⟩, then

a · b = a1b1 + a2b2 + a3b3

. . . . . .

Algebraic properties of the dot product

FactIf a, b and c are vectors are c is a scalar, then

1. a · a = |a|22. a · b = b · a3. a · (b + c) = a · b + a · c

4. (ca) ·b = c(a ·b) = a · (cb)

5. 0 · a = 0

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Geometric properties of the dot product

FactThe projection of b onto a is given by

proja b =a · b|a|2

a

FactThe angle between two nonzero vectors a and b is given by

cos θ =a · b|a| |b|

,

where θ is taken to be between 0 and π.

. . . . . .

More geometric properties of the dot product

FactThe angle between two nonzero vectors a and b is acute if a · b > 0.obtuse if a · b < 0, right if a · b = 0. The vectors are parallel ifa · b = ± |a| |b|.

◮ b is a positive multiple of a if a · b = |a| |b|◮ b is a negative multiple of a if a · b = − |a| |b|

. . . . . .

Other uses of the dot product

◮ similarity◮ Work

W = F · d

. . . . . .

The Cross ProductLearning objectives for Section 9.4

◮ To understand and be able to use the definitions of the crossproduct to find a vector that is orthogoal two vectors. Giventwo vectors a = ⟨a1, a2, a3⟩ and b = ⟨b1, b2, b3⟩,

a×b = (|a||b| sin θ)n = ⟨a2b3−a3b2, a3b1−a1b3, a1b2−a2b1⟩,

where n is a unit vector perpendicular to both a and b.◮ To understand and be able to apply properties of the dot

product, especially a × b = −b × a.◮ To understand that two vectors are parallel if and only if their

cross product is zero.◮ To be able to use vectors and the cross product in applications.

. . . . . .

Definition of the cross product

DefinitionGiven vectors a and b in space, the cross product of a and b isthe vector

a × b = |a| |b| (sin θ) n,

where n is a vector perpendicular to a and b such that (a, b, n) is aright-handed set of three vectors.

In components, if

a = ⟨a1, a2, a3⟩= a1i + a2j + a3k

b = ⟨b1, b2, b3⟩= b1i + b2j + b3k

Then

a × b = (a2b3 − b2a3)i + (a3b1 − b3a1)j + (a1b2 − b1a2)k

= ⟨a2b3 − b2a3, a3b1 − b3a1, a1b2 − b1a2⟩

. . . . . .

Determinant formula

This is only to help you remember, in case you’ve seen determinantsof 3 × 3 matrices:∣∣∣∣∣∣

i j ka1 a2 a3

b1 b2 b3

∣∣∣∣∣∣ = i

∣∣∣∣a2 a3

b2 b3

∣∣∣∣ − j

∣∣∣∣a1 a3

b1 b3

∣∣∣∣ + k

∣∣∣∣a1 a2

b1 b2

∣∣∣∣= (a2b3 − b2a3)i − (b3a1 − a3b1)j + (a1b2 − b1a2)k

= a × b

. . . . . .

Algebraic Properties of the Cross Product

If a, b, and c are vectors and c is a scalar, then

1. a × b = −b × a

2. (ca) × b = c(a × b) = a × (cb)

3. a × (b + c) = a × b + a × c

4. (a + b) × c = a × c + b × c

. . . . . .

Geometric Properties of the cross product

◮ a × b = 0 exactly when a and b are parallel.◮ The magnitude of the cross product a × b is the area of the

parallelogram with sides a and b.

. .a

.b .|b| sin θ

.θ

. . . . . .

Applications of the cross product

◮ Area of a parallelogram◮ Torque τ = r × F◮ Volume of a parallelipiped (scalar triple product)

V = |a · (b × c)|

. . . . . .

The scalar triple product in actionExampleFind the volume of the parallelipiped spanned by a = ⟨2, 0,−3⟩,b = ⟨1, 1, 1⟩, and c = ⟨0, 4,−1⟩

Solution

.

.a

.b

.c

The scalar triple product can becalculated by∣∣∣∣∣∣

2 0 −31 1 10 4 −1

∣∣∣∣∣∣ = −22

It’s negative because the triple(a, b, c) is left-handed. So thevolume is 22.

. . . . . .

Equations of Lines and Planes I

◮ To understand and be able to represent a line ℓ in R3 using◮ the vector equation

r = r0 + tv

◮ the parametric equations

x = x0 + at y = y0 + bt z = z0 + ct

◮ the symmetric equations

x − x0

a=

y − y0

b=

z − z0

c

. . . . . .

Equations of Lines and Planes II

◮ To be able to represent line segment in R3 from r0 to r1,

r(t) = (1 − t)r0 + tr1,

where 0 ≤ t ≤ 1.

◮ To understand and be able to represent a plane in R3 given anormal vector n = ⟨a, b, c⟩ and a point P0 = (x0, y0, z0) in theplane,

n · ⟨x − x0, y − y0, z − z0⟩ = 0

ora(x − x0) + b(y − y0) + c(z − z0) = 0.

. . . . . .

Equations of Lines and Planes III

◮ To understand and be able to calculate the distance D from apoint P1 = (x1, y1, z1) to the plane ax + by + cz + d = 0,

D =|ax1 + by1 + cz1 + d|√

a2 + b2 + c2.

. . . . . .

Applying the definition

ExampleFind the vector, parametric, and symmetric equations for the linethat passes through (1, 2, 3) and (2, 3, 4).

Solution

◮ Use the initial vector ⟨1, 2, 3⟩ and direction vector⟨2, 3, 4⟩ − ⟨1, 2, 3⟩ = ⟨1, 1, 1⟩. Hence

r(t) = ⟨1, 2, 3⟩ + t ⟨1, 1, 1⟩

◮ The parametric equations are

x = 1 + t y = 2 + t z = 3 + t

◮ The symmetric equations are

x − 1 = y − 2 = z − 3

. . . . . .

Planes

The set of all points whose displacement vector from a fixed point isperpendicular to a fixed vector is a plane.

.

.x

.y

.z

.r0

.n

. . . . . .

Equations for planes

The plane passing through the point with position vectorr0 = ⟨x0, y0, z0⟩ perpendicular to ⟨a, b, c⟩ has equations:

◮ The vector equation

n · (r − r0) = 0 ⇐⇒ n · r = n · r0

◮ Rewriting the dot product in component terms gives the scalarequation

a(x − x0) + b(y − y0) + c(z − z0) = 0

The vector n is called a normal vector to the plane.◮ Rearranging this gives the linear equation

ax + by + cz + d = 0,

where d = −ax0 − by0 − cz0.

. . . . . .

ExampleFind an equation of the plane that passes through the pointsP(1, 2, 3), Q(3, 5, 7), and R(4, 3, 1).

SolutionLet r0 =

−→OP = ⟨1, 2, 3⟩. To get n, take

−→PQ ×−→

PR:

−→PQ ×−→

PR =

∣∣∣∣∣∣i j k2 3 43 1 −2

∣∣∣∣∣∣ = ⟨−10, 16,−7⟩

So the scalar equation is

−10(x − 1) + 16(y − 2) − 7(z − 3) = 0.

. . . . . .

Distance from point to line

DefinitionThe distance between a point and a line is the smallest distancefrom that point to all points on the line. You can find it by projection.

...P0

..Q

.v

.b

..θ

.projv b =b · vv · v

v

.

∣∣∣∣b − b · vv · v

v

∣∣∣∣

. . . . . .

Distance from a point to a plane

DefinitionThe distance between a point and a plane is the smallest distancefrom that point to all points on the line.

...P0

..Q

.n

.b.|n · b||n|

To find the distance from the a point to a plane, project thedisplacement vector from any point on the plane to the given pointonto the normal vector.

. . . . . .

Distance from a point to a plane II

We have

D =|n · b||n|

If Q = (x1, y1, z1), and the plane is given by ax + by + cz + d = 0, thenn = ⟨a, b, c⟩, and

n · b = ⟨a, b, c⟩ · ⟨x1 − x0, y1 − y0, z1 − z0⟩= ax1 + by1 + cz1 − ax0 − by0 − cz0

= ax1 + by1 + cz1 + d

So the distance between the plane ax + by + cz + d = 0 and the point(x1, y1, z1) is

D =|ax1 + by1 + cz1 + d|√

a2 + b2 + c2

. . . . . .

Functions and surfacesLearning objectives for Section 9.6

◮ To understand and be able to represent a function of twovariables, z = f(x, y), graphically. Section 9.6)

◮ To understand and be able to use the trace (cross-section) of afunction. Section 9.6)

◮ To understand and be able to determine the domain of afunction z = f(x, y). Section 9.6)

◮ To understand and be able to represent a selection of quadricsurfaces graphically.

. . . . . .

function, domain, range

DefinitionA function f of two variables is a rule that assigns to eachordered pair of real numbers (x, y) in a set D a unique real numberdenoted by f(x, y). The set D is the domain of f and its range isthe set of values that f takes on. That is

range f = { f(x, y) | (x, y) ∈ D }

. . . . . .

Example

ExampleFind the domain and range of f(x, y) =

√xy.

Solution

◮ Working from the outside in, we see that xy must be nonnegative,which means x ≥ 0 and y ≥ 0 or x ≤ 0 and y ≤ 0. Thus the domainis the union of the coordinate axes, and the first and third quadrants.

◮ The range of f is the set of all “outputs” of f. Clearly the range of f isrestricted to the set of nonnegative numbers. To make sure that wecan get all nonnegative numbers x, notice x = f(x, x).

. . . . . .

Solution continued

Here is a sketch of the domain:

.

. . . . . .

Traces and surfaces

ExampleDescribe and sketch the surfaces.

(i) z = 4x2 + y2

(ii) z = 4 − x2 − y2

(iii) z2 = 4(x2 + y2)

(iv) x = 2y2 − z2

(v) x2 + y2 − 9z2 = 9

(vi) y2 − 9x2 − 4z2 = 36

(vii) x2 + 4y2 + 2z2 = 4

(viii) y2 + z2 = 1

. . . . . .

Sketch of z = 4x2 + y2

The trace in the xz-plane (y = 0) is the parabola z = 4x2. The tracein the yz-plane is the parabola z = y2. The trace in the xy-plane is thecurve 4x2 + y2 = 0, which is just a point (0, 0). But the trace in theplane z = k (k > 0) is the ellipse 4x2 + y2 = k. So we get ellipsescrossing the parabolas, an elliptic paraboloid.

-1.0

-0.5

0.0

0.5

1.0

-2

-1

0

1

2

0.0

0.5

1.0

1.5

2.0

. . . . . .

Sketch of z2 = 4(x2 + y2)

The trace in the xz-plane is the equation z2 = 4x2, or z = ±2x. Thetrace in the yz-plane is the equation z = ±2y. Traces in the planez = k are circles k2 = 4(x2 + y2). So we have a cone.

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

. . . . . .

Sketch of x = 2y2 − z2

The trace in the xy-plane is a parabola x = 2y2. The trace in thexz-plane is a parabola opening the other direction: x = −z2. Tracesx = k give hyperbolas. So we have a hyperbolic paraboloid.

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

. . . . . .

Sketch of x2 + y2 − 9z2 = 9

The x2 + y2 piece tells us that this is a surface of revolution, wherethe z-axis is the axis of revolution. If we find the xz-trace, we getx2 − 9z2 = 9, a hyperbola. We get a hyperboloid of one sheet.

-4

-2

0

2

4-4

-2

0

2

4

-2

-1

0

1

2

. . . . . .

Sketch of y2 − 9x2 − 4z2 = 36Traces in the xy and yz planes are hyperbolas. Traces in the planey = k are ellipses 9x2 + 4z2 = k2 − 36. We have a hyperboloid oftwo sheets.

-2

0

2

-10

-5

0

5

10

-4

-2

0

2

4

. . . . . .

Sketch of x2 + 4y2 + 2z2 = 4

The traces in the planes x = 0, y = 0, and z = 0 are all ellipses. Weget an ellipsoid.

-2-1

0

1

2-1.0

-0.5

0.00.5

1.0

-2

-1

0

1

2

. . . . . .

Traces and surfaces

ExampleDescribe and sketch the surfaces.

(i) z = 4x2 + y2 ellipticparaboloid

(ii) z = 4 − x2 − y2 ellipticparaboloid

(iii) z2 = 4(x2 + y2) cone(iv) x = 2y2 − z2 hyperbolic

paraboloid

(v) x2 + y2 − 9z2 = 9hyperboloid of one sheet

(vi) y2 − 9x2 − 4z2 = 36hyperboloid of two sheets

(vii) x2 + 4y2 + 2z2 = 4 ellipsoid

(viii) y2 + z2 = 1 cylinder

. . . . . .

Cylindrical and Spherical Coordinates ILearning objectives for Section 9.7

◮ To understand and be able to apply the polar coordinate system inR2, and to understand the relationship to rectangularcoordinates:

x = r cos θ y = r sin θ

r2 = x2 + y2 tan θ = y/x,

where r ≥ 0 and 0 ≤ θ ≤ 2π.

. . . . . .

Cylindrical and Spherical Coordinates IILearning objectives for Section 9.7

◮ To understand and be able to apply the cylindrical coordinatesystem in R3, and to understand the relationship to rectangularcoordinates:

x = r cos θ y = r sin θ z = z

r2 = x2 + y2 tan θ = y/x z = z,

where r ≥ 0 and 0 ≤ θ ≤ 2π.

. . . . . .

Cylindrical and Spherical Coordinates IIILearning objectives for Section 9.7

◮ To understand and be able to apply the spherical coordinatesystem in R3, and to understand the relationship to rectangularcoordinates:

x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ

ρ2 = x2 + y2 + z2,

where r ≥ 0, 0 ≤ θ ≤ 2π, and 0 ≤ φ ≤ π.

. . . . . .

Why different coordinate systems?

◮ The dimension of space comes from nature◮ The measurement of space comes from us◮ Different coordinate systems are different ways to measure

space

. . . . . .

Vectors and the Geometry of Space

..x

.y.r.θ

Conversion from polar tocartesian (rectangular)

x = r cos θ

y = r sin θ

Conversion from cartesian topolar:

r =√

x2 + y2

cos θ =xr

sin θ =yr

tan θ =yx

. . . . . .

Cylindrical CoordinatesJust add the vertical dimension

.

.x

.y

.z

..

.r.θ

.(r, θ, z)

.z

.x.y

.z

Conversion from cylindrical tocartesian (rectangular):

x = r cos θ y = r sin θ

z = z

Conversion from cartesian tocylindrical:

r =√

x2 + y2

cos θ =xr

sin θ =yr

tan θ =yx

z = z

. . . . . .

Spherical Coordinateslike the earth, but not exactly

.

.x

.y

.z

..(r, θ, φ)

.

.ρ.φ

.θ.x

.y

.z

Conversion from spherical tocartesian (rectangular):

x = ρ sin φ cos θ

y = ρ sin φ sin θ

z = ρ cos φ

Conversion from cartesian tospherical:

r =√

x2 + y2 ρ =√

x2 + y2 + z2

cos θ =xr

sin θ =yr

tan θ =yx

cos φ =zρ

. . . . . .

Outline






. . . . . .

Vector Functions and Space CurvesLearning objectives for Section 10.1

◮ To understand and be able to apply the concept of avector-valued function,

r(t) = ⟨f(t), g(t), h(t)⟩ = f(t)i + g(t)j + h(t)k.

◮ To be able to apply the concepts of limits and continuity tovector-valued functions.

◮ To understand that a curve in R3 can be representedparametrically

x = f(t) y = g(t) z = h(t)

or by a vector-valued function

r(t) = ⟨f(t), g(t), h(t)⟩

. . . . . .

ExampleGiven the plane curve described by the vector equation

r(t) = sin(t)i + 2 cos(t)j

(a) Sketch the plane curve.

(b) Find r′(t)

(c) Sketch the position vector r(π/4) and the tangent vectorr′(π/4).

. . . . . .

Solution

r(t) = r(t) = sin(t)i + 2 cos(t)j

r′(t) = cos(t)i − 2 sin(t)j

t r(t)0 2j

π/2 iπ −2j

3π/2 −i2π 2j

. .x

.y

.r(π/4)

.r′(π/4)

. . . . . .

Derivatives and Integrals of Vector FunctionsLearning objectives for Section 10.2

◮ To understand and be able to calculate the derivative of avector-valued function.

◮ To understand and be able to apply the basic rules ofdifferentiation for vector-valued functions.

◮ To understand and be able to find the definite integral of avector-valued function.

. . . . . .

Derivatives of vector-valued functions

DefinitionLet r be a vector function.

◮ The limit of r at a point a is defined componentwise:

limt→a

r(t) =⟨

limt→a

f(t), limt→a

g(t), limt→a

h(t)⟩

◮ The derivative of r is defined in much the same way as it isfor real-valued functions:

drdt

= r′(t) = limh→0

r(t + h) − r(t)h

. . . . . .

Rules for differentiation

TheoremLet u and v be differentiable vector functions, c a scalar, and f areal-valued function. Then:

1.ddt

[u(t) + v(t)] = u′(t) + v′(t)

2.ddt

[cu(t)] = cu′(t)

3.ddt

[f(t)u(t)] = f′(t)u(t) + f(t)u′(t)

4.ddt

[u(t) · v(t)] = u′(t) · v(t) + u(t) · v′(t)

5.ddt

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

6.ddt

[u(f(t))] = f′(t)u′(f(t))

. . . . . .




(b) Find r′(t)


. . . . . .

Solution



t r(t)0 2j

π/2 iπ −2j

3π/2 −i2π 2j

. .x

.y

.r(π/4)

.r′(π/4)

. . . . . .




(b) Find r′(t)


. . . . . .

Solution



t r(t)0 2j

π/2 iπ −2j

3π/2 −i2π 2j

. .x

.y

.r(π/4) .r′(π/4)

. . . . . .

Integrals of vector-valued functionsDefinitionLet r be a vector function defined on [a, b]. For each whole numbern, divide the interval [a, b] into n pieces of equal width ∆t. Choose apoint t∗i on each subinterval and form the Riemann sum

Sn =n∑

i=1

r(t∗i )∆t

Then define∫ b

ar(t) dt = lim

n→∞Sn = lim

n→∞

n∑i=1

r(t∗i )∆t

= limn→∞

[n∑

i=1

f(t∗i )∆ti +n∑

i=1

g(t∗i ) ∆tj +n∑

i=1

h(t∗i )∆tk

]

=

(∫ b

af(t) dt

)i +

(∫ b

ag(t) dt

)j +

(∫ b

ah(t) dt

)k

. . . . . .

FTC for vector functions

Theorem (Second Fundamental Theorem of Calculus)If r(t) = R′(t), then ∫ b

ar(t) dt = R(b) − R(a)

Example

Given r(t) = ⟨t, cos 2t, sin 2t⟩, find∫ π

0r(t) dt.

Answer ⟨∫ π

0t dt,

∫ π

0cos 2t dt,

∫ π

0sin 2t dt

⟩=

⟨π2

2, 0, 0

⟩

. . . . . .

Arc Length (not Curvature)Learning objectives for Section 10.3

◮ To understand and be able to apply the arc length of a curve,r(t) = ⟨f(t), g(t), h(t)⟩, over an interval a ≤ t ≤ b,

L =

∫ b

a|r′(t)| dt =

∫ b

a

√[f′(t)]2 + [g′(t)]2 + [h′(t)]2 dt.

◮ To understand and be able to apply the arc length function of acurve, r(t) = ⟨f(t), g(t), h(t)⟩, over an interval a ≤ t ≤ b,

s(t) =

∫ t

a|r′(u)| du =

∫ t

a

√[f′(u)]2 + [g′(u)]2 + [h′(u)]2 du.

. . . . . .

Length of a curve

Break up the curve into pieces, and approximate the arc length withthe sum of the lengths of the pieces:

. .x

.y

. . . . . .

Length of a curve

Break up the curve into pieces, and approximate the arc length withthe sum of the lengths of the pieces:

. .x

.y

.L ≈n∑

i=1

√(∆xi)2 + (∆yi)

2

. . . . . .

Sum goes to integralIf ⟨x, y⟩ is given by a vector-valued function r(t) = ⟨f(t), g(t), ⟩ withdomain [a, b], we can approximate:

∆xi ≈ f′(ti)∆ti ∆xi ≈ g′(ti)∆ti

So

L ≈n∑

i=1

√(∆xi)2 + (∆yi)

2 ≈n∑

i=1

√[f′(ti)∆ti

]2+ [g′(ti)∆ti]

2

=n∑

i=1

√[f′(ti)

]2+ [g′(ti)]

2 ∆ti

As n → ∞, this converges to

L =

∫ b

a

√[f′(t)]2 + [g′(t)]2 dt

In 3D, r(t) = ⟨f(t), g(t), h(t)⟩, and

L =

∫ b

a

√[f′(t)]2 + [g′(t)]2 + [h′(t)]2 dt

. . . . . .

Example

ExampleFind the length of the parabola y = x2 from x = 0 to x = 1.

SolutionLet r(t) =

⟨t, t2

⟩. Then

L =

∫ 1

0

√1 + (2t)2 =

√5

2+

14

ln∣∣∣2 +

√5∣∣∣

. . . . . .

Motion in Space: Velocity and AccelerationLearning objectives for Section 10.4

◮ To understand the first derivative as the velocity of a curve andthe second derivative as the acceleration of a curve and be ableto apply these definitions.

. . . . . .

Velocity and Acceleration

DefinitionLet r(t) be a vector-valued function.

◮ The velocity v(t) is the derivative r′(t)◮ The speed is the length of the derivative |r′(t)|◮ The acceleration is the second derivative r′′(t).

. . . . . .

ExampleFind the velocity, acceleration, and speed of a particle with positionfunction

r(t) = ⟨2 sin t, 5t, 2 cos t⟩

Answer

◮ r′(t) = ⟨2 cos(t), 5,−2 sin(t)⟩◮

∣∣r′(t)∣∣ =√

29◮ r′′(t) = ⟨−2 sin(t), 0,−2 cos(t)⟩

. . . . . .

Parametric SurfacesLearning objectives for Section 10.5

◮ To understand and be able to apply the concept of a parametricsurface. A parametric surface may be defined by a vectorequation,

r(t) = x(u, v) i + y(u, v) j + z(u, v) k

or by a set of parametric equations

x = x(u, v) y = y(u, v) z = z(u, v),

where u and v are variables with a domain D contained in R2.◮ To understand and be able to represent surfaces of revolutions

parametrically.

. . . . . .

Surfaces with easy parametrizations

◮ graph◮ plane◮ sphere◮ surface of revolution

. . . . . .

Parametrizing a graph

ExampleParametrize the surface described by

z = x2 + y2, −2 ≤ x ≤ 2, −3 ≤ y ≤ 3

SolutionLet

x = u, y = v, z = u2 + v2

on the domain −2 ≤ u ≤ 2, −3 ≤ v ≤ 3.

. . . . . .

parametrizing a sphere

ExampleFind a parametrization for the unit sphere.

SolutionUse spherical coordinates. Let:

x = cos θ sin φ, y = sin θ sin φ, z = cos φ

The domain of parametrization is 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π.

. . . . . .

Parametrizing a surface of revolution

ExampleFind a parametrization of the surface described by

x2 − y2 + z2 = 1, −3 ≤ y ≤ 3

SolutionThe surface is the graph of z2 − y2 = 1 revolved around the y-axis. So let

x =√

1 + y2 cos θ, y = u, z =√

1 + y2 sin θ

The domain is −3 ≤ u ≤ 3, 0 ≤ θ ≤ 2π.

. . . . . .

Outline






. . . . . .

Functions of Several VariablesLearning objectives for Section 11.1

◮ To understand functions of several variables and be able torepresent these functions using level sets.

. . . . . .

A contour plot is a topographic map of a graph

Consider the graph z =√

x2 + y2 Intersect the cone with planesz = c and what do you get? Circles. A contour plot shows evenlyspaced circles.

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

-2

0

2

-2

0

2

0

1

2

3

4

-2

0

2

. . . . . .

The paraboloid

ExampleGraph z = x2 + y2.

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

-2

0

2

-2

0

2

0

5

10

15

-2

0

2

. . . . . .

The hyperbolic paraboloid

ExampleGraph z = x2 − y2.

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

-2

0

2

-2

0

2-5

0

5

-2

0

2

. . . . . .

Plotting a Cobb-Douglas function

ExamplePlot z = x1/2y1/2.

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3

0

1

2

3 0

1

2

3

0

1

2

3

0

1

2

. . . . . .

Utility Functions and indifference curves

◮ If u is a utility function, a level curve of u is a curve along whichall points have the same u value.

◮ We also know this as an indifference curve

. . . . . .

Limits and ContinuityLearning objectives for Section 11.2

◮ To understand and be able to apply the concept of a limit of afunction of several variables.

◮ To understand and be able to apply the definition of continuityfor a function of several variables.

. . . . . .

DefinitionWe write

lim(x,y)→(a,b)

f(x, y) = L

and we say that the limit of f(x, y) as (x, y) approaches (a, b) isL if we can make the values of f(x, y) as close to L as we like by takingthe point (x, y) to be sufficiently close to (a, b).

. . . . . .

easy limits

◮ lim(x,y)→(a,b)

x = a


y = b


c = c

. . . . . .

Like regular limits, limits of multivariable functions can be◮ added◮ subtracted◮ multiplied◮ composed◮ divided, provided the limit of the denominator is not zero.

. . . . . .

Limit of a Polynomial

ExampleFind lim

(x,y)→(5,−2)(x5 + 4x3y − 5xy2)

Solution

lim(x,y)→(5,−2)

(x5 + 4x3y − 5xy2) = (5)5 + 4(5)3(−2) − 5(5)(−2)2

= 3125 + 4(125)(−2) − 5(5)(4)

= 2025.

. . . . . .

Limit of a Rational Expression

ExampleCompute

lim(x,y)→(1,2)

x2

x2 + y2 .

Solution

lim(x,y)→(1,2)

x2

x2 + y2 =(1)2

(1)2 + (2)2

=15

. . . . . .

What can go wrong?

The only real problem is a limit where the denominator goes to zero.

◮ If the numerator goes to some number and the denominatorgoes to zero then the quotient cannot have a limit.

◮ If on the other hand the numerator and denominator both go tozero we have no clue.

. . . . . .

Showing a limit doesn’t exist

TheoremSuppose lim

(x,y)→(a,b)f(x, y) = L. Then the limit of f as (x, y) → (a, b) is L

along all paths through (a, b).

There are two contrapositives to this statement:◮ If there is a path through (a, b) along which the limit does not exist,

the two-dimensional limit does not exist◮ If there are two paths through (a, b) along which the limits exist but

disagree, the two-dimensional limit does not exist

. . . . . .

Continuity

DefinitionA function f of two variables is called continuous at (a, b) if

lim(x,y)→(a,b)

f(x, y) = f(a, b).

We say f is continuous on D if f is continuous at every point (a, b)in D.

. . . . . .

Partial DerivativesLearning objectives for Section 11.3

◮ To understand and be able to apply the definition of a partialderivative.

◮ To be able to compute partial derivatives.◮ To understand and be able to apply Clairaut’s Theorem. If f is

defined on a disk D that contains the point (a, b) and thefunctions fxy and fyx are continuous on D, then

fxy(a, b) = fyx(a, b).

◮ To understand the idea of a partial differential equation, and tobe able to verify solutions to partial differential equations.

. . . . . .

DefinitionLet f(x, y) be a function of two variables. We define the partial

derivatives∂f∂x

and∂f∂y

at a point (a, b) as

∂f∂x

(a, b) = limh→0

f (a + h, b) − f (a, b)h

∂f∂y

(a, b) = limh→0

f (a, b + h) − f (a, b)h

In other words, we temporarily treat the other variable as constantand differentiate the resulting one-variable function as in Calculus I.

. . . . . .

ExampleLet f(x, y) = x3 − 3xy2. Find its partial derivatives.

SolutionWhen finding

∂f∂x

, we hold y constant. So

∂f∂x

= 3x2 − (3y2)∂

∂x(x) = 3x2 − 3y2

Similarly,∂f∂y

= 0 − 3x(2y) = −6xy

. . . . . .

Second derivatives

If f(x, y) is a function of two variables, each of its partial derivativesare function of two variables, and we can hope that they aredifferentiable, too. So we define the second partial derivatives.

∂2f∂x2 =

∂

∂x

(∂f∂x

)= fxx

∂2f∂y ∂x

=∂

∂y

(∂f∂x

)= fxy

∂2f∂x ∂y

=∂

∂x

(∂f∂y

)= fyx

∂2f∂y2 =

∂

∂y

(∂f∂y

)= fyy

. . . . . .

Don’t worry about the mixed partials

The “mixed partials” bookkeeping may seem scary. However, we aresaved by:

Theorem (Clairaut’s Theorem/Young’s Theorem)If f is defined near (a, b) and fxy and fyx are continuous at (a, b), then

fxy(a, b) = fyx(a, b).

The upshot is that we needn’t worry about the ordering.

. . . . . .

Example (Continued)Let f(x, y) = x3 − 3xy2. Find the second derivatives of f.

SolutionWe have

fxx = (3x2 − 3y2)x = 6x

fxy = (3x2 − 3y2)y = −6y

fyx = (−6xy)x = −6y

fyy = (−6xy)y = −6x

Notice that fyx = fxy, as predicted by Clairaut (everything is apolynomial here so there are no concerns about continuity). The factthat fxx = fyy is a coincidence.

. . . . . .

Partial Differential Equations

DefinitionA partial differential equation (PDE) is a differential equationfor a function of more than one variable. So the derivatives involvedare partial derivatives.

ExampleLet f(x, y) = x3 − 3xy2. Show that f satisfies the Laplace equation

∂2f∂x2 +

∂2f∂y2 = 0

SolutionWe already showed that fxx = 6x and fyy = −6x. So fxx + fyy = 0.

. . . . . .

Tangent Planes and Linear Approximations ILearning objectives for Section 11.4

◮ To understand the concept of a tangent plane to a surfacez = f(x, y) and to be able to compute the equation of tangentplanes

◮ To understand and be able to find a linear approximation to afunction z = f(x, y).

. . . . . .

Tangent Planes and Linear Approximations IILearning objectives for Section 11.4

◮ To understand the concept of a tangent plane to a parametricallydefined surface

r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k

and to be able to compute the equation of tangent planes.

. . . . . .

Tangent Planes and Linear Approximations IIILearning objectives for Section 11.4

◮ To understand and be able to find the differential to a functionz = f(x, y),

dz =∂z∂x

dx +∂z∂y

dy

To be able to use the differential to estimate maximum error.

. . . . . .

DefinitionIf f is a function and P0 = (x0, y0, z0 = f(x0, y0) a point on its graph,then

◮ The linearization of f near (x0, y0) is the function

L(x, y) = f(x0, y0) +∂f∂x

(x0, y0)(x − x0) +∂f∂y

(x0, y0)(y − y0)

◮ The tangent plane to the graph at P0 is the graph of L

The linearization is the best possible linear approximation to afunction at a point.

. . . . . .

ExampleFind an equation of the tangent plane to the surface z = y cos(x − y)at (2, 2, 2).

SolutionWe have

∂z∂x

= −y sin(x − y),∂z∂y

= cos(x − y) + y sin(x − y)

So∂z∂x

∣∣∣∣(2,2)

= 0,∂z∂y

∣∣∣∣(2,2)

= 1

Therefore the equation of the tangent plane is

z = 2 + 0(x − 2) + 1(y − 2) = y

. . . . . .

ExampleThe number of units of output per day at a factory is

P(x, y) = 150[

110

x−2 +910

y−2]−1/2

,

where x denotes capital investment (in units of $1000), and y denotesthe total number of hours (in units of 10) the work force is employedper day. Suppose that currently, capital investment is $50,000 and thetotal number of working hours per day is 500. Estimate the change inoutput if capital investment is increased by $5000 and the number ofworking hours is decreased by 10 per day.

. . . . . .

Solution

∂P∂x

(x, y) = 150(−1

2

) [110

x−2 +910

y−2]−3/2 (

−210

)x−3

= 15[

110

x−2 +910

y−2]−3/2

x−3

∂P∂x

(50, 50) = 15

∂P∂y

(x, y) = 150(−1

2

)[110

x−2 +910

y−2]−3/2 (

910

)(−2)y−3

= 15[

110

x−2 +910

y−2]−3/2

x−3

∂P∂y

(50, 50) = 135

So the linear approximation is L = 7500 + 15(x − 50) + 135(y − 50)

. . . . . .

Solution, continued

So the linear approximation is

L = 7500 + 15(x − 50) + 135(y − 50)

If ∆x = 5 and ∆y = −1, then

L = 7500 + 15 · 5 + 135 · (−1) = 7440

The actual value isP(55, 49) ≈ 7427

So we are off by13

7427≈ 1.75%

. . . . . .

Tangent planes to parametrized surfaces

◮ If r(u, v) parametrizes a surface S, and P0 = r(u0, v0), then thevector

ru(u0, v0) = limh→0

r(u0 + h, v0) − r(u0, v0)

h

is tangent to S at P.◮ If r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, then

ru(u0, v0) =∂x∂u

(u0, v0)i +∂y∂u

(u0, v0)j +∂z∂u

(u0, v0)k

◮ We have another tangent vector

rv(u0, v0) =∂x∂v

(u0, v0)i +∂y∂v

(u0, v0)j +∂z∂v

(u0, v0)k

◮ Their cross product is normal to the tangent plane.

. . . . . .

ExampleFind an equation of the tangent plane to the parametric surface

r(u, v) = u2 i + 2u sin v j + u cos v k

at the point u = 1 and v = 0.

SolutionThe point we are finding the tangent plane at is r(1, 0) = ⟨1, 0, 1⟩. Thetangent plane is spanned by the two vectors ru(1, 0) and rv(1, 0):

ru(u, v) = ⟨2u, 2 sin v, cos v⟩ =⇒ ru(1, 0) = ⟨2, 0, 1⟩rv(u, v) = ⟨0, 2u cos v,−u sin v⟩ =⇒ rv(1, 0) = ⟨0, 2, 0⟩

So a normal vector to the tangent plane is

ru(1, 0) × rv(1, 0) = ⟨2, 0, 1⟩ × ⟨0, 2, 0⟩ = ⟨−2, 0, 4⟩

This means an equation for the tangent plane is

−2(x − 1) + 4(z − 1) = 0.

. . . . . .

Differentials

This is really just another way to express linear approximation.Define new variables dx = ∆x, dy = ∆y, dz. Then the equation forthe tangent plane through (x0, y0, z0 = f(x0, y0)) is

dz =∂f∂x

dx +∂f∂y

dy

And the concept that this is a good linear approximation is expressedas

∆z = f(x + dx, y + dy) − f(x, y) ≈ dx

when dx and dy are “small enough”

. . . . . .

ExampleIf z = 5x2 + y2 and (x, y) changes from (1, 2) to (1.05, 2.1), comparethe values of ∆z and dz.

SolutionWe have

dz = 10x dx + 2y dy

When x = 1, y = 2, dx = 0.05, and dy = 0.1, we get dz = 0.9. On theother hand

z(1.05, 2.1) = 9.9225

So ∆z = 0.9225. The difference is 0.0225, or 2.4%.

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