Math 127 Introduction and Review - University of Kansass890m022/Math127_F18/Intro_and_Review.pdfux, pressure, velocity, and acceleration. Geometric Representations of Vectors: a segment

Math 127

Introduction and Review

(1) Recap of Differential Calculus and Integral Calculus

(2) Preview of Calculus in three dimensional space

(3) Tools for Calculus 3

MATH 127 Introduction to Calculus III The University of Kansas 1 / 22

Differential Calculus

Types of problems in Differential Calculusf : R→ R , y = f (x)

Slope of the tangent line (rate of change) at (a, f (a)) is f ′(a)

Area under the graph of y = f (x) between two x values is

∫ b

af (x) dx


Differential Calculus

Optimization Arc Length∫ b

a

√1 +

(dy

dx

)2

dx


Multivariable Calculus

Scalar Valued Functionsf : R2 → R, z = f (x , y), w = g(x , y , z)

Examples1 z = x2 + y2

2 w = exyz

Partial Derivative Line Integral


Multivariable Calculus

Optimization Volume


Vector Valued Functions

f : R→ R2, r(t) = 〈x(t), y(t), z(t)〉

f : R2 → R3, r(u, v) = 〈x(u, v), y(u, v), z(u, v)〉


Vector Field

f : R2 → R2 or f : R3 → R3


Vectors

Definition

A vector is a geometric quantity described by magnitude (length) anddirection. A scalar is a constant in R which has no direction and ismagnitude only.

Example: Force, flux, pressure, velocity, and acceleration.

Geometric Representations of Vectors: a segment AB from A to Bwith an arrow beginning at A and ending at B.

A - Initial Point B - Terminal Point

Notation: ~v =−→AB or v= ~AB

Only the magnitude and direction are relevant to a vector. It’s location isNOT relevant. Vectors are invariant under translation.


Vector Addition and Scalar Multiplication

Vector Addition

Vectors ~u and ~v are summed throughthe Parallelogram Law.

Scalar Multiplication

When a vector ~v and a scalar c aremultiplied, the direction remainsunchanged and the magnitude of ~v ismultiplied by c .


Cartesian Representation of Vectors

Components of a Vector ~v

Position the initial point of the vector ~v at the origin of the Cartesiancoordinate system. The coordinates of the terminal point of ~v will be thecomponents of ~v .

~v =−→OP = 〈a, b, c〉

In general, if ~v =−→AB where

A(x1, y1, z1) and B(x2, y2, z2) then

~v = 〈x2 − x1, y2 − y1, z2 − z1〉


Vector Addition and Scalar Multiplication

If ~v = 〈a1, b1, c1〉 and ~u = 〈a2, b2, c2〉 and k is a scalar, then

(I) ~v + ~u = 〈a1 + a2, b1 + b2, c1 + c2〉 = ~u + ~v

(II) k~v = 〈ka1, kb1, kc1〉(III) ~v − ~u = 〈a1 − a2, b1 − b2, c1 − c2〉 = − (~u − ~v)

The zero vector is ~0 = 〈0, 0, 0〉.~i = 〈1, 0, 0〉 ~j = 〈0, 1, 0〉 ~k = 〈0, 0, 1〉

Algebraic Properties

〈a, b, c〉 = a~i + b~j + c~k

~v + ~u = ~u + ~v

(~v + ~u) + ~w = ~v + (~u + ~w)

~v +~0 = ~v = ~0 + ~v

~v − ~v = ~0

k (~v + ~u) = k~v + k~v

(k1 + k2)~v = k1~v + k2~v

(k1k2)~v = k1(k2~v)MATH 127 Introduction to Calculus III The University of Kansas 11 / 22

Magnitude of a Vector

Magnitude

The magnitude (or length) of a vector ~v = 〈a, b, c〉 is

|~v | =√a2 + b2 + c2

If ~v =−→AB with A(x1, y1, z1) and B = (x2, y2, z2), then

|~v | =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

The zero vector ~0 = 〈0, 0, 0〉 is the only vector with magnitude zero.

Unit Vector

A unit vector is a vector of magnitude one.

If ~v = 〈a, b, c〉 6= ~0, then the unit vector with the same direction of ~v is

−→U ~v =

~v

|~v |=

1√a2 + b2 + c2

〈a, b, c〉


In addition to the vector addition and scalar multiplication, there are otheroperations among vectors. In this course we will discuss two types ofvector products:

(I) The Dot Product,

(II) The Cross Product.

The dot product of two vectors ~v and ~u is thenumber given by

~v · ~u = |~u| |~v | cos (θ)

where θ is the acute angle between the vectors ~vand ~u.


Orthogonal

Two vectors ~v and ~u are orthogonal to each other, denoted ~v ⊥ ~u, if andonly if θ = π

2 if and only if ~v · ~u = 0.

~v · ~u is positive if and only if 0 ≤ θ < π2 ,

~v · ~u is negative if and only if π2 < θ ≤ π

Dot Product in Component NotationSuppose that ~v = 〈a1, b1, c1〉 and ~u = 〈a2, b2, c2〉.

Recall the Cosine Law

|~v − ~u|2 = |~v |2 + |~u|2 − 2 |~v | |~u| cos (θ)

= |~v |2 + |~u|2 − 2~v · ~u

~v · ~u = a1a2 + b1b2 + c1c2


Example: Find the angle between ~v = 〈−1, 2, 0〉 and ~u = 〈3, 1, 7〉.Solution:

~v · ~u = |~v | |~u| cos (θ)

cos (θ) =~v · ~u|~v | |~u|

=(−1)(3) + (2)(1) + (0)(7)√

(−1)2 + (2)2 + 02√

32 + 12 + 72=−1√295

θ = arccos

(−1√295

)>π

2


Properties of Dot Product~v · ~u = ~u · ~v~u · (~v ± ~w) = ~u · ~v ± ~u · ~w(k~u) · ~v = k (~u · ~v)

~0 · ~v = 0

~v · ~v = |~v |2

Example:

Suppose that−→OR ⊥

−→PQ and

−→OS ⊥

−→PQ, show that

−→RS ⊥

−→PQ.

Solution: Since the vectors are orthogonal,

−→OR ·

−→PQ = 0 =

−→OS ·

−→PQ

As−→RS =

−→OS −

−→OR we have

−→RS ·

−→PQ =

(−→OS −

−→OR)·−→PQ =

−→OS ·

−→PQ −

−→OR ·

−→PQ = 0


The cross product of two vectors ~v and ~u, denoted ~v × ~u, is the vector

~v × ~u = (|~v | |~u| sin (θ)) ~n

where

(i) θ is the angle between ~v and ~u,

(ii) ~n is the unit vector, so that ~n ⊥ ~v , ~n ⊥ ~u, and the vectors (~v , ~u, ~n)satisfies the right-hand rule.


~v × ~u = (|~v | |~u| sin (θ)) ~n

Properties of the Cross ProductIf ~v and ~u are parallel, then ~v × ~u = ~0.

(~v × ~u) ⊥ ~v and (~v × ~u) ⊥ ~u.

~v × ~u = −~u × ~v

|~v × ~u| is the area of the parallelogram determined by ~v and ~u whichis twice the area of the triangle with edges ~v and ~u.


Cross Product in Component Form

~i × ~j = ~k ~j × ~k = ~i ~k × ~i = ~j

If ~v = 〈a1, b1, c1〉 and ~u = 〈a2, b2, c2〉 then we can distribute over thecross product:

~v × ~u =(a1~i + b1~j + c1~k

)×(a2~i + b2~j + c2~k

)= (b1c2 − b2c1)~i − (a1c2 − a2c1)~j + (a1b2 − a2b1)~k

Alternatively, you can consider the determinant of the matrix:

~v × ~u =

∣∣∣∣∣∣~i ~j ~ka1 b1 c1a2 b2 c2

∣∣∣∣∣∣=

∣∣∣∣ b1 c1b2 c2

∣∣∣∣~i − ∣∣∣∣ a1 c1a2 c2

∣∣∣∣~j +

∣∣∣∣ a1 b1a2 b2

∣∣∣∣ ~k= (b1c2 − b2c1)~i − (a1c2 − a2c1)~j + (a1b2 − a2b1)~k


Example:

(I) Find a vector which is orthogonal to the plane through the pointsP(2, 0,−3), Q(3, 1, 0), and R(5, 2, 2).

(II) Find the area of the triangle with vertices at P, Q, and R.

Solution: Let ~u =−→PR = 〈3, 2, 5〉 and ~v =

−→PQ = 〈1, 1, 3〉.

(I) ~u × ~v =

∣∣∣∣∣∣~i ~j ~k3 2 51 1 3

∣∣∣∣∣∣ = 〈1,−4, 1〉 is

orthogonal to the plane.

(II) The area of the triangle is12 |~u × ~v | = 3

√2

2


Example:

1 Find the algebraic equation of the plane containing the points (1,0,1),(0, -1, -1), and (1, -1, 0).

2 Find the parametric equations of the line perpendicular to the planein (1), passing through the point (1,0,1).

Solution:Let P = (1, 0, 1), Q = (0,−1,−1) and R = (1,−1, 0)

⇒ u = ~PQ = 〈−1,−1,−2〉 and v = ~PR = 〈−1, 0,−1〉

u and v lie in the required plane. The normal vector of that plane is

u× v =

~i ~j ~k−1 −1 −2−1 0 −1

= 〈1, 1,−1〉

⇒ The equation of the plane is 〈1, 1,−1〉 · (〈x , y , x〉 − 〈1, 0, 1〉) = 0⇒ (x − 1) + (y − 0)− (z − 1) = 0 or x + y − z = 0


Example continued

Vector Equation: r(t) = (1, 0, 1) + t〈1, 1,−1〉 = 〈1 + t, t, 1− t〉

Parametric Equations:x(t) = 1 + ty(t) = tz(t) = 1− t

Symmetric Equations: x − 1 = y = 1− z


Documents

Math 127 Introduction and Review - University of Kansass890m022/Math127_F18/Intro_and_Review.pdfux, pressure, velocity, and acceleration. Geometric Representations of Vectors: a segment