Bear H.S. Understanding Calculus

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192

Understanding Calculus

Figure

31.12

(

IIFII-'-+-

=100

2 2

IIF211

= 200/(1 +V3)

73.2(lbs),

89.7(lbs).

As a check we calculate

IHIII

= 89.7 cos 45° = 63.4;

IIV11I

= 89.7 sin 45° = 63.4;

IIH

211

= 73.2 cos 30° = 63.4;

IIV211

= 73.2 sin 30° = 36.6.

The horizontal components have equal magnitude,

HI= - H

,and the sum

the vertical components

100

IIV

V211

100

PROBLEMS

31.1 Write

in terms

ofi

and j and find the norm

IIPQII

(a) P = (2, 4), Q=

(-1,

(b) P

(0,0),

Q=(3,

-2)

(c)

P=(5,2),Q=(I,-4)

(d) P = (1, 7), Q= (6,

-5)

31.2 Find the point R = (x,y) on the segment from P = (3, 1) to Q=

(-1,

5) so that R is

the

way from

P to

31.3 Find the coordinates

the point R on the line through P = (2, I) and Q= (5, 0) so that P

and R are on the same side

and the distance from Qto R is twice the distance from

QtoP

31.4 Find a unit vector A which is parallel to i - 2j.

31.5 Find a unit vector which is parallel to a line with slope

31.6 Find a unit vector parallel to a line with slope 3.

31.7 Find a unit vector which is parallel to a line which makes an angle 8 with the x-axis.

Chapter31 •

Vectors

V = 200 J

()= ?

IIF II = ?

193

50 i

200

Figure 31.13

31.8 What is the slope

ofa

lineparallelto v =3i + 7j?

31.9 Checkthat the line 3y =4x + 4 goesthroughthe point (2, 4).

(a) Finda unit vectorparallelto the line.

(b) Findthe twopointson the line whichare a distance5 units from(2, 4).

31.10 The segmentfrom (I, 1)to (2, 3) and the segmentfrom (I, I) to (4, 2) are adjacentsides

of a parallelogram. What is the fourthvertex?

31.11 Let A, B, C, D, E, F, G, H be vectorsalongconsecutive sides of a regularoctagonwith

sidesoflength

1. Let A = i, B =

i -

j, C =

-j,

etc.

(a) Drawthe figureand write

D, E, F, G, and H in terms

ofi

and j. Checkthat the

sum

isO.

(b) WriteD, E, F, G, and H in terms of A and B.

31.12 In Figure 31.I I it appears that the three medianvectors intersect in a point. Showthat

this is the case,and the point is

3of the way alongeach. Hint:It sufficesto showthat for

instance

fiQ

fiR

3R.

(Thepoint of intersection is calledthe centroid of the tri-

angle.)

31.13 A bucketof wetconcrete

weighing

200 lbs. hangsfroma crane. A controlrope puts a 50

lb. horizontal force on the bucket, pulling the suspension cable

away

from the vertical

(Figure

31.I 3). What is the tension

IIFII

in the cable,and whatangle does the cablemake

with the vertical?

31.14 An airlinermaintains a headingof 60

(60

east of north). The plane's speedthroughthe

air is 500mph, andthe wind at cruisingaltitudeis 100mph

toward

the east.What is the

plane's true course (degreeseast of north) and what is its speed overthe ground?Hint:

The plane's velocitythroughthe air is 500(sin 60

i +cos 60

j) and the wind's velocity

is 100

The

Dot

Product

The dot product of two vectorsA and B is define by

A . B

IIAIIIIBII

cos

(J,

where (Jis the angle betweenA and B. The dot product is sometimes also called the scalar

product

sincethe result is a

number-a

scalar.

It is clear from the definition that B . A is the same as A . B, so the dot product is a

commutative operation. It is also easy to see that scalars behavenicely with the dot product,

so that

(aA) · B =a(A . B) =A . (aD).

IfC

is a unit vector,

IICII

= 1,then A . C =

IIAII

cos (Jis the lengthof the projectionof A

on C (Figure32.1a).This projectionis a signednumber,so if

< (J S 'IT with cos (J< 0, then

A . C is a negativenumber (Figure32.1b).

FromFigure32.2we see that the projectionof A +

Bon

C is the sum of the projections

A on C and B on C. Therefore,if C is a unit vector,

(A + B) . C = A . C + B . C.

(32.1)

If C is not a unit vector, then both sides of (32.1) are multipliedby the length of C, so (32.1)

holds for any vector C. Sincethe dot productis commutative we also have from (32.1)

C . (A

+B) =C . A +C . B =A · C +

B·C.

Two

vectors A and B are perpendicular if the angle betweenthem is !, so cos (J=

HenceA and B are perpendicularif and only if A . B =

Of course if A =0 or B =0 then

A . B = 0, so we agree for convenience that the zero vector is perpendicularto every

vector.

In particular, i andj are not only unit vectors,they are perpendicularto each other,so

i·i=j·j=1

i . j =J: i =

(32.2)

195

196

I-IIA

cos

--I

(a)

Understanding Calculus

- - - _.-1

.'--

I-IlAllcose-l

(negative)

(b)

Figure 32.1

(A + B) • C =A • C + B • C

Figure

32.2

use

(32.2)

calculate

A·

B when A and B are

expressed

in terms of i and i. If

ali

+a2i andB =bli +b2i, then

A . B =

(ali

+ a2i) · (bli +b2i)

=a.b, i · i +a

i · i +a

i . i +a

i . i

alb

•

If A =

ali

+a2i, thenB =

-a

i +

ali

is perpendicular to A since

A · B

(ali

+a2i) .

(-a

i +

ali)

-a

If A =aIi +a2i is a unit

vector,

= 1,thenof

course

B =

-a

i +aIi is alsoa unit

vector.

HereB is obtained byrotating A

through

90°in the

positive

direction

(Figure

32.3).

EXAMPLE 32.1

Findthe angle 8betweenthe vectorsA =5i + 2j and B = 3i - j.

Solution

Since

IIAII

= V29 and

IIBII

vTO,

the vectors

Aand

B are unit vec-

tors and their dot productgivesthe cosine:

Chapter 32 • The Dot Product

Figure 32.3

1 1 1

cos 8

V29

A .

\7iQB

= Vi90 A . B

cos 8 =

:vi9O

(5i +2j) . (3i - j)

1 13

:vi9O

·3

+2(-1)) =

Vi9O.

From the calculator we get 8

40.2

or 8

.702 radians.

197

EXAMPLE

32.2

Find the angle between the line through P = (1, 3) and Q= (6, 1), and the line through R =

(-1,3)

and

S = (2, 0). (Figure 32.4).

Solution

~ ~

PQ and RS are vectors parallel to the lines, and

PQ=

5i-2j

RS=3i-3j.

-2

Figure 32.4

198 Understanding Calculus

Wefind unit vectorsalongthese vectorsand take the dot product:

1 1

cos 8

= V19 (5i -

2j)·

v18

(3i - 3j)

1 21

~(15+6)=~.

The calculatorgives 8

23.2° or 8

.405 radians.

EXAMPLE

32.3

Finda unit vectoru parallelto A = 4i + 3j and a unit vectorv perpendicular to A. Write a generalvec-

tor B

= xi +yj in the form au +by.

Solution

Since

IIAII

= 5, we can take

143

u =

-(4i

+3j) =

-i

-j

5 5

A unitvectorperpendicular to A and u is

The component of B along u will havelengthB . u and the component of B along v will have length

B . v. Therefore

B=xi+yj

= (B . u) u + (B . v) v

=(x i+Y

fj)

U +(x i+y

(-fi

~j)

(~x+

~y)

U +

(-~x+

~y)

The kind of calculation in Example3 is useful if you wantto introduce a newcoordinate system

with the new axes rotated through an angle 8. (Figure

32.5). Let x and y denote the coordinates of a

point in the rotated

system.

As we sawin Example3, the u-component of

i5P

(i.e., the x-component)

will havelengthOi": u. Hence

x=i5P·u

= (xi +yj) . (cos 8 i +sin 8 j)

= x cos 8 +Y sin 8.

Similarly,

y=i5P·v

= (xi +yj) . (-sin 8 i +cos 8 j)

-x

sin 8+Y cos 8.

Here are the rotationequations:

x= x cos 8+Y sin 8

y =

-x

sin 8+ Y cos 8.

(32.3)

Chapter32 • TheDot Product

Figure 32.5

199

Sincethe x andy coordinates are obtained fromthe X,y

coordinates

by rotating

through

the angle-8,

weuse

(32.3)

with

-8

in placeof 8to get

x=xcos

8-ysin

y=xsin

8+ycos

EXAMPLE

32.4

Show

thatthe curvexy = 1is the

hyperbola

!(x

y2)

= 1 rotated

through

45°.

Solution

If the coordinate systemis rotated45°the newcoordinates satisfy

V2 V2

x =

- -

y-2-

V2 V2

y=x-

+Y-2-·

WegettheX,y equation for the curveby substituting the

above

formulas

inx andy forx andy:

xy=

(

v'2

)(x

V2 +

V2)

= 1

2 2 2 2 '

1 1

"29=

(32.4)

recognize

the last equation as thatof a

hyperbola,

soxy = 1 is this

hyperbola

rotated

through

45°.

As a final application, which

shows

dramatically how vectornotation simplifies manycalcula-

tions,we

derive

the law of cosines. LetA and B be two

vectors

fromthe

origin,

with

IIAII

=a,

IIBII

=b.

Let C=A - B be thethirdsideof the triangle, with

IICII

=c.

(Figure

32.6).

Then

C . C = (A - B) . (A - B),

C . C = A . A +B . B - 2A . B,

= a

+ b

2ab cos 8.

Formula

(32.5)

is the lawof

cosines.

(32.5)

200

Understanding Calculus

Figure 32.6

PROBLEMS

Findthe anglebetweenA and B, and find the lengthof the projectionof A on B.

32.1

A = i +j ; B =

-i

+j.

32.2 A =3i - 2j B =5i +j.

32.3 A =

-2i

+j ; B =3i +4j.

32.4 A =6i +2j ; B = 12i+5j.

Findthe vectorA = ai +hj whichhas the givenproperties.

32.5

A makesan angleof 45° with the

X-axis,

and

IIAII

= 3.

32.6 A is a unitvectoralongthe line

y = 3x - 2.

32.7

Ais perpendicular to 2i +j and

IIAII

= 2.

32.8 Find the angle betweenthe line through (0, 0) and (3, 4) and the line through(3, 4) and

(5,0).

32.9 Findthe anglebetweenthe linesy =2x+3 andy =

-x

+ 1.

32.10 Let

A =2i +2j and C =4i +j. WriteA as the sumAl +

whereAl is parallelto C and

is perpendicular to C.

32.11 Let

Xand Ybe

axesof a rotatedcoordinate system,and let u =

vb-

i +

j be a unit

vectoralongthe

X-axis.

(i) Whatwas the angleof rotation? _

(ii) Let v be the unit vectoralongthe Y-axis. Findxandy suchthat 6i - 5j = xu + yv.

32.12

Verify

the identity: IIA+ BII

-IIA

- BII

= 4A . B.

32.13 Let A and B be perpendicular, and C = A +B. Calculate(A + B) . (A +B) to verifythe

Pythagorean

Theorem:

IICII

IIAII

IIB11

•

32.14 Show that every angle inscribed in a semicircle is a right angle. Hint: In Figure 32.7,

writeC and D in terms of A and B and showthat C . D

-A

Figure 32.7

Lines and Planes in Space

Tointroduce coordinates into 3-space, weput another

axis-the

z-axis-perpendicular to the

xy-plane

asshowninFigure33.1.Thepointsinthexy-plane nowhavecoordinates (x,y, 0),and

the point (x, y, z) is

Izl

units up or downfrom (x, y, 0). The three coordinate planes-the xy-

plane,the

xz-plane,

and theyz-plane--divide3-spaceup intoeightoctants. Thefirst octant

consists of the points(x,y, z) withpositive coordinates, andthe otheroctants are

unnamed.

An equation of the form F(x,y, z) = 0 or z = fix, y) generally represents a surface in

space.

For

example,

the equation z =1is the equation of a planeparallelto thexy

plane.

(Fig-

ure

33.2).

The equation x =z is the plane whichmakes an angle of 45° with the xy andyz

planes.

(Figure

33.3).

Vectors

in space are again represented by directedline

segments.

We add a new unit

vector,

k, along the z-axis, so any vectorin spacehas a representation of the form ai + bj +

ck. In particular, the vectorR fromthe originto the point(x,y, z) is

R = xi + yj + zk,

andthe magnitude or

norm

of R is

(Figure

33.1)

IIRII

+Z2.

Vector

addition, and multiplication by a

scalar,

and the dot product

behave

just as for

plane

vectors.

Thusif A =ali +a

j +a

and B =

bli

j +b

3k,

then

= cali +ca

i +ca

A +B = (a

i +(a

)

j +(a

)

A · B =

IIAllllBllcos

(J=

a.b

; +a

•

The straightlinethrough

=(x

Yo,

zo)

andparallelto A =

+mj +nk canbe repre-

sentedby the vectorfunction

R(t) = OP

+ tA

=xoi+yoj +zok + t(t i +m j +n k)

=(x

tt)

i +

(vo

+tm) j +

(zo

+tn) k.

201

202

Understanding

Calculus

,z)

J'E.:-----------j---;---

Figure

33.1

z = l

Figure 33.2

The plane x = z

Figure

33.3