Hints and Answers to Exercises

Exercises 1.3

1 r;;- l 3 1.1. (a) 11, (b) - 7 , (c) -4, (d) v5, (e) 2(b- a), (f) 5•

(g) t· (h) 1, (i) 1.

1.2. Terminal speed 8 m s- 1, 1.24 seconds.

1.3. Use f(x Z h) - f(x) and Iet h ~ 0, so

1.4.

(x + h )2 - x2 x2 + 2hx + h2 - x2

(a) h = h = 2x + h ~ 2x

as h ~ 0.

(b) cos(x + ~) - cosx

cosx cosh - sinx sinh - cosx h

COSX - hsinx - COSX

h

cos(x + h) - cosx . for small h. So h ~ -smx as h ~ 0.

ex+h - ex exeh - e [eh - 1] (c) h = h = e' -h- ~ ex as h ~ 0 since

eh = 1 + h for small h.

1

(ct)~--:x

h ~ 0.

x-x-h hx(x + h)

1 1 x(x + h) ~ -:xz as

_!(y") = lim [[y(x + ~x)]" - [y(x)]") dx at-->0 ~

= lim [ [y(x + ~)]" - [y(x)]" . ~y j at-->0 ~y ~

from which the result follows, provided ~y and ~x tend to zero jointly in a uniform manner, and all the Iimits exist.

1.5. For this function, the derivative f(x) has the definition

160

f'(x) = [ sin x 2x

X< 0 x2:0

This implies that f(x) is continuous at the origin as both sin x and 2x are zero there. The second derivative, however, is as follows:

f"(x) = ~~OS X X< 0 X 2: 0

which is not continuous at the origin. Therefore f'(x) does not exist at the origin.

1.6. (a) ; , min.; (b) 0, max.; ±~/6 both min.; (c) 0, min.;

(d) -.[2: min.; f2, max.

1.7. The function tan x is discontinuous at x = ~· therefore the

conditions of Rolle's Theorem are violated.

1.8. This is Rolle's Theorem in slightly more precise terms than given in Example 1.8. The explanation is similar.

1.9. Rolle's Theorem: If a particle retums to its initial position, there is a time when its speed is zero.

Mean Value Theorem: The average speed is attained at least once by the object.

1 1 1 1 1.10. (a) x - 3Tx3 + 51 x 5; (b) 1 - 21 x2 + 41 x4;

1 1 1 1 (c) x - 2x2 + 3x 3; (d) 1 + -zx - 3x2 •

1.11. The nth derivative of a polynomial is zero. The definition of a polynomial is P(x) = a0 + a1x + a~ + · · · + a._x" for some positive integer n. This is precisely the form of a truncated Maclaurin Series. A polynomial is thus its own Maclaurin Series, with

1 ak = !JP<kl(O) for k = 1, 2, ... , n.

1.12. (a) xlnx = (x - 2 + 2)ln(x - 2 + 2) = (u + 2)ln(u + 2) where u = x - 2. Expanding in terms of u then writing u = x- 2 gives

- ~ ( -1)k k xlnx - 2ln2 + (1 + ln2)(x - 2) + Lk(k _ 1)2k-l (x - 2)

k~O

(b) c1 - 2x)-3 = s-3 L<k + 2)(k + 1) 2~:~ cx + 2)k k~O

(write u = x + 2).

( c) sin x = sin(x - 1r + tr) = - sin(x - tr) hence

. ~ ( _ 1)k+l(x _ 7r)2k+l sm x = L (2k + 1)!

k~O

Note, always use the easiest method of expanding a func­tion (by calculus, binomial theorem, using known series etc.). Theory then teils us that the series found is the Taylor's Series since it is unique.

1.13. [61 is evaluated most straightforwardly by considering

j6l = ~ = 8 ( 1 - i4 ) 112 and expanding using the

binomial theorem. Whence

j6l = 8 1 - 128 - 8 64 - 16 64 - ... ( 3 1 ( 3 )2 1 ( 3 ) 3 )

that is j6l = 7.81025

8 1 25 1.14. (a) 9 , (b) 0, (c) 9 , (d) -1, (e) 0, (f) - 49, (g) a - b,

(h) in a (recall that ! (ax) = ax!n a in using L'Höpital's

Rule), (i) -1 (it is necessary to use L'Höpital's Rule three

times here), (j) -21 (in this example, use that if lim f(x) = C

x_.L

then !im !JW = .jC), (k) 0, (l) 1, (m) 1 (take Iogs and use x_.L

that if !im fix) = C then lim(lnfix)) = ln(C)), (n) 5 (L'Höpital's x~L x~L

Rule is not essential here, but take Iogs if you want to use it).

1.15. The Maclaurin Series for the two parts of the numerator are

sin(tanx) = x + ix3 - 410 x 5 - 1~~8 x7 + ... , and

tan(sinx) = x + ix3 - 4~x 5 - ~gJ0x7 +

The first three terms of each are identical, so when they are

X? subtracted we obtain sin(tanx) - tan(sinx) = - 30 + ....

Hence the limit in the question has the value - 3~ . lt is a

brave student who attempts this by hand; I used computer algebra.

1.16. Start with the calculator's best estimate as first guess. If you started with [i = 1.4142136, then the Newton-Raphson method applied to the equation f(x) = x 2 - 2 yields.

(1.4142136)2 - 2 x 1 = 1.4142136 - 2 X 1.4142136 = 1.41421356

an improvement. Similarly, for fland % the start values are 2.6457513 and 1.8171206 and these are improved after one step of the Newton-Raphson method to 2.645751311 and 1.817120593 respectively.

1.17. With f(x) = x 2 - a, f'(x) = 2x so the Newton-Raphson method yields

- f(xn) - x2 - a . . xn+l - xn - f'(xn) - xn - 2x:-- which gives the result.

1.18. The integral is equal to the Iimit of the function } :t ~: k~l

as n ~ oo which, using the summation given Ieads to the limit

. { 1 n2 1 } 1 hm - - - (n + 1)2 = -. n ... ~ n 4 n3 4

1 7C 1 1 2 1 2 1. 9. (a) 4 - 2 ln2, (b) 2 X 1n X - 4 X + C,

xe' 1 (c) - x + 1 + e' + C (choose v = (x + 1)2 , u = xex).

f 1 n- 1 In = cosnxdx, then In = - cos"- 1 x sinx + -- In_ 2, also n n 1.20.

I 1 3• 3. 3 c d 4 = 4 COS X SlllX + g COSX SlllX + g X + , an

I 1 4 • +4 2 • +8. C 5 = 5 COS X SlllX 15cos X SlllX 15s1nx + .

1.21. In = f sin"xdx, then In = - ~sinn-t X COSX + n -n 1 In-2•

I 1 . 2 2 d 3 = - 3 Sill x cosx - 3 cosx an

I 1.4 4.2 8 c 5 = - S Sill X COSX - 15sill X COSX -lSCOSX + .

1.22. (a) ~ ' (b) In(~), (c) 1 , (d) -2, (e) 2~b ·

16 1.23. 3 7C.

3 1.24. 5 h.

Exercises 2.3 2.1.

2.2.

2.3.

2.4.

2.5.

2.6.

fx = 4x3 + 15x2y + Si, /y = 5x3 + 15xl + 4i, fxx = 12x2 + 30xy, fYY = 30xy + 12l, /xy = 15(x2 + l). fx = (1 + X + Y + z)e(x+y-z),fy = (1 +X + y + z)e(x+y-z), f, = (1 - X - y - z)e(x+y-z),fxy = (2 + X + y + z)e(x+y-zl, J,x -(x + y + z)e(x+y-zl, /y, = -(x + y + z)e<x+y-z), fxx = (2 + X + Y + z)e(x+y-z),fyy = (2 +X+ y + z)e(x+y-z), fzz = (x + Y + Z - 2)e(x+y-z).

dU dV dU dV (a) - = 2.x = - - = -2y = - -· ~ ~·~ ~· dU dV dU . _ dV .

(b) dx = ex cosy = dy , dy = -e' silly = dx'

(c) dU X dV dU y dV dx = x2 + l = dy ' iJy = x2 + y2 = - dx ;

(d) dU l - X2 dV dU -2xy dV dX (x2 + l)2 - dy ' dy = (x2 + l)2 dx ·

That each u and v above is also harmonic is easily checked. df df dX = 2Bxl + 4Ax3 , dy = 2Bx2y + 4Ci, hence

df df x- + y- = 2Bry2 + 4Ax4 + 2Brl + 4Cl = 4f

dX dy as required.

dX dy . dX . dy dR = cosO, dO = RsillO, dO = -RsillO, dR = sinO, thus

dX dy dX dy 2 . 2

dR dO - dO dR = Rcos 0 + Rsill 0 = R.

dP T dP T dP k dV = -kv2, so v dV = -kv = -P. Also dT v

dP T dP dP therefore T dT = k V hence V dV + T dT = 0.

161

2.7.

2.8.

2.9.

dz Using dt dZ dy dZ dx - - +- - we get

dy dt dx dt dz 2x 2y . dt - (x _ y)2 cost + (x _ y)2 smt

hi h · dz 2 U · d" d"ff · w c 1s - = ---2 . smg 1rect 1 erentatwn, we get dt (x - y)

dz d (cost + sint) 2 . - = - = after usmg the dt dt cost - sint ( cost - sint)2

quotient rule and a little trigonometry. The results are the same.

2

Differentiating the given function, fx = ( y )2 and x+y

• r o·ff . . . . • 2y2 Jy = ( ) 2 . 1 erentlatmg agam g1ves Jxx = - ( )3 , x+y x+y

2r _ 2xy 2 /yy = (x + y)3 and fxy - (x + y)3 .Thus, X fxx + 2xyfxy

- 2rl + 2xy · 2xy - 2x2y2

(x + y)3 = 0.

With '" = F (I) dl/J = - 2 F and dl/J = 1. F' from which 'I' X ' dX X1 dX X

the result follows.

2.10. Writing x = ~ + 11 and t = - ~ 17, using tlte chain rule gives

d dd d d h h .. d~ = ~' dT/ = C dX- af' W ence t e gtven equat10n

d1l/J d1l/J . . d1l/J C dX1 - dXdt = 0 Imphes d~dT/ = 0 SO that

l/J = f@ + go(T/) = f(x + ct) + g(t).

2.11. Differentiating r 2 = ~ + ~ + ... + r. yields 2r ~r = 2x, OX;

. ~~ dr ~ x1 r1 for al11 = 1, 2, · · ·, n. Thus ~· :;- = ..t..J -' =- = r.

i=t oxi i=I r r

dw wyz cos xyz dW wxz cos xyz 2·12· dX 2w + sin xyz' dy 2w + sin xyz '

dW wxy cos xyz dz 2w + sinxyz ·

2.13. Using the chain rule, ;~ = df df

e"secv ax + e"tanv oy and

df = dU

df df e"secv tanv - + e"sec2v -

dx dy

Hence (:~ r + cos2v( ~~ r =

(sec2v - tan2v){ ( ~~ r -( ~; rJe2" from which the result

follows. 2.14. This exercise is quite difficult. Use the chain rule and keep

a clear head!

162

dl/J dl/J dl/J dl/J dl/J dl/J . - = - + y - and - = - + x- from wh1ch dX dU dV dY dU dV

dl/J dl/J dl/J . . . dl/J x- - y- = (x - y)- (ehmmatmg -). ax ay au av Expanding the operator (x:x - y~)(x ~! - y ~~)

( x :x - y :y ){ (x - y) ~~} gives, taking care with the

products:

X2l/Jxx - 2xyl/Jxy + ll/Jyy + xl/Jx + Yl/Jy = xcf>. + Ylflu + (x - y)2l/J •• using the suffix derivative notation. However, using the first two results:

dl/J dl/J dl/J dl/J X dX + y dy = (X + y) dU + 2XY dV'

dl/J dl/J dl/J and - - - = (x - y)-. Thus

dX dy dV

X2l/Jxx - 2xyl/Jxy + ll/Jyy = -(x + y)l/J. -2xyl/Jv + (x + y)l/J. + (x - y)2l/J ••

= -2xyl/Jv + (x - Y) 2l/Juu

Multiplying this by (x - y) yields (x- y)(x2l/Jxx - 2xyl/Jxy + ll/Jyy) -2xy(x- y)l/Jv

+ (x - Y) 3l/Juu

2xy( lflx - if}) + (X - Y) lfluu

hence if l/J obeys the equation given in the question, we must have, (x - y)3l/J •• = 0. Integrating this twice with respect to u yields l/J = (x + y)f(xy) + g(xy) where fand g are arbitrary functions.

+ y2

o(u, v) 2.15. The Jacobian -:-.-- is given by

o(x, y)

(1 - xy)2 (1 1

- xy)z = 0 1

hence there is a functional relationship between u and v. To see what this is, write (} = tan- 1x, lfJ = tan- 1y, then u = (} + l/J, and

tan (} + tan cp tan u =

1 - tane tanl/J X+ y --- = v, so v = tan u. 1 - xy

2.16. From V = 11th(b2 + ba + a 2) we get by logarithmic

differentiation:

2.17.

dV V

b 2+-

a da -~-~~~ - +

b ( b )2 a + ~+ ~

and using the numbers in the question gives

dV 1 ( 8 10) V = 100 3 + 9 + ? = 5.32 per cent.

1 • dA da db A = -zabsm C, whence A = a + b + cot CdC from

which the maximum error is 12.7 per cent. In this exam­ple, if C is very small the area is very sensitive to changes in its value. This is very important in numerical analysis where it is related to ill-conditioning.

f~ n' 2.19. t"e -s'dt = -n:i:-i for all positive integers n.

0 s

Exercises 3.3

3.1.

3.2.

3.3.

3.4.

x2y2 (a) exy = 1 + xy + ll + ... by direct expansion. The

question only asks for second-order terms, therefore exy = 1 + xy suffices. About the point (0, 1) we have

and about the point (1, 0) we have

Finally, about the point (1, 1) we have

e"Y = e + e(x - 1) + e(y - 1) + te[(y- 1)2 + 4(x- 1)(y- 1) + (x- 1)2] + · · ·

(b) Using Taylor's Series directly gives, about (1, 0),

cosec(x + y) = cosec(l) - ((x- 1) + y)cosec(1)cot(l) + t((x - 1)2 + 2(x - 1)y + y2)(cosec(1)cot2(1) + cosec3(1)) + · · ·

and about (0, 1),

cosec(x + y) = cosec(1) - (x + (y - 1))cosec(1)cot(1) + t(x2 + 2x(y - 1) + (y - V)(cosec(l)coe(l) + cosec3(1)) + · · ·

h2 + _4 + ... u!

where the h's are the variables. (a) (2, 6) minimum, (b) (0, 0) saddle, (1, 1 ), ( -1, -1) both maxima.

(c) f = -xye-1<x'+J'>, so fx = y(x2 - l)e-1<x'+y'l,

f, = x(y2 - l)e-~<x'+J"l, fxx = xy(3 - x2)e-~<x'+y'l,

JYY = .xy(3 - y2)e-1Cx'+J'>, and fxy = (x2 - 1)(1 - y2)e-1<x'+J'l.

This Ieads to the following extrema: (0, 0) saddle, ( 1, 1) minimum, ( -1, 1) maximum, (1, -1) maximum, (-1, -1) minimum. (d) Extremum whenever x = 2y (x * 0).

By writing x = 2t + e cos&, y = ~ + e sin& and expanding

x> + 2y> the functionfix, y) = - ( )2 in powers of e, we obtain: X+ y

1(2t + e cose, ~ + e sine) - 1(2t, ~) =

- e>2 [cos28 + 2sin28 + 10 (cos& + sin8)2] 9t 3

and this is < 0 for all 8. Hence x = 2y is a line of maxima. (e) f(x, y) = ex cosy has no extreme values.

3.5. fix, y) = sin.xy, sofx = ycos.xy,f, = xcos.xy. Lines of extrema along reetangular hyperbolae xy = c where

2n+1 . u· h .. c = --2- n, n an mteger. smg t e parametensatlon

X = t + ecos8, y = l + esin8 it is shown that t

n . h . 7r 5n 9n d 1 ,x, y) = sm.xy as maxtma at .xy = 2· 2• 2• ... an

jir ) . h . . 3n 7n 11n ,x, y = sm xy as mtmma at xy = 2, 2 , 2

3.6. If f(x, y) = 4 - -Yx2 + l then, by inspection, 4 is a maxi­mum value occurring at the origin (0, 0). This function rep­resents a cone with the vertex at the origin at which de­rivatives do not exist, therefore normal tests for extrema fail.

3.7. f(x, y) = sinx siny has extrema wherever either 2k + 1 21 + 1 x = --2- n, y = - 2- n, or x = rn, y = sn where k, I,

r, s are integers. This Ieads to the following classification: 2k + 1 21 + 1 . . ·r (k I) . x = --2- n, y = --2- 1r 1s a max1mum 1 + 1s

odd and a minimum if (k + I) is even, x = rn, y = sn is always a saddle point for all r and s.

3.8. f(x, y) = x4 - 2(x - y)2 + lleads to extrema wherever the following two equations hold: x3 = x - y and ,1:3 = y - x, h~ce the three tuming points are (0, 0), (--fi, - v2) and ( ---fi, .Y2). Usi!tg th~usual proc~ur~ it is Straightforward to deduce that (.Y2, -.Y2) and ( -.Y2, .Y2) are both minima. However, at the origin the second derivatives of f(x, y), fxx, f,y and /xy all have the value 4 so fxxfyy = f~. However, near the ori­gin, it is quite easy to deduce that f = -2e>(cos8 - sin8)2 < 0 by writing x = ecos&, y = esin&. Hence the origin is a maximum.

3.9. Extreme values are at the five points (0, 0), (a, a), (- a, a), (a, -a) and (-a, -a). The last four are all saddle points, since fxx = f,Y = 0 yet f~ = sin2 x sin2 y > 0. At the origin, all the second derivatives of f are zero, so we put x = ecos&, y = esin& to obtain the expansion

j(ecos&, esin&) = 1 + 2~ cos48 + O(l), whence (0, 0) is

a saddle point too. 3.10. With f(x, y, z) = xyz(1 - x - y - z), the first derivatives

are zero under the following conditions:

- 2x - y - z = 0 or yz = 0

- x - 2y - z = 0 or xz = 0

- x - y - 2z = 0 or yx = 0

If either x, or y or z is zero then either y + z = 1, or x + z = 1 or y + x = 1. In any of these cases, f(x, y, z) = xyz(l - x - y - z) = 0. On the other hand, if none of x, y, z is zero, the only solution of the three conditions is

1 . h. h x = y = z = 4 m w 1c case

fix, y, z) = xyz(l - x - y - z)

= i . i . i ( 1 - i -} -} ) = 2~6 , which is obviously a maximum.

163

3.11. x = 1, y = 0 givesf = 2, a minimum. x = 0, y = 1 gives f = 4, a maximum.

3.12. The equations ~ 1rbc + A = 0, ~ n:ac + A = 0,

1~rab + A = 0 result from differentiating the Lagrangian

H = 1 n:abc + A(a + b + c - 1) and putting each deri­

vative equal to zero. These imply a = b = c, that is the ellipsoid is a sphere.

3.13. With H = m-2h + A(27rr2 + 2m-h - 1), differentiation

gives r = 2h, whence h = . b, r = ~- This gives a '1127r '137r

volume of . ~ , obviously a maximum. 6~37r

3.14. With the box (cuboid) with dimensions 2x, 2y and z the Lagrangian is H = 4xyz + A(z + --.fx2 + y2 - 9) from

which y = x, z = .Jz .The equation of the cone then implies

that z + --.fx2 + l - 9 = 0 ==? z + xfi = 9 so that

x = 9 = 3..f2 = y, z = 3. Thus the dimensions are ...J2+_l_

...J2 3fi, 3...f2 and 3, giving a volume of V = 216.

3.15. The Lagrangian for this problern is

--./di + x 2 --.fd~ + l H = + + A(x + y - a).

VI V2 lt is essential to eliminate the angles as these depend on x and y. The conditions

aH = .! (d2 + x2flt2 . 2x + ;., = 0 ax 2 I VI

aH = l (d2 + /)-112 . 2x + ;., = 0 hold ax 2 2 V2 '

from which x = y or sin81 = sin82 V --.fd2 + X2 v2--.fd2 + y2 V V I I 2 I 2

3.16. If the corridor has width a and b then the length of the

rod is I = bcosec8 + asec8 so;~ = 0 implies tan3 8 = b/a.

Hence I = (a213 + b213)312.

Exercises 4.3

4.1. f(x, y) = 4x2 - 4xy + 2l, so Vf = (-~ ~ t). The line

direction is x 1 = ( 2 - 4;., )· Evaluating Vf on this line 3- 4.1c

then demanding that x1 · Vf is minimal gives -32 + 64.1c

= 0 ==? A = ~ whence x1 = ( ~ )· Repeating this pröcess

. ( 4.1c ) d ;., 1 h . ( 0.4 ) . g1ves x2 = 1 _ 4;., an = 10 , so t e pomt 0_6 1s

the result of two iterations.

4.2. f(x,y) = -(x - 1)4 - (x - y) 2,

164

__ ( -4(x - 1)3 - 2(x - y)) so Vf

2(x - y)

Starting the first iteration with (0, 0), we get to (0.41025, 0) with A = 0.10256 after the second iteration. The third iteration leads to (0.41025, 0.41025) via A = 0.5.

4.3. Putting g = Vf in the Taylor Polynomial gives the result.

4.4. f(x, y) = x4 - 2xy + (y + 2)2, starting at ( ~ , 1) leads to the

following formulae:

_ ( 4x3 - 2y ) _ (12x2 -2) ( 1 ) Vf - 2y _ 2x + 4 , G - 0 2 . At z• 1 we

therefore have Vf = ( -t )· G = ( ~

-I 1 ( 2 2) and G = 6 0 3 , whence x 1 =

This means that the Newton-Raphson method has actually taken us further away from the actual minimum. There is therefore little point in carrying on with the method (see the answer to the next exercise).

4.5. Using software, the minimum value of f(x, y) = x4 - 2xy + (y + 2)2 is x = -1.165373, y = -3.165373

4.6. f(x, y) = x4 + xy + l + 2y, the Newton-Raphson method gives the results:

( 1 ) ( 0.7826) ( 0.7058) ( 0.6960) Xo = -2 ' XI = -1.3913 ' Xz = -1.3529' X3 = -1.3480

4.7. f(x, y) = x2 + Si + z2 + 2xy + 4yz + 8y - 2z

Vf = (2x + i~//~z + 2) and G = ( i 1~ ~) which 4y + 2z - 2 0 4 2

is of course constant.

lnverting G gives G-I = ! ( -i -/ -~ )· whence 2 -2 7

the exact minimum.

4.8. U•ing lho DFP molhod g;,., ( :; ) U•ffig lho molhod of

steepest descent gives

( 2(x - y) + ~ (x + y + 1 )3 )

V!= -2(x- y) + ~(x + y + 1)3

( 0 ) ( ! ) (-! .lc) At O ,Vf= ; ,X1 _;A

1 ( (1 - kW) so that on x1, Vf = - 4 . Demanding that o- kW

xc Vf ~ 0 impHod ~ l wh'"'"• ~ ( :!). lho ox.ol

result. Using the Newton-Raphson method gives

(-0.1667) df 00123 f . . d(-0.2778) _ 0_1667 an = . a ter one 1terat1on, an _ 0_2778

and f = 0.0024 after two. 4.9. A start vector that rnakes the residual r as close to 0 as

possible is the choice a = 4 and 4eh = 2 so b = ln(0.5) = -1. So choose (a, b) = (4, -1). The DFP rnethod then gives the following results:

Step a b rrr 0 4 -1 1.1773 1 4.0748 -0.5652 0.1779 2 3.9388 -0.5219 0.1490 3 3.8903 -0.5349 0.1401 4 3.8884 -0.5349 0.1401 5 3.8884 -0.5349 0.1401

that is, convergence after 4 steps. 4.10. Denoting D2 by (0, 0) and S by (x, y) then D 1 is (0, 6) and

D 3 is (4, 0). We thus rninirnise the function f(x, y)

= --J x2 + y2 + --J (x - 4)2 + l + --J x2 + (y - 6f The nurnerical procedure is set up as follows: First of all, we calculate VJ, then set H0 = I and y1+1 V/1+1 - VJ,. For this problern

- ( x(x2 + y2)-1!2 + (x - 4)((x - 4)2 + y2)-I/2 Vf- + x(x2 + (y _ 6)2)-112

y(x2 + yz)-112 + y((x _ 4f + l)-112 ) + (y _ 6)(x2 + (y _ 6)2)-112

The step length h is our decision, whence x1 = h - x0.

We have V/0 = ( = ~) thus we can find y1 = V/1 - V/0

(after finding V/1 frorn x1 = h - x0). Then we either use the DFP algorithrn

or the more cornplex BFGS algorithrn

h,yf + Y1hf hiyi

with i = 1, 2, ... the right-hand side of each being known for i = 1. The solution to this soakaway location problern is (1.06566, 0.95725).

4.11. The DFP algorithrn works reasonably well when applied to Rosenbrock's function lOO(y - x2)2 + (1 - x?. Steepest descent and Newton-Raphson fail to converge because of the curved shape of the valley in which the rninirnurn lies. The 12th, 13th and 14th steps give the Co-ordinates

0.9971091 0.9941688 0.9999877 0.9999816 1. 00000 16 1. 0000031

obviously very close to (1, 1). 4.12. (a) The Hessian rnatrix for Powell's function is singular,

therefore no rnethod works weil, even though (0, 0, 0, 0) is an obvious minimum. (b) Fleteherand Powell's helix function has a helical-shaped valley leading to the point (1, 0, 0). (c) Wood's function presents problerns because there is a saddle very close to the minirnurn at (1, 1, 1, 1 ).

4.13. With L(x, y, A) = x2 + l + A(xy - 3), zero partial derivatives of L soon reveal that A = 2, x = y = f3 and f = 6.

4.14. The best way to tackle this problern is to calculate VJ, evaluate it at the point and then check that this point is stationary by cornputing certain gradients. First of all, it is

easy to check that the point ( ~. h ~) lies on both constraints.

Th<n if w< d<fin< V< ~ (1 ~ ) wh«< '"h <olumn

consists of the three partial derivatives of the two constraints, then a feasible direction is s where s · Vc = (0, 0). To find s takes us into the practice of finding eigenvectors ( see

the Fact Sheet, section 4.1). At the point (~. f, ~). s = k(l, t -1) where k is any constant. lt is now easy

to ohock th" V[ ~ ( }~ ) .ati'fi" ' · V[ ~ 0. Thi' mo.m

that the point (~. +. ~) is indeed an extrernurn. lt is in fact a

rninirnurn, which is rnost easily verified frorn first principles. 4.15. The equivalent equality constraint rnodel is:

rninimise f(x) = x2 - 2xy + 2l + 6x + 7y

subject to: 25 - x2 - y2 - u2 = 0

-x- 3y - v2 = 0

where u2 and v2 are surplus variables. Using software, the solution to this problern is (x, y) = ( -3, -4). [The topic of inequality constraints is a great deal rnore cornplex than implied here, but is well outside the scope of this Work Out.]

4.16. Salutions are (x, y, z) = (0.4, 1, -0.6). The Newton-Raphson rnethod is exact for quadratic problerns, hence it works very well here.

Exercises 5.3 5.1.

5.2.

(a) 6i + 6j; -4i + 4j, (b) 2i; -2j, (c) 2i - 2k; -2j, (d) -i + 4j - k; 5i - 2j + 5k, (e) i + j; i - j + 2k. Direction cosines are as follows:

1 5 (a) ill' ill;

5 1 {26'{26

1 1 (b) {2' {2'

1 1 {2';}2

1 1 __ 1 __ (c) {3' - {3, {3'

_1 _1 {3' {3'

(d) l l. ~-3, 3, 3,

1 1 (e) -{2' 0, {2'

1 1 1 ---{3'{3' {3

1 1 {2' 0,-{2.

_1

--J3

165

5.3. a = -2. ß = 1. r = -3 5.4. A = 40.64°, B = 49.36°, C = 90°.

5.5.

5.6. 5.7.

5.8.

7r 37r 8 = 2' 2· -1, -3, 1. (a) 67.38°, (b) 90°, (c) 70.5°, (d) 125.26°, (e) 90°.

1 (a) p = - 2 , (b) p = 0 or 4.

5.9. j + k. 5.10. From the definition of scalar product,

1\ · f 2 = lf111f21 cos (a + ß) = cos(a + ß). However, using components, f 1 • f 2 = (cosß, sinß) · (cosa, -sina) = cosa cosß + sinasinß, whence cos(a + ß) = cosacosß + sinasinß. Similarly, f 1 X f 2 = lf111f21 sin(a + ß)k

I i j k I cosa - sina 0 , whence sin( a + ß) cosß sinß 0

sina cosß + cosa sinß. 5.11. a X b = 4i + 3j - 5k, b X a = -4i - 3j + 5k,

(a - b) X (a + b) = 8i + 6j - lOk. (a - b) X (a + b) = a X a - b X a + a X b - b X b = 2a X b.

5.12. (a) 3,

(b) F · (1, 2, 3) = _12_ -./ 1 + 4 + 9 {14'

( ) Th k d . F · (2, 5, 6) 40 c e wor one ts -./ 4 + 25 + 36 -./ 65 .

5.13. Using the notation of Figure 5.15, A is i + k, B is j + k, C is i + j and D is i + j + k.

B Ar----------::.., D

/ /

/ /

~- .. // i +i + k

/

/ /

/

/

/ /

/

c

Figure 5.15 A cube and its diagonals.

That i + k + j + k + i + j = 2(i + j + k) proves the result.

5.14. a = (1, -2, 1), b = (0, -2, 3) whence b - a = ( -1, 0, 2) and the line has equation r = (1 - A, -2, 1 + 2A). The plane that contains the points (0, 0, 0), (2, 4, 1) and (4, 0, 2) has equation r X (2, 4, 1) · (4, 0, 2) = 0, the condition that three vectors are coplanar. This expands and simplifies to x = 2z. The value of A, corresponding to a

point in this plane is thus 1 - A, = 2 + 4A, or A, =

The point of intersection is thus (1· -2,1 )-166

1 5'

5.15. The result follows from the formula ~absin C for the area of MBC which is

~IPQIInlsin(} = ~IPQ X nl (see Figure 5.16).

Q

R

Figure 5.16 The triangle PQR.

1 5 (a) 213, (b) 2 {5.

5.16. Given the vertices as position vectors, the sides are:

AB = (-3, 0, -3), BC = (4, 2, 4) and cA = (1, 2, -7).

Whence cos A

~ ~ AB· AC

IÄh IIACI 1 -3-./ 3, and similarly,

1 _r;: ~ ~ cos C = "3 '16. As AB · BC

right angle.

5.17. 1 (5i - 3j + 4k). -./50

-12 + 12 0, B is a

5.18. The sides are a - b, b - c, c - a so the area is 1 . 2 (a - b) X (b - c) whtch expands to

1 . 2 (a X b + b X c + c X a) as reqmred.

5.19. Since fi bisects ä and 6 we have from the vector addition law, ä + 6 = kn. Take the scalar product with fi to give fi · ä + fi · 6 = k. Take the scalar product with ä then with 6 to give the two equations 1 + ä · 6 = kä · fi, and 1 + ä · 6 = k 6 · fi, so ä · fi = 6 · fi, and k = 2(ä · n). Substitut­ing for k gives the result.

5.20. a X (b X a) = a 2b - (a · b)a, using the vector triple prod­uct. Whence (a X (b X a)) · b = a 2b2 - (a · b)2, 'dotting' with b. The left-hand side can be interpreted as a scalar triple product so b · a X (b X a) = b X a · (b X a) and this is (b X a)2 = (a X b)2 which proves the result.

b ·X C 5.21. Given x2 + a + a = 0 we complete the square to give

(X + 2~r -4~2 + ! = Ü, and taking the square root

• b + ( c b2 )l/2 A h A • b' gtves x = - 2a -; - 4a 2 e w ere e ts an ar 1trary

unit vector.

5.22. The equation of the line in parametric form is: r = (2 + 6t)i + (-3 + 2t)j + ( -1 + 3t)k from which

X - 2 y + 3 Z + 1 . t = - 6- = - 2- = - 3- as reqmred. A corresponds

to t = 0 and an arbitrary point P corresponds to (2 + 6t, -3 + 2t, -1 + 3t). For AP tobe 14 we require 36t2 + 4t 2

+ 9t2 = (14)2, and so t = :±:2 giving the two points (14, 1, 5) and (-10, -7, -7).

5.23. Figure 5.17 shows the set-up.

z

y

Figure 5.17 The line I is parallel top; the dotted line is the required perpendicular distance.

a = (1, 4, -2), b = (3, 1, -4) and p = (6, -2, 3), so a line joining an arbitrary point on the line I to A is r - a where r = b + tp. This is to be perpendicu1ar to p which implies

(a-b)·p 12 (r - a) · p = 0 so t = p2 49 using the data.

Thus, the required distance is

12 I c2. -3, -2) - 49 (6, -2. 3) I = 3.71

5.24. In vector form the planes are r · (2, 3, -4) = 1 and r · (3, 1, -2) = 2. The vector perpendicular to the line of intersection is (2, 3, -4) X (3, 1, -2) = (-2, -8, -7). The plane is thus r · (-2, -8, -7) = - c and to pass through (2, -4, 5), c = 7, which gives the result.

5.25. The intersection of the two planes is given by 22y = 54 - 14z = 209 - 77x which can be considered to be the equation of the line of intersection. Writing the two planes in vector form r · (3, -1, 1) = 12, and r · (1, 4, - 2) = -5 the perpendicular plane is r·(3, -1, 1) X (1,4, -2) = kandifthisistopassthrough the point (1, 2, -1), its equation is -2x + 7y + 13z = -1.

5.26. The cube of side a is shown in Figure 5.18 together with the diagonals FG and AE.

E

A

Figure 5.18 The cube of side a, the diagonals FG and AE, and the shortest distance between them, p.

Label the co-ordinates as follows: F is (a, 0, 0), A is (0, a, 0), G is (0, 0, a), B is (a, a, 0), D is (0, a, a), E is (a, 0, a) and C is (a, a, a). From this we see

~ ~ ~ ~ that FG = FO + OG = -ai + ak and EA = aj - ai - ak. A line perpendicular to both is

~ ~ (FG) X (EA) = -a2(i + 2j + k), so a unit vector in this

direction is ~ (i + 2j + k) . xt is the vector -aj + ai so

the required perpendicular distance is

~ (i + 2j + k).Äf' = .{6. As this must be positive, the

distance is arJ6.

Figure 5.19 shows the triangle OEA.

a

A

Figure 5.19 The triangle OEA and the shortest distance p.

5.27.

5.28.

5.29.

5.30.

The equations of the two diagonals are as follows:

~ ~ ~ ~ OA + t AE = (0, a, 0) + t(a, -a, a) and AF + sFG = (0, -a, 0) + s( -a, 0, a). The centre of the sphere is, say, (b, b, b). It touches the

plane x + y + z = a at the point (~, ~, ~)· However,

the distance to the centre of the sphere from the origin is b13 which means that the radius of the sphere must be

13 _r;; . a-13 a 3 - bv3 from wh1ch b = (1 + '13) so the centre of the

sphere is at the point% (3 -13)(1, 1, 1).

By taking the scalar product with v derive the expression 2

V· r = yv · r' + yv 2t; and using 1 + y2 V2 = y2 gives c

t = yt' + r V ;t . Substituting for V • r and t in the expression

r' = r + [r ;2 1 v · r yt ] v gives after manipulation

r = r' + [y ~2 1 V· r' + yt'] V.

If a and b are the position vectors of the ends of a diam­eter, AB say, of a sphere, then r - a and r - b together with AB form a triangle APB where P has the position vec­tor r. The condition that P describes a sphere is the condi­tion that the angle APB is a right angle, that is (r - a) · (r - b) = 0. The lines are r = (9, 0, 5) + t(-4, 1, 1) = a + tb and r = (7, 8, 0) + s(2, 1, 1) = a' + sb'. Using Examples 5.21 and 5.22 gives the perpendicular distance as

b' X b 1 14 (a'- a) lb' X bl = (-2, 8, -5)ill (1, 1, -3) =TI·

167

Exercises 6.3 6.1.

6.2.

Using the definition of derivative, we have

_i (1/JF) = lim{I/J(t + .M)F(t + .M) - 1/J(t)F(t)}o Adding and dt l>t--.0 t::.t

subtractinl?- the term 1/J(t + t::.t)F(t) from the numerator then taking the Iimit gives the resulto

F 0 dF = F dF 1 + F dF2 + F dF3

dt I dt 2 dt 3 dt

= _!__ ~ (F2 + F 2 + F2 ) = F d~ 2 dt I 2 3 df

6.3. ~~ = :r (r X m ~n = r X~ (m ~~) + ~~ X m ~~ the second term of which is clearly zeroo Hence

dL d2r dt = r X m dt2 = r X F = ro So r X F = 0 ~ L = const.

6.4. V = (2t - 6)i + (3f - 2t)j f2 + 2k,

6.5

6.6.

a = 2i + (6t - 2)jfi,

v 0 a = (t - 1)(18t2 + 6) = 0 when t = 1.

~[ r x a] = (roa)(r X a)- (r X a)(roa) dt r 0 a (r 0 a)2 0

(a) r = ~at 2 + At + 8 where A and 8 are arbitrary constant vectorso

d2r b X a (b) Convert to df2 = Jla + -a-2- (see Example 5017),

then integrate to give r = !llat2 + b 2: 2 a t2 + Ct + D where

C and D are arbitrary constant vectorso

6.7. (a) T = (2t, 3t2 -1, 0) , --J9t 4 - 2t2 + 1

(b) T = (a cost, - a sint, t),

--Ja2 + b2

(c) T = (1, 2t, 3t2)

--J9t4 + 4f + 6.8. Written in the form r = (1, 1, 1) + A(l, 2, 1) and

r = (0, 0, 2) + JL(2, 1, -4) the scalar product (2, 1, -4) 0 (1, 2, 1) = 2 + 2 - 4 = 0 shows that the lines are at right angleso

6.9. !~ = (a cost, -asint, b) = (y, -x, b) as requiredo

6.10. For a plane curve, since T and N are in the plane of the curve, B is a constant vector directed perpendicularly to the plane of the curveo The third Serret-Frenet formula

dB = -TN thus implies -TN = 0, from which T is zeroo ds

6.11. Using the method of Example 608 we obtain:

T = -- smt, - cost, - , 1C = 2 , , ( a 0 a b) a c c c c

168

N = ( -cost, -sint, 0),

B = (~ sint, -~ cost, ~ ) and T = _!!_ c c c c2

where c2 = a2 + b2o In this example, t = !.. 0

c

6.12. First of all ~~ = (682, 68, 3)

whence ~ = I~ I = 3(484 + 482 + 1)112 = 3(282 + 1) d8 d8

as requiredo The method of Example 608 then gives the following results:

t = (282 1+ 1) (282, 28, 1), I( = ~ (282 + 1)-2,

A 1 N = (282 + 1) (28, 1 - 282, -28),

A 1 2 8 ( 1 28 282) d ". --3 (282 + 1)-20 = (282 + 1) - ' ' - an , =

From which IICI = I Tl = ~282 + 1)--20

6.13. r = (2acost, 2asint, bt2),

r = ( -2asint, 2acost, 2bt) and

f = ( -2acost, -2asint, 2b)

from which r 0 f = 4b2t follows at onceo Taking cross prod­ucts, we obtain

r X f = (4abcost + 4abtsint, - 4abtcost + 4absint, 4a2)

and I r X f I = ,) 16a2b2 + 16a2b2t2 + 16a4

= 4a[a2 + b2 + b2t2]1/20

Finally, T = ~ = (-asint, acost, bt)o lrl --Ja2 + b2f

6W ~ 0 V) • • V = dt = (U smOJt, U COSOJt, ,

v X 8 = (BU coswt, -BU sinwt, O)o Also,

m dv = m( wU coswt, - wU sinwt, O)o Thus dt

m !; = q(v X 8) provided mw = qBO

6.15. r = (x, y, z) thus :: = (x', y', z') = t by definitiono Hence

d2r dT 0 - = (x" y" z") = - but the nght-hand side is ICN ds2 ' ' ds'

using the first of the Serret-Frenet formulaeo Taking modulus

thus gives 1C = ,J (x")2 + (y")2 + (z")2 as requiredo 6.16. This problern demands extensive use of the Serret-Frenet

formulaeo

dr , d2r dT , - = T and - = - = ICN 0 Also, ds ds2 ds

d3r ctN diC A diC A A

ds3- 1(- + -N = -N- ICT8o ds ds ds

f h. . h dr T. . h Taking the scalar product o t 1s w1t ds = giVes t e

result. . ar ar .

6.17. Usmg dr = au du + Clv dv g1ves

dr . dr = ds2 = ( ~:du + ~: dv ) . ( ~:du + ~: dv )

= ar . ar du2 + 2 ar . ar dudv + ar . ar dv2 au au au av av av

which shows the first part.

ar ar . If u and V are orthogonal, then au . av = 0 and F iS zero.

ar ar 0 h" h . 1" h d If F is zero, then - . "'> = , w 1c 1mp 1es t at u an v au av

are orthogonal.

Exercises 7.3

7.1. VI/> = 2x(l + z2)i + 2y(z2 + x2)j + 2z(x2 + l)k, so the

unit normal at (1, 1, 1) is ~ (1, 1, 1) and that at (1, 0, -1)

7.2.

is .A (1, 0, -1). It is just coincidence that the unit normals

have the same form as the points themselves, this being a consequence of the form of V 1/J. The normal at (a, 0, -a) is n1 = 2a3i - 2a3k and the normal at (a, 0, a) is n2 = 2a3i + 2a3k. It follows immediately that n1 • n2 = 0 thus the normals are perpendicular.

(a) Cli/J = nx"-1, Cli/J = ny"-1 and Cli/J = nz"-1, so the result ax ()y az follows.

(b) Cli/J = ax"-1/zc, Cli/J = bx"/-1zc and Cli/J = cxalzc-1 ax ()y az from which the result follows.

7.3. At (a, b, c), Vif> = ~ i + ~ j + ~ k from which the direction

cosines are k(a, b, c) where k = (:2 + ;2 + :2 r12•

7.4. This result follows from the use of the quotient rule on the components

7.5.

7.6.

v(j_) = i.(1_) i + i.fj_)j + i.Jj_) k. "' ax "' dy\"' az \ "'

(a) 1/J = 1nxyz,

(b) 1/J = sin xyz,

(c) 1/J = x2iz4,

(d) 1/J = xy + ix2iz.

_ P1X + PzY + P32 Use 1/J - (r + y2 + 22) 312 and differentiate to give:

E = -VA> = -!!. + 3(p. r)p -r r3 rs

13 7.7. (a) m'

7.8.

7.9.

7.10.

7.11. 7.12.

7.13.

7.14.

23 (b) - --!45'

() ~ c {45"

Maximum change is along the direction VT = aT0(1 + cz + by)eaxj + bT0eaxj + cT0eaxk which at the origin is T0(a, b, c).

The unit normal is .b (-2, I, 3), found through VI/> in the ~13

usual way. The tangent plane is r · (- 2, 1, 3) = c where c is a constant. So the plane passing through the point (1, -3, 2) is -2x + y + 3z = 1. The angle between the surfaces is also the angle between the normals to the surfaces. We check that the point ( 1, 2, -1) lies on both surfaces, then calculate both normals: Vif> = (6x, -2y, 2), Vlf/ = (y2z - 3, 2xyz, xy2 + 2z). At the point (1, 2, -1) the normals are (6, 4, 2) and (1, -4, 6) whence the angle is

cos- 1 [{56 2 ml = 87.9°.

V . F = y + z, V x F = -yi - xk and V(V · F) = j + k. (a) 4xz - 4xyz + 6yz, (b) 2xy - xz + 2yz, ( c) a 1yzexyz + ar:zexyz + a3xyexyz. (a) (2z4 + 2x2y)i + (3xz2 + 4xyz)j - 4xyzk, (b) 0, since 1/JV 1/J = V( tf), or use the formula V X (aA) = aV X A + Va X A with a = 1/J, and A = Vif>. (c) 0, using the determinant for expanding curl. (a) Using the determinant for curl

j k

V X (I/JA)= a a ax ()y :z = i ( ~ (ql<\3) - ! (ql<\z)) + · · ·

I/JA1 1/JAz I/JA3

The identity follows by expanding the derivatives using the product rule then grouping the 1/J terms and the A terms. (b) This is a particularly nasty formula to derive. The only practical way without recourse to computer algebra is to note that the i component of the left-hand side is

:x (A 1B 1 + Aj32 + A3B3), then to show that this is also

the i component of the right-hand side. This is done term by term:

Adding these together reveals that all but the x derivatives cancel, and further these x derivatives combine to form

:x(A1B1 + Af12 + A3B3), the left-hand side. We can argue

169

by symmetry that the j and k terms behave similarly, hence the identity is established.

(c) This is done as part (b). The left-hand side has i component

The right-hand side has the four terms

= i(B~:x + B2;y + B3:2r1 - i(A~:x + A2;y + A3;z ~~

-iB1 (()AI + ()A2 + ()A3) + iA1 (()BI + ()B2 + ()B3) + ... ax ay az ax ay az

Once again, many terms cancel and the two i terms are shown tobe the same. Symmetry then establishes the identity. Put A = 8 = u in part (b) to give V(u · u) = 2(u · V)u - 2u X V X u from which (u · V)u = V(~u2) - u X V X u. For irrotational ftow, the second term on the right is zero and so the potential is tu2·

7.15. Since V· (VI/> X VIJI) = VIJI· V X VI/J- VI/>· V x VIJI = 0 the combination V 1/J X V lJI is solenoidal and the result is established.

7.16. From the definition of divergence and curl we can imme­diately write that

di/J di/J di/J ax ay az

VI/J· VIJI X Vn = dljl dljf dljf --

ax ay az an an an ax ay a;-

= d(I/J, "'' n) a(x, y, z)

Hence if 1/J, 'I' and n are functionally related, this determinant must be zero. This is equivalent to the three gradients V 1/J, VIJI and Vn lying in the same plane.

7.17. By direct differentiation

so that

7.18. Substitution and use of the rules of differentiation establish that the given forms of E and H satisfy Maxwell's equations as stated.

7.19. (See 7.18). 7.20. Demanding that the ftow satisfies V· u = 0 implies

a + l = 0. Demanding that the ftow is also irrotational, that is V X u = 0 gives k - b = 0. Hence

u = (ax + by)i + (bx - ay)j di/J i + di/J j. ax ay

Integration gives the result. 7.21. Differentiation gives:

ar ß' 0 ß.' + 0 ak ()p = -cosa COS I - cosa Slll ~ Slll

170

ar 0 ß' 0 ° ß.' ak ()a = psma COS I + psmasm ~ + p COS

-(a - pcosa)sinßi + (a - pcosa)cosßj

Hence h1 = I ~ I = I, h = I ~ I = p and ap 2 aa

h3 = I ~~ I = (a - pcosa). Also,

el = -cosacosßi - cosasinßj + sinak

e2 = sinacosßi + sinasinßj + cosak

e3 = -sinßi + cosßj

are easily tested to be orthogonal. 7.22. In orthogonal curvilinear co-ordinates curl takes the form:

h1e1 h2e2 h3e3

VXA a a a

dU] au2 au3 so

hiAI h~2 h03

_Q_ _Q_ _Q_ aul au2 au3

a a a dU1 au2 au3

V· (V X A)

hiAI h02 h3A3

which is zero for the same reason as it is in Cartesian co-ordinates.

h1e1 h2e2 h3e3

a a a V x VI/>= au; dU2 dU3 = 0 by directly multiplying out.

di/J iJI/J iJI/J Tu; au2 au;

7.23. V· (1/Jep) = ( I ) aa (1/Jp(a - pcosa)). So if this p a - pcosa p is zero, we have 1/Jp(a - pcosa) = f(a) where the right­hand side is an arbitrary function of the co-ordinate a.

Hence 1/J = f(a) . On p = a, 1/J = (sin2 ta)-I, p(a - pcosa)

I

f(a) _ 1 _ a · 2asin22a _ 2 hence ( ) - -. -2-1- or f(a) - . 21 - 2a. a a - acosa sm .:,p sm .:,p.

Whence the solution is 1/J = 202 p(a - pcosa)

Exercises 8.3

8.2. (a) 2, (b) 27?-, (c) 27?-. (b) and (c) are equal because the end points are the same and the integral is independent of the path since r = V <tr2).

8.3. (a) 2, (b) 2. 1/J = x2y + z2•

8.4. (a) 185 i + fs j + 5~67 k, (b) 20 ~53fi'

() 9-2'1/2._ s.+ 2'1/"2-1k C 18 I 9 J 2 .

8.5. J c A.dr = J/ 1dx + A2dy + A3dz = [A 1x + AzY + A3z]c

since A is a constant vector. This is zero since x, y and z have the same values at both 'ends' of C.

8.6. J /ds = (0, 0, 2nb(a2 + b2) 112) and

J / X dr = (0, 2nab, 27ra2).

8.7. -61t.

8 8.8. (a) 3, (b) 4.

8.9. a4

- 2 nk.

_1- J · I dr (a) 2na, (b) 4, (c) 2 + '12. In fact c T · dr = ds · dr = J c ds which is the perimeter of C. C

8.10.

8 11 Th . 1 . 8 _ I zn -_z_,(=-bc_o-:s_e_i_+---=a_s_in::-8-'j'-'-)_--=--=-a=-b_k d (} . . e mtegra ts - . (a2cos28 + b2sin28 + z2) 112

0

Exercises 9.3

9.1. (a) t, (b) t, (c) 2 sin1 - sin2; the fact that 0 ~

J J sin(x + y) dxdy ~ 1 follows since the modulus of the D

integrand is always less than one. 4

(d) -s·

9.2. (a)

y

2

1.5

0.5

Figure 9.21

2 27'

0.5 1.5 2 X

(b)

y

3

2.5

2

1.5

0.5

Figure 9.22

0.2

9 280.

(c)

Figure 9.23

(d)

Figure 9.24

..2.. 70.

0.4 0.6 0.8 1.4 X

y

X= 2y

X

0.5 1.5

171

9.3. (a)

y

4

3

2

Figure 9.25 I yl!4

f J f(x, y)dxdy. Jo yll2

(b)

y

1.2

0.8

0.6

0.4 Region of Integration

0.2

0.2 0.4 0.6 0.8

Figure 9.26

JIJI f(x, y)dxdy. 0 yl/2

(c)

y

2 X

1.2 1.4 X

Figure 9.27 The triangular region of integration and two typical vertical strips.

r r f(x, y)dydx + r rf(x, y)dydx. -1 - X 0 X

172

(d)

12.5 y

10

7.5

5

2.5

0

-2.5

X= I

Figure 9.28 The shaded region is the domain of the integral, and three typical horizontal strips are shown.

J-JJ3 Jl J3 J9J3 f(x, y)dxdy + f(x, y)dxdy + f(x, y)dxdy. - 3 - y - 1 I I "Y

9.4. (a)

Figure 9.29 The region of integration (shaded). Integration with respect to y first.

(b)

y

0.2 0.4

Figure 9.30 The region of integration D, and a typical vertical strip .

...J2 2

I I ( 3 + 1) 9.5. (a) J J sin ~ dydx

0 -.rx fiJ/ ( 3 + 1) Jo 0 sin ~ dxdy

i<cos-}- cos1),

(b)JIJI kdydx = flf -fj-v X3 dxdy = l(..JZ- 1), Oxzx4+yl ooX4+yl 8

(d) J~ { x2e'4 dydx = L J: x2e'4 dxdy = / 2 (e - 1 ).

9.6. The required area is 4 X shaded area

Figure 9.31 The lemniscate of Bemoulli with the region of integration shaded.

9.7

f1t/4J 3cos29 9 = 4 Rd.RdO = - n.

0 0 4

1L 12"

n/2

9.8. The integral becomes f 2cos2e 2 d0 = __!__ (67t - 20), Jo (1 + cos O) 3

either by computer algebra or by using the substitution

t = tan!O.

rt/4 I

9.9. The integral becomes J J 21nRd.RdO = - ~-o 0

9.10. (a) For this Iransformation the Jacobian

d(r, 8) 1 h h . 1 . d(x, y) = 4(cos0sinO)II2, ence t e mtegra IS

rt/2 I rt/2

f f (rcos0)3tz (l - rz) drdO = 2_I cose dO = _!. Jo Jo 4(cos0sin0) 112 45 0 ..JsinO 45

f2nfb (b) The integral is 2RlnRd.RdO = 7t(b2 - a2).

0 a

(c) For this transformation the Jacobian aa(u, v) = X ~ Y, (x, y) x

I 2- y

hence the integral becomes f f X + y e+r dxdy =

o ' r

JJ: e"dudv = e2 - 1, and the triangle in the x-y plane

transforms to the reetangle in the u-v plane as shown in Figure 9.32.

x- yplane u - vplane

y V

Y=X X+y=2

Figure 9.32 The shaded areas map to each other.

9.11. Using Green's Theorem the integral becomes

rr 4ydydx = 2a3

9.12. Using Green's Theorem

u

U=2

fc e'sinydx + e'cosydx = J L (e'cosy - e'cosy)dxdy = 0.

In terms of a potential, if e'siny = ~~ and e'cosy = ~: then

<fJ = e'siny and fc V<fJ · ds = 0.

9.13. Integrate this directly since

8 53 9.14. (a)3, (b) 104, (c) 105 .

9.15.

The grinder Y - \X

(at the back)

Figure 9.33

z

i lf Vif. 2-x-y 2 dzdydx

0 X 0

11 20"

The plane x + y + z = 2

X

173

9.16. 61t - 2.

9 17 27ta3 ( 1 - ) d h - !E h I . ~ 3 • • 3 cosa , an w en a - 2 , t e vo ume IS 3 1ta .

16 -9.18. 5 n>/2.

9.19. The formulae follow immediately from writing r

plane polar co-ordinates.

(a) For the cardioid, x = ~ 1ta, y = 0 (by symmetry).

(b) For the quarter lemniscate, computer algebra is useful

for evaluating the two integrals J "14cos31228cos8d8 and

fnM o 0

cos312 28sin8d8; the results give

x = i 1ta, y = (~ ln(l + >12) + 7~2)a. 9.20. The volume is given by the triple integral

{3[" ~ f f RdRd8dz = 4n>/3. -f3 0 I

9.21. aa2r(cos8sin8)"- 1•

Th . I " 1 J~ f~ ~ d d _l 312 9.22. e mtegra transtorms to 2 -~ -~ 1 + vz u v - 2 1t .

9.23. The moment of inertia is given by the triple integral

f2nfhfw/h 3 0 0 0

R3dzdRd8 = 10 ma2 where m is the mass of the cone.

9.24. The volume of the ice crearn cone is given by the integrals

f 2nf f3!2f ztl3 f 2n1l f-M RdRdzd8 + RdRdzd8

0 0 0 0 3/2 0

~ 1t(2 - '>/3).

Similarly, the moment of inertia is the sum

f 2nf..f3!2 zf'i3 f2nfl J.,fhi i pR3dRdzd8 + pR3dRdzd8 where p 00 0 Of3/0

is the density which is messy but Straightforward to evaluate. In terms of the mass of the ice crearn cone m, the answer is

286 - 147>/3 320(2 - >13) m.

Exercises 10.3

10.1.

10.2.

10.3.

174

2na2 81 (a) 108, (b) - 3-, (c) 2.

8 (a) 3 7ra3, (b) 4na3, (c) 3a3, (d) 3207t.

(All of these answers can be verified using Gauss's Flux Theorem, see Chapter 11.) Since z - f(x, y) = 0 is the equation of the surface S, we deduce that the unit normal to S takes the form

A ( dj. dj. ) I ( df) 2 ( df) 2

n = - (fx I - ay J + k dX + dy + 1.

Using the fact that area = f dS = f n · dS = f ~ s s R I" . kl

then gives the result.

10.4.

10.5.

10.6.

10.7.

10.8.

10.9.

Use of projection gives

Jt · dS = L (2x2 + 2y2 + z)dR = J) :(R2 + 1)RdRd8

31t = 2· where we have z = 1 - x2 - y2 on R.

Using direct evaluation means parameterising the sphere which Ieads to

fs<x + y + z )dS =

f n/2f n/2 0 0

(sin8cosA. - sin8sinA. + cosA.) sin8d8dA.

which is evaluated straightforwardly to be ~1t . Evaluating the same integral as is in Ex:ercise 10.5 but by projection Ieads to

fs<x + y +z)dS = t<x + y +z)c~e

J "'zf I(R(cos8 + s~8) + 1 \lldRdO. o o >~1 - R r

This is evaluated reasonably straightforwardly to be ~1t . This method is Ionger than that of the last example. 4nA., hence independent of the radius of the sphere. 41t}1

------;1 hence no Ionger independent of the radius of the sphere.

We use the result that dS = h.hvdudv where h. = I~: I and

hv = ~~:1· (See Chapter 7.) ~: = cosvi + sinvj, and

dr dV

- usinvi + ucosvj + bk whence

10.10. When the rain comes straight down, the calculation is as follows:

F . n = ~ and n . k = ~ thus fs F . dS = L dxdy = 1t.

If the rain is at a slant, the calculation proceeds in this fashion:, ,

F = - *- ~. therefore F . n = ->/I (~ - i). whence

f F · dS = -f _,L_ (.:!. - !..) I. dR = s R >12 r r z

1 J2"J 1 Rcos8 + Rsin8 RdRdB h. h h 1 - ...[2 0 0 R w 1c as a va ue

n;>/2. This gives some insight into the inaccuracy of home­made rain gauges.

10.11. The area of a curved survace is A(S) = fs dS. Using re­

sults from curvilinear co-ordinates (Chapter 7), we have

ldr drl dr dX dy

dS = ~ X ~ dudv, whence since du = du i + du j

dZ dr dX dy dZ + - k and - = - i + - j + - k we have

dU dV dV dV dV

dr dr X

dU dV

i j k dX dy dZ dU dU dU

dx dy dz dV dV dV

d(y, z) . d(z, x) • d(x, y) a(u":-;;) I + ~ J + ~ k and so

ldr X drl = ~(d(y, z)\2 + (d(z, x))2 + (d(x, y))2 from du dV d(u, v)} d(u, v) d(u, v)

which the result follows.

(a) For the cone, ~: = cosvi + sinvj + k and g~ -usinvi + ucosvj and so (by evaluating the cross product and taking the modulus, it's quicker than using Jacobians)

we have that the area of the cone is thus fs dS

J~ C u~2dvdu = rt~2.

(b) For the helicoid, ~: = cosvi + sinvj and ~: =

-usinvi + ucosvj + k so proceeding as before, the area of the helicoid is given by

fsdS = f~ c~v2 + ldudv = 2rtJ~~v2 + ldv =

n<~2 + ln(l + ~2)).

10.12. Writing x = u, y = v, z = f(u, v) as a valid parameterisa­

tion of the surface S, we can derive the normal ~: X ~: as

df. df. . - du I - dv J + k. The result then follows smce

fl · dS = L F · (~ X fv) dxdy upon writing u = x and

V= y.

Exercises 11.3 11.1. (a) 108rt, (b) 3a3•

11.2. For F = r the ftux consists of the two parts:

L"'zdS + Lmd (x2 + i)dS = 2 X 1t + 2 X 2rt= 6rt. part

Using the Divergence Theorem, V · r = 3 so

L F · dS = LV · FdV = 3 X Volume 6rt.

The second method is easier.

11.3. Wehave already shown that L (x2i + lj + z2k) · dS = 1t

where S is the surface of the sphere x2 + l + z2 = 1.

We thus need to find LV · (x2i + lj + z2k)dV where V

is the sphere r + l + z2 :s: 1. Using spherical polar CO­

ordinates this volume integral becomes

JJ:J:n2r(cos.Asin8 + sin8sin.A + cos8) r2sin8drd8d.A

= ·M:J:(cosA.sin8sin8 + sin28sinA + sin8cos8)d8d.A

= i( rt(sin .?.. + cos A) dA = rt. Gauss's Theorem is

thus confirmed.

11.4.

11.5.

11.6. 11.7.

11.8.

11.9.

11.10.

Using the Divergence Theorem, V · F = 7x2z2 , so the ftux is

h. h . 1t w IC mtegrates to 6 . If F = V cp then V · F = V2cp = 0 since cp is harmonic. If

V · F = 0, then fvv · FdV = 0 = fsF · dS for all closed

surfaces S. Thus F · dS is an exact differential, dc/J say. We thus have F · dS = F 1dx + F2dy + F3dz = dc/J. Using the chain rule thus gives

F1 = dc/J Fz = dcp F3 = dc/J so F = V" dX ' dy' dz .,.,.

This needs no further comment. Let F = <f>a where cfJ is a scalar and a is an arbitrary but constant vector. Using the vector identity V· F = V· (c/Ja) = a · Vcp, Gauss's Divergence Theorem

becomes fvv · FdV = fva · VcpdV = a · fvVc/JdV which is

equal to fsF · dS = fsac/J · dS = a · fsc/JdS. We thus have

a · [fvVc/JdV - fsc/Jds] = 0 and since a is arbitrary, the

result follows.

Suppose that the function cfJ has an extremum at the point C, and take the origin of co-ordinates to be at C. Sur­round C by a small sphere of radius e which lies entirely within the volume V. The fact that cfJ is harmonic together with Gauss's Theorem applied to the sphere then implies

f2"f" dc/J 0 0 dr sin8d8d.A = 0. Multiplying this result by dr and

integrating from r = 0 to r = e yields rar f~ ~~ sin8d8dAdr = 0 or r J~(c/J(e) -cp(O))sinectect.A

= 0. Now, c/J(O) is the extreme value that does not depend

Oll either e or A, hence J:nJ: c/J(e)sinectect.A = 4rtc/J(O). Mul­

tiplying this by E? then integrating between 0 and e gives

c/J(O) = -v1 J cfJdV where V, denotes the small sphere. This e v,

expression implies that c/J(O) is the average value over the sphere. It cannot therefore be an extreme value and the result is proved. Let A = a X F where a is a constant vector, we have the identity V(a X F) = F · V X a - a · V X F = -a · V X F. Apply Gauss's Divergence Theorem to obtain

-a -fvv x FdV = fsa X F · dS = a · fsF X dS using a

property of the scalar triple product. The result follows since a is an arbitrary vector. (a) With F = (2x - y)i - yz2j - izk, V X F = k,

whence fsv X F · dS = fsk · dS = Ldxdy = rta2, where

R is the disc r + i :s: a2, being the projection of the hemi­

sphere on to the x-y plane. Directly, we have J F · dr

Jz. c = 0 ( -2a2cos8sin8 + a 2sin28)d8 = rta2•

(b) F = xyi + yzj + zxk, V X F = 6k so that

J V X F · dS = 6f k · dS = 6f dxdy = 6 x ~ = 3rt. S S R 4 2

175

F 3 · e· 3 e· 1 k d d Directly, = - 2 sm 1 + 2 cos J + 4 an r

= -1sin0dei + -i'cosOd~, whence

fc F · dr = 2n (- ~) = - 3; .

11.11. For this problem, an attempt to draw the curve C has been made in Figure 11.6.

z

X

Figure 11.6 The intersection of the plane z + y the cylinder x2 + i = b2 in the curve C.

176

y

V X F = -yi - zj - xk, but the area over which the surface integral is taken is shown shaded in Figure 11.6, and this is a 'sawn off' cylinder. The use of projection is not possible since the cylinder is wrapped around the z-axis. Instead we use parameterisation. The unit normal to the

curved part of S is ft = i (xi + yj), and noting that the

contribution to J. V X F · dS from the flat disc on the z = 0

plane is zero, w: have that LV X F · dS = - tfsy(x + z)dS.

On S, y = bsint, x = bcost, z = z so that dS = bdzdt and

1 i r2x f aLbsint so -b /(x + z)dS = - J0 0 (b2sintcost + zbsint)dzdt

= na2b2• This can be checked by direct evaluation of

J/· dr. On C

y = bsint, x = bcost, z = a2 - bsint and the integrand F · dr is given by F · dr = -b3sin2tcostdt + b2sintcost(a2 - bsint)dt

-b2cos2t(a2 - bsint)dt

and this integrated from 0 to 2n is clearly -a2b2J:" cos2tdt

= -na2b2• The negative sign results from the direction that C is traversed. In this case, direct evaluation is perhaps easier!

11.12. P = E X H thus using the identity V · (E X H) = H ·V X E - E ·V X Hin Gauss's Divergence Theorem

gives fs E X H · dS = LV · (E X H)dV, and expanding

the integrand then using the two Maxwell's equations V X E = 0 and V X H = J gives the result.

Appendix A: Conjugate Harmonie Functions

In this appendix an outline is given of the topic of conjugate function theory. This subject belongs to Chapter 7, and some of the problems in that chapter are put into context by it. Consider a vector field F that satisfies the field equations:

VXF 0 V· F 0.

From the first of these, F is conservative (or irrotational) so the existence of a potential 1/J such that

F = VljJ

is assured. From the equation V · F = 0, 1/J is harmonic, that is

V 2 1/J = 0

If, in addition, it is assumed that 1/J is a function only of the two variables x and y, and not depend­ent on z, with F = F1i + F 2j, we have

This is satisfied by the function lfl where

p 1 = Olfl F = _ Olfl ay ' 2 ax

provided lfl has continuous second-order partial derivatives. The equation V X F = 0 then implies

j k j k

a a a a a a ox oy oz OX oy oz

0

Fl Fz 0 olfl _ olfl 0 oy ox

or

Hence we have the following equations valid for 1/J and vr.

together with

177

178

These last two equations arise from the two forms of the components of the vector field F:

F _ F . + F . _ al/J . + al/J • _ alfl . alfl . -I j--1 -j--1--J I 2 ax ay ay ax

and are often referred to as the Cauchy-Riemann equations. The functions lfJ and 1f1 that are both harmonic and obey the Cauchy-Riemann equations are called conjugate harmonic functions. Con­jugate harmonic functions play a key role in inviscid fluid dynamics where l/J is the fluid potential and 1f1 is the stream function. They are also encountered in electromagnetism where the lines 1/>(x, y) = const. are equipotential lines and the lines lfl(x, y) = const. are the electric or magnetic force lines.

The sets of lines l/J(x, y) = const. and lfl(x, y) = const. intersect at right angles, as can be seen from the following argument:

The normal to the line l/J(x, y) = const. is of course Vl/J, and the normal to the line lfl(x, y) = const. is of course Vlf/. Now,

al/J ~ + al/J ~ ax ax ay ay

0

~ al/J ay ax

Therefore the normals are at right angles and so the tangents to the curves must be orthogonal. Forthose familiar with complex variables, functions of a complex variable z = x + iy, i = R

give a rich source of conjugate harmonic functions. This is because the real and imaginary parts of a regular complex function are conjugate harmonic functions:

f(z) = f(x + iy) = l/J(x, y) + ilfl(x, y)

The proof of this is outside the scope of this text, but the following table gives some conjugate harmonic functions generated in this way:

f(z) 1/>(x, y) lfl(x, y) zz xz- i 2xy

X __ Y_ z xz + i xz + i e' excosy exsiny

sinz sinx coshy cosx sinhy

Finally, some mention is made of conjugate harmonic functions in plane polar co-ordinates. In plane polar co-ordinates (r, 0), the Cauchy-Riemann equations are:

1 alfl 1 al/J ~ ao' ;: ae - alfl

ar Examples of conjugate harmonic functions in plane polar co-ordinates are: r"cosnO, r"sinnO (n a constant); lnr, 0.

Appendix B: Vector Calculus

8.1 Vector ldentities

In this appendix a Iist of common vector identities is given. It is assumed that 1/J and lfl are scalar­valued functions with continuous second-order partial derivatives and that F and G are vector­valued functions with continuous second-order partial derivatives:

V(i/J + lfl) = Vi/J + Vlfl V · (F + G) = V · F + V · G V X (F + G) = V X F + V X G V · (1/JF) = (Vi/J) · F + 1/JV · F V X (1/JF) = (Vi/J) X F + 1/JV X F V · (F X G) = G · V X F - F · V X G V X (F x G) = (G · V)F - G(V · F) - (F · V)G + F(V · G) V(F · G) = (G · V)F + (F · V)G + G X (V X F) + F X (V X G) V X (Vi/J) = 0 V· (V x F) = 0 V X (V X F) = V(V · F) - V2F

where in any co-ordinate system other than Cartesian, the last equation is to be taken as the defini­tion of V2. V2 On the other hand has been defined in terms of Orthogonal curvilinear CO-ordinates in Chapter 7 (see the next section).

8.2 Curvilinear Co-ordinates

If e1, e2 and e3 are orthogonal unit vectors in the directions defined by mutually orthogonal CO­ordinates u, v and w then:

so that the quantities: h1, h2, h3 and e1, e2, e3 define particular curvilinear systems. The quantities V 1/J, V · F, V X F and V21/J are given in terms of curvilinear co-ordinates as follows:

h1F 1 h2F2 h3F 3

v21/J _ 1 Ia [!!h. oi/J] + a [!!.A oi/J] + a [~~]} hlh2h3 l dU hl dU dV h2 dV dW h3 dW

179

180

The foilowing table gives some h1, h2, h3 for weil known and perhaps less weil known co-ordinate systems:

Title

Cylindrical Polars (R, 8, z)

Spherical Polars (r, 8, A)

Parabolic Cylinder (u, v, z)

Paraboloidal (u, v, l/>)

Elliptic Cylinder (u, v, z)

Bipolar (u, v, z)

aVsin2u + sinh2v a

(cosh v - cos u)

R

r

aVsin2u + sinh2v a

(cosh v - cos u)

r sin 8

uv

In the above table, for the elliptic cylinder co-ordinates, ( -a, 0) and (a, 0) are the foci of the ellipses (u constant) and hyperbolae (v constant) that constitute the co-ordinate system. For the bipolar co-ordinates, the points ( -a, 0) and (a, 0) are the centres of the two systems of circles that form the co-ordinates v = constant; the u co-ordinates are then circles that pass through these points.

Bibliography

In the course of producing this text, several published works have been consulted but none have been specifically cited. Below is a by no means exhaustive Iist of books that the reader might find relevant to the study of Advanced Calculus. The comments that follow each are my own personal opinion.

Durrant, A. V. (1996) Vectors in Physics and Engineering, 288pp. Chapman & Hall, London. Good text for those who like the engineering approach. The examples favour units and numbers.

Dyke, P. P. G. (1995) Mechanics (Work Out), 215pp. Macmillan, Basingstoke, UK. The companion to the present text with a little overlap in Chapter 6. Hard not to recommend it!

Etgen, G. J. (1995) Calculus (one and several variables) Salas and Hille, 1370pp. John Wiley, New York. This is one of many very thick calculus books that are brilliantly produced with wonderful dia­grams, many in colour and in 3-D. This one is my personal favourite and is a seventh revision of a classic text by Salas and Rille. lt is very good for the preliminary calculus of Chapter 1 (and before), but do not pay full price for it!

Gilbert, J. (1991) Guide to Mathematical Methods, 309pp. Macmillan, Basingstoke, UK. This is here because it contains my favourite treatment of max. and min. of functions of two variables. Also covers the material of Chapters 2, 5 and 9.

Hirst, A. E. (1995) Vectors in 2 or 3 Dimensions, 134pp. Amold, London. A title in the 'Modular Mathematics' series, therefore brief. The treatment is morepure than the present text, but it is readable.

Lewis, P. E. and Ward, J. P. (1989) Vector Analysis for Engineers and Scientists, 406pp. Addison­Wesley, Wokingham, UK. A more mathematical approach than Durrant and proceeding to a higher Ievel. Highly recom­mended; slightly more applied than Marsden and Tromba.

Marsden, J. E. and Tromba, A. J. (1976) Vector Calculus, 655pp. W. H. Freeman, New York. This is an excellent textbook that covers almost all of the topics (but not optimisation). There are later reprintings and perhaps a new edition. Highly recommended, but be prepared for some rigour.

Spiegel, M. R. (1959) Vector Analysis, 225pp. Schaum, McGraw-Hill, New York. Perhaps the direct competitor to Spiegel (1963), this has dated but is still an excellent source book for both solved and unsolved problems.

Spiegel, M. R. (1963) Advanced Calculus, 384pp. Schaum, McGraw-Hill, New York. This is the standard 'problem solver'. Now a little dated, and covering a wider class of problems than the present text (such as Series, Fourier Series, Complex Variable). Worth looking at.

181

182

Index

Aceeieration 79, 86 Ampere's circuit law 158 Angular momentum 90 Approximation to a root of an equation 10,

11, 16 Approximation to error 26, 27 Are length 79, 82 Area expressed as a line integral 153 Area of a triangle 166

Binomial theorem 5 Binormal 79, 82 Bio-Savart Law 111, 112 Broyden-Fletcher-Goldfarb-Shanno method

(BFGS) 50, 51, 54, 58, 59

Cardioid 153, 154 Catenary 116 Cauchy-Riemann equations 32, 178 Centre of mass 14, 15, 131 Centripetal acceleration 86, 90 Chain rule 18, 82, 92, 110 Circulation 116 Complex variables 178 Components 62, 102 Conservative 108, 109

function, fields 99 Constrained optimisation 44, 52 Constraints 35 Continuity 1, 12 Continuous 1 Contour integral 108 Coriolis acceleration 89, 90 Cross product see Vector product Curl 91, 98, 100, 102 Curvature 79, 82 Curvilinear co-ordinates see Orthogonal

curvilinear co-ordinates Cylindrical polar co-ordinates 92, 103, 111,

133, 136

Davidon-Fletcher-Powell method (DFP) 50, 51, 54, 57, 59

Dei see Gradient, Divergence and Curl Derivative 1, 5, 7 Derivative of products see Chain rule Determinant 19, 25, 35, 49, 50 Differentiability 1 Differential 19 Differentiation, rules of 6 Differentiation under the integral sign 20, 27,

156

Direction cosines 62, 64 Directional derivative 91, 95 Discriminant 40 Div see Divergence Divergence 91, 96, 97, 102 Domain 117 Double integrals 117

Eigenvalue 50, 81 Eigenvector 50, 81 Electric field 106, 158 Equipotentials 178 Errors 20, 26, 27 Euler's Theorem (for homogeneous

functions) 19, 30 Explicit 10 Extrema 34 Extreme values 11

Faraday's law of electromagnetic induction 158

First Mean Value Theorem 7, 8, 27 Flux 96, 146 Folium of Descartes 153, 154 Frenet formulae see Serret-Frenet formulae Function 1 Functional relationship 25, 26

Gauss 's Divergence Theorem 148, 153, l55

Gauss's Law (electromagnetism) 158 Gauss's Theorem (for a closed surface) 157 Generalised polar co-ordinates 138 Grad see Gradient Gradient 91, 92, 101 Green's Second Theorem 148, 152 Green's Theorem in the Plane 126, 127, 137,

151, 173

Harmonie functions 32, 155, 177, 178 Helicoid 147 Hertzian vector 106 Hessian matrix 51, 56 Homogeneous function 20

Ill conditioning 162 Image ( of a mapping) 122 Implicit 10 Improper integrals 14 Indeterminate values 3 Integration 2, 12, 13 Integration by parts 13

Integration under the integral sign 31 Irrotational 98, 99, 106, 107, 170 lterated integrals see Double integrals and

triple integrals

Jacobian 19, 32, 101, 118, 134

L'Höpital's Rule 3, 9, 10 Lagrange multipliers 35, 43, 52, 55 Lagrangian 52, 56 Laplace's equation 22, 23 Laplacian 81, 104 Least squares estimate 45 Left-hand limit 3 Leibniz's Rule 20, 27, 31, 32 Lemniscate 137, 173 Limits 1, 2, 3, 4 Line, equation of (see also Straight line) 72,

73, 77 Line bound vector 62, 63, 73 Line integral 108 Linear dependence 76 Logarithmic differentiation 27

Mactaurin 's Series 3 Magnetic field 106, 111, 158 Magnetic induction 112, 116 Matrix 49 Maximum 34 Maxwell's equations 106 Minimum 34 Moment of inertia 119, 135

Newton's Second Law 79, 87, 90 Newton-Raphson method 10, 11, 50, 51, 55,

56, 57 Numerical methods 58, 59

Operational research 50 Optimisation 50, 59 Orthogonal curvilinear co-ordinates 92, 101,

179

Parabola 83 Paraboloid 140 Parameterisation 83, 92, 109, 115, 139 Partial derivative 18, 21, 78 Particle mechanics 86 Path 79 Pathologica1 (curve) 94 Penalty functions 61 Pinching theorem 4 Plane, vector equation 67, 96 Plane polar co-ordinates 118, 123, 124, 127,

137 Polar co-ordinates 22 Position vector 67, 72 Potential function 99 Poynting's vector 146, 159 Principal normal 79, 82 Projection 139, 140

Quadratic form 35

Quasi-Newton methods 57

Radius of curvature 82 Reduction formula 16, 17 Removable singularity 3 Repeated integrals see Double integrals and

Tripie integrals Reversal of limits 120, 121 Right-hand limit 3 Rolle's Theorem 7, 8 Rotating co-ordinates 88

Saddle point 38, 41 Scalar 62 Scalar functions 91 Scalar potential 106, 108, 114 Scalar product 62, 65 Scalar triple product 63, 71, 72 Serret-Frenet formulae 79, 82, 83, 168 Singular (transformation) 134 Singularity 3 Skew lines 72, 73 Solenoidal 97 Sphere, vector equation 67 Spherical polar co-ordinates 92, 100, 104,

145 Square roots (calculation of) 16 Step length 55, 56 Stokes' Theorem 127, 148, 151 Straight line (equation of) 96 Surface integrals 139

Tangent 79, 82 Taylor polynomial 37, 50 Taylor's Series 3, 35, 56 Taylor's Theorem 9, 34, 35, 36, 37, 53 Tolerance 59 Toroidal co-ordinates 107 Torque 90 Torsion 79, 82 Torus 143 Total differential 20 Transformation 101, 118, 131, 137 Transpose 50 Tripie integrals 117, 130

Undetermined multipliers 35, 43 Uniqueness theorem 152 Unit tangent 115 Unit vector 62

Vector 49, 62 Vector analysis 62 V ector differential operator 91 Vector equations 69, 70, 71, 77 Vector functions 78, 91, 148 V ector identities 91 , 97, 99, 100 V ector potential 106 Vector product 62, 68, 69 Vector triple product 63, 75, 76, 166 Velocity 79, 86

Work done 77, 108, 114, 115

183