Given P(1,-4,-3), convert to cylindrical and spherical values;

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Iris Lambert
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1 CHAPTER 1 Poblems Pob. 1.1 Pob. 1.2 () Given P(1,-4,-3), convet to cylindicl nd spheicl vlues; 4

2 x y = + = + = = 1 ( 4) y 1 4 = tn = tn = x 1 P(,, ) = (4.123, , 3). Spheicl : x y = + + = + + = = tn = tn = P (,, ) = P(5.099, , ). (b) (c ) 1 y 1 0 = 3, = tn = tn = 0 x 3 o Q(,, ) = Q(3,0,5) o = = 5.831, = tn = tn = o o Q (,, ) = Q(5.831,30.96,0 ) = = 6.325, = tn = o R(,, ) = R(6.325,108.4,0) = = 6.325, = tn = tn = 90 0 o o R (,, ) = R(6.325,90,108.4 ) o o o Pob

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4 Pob. 1.4 () x = cos, y = sin, 2 V = cos sincos + sin (b) U = x + y + + y = + sin sin + 2 cos = [1+ sin sin + 2 cos ] Pob

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9 Pob. 1.6 () x + F cos sin 0 y F sin cos 0 = + F F F F 1 = + = + + [ cos sin ] ; 1 = [ cossin+ cossin ] = 0; + = F = ( + 4 ). + ; In Spheicl: x sin cos sin sin cos F y F coscos cossin sin = F sin cos F = sin cos + sin sin + cos = sin + cos ; 4 4 F = sin cos cos + sincos sin sin = sincos sin ; F = sincos sin + sinsincos = 0; F = (sin + cos ) + sin (cos ). 12

10 (b) 2 x + G cos sin 0 2 y G = sin cos 0 + G G G G 2 3 = [ cos + sin ] = ; + + = 0; = G = ( + ). + ; Spheicl : Pob G = ( xx + yy + ) = = sin sin 13

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13 Pob. 1.8 () D cos sin 0 0 D = sin cos 0 x + D D = ( x + ) sin = ( cos + ) sin D = ( x+ ) cos = ( cos + ) cos D = ( cos + )[sin + cos ] Spheicl : Since D = D = 0, we my leve out the fist nd x thid column of the tnsfomtion mtix. Thus, D... sinsin... 0 D... cossin... x = + D... cos... 0 D = ( x+ )sinsin = (sincos+ cos )sinsin. D = ( x+ )cossin = (sincos + cos )cossin. D = ( x+ )cos = (sincos+ cos ) cos. D= (sincos+ cos )[sinsin + cossin + cos ]. (b) Cylindicl: E cos sin 0 y x E sin cos 0 = xy E x E = y x + xy ( )cos sin = (sin cos ) cos + cos sin 2 = cos 2 cos + sin cos. 2 E = y x + xy ( )sin cos 2 = cos 2 sin + sin cos. 2 E = x = cos. 2 E = cos ( sin cos 2 ) + sin ( cos + cos 2 ) + ( cos ). 16

14 In spheicl: E E = y x + xy + x E sincos sinsin cos y x E coscos cossin sin = xy E sin cos 0 x ( )sincos sinsin ( )cos ; but x= sincos, y = sinsin, = cos ; 2 3 = sin (sin cos + (sin cos cos ) cos ; )cos + sin cos sin cos E = ( y x )coscos + xy cossin ( x )sin ; 3 2 = sin cos2coscos + sin cos sin cos (sin cos 2 cos 2 ) sin ; E = ( x y )sin + xy cos 3 = sin cos2sin + sin cos sin cos ; [ sin cos 2 cos sin cos sin cos (sin cos cos ) cos ] E = [ sin cos 2coscos + sin cos sin cos sin (sin cos cos )] 3 [ sin cos 2 sin sin cos sin cos ] Pob. 1.9 () H cos sin 0 3 H sin cos 0 2 = H H = 3cos + 2sin, H = 3sin + 2cos, = 4 H H = (3cos + 2sin ) + ( 3sin + 2cos ) 4 o (b) At P, = 2, = 60, = 1 H = (3cos60 + 2sin60 ) + ( 3sin60 + 2cos60 ) 4 o o o o = 17

15 Pob () 18

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17 Pob () = 2 x + y + = +. 1 = tn ; =. o = x + y = sin cos + sin sin. = sin ; = cos ; =. (b) Fom the figues below, - cos sin sin ( ) = sin + cos ; = cos sin ; = Hence, sin 0 cos = cos 0 sin Fom the figues below, = cos + sin ; = cos sin ; =.. 20

18 sin ( ) cos cos sin sin cos cos sin = Pob

19 Pob Using eq. (1.32), d = cos( 2 1) + ( 2 1) o o 2 = (5)(10)cos(60 30 ) + ( 4 2) = 74.4 d = 74.4 = Pob () d = ( 6 2) + ( 1 1) + ( 2 5) = 29 = (b) d = ()()cos 3 5 π + ( 1 5) = 100 d = 100 = 10 22

20 (c) d π π π π π π = (10)(5)cos cos 2(10)(5)sin sin cos(7 ) π π π π o = (cos cos sin sin ) = cos75 = d = = Pob A x Ay A cos sin 0 A = sin cos 0 A A x y 0 x + y x + y y x = 0 x + y x + y A A A Ax sincos coscos sin A = Ay sin sin cos sin cos A A cos sin 0 A x x y x + y + x + y x + y + x + y y y x = x + y + x + y x + y + x + y x + y 0 x + y + x + y + A A A 23

21 Pob

22 Pob () A B = (5,2, 1) (1, 3,4) = 5 (b) (c ) AxB= = A B 5 cosab = = = AB = AB o 25

23 (d) n AB x (5, 21, 17) (5, 21, 17) = = = AB x = ( A BB ) 5B (e) AB = ( A B) B = = = ` 2 B 26 Pob

24 Pob () At Q, = 10, = π / 2, = π / 3 x = sin cos = 10sin π / 2cos π /3 = 10(1)(1/ 2) = 5 y = sinsin = 10sin π / 2 sin π / 3 = 10(1)( 3 / 2) = 8.66 = cos = 10 cos π / 2 = 0 Q(x,y,) = Q(5,8.66,0) (b) At P, = = o cos = = = = cos = x 3 o = = 0.6 = x + y 5 o o P (,, ) = P (5.385,68.2, ) (c ) d = ( x x ) + ( y y ) + ( y y ) = = p Q p Q p Q d = Pob () An infinite line pllel to the -xis. (b) Point (2,-1,10). (c) A cicle of dius sin = 5, i.e. the intesection of cone nd sphee. (d) An infinite line pllel to the -xis. (e) (f) A semi-infinite line pllel to the x-y plne. A semi-cicle of dius 5 in the y- plne Pob

25 Pob ( ) At T, x = 3, y = 4, = 1, = 5,cos = 5 3 A= 0 5(1)( ) + 25(1) 5 = = 26, sin = cos = 26, B = 26( ) + 2( 26) 5 26 =

26 ( b) In cylindicl coodintes, B B B sin cos B = B cos sin = 15.6 sin = 15.6( ) = = 10, B = 15.6 cos = B(,, ) = ( 15.3,10, 3.059) AB = ( A B) B = ( A B) B 1 ( )( 15.3,10, 3.059) = B = () c In spheicl coodintes, A sin 0 cos 0 A = cos 0 sin 3 A A = 25cos = = A = 25 sin = 25( ) = A = 3 A B = = ± A B AxB = = ± + (

27 Pob At P( 02,, 5), = 90 ; B B B x y cos sin 0 = sin cos = B = 5 3 x y B B B ( ) A+ B = ( 2410,, ) + ( 1, 5, 3) = + 7 x y A B 52 ( b) cosab = = AB AB = cos ( ) = A B 52 () c AB = A B = = = B

28 Pob cossin 2 G= cos x + y+ (1 cos ) sin = cos + 2 cot sin + sin x y 2 G sin cos sin sin cos cos = G cos cos cos sin sin 2 cot sin 2 G sin cos 0 sin 3 G = sincos + 2cossin + cossin G G 3 2 = sincos + 3cos sin 3 = coscos + 2cotcossin sinsin 2 = sincos + 2cotsincos 3 2 G = [sin cos + 3 cos sin ] 3 + [ cos cos + 2 cot cos sin sin sin ] + sincos (2cot cos ) 31

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4 MATHEMATICS IV MARKS. If + + 6 + c epesents cicle with dius 6, find the vlue of c. R 9 f c ; g, f 6 9 c 6 c c. Find the eccenticit of the hpeol Eqution of the hpeol Hee, nd + e + e 5 e 5 e. Find the distnce