::::l<r/ L- 1-1>(=-ft\ii--r(~1J~:::: Fo. l. AG -=(0,.2,L}> M - &-c ==- < ) I) ~..-.::.1 ( \ I 0. /:rf!:,-t- f1c =- <I _,, -2...

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Math 3298 Exam 1 NAME: SCORE: l. Given three points A(I, l, 1), B(l,;2, 3), C(2, - l, 2). (a) Find vectors AD, AC, nc. (b) Find AB+ DC, AB - AC, and 2AD. -->,,. /:rf!:,-t- f1c =- <I _,, -2...J I) ~..----::. M - &-c ==- <--1.1 3) I)..) l. AG -=(0,.2,L}> (c) Find length of AB and area of the parallelogram spanned by AB, AC. ~..-.::.1 ( 1 1 1-\ I 0 (}:rev..- = I A ~ >'- kc. -=- ( _ v 1 J -, ::::l<r/ L- 1-1>(=-ft\ii--r(~1J~:::: Fo 2. Given two unit vectors u, v. The angle between them is 7r/4. What is u v? 3. Find the equation of a line passing point P(l, 1, 5) and parallel to another line whose equation is x = l + t, y = 3 - St, z = 9 + 2t. Jv;( v = <. l / a/ r-. v >

4. find equation of the plane which pa:;s through point P(2, 1, L) n.nd parallel to another plane x + 2y - z = o. ~ ~\ JJ< n ~(1,, 2-) -\> LI ( x - v) -+ z..{ 'd-- I ) I. ( i' ~1) ~ 5. Fin<l the point in which the line x = l - t, y = l + 3t, z = t intersects the plane 2x -y + z = 2. (n- t-k. 1 \J-.e~\qn_; 2(1-t).._ (I 1-5 f:) t f- ~ 1- --~ t-==- i.- L+ l ~ 0 ><---- " ~ -- i poi..\.- C:.L- 1:;--= - 4-6. Make a rough sketch of tlie quadratic surfaces, and give their names (i) z = 4x 2 + 9y 2. (ii) x2 + y2 - z2 = 0. :/::, l / 7. ind a vector function r(t) = (x(t), y(t), z(t)) for the curve which is the intersection of surfaces 4 = x 2 + y 2 and z 2 = x 2 + y 2.

8. Suppose a =f. 0. If ax (b - c) = 0, can we claim that b = c? If Yes, prove it. If 110, give a.11 example to support your conclusion. U. A particle moves with position function r(t) = (cosl,sint,t). Find its velocity, speed, an<l acceleration. vb < = Sv\,,-\-,.-- ('.,o-yt,, 1/ / / \ n x ~ 11 L /;-) lta T ~ / ~ I? c.."' ~ -k'- v,_ "'- t ~ I s~itj ~ l (sf (I ~ (: ~ :_~ ; ~ r; k -::: ( -~ f- -;-!;v l.;f- ' 0 ) 11. particle starts at the origin with initial \I locity v(o) = (0, 0, 0). Its acceleration is a(t) = (2, 1, 0). Find the distance it travels from t = 0 to t = 1. Vu-,= ) ' ~(h~(;- ~ ) \>-,I, o).iv~ ~ <!. = (lk, -t ' o) + c::...:... ~ ~ S"vL...>- \} Ci. ; -:::.- Ct> ; o ) ~ "=3> G -::- ( o; li 1 o> ~ \J (:h ~z)/t '- {;- / D> \/ lh ~ J~t)"'+-t 1---t l)~ ~ (? ~ I S:::: ( fr\;-~\-- -:; Ji-t-vl Jo v o I

L2. Fi11d the local max am! min values a11d :-;addle poi11t::; of the f1111ctio11 J(.c, y) =.c 1 +xy+ y 1 1-.i:. 2- { '----- ~- *' u = ( 0, L3. (i) Find V' J for J(x, y, z) = :-;in x ~ os ~ ez. (ii) Find the directional derivative of J along 72). (3) At point (rr, Jl', 0), find the direction which will yield the largest directional derivative of J c-- -c-- :t;- \7 f -= (l<s-\x Of\'(} e / - s-. h.;< s,k ~ ~ ; S-vtv:>< (,../) 'O.e, ) j) ~ J_?--.L ~. 7(; IA f -::.. \7 f. la -::. r1-))~ K (ra d - ~ -f rv S -..ivx (fa. D e = it 6<5 j ('_ t-(c.ri,x + <;,), x) "Tf, <f '" )J < ~ lig(- ( t-1) (-I) ~, 0 ' 0 > ~ (I / D, 0 > j)j, r rul M"-'XI ~" la = i1}~(-if,of, o) = < \ 1 D, 0,) 14. Find absolute max and min off (x, y) = x + y - xy on the closed region D bounded by the triangle with vertices (0, 0),(0, 2),(4,0) '. ('. f \\, \f( D I ), =/ I, l ) -::c I + I - I ~

Math 3298 Exam 2 NAME: SCORE: 1. Evaluate the iterated integral r212y Jo Y x 2 ydxdy. 2. Sketch the region D of the double integral associated with the iterated integral in last problem. Then change of order of the iterated integral to dydx. Evaluate it. ~ ) ~ f: \. J-A )74 (clyd# j) 0 f"i ~ -- "'f 1\ I\ - 4'. l r 1.- ;:; - 1

4. A plate lies on the region D bounded between the circles x 2 + y 2 = 11 and x 2 + y 2 = 4. The density is p(x, y) = h Find the mass of the plate. Hint: switch to polar coordinates. x +y -L)\ 2)T ( z._ 2, Ry,/wo\,(9. 5. Find the volume of the solid under the surface z = 9 - x 2 - y 2, and over the rectangle D = [O, 2] x [O, 2]. ~ ~ \'J-1<'-1") JA ~ \, \. (1-~~~')h ""7< j) l.- 1... 1..-- ~ \ 0 ( 9 '6 - )<"'" -f J' ) \ J.-,. - )}c~-:) - t)e - ~ 'l- Ji- } 6. Find the surface area of the paraboloid z = x 2 + y 2 between z = 1 and z = 4. _j) 3 2.--ll J_ ( I -+ ~ ~ 1-) 2. \ i.

7. Given the following double integrals f 1 Jx xydydx + r.,/2 rx xydydx + J 2 11;../2 ~ 11 lo.,/2 lo Sketch all three regions of integration on one XY plane. { J 4 -x 2 xydydx. -- 8. Use an easy way to calculate the integral in last problem. 9. Rewrite the integral 1 1-1 x2 lo rl-y J(x, y, z)dzdydx as an iterated integral in the order dxdydz. Sketch the region first. /

10. Calculate the triple integral j j f 2 1 dv by converting it to spherical coordinates. }EX +y 2 +z 2 Here E is the region between sphere x 2 + y 2 + z 2 = 1 and x 2 + y 2 + z 2 = 4. \L: ~~ \ L t1-. ( S '-k c\( Jc~ ctr C ~~ s, ~4 - f \ ~ l~ ct~ - f ~ ' If (;v) ~ \ cj1t ~ ~ ~ Yi\ ( l + I) - ~If

Math 32D8 Exam [[I NAME: SCORE: l. Compute the line in tegral 1 yz cos xcl.s, c C : :i; = t, y = 3 cost, z = :3 sin t, 0 ~ t ~ 1r. 2. Use Green's Theorem to evaluate k F cir = k J l + x:lclx + 2xycly where C is the triangle with vertices (0, 0), (1, 0), an<l (1, 3) with positive orientation.

:.l. Find a parametric equation for the curve!j = sin x starting from (0, 0) and ending at (7r, 0). Specify the parameter region. '' Find a parametric equation for the surface which is the pnrt of the sphere x 2 + y 2 + z 2 = 4 and above the cone z = j x 2 + y 2. Specify the parameter region. x ~ 2- (' 0 \} $ "-- f d ~ 0~4)- S Vf c 9 / q ) E-j) ~ [ G) vrr) X ( d, -l j t: ~ ~~1 5. Given a parametric surface )- z_ 6L= tt ~ :C -:::fil r ~ ~- 1 ~\ == ~ r (u,v) = (ucosv,usinv,sinu), 'U E [-Jr, Jr], v E [0, 27rj. Find its Cartesian equation. 2..- 'l-- '},.- 1--- 7, \ v '){ -+J = t.t ~ \J.+\j\ "'\!\ \) 6. Given a vector field F = (p(x, y, z), Q(x, y, z), R(x, y, z)). Define curlf and divf. -> 9rz w JL.d.R ~ _ ll> c(a,'< \ f -:::: ~ - o-c ) 0 -c - ())I). / 'O )\ 60 t

7. Suppose that yon know that J(:c, y) = e.i: +.1: 2 y'2 a nd 'V f = F = (c'" + '2.c.1/, 2.c 2 y). Eval11ate the li11c integral ( F dr where C: is r(l) = (l +sin rct, '2t +cos rct), O :::; t :::; 1. le L~-Jy " f(yl1l]-f(ylo\) =:,f ( f) I) - f ( o 1 1) 8. Evaluate the surface integral j h F ds, where F(x, y, z) = (xz, - 2y, 3x), S: r(u,v) = (2s in ucosv,2s in usi ~ v,2cosu) au<l 0 :::; u ~ 71', and 0 :::; v :::; 27!'. ) J ~~r-.js s F-. (j If i '(.i:) J.A - ~) <4»lu. C,.,,U."" I/ I - (j. S.\.I<_". v' G ~ 'kvl "''">. v

F = (l11y 1-'2x!i,3. i:~.1/ 1-:!.:) y is conservati\'p. If so, fi11cl the potential fu11ctio 11 f (:i:, y) such that F = 'V f. P 1 ~ df'.l f L :::: {V-~ -t ))('0 ~-= V\ -t b '/.. J -..) u & == 3 x~i--r ~ ~ ~ ::: l t t- J ir-\- > r ~~iv-t_ fu<, ~l "" ) (.e"' ~-+v\<'.'j') J..1< = lz.q"' ~ -+ i<. ~ ~" -+ ~ { ~ ),... ol- v i. i. I 1!,.,_, ;L ~ qr ft = 1 + j x ~ + d { ~ ) -= Cl -;;= 3 i- -0 t 'a -( u l 'J ) ~ 0 ~ cri~)~lj. 10. Find the surface area of the upper half sphere :i: 2 + y 2 + z 2 = 4 and z ~ 0 by surface integral. s ; 0( (! 1f \ c=q [0~,\-Cf ' 2. s.\,, ~),\cf " AC'>)= ))1Js = c):~y6 n~\clt\ 2.(,U\{\J >., tj. E:[ 0, ), TI] ' p E-[o,~ l "' \": \ ~~ -23,'Ml s, ve. ~ It ' '1, 0 > ){ < µ..;~ un1 u... ~ ~ :-'--'.-.~A"" ;r -=: [ \~ } 51~ q jfj cl Cf ' $-TI ( - Gr> <I')\: :=: g Tl

t l l. Use th0 Lagnu1µ;p M11lt.ipli<'r nwt.hod to find t.h1 ~ whose s111n is 100. maxirn11n1 procl11rt. of thrc'c' posit.ivc' mtmlwrs ~~' 1 h -:: )( ~ i ~t< X~c ~,t-- ~ ~ --t' ~ -=--{,(JO,X. 1 - =t--?---v 12 3 L~~ ~I 4t_,_,, ""-" ~ ~?; =\ l (L_ Xe ~ ~., Q2 Xb ~ -;\-I Q X-t-~ -t l =(ITT> G x. t: ~ 1 l-- -:) re )( -J) = o? x ~ ~ 't:=-o w,~l\ ~\J-.l(_ X~t-=--V vw~ ~14-6 ( - 'fil- =- )( ~ ;> )( ( l- - ~ ) ~ ~ ~ ~ -l- ')( =-1> W "vi,-(!r V "l 'l( ~ ~ ~ 0 I VU. f- 'i;vu'-')<. ')( -=: ~ -=- r e (QI) X=~=:i~3 ( (J1j 0 tl lrb ~7