thq_01-key

# thq_01-key - Exam 16s MATH 2451 thqf01 TDC Use Course...

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Unformatted text preview: Exam : 16s MATH 2451 thqf01 TDC. Use Course : Multivariable Calculus Duration : 15 min Due Date : 2016-01-21 Instructor : McCary “5/ (First Name) (Last Name) 2016-01-14 14 : 15 1. (10 points} The unit circle 81 C R2 is: 5‘1={(:€a:ar}|:€2 + 92 =1} {a} Show that 81 is not a subspace of R2 by ﬁnding two points in 81 whose sun] is not in Si. 13?, é+<-e’;>=3;é 6’ (b) Are there any two points in 81 whose sun] is again in 81'? Find the conditions which 55,5 E 81 must satisfy so that ('1' + b E 81, and draw a sketch of a particular ('1' and b which satisfy these conditions. AMBER/i: 6%,“: mm m an. \i_\ M 9? (Ia/P -— //z+z//“’ -— z A / / (m, ‘ 4 ﬂaw/ﬂ 1 =//z//a+m.2+/w//° 1 ' 1+ eE-HH —1 - Mt = 3.13 a —//a///Iw/ ML w — £5 = more) 371 : 9 2. (1(Jp01nts} (a) Suppose points A, B and C are vertices of a triangle. Find E? + + and provide a sketch. 5 a a V? a; O /f N‘. 1 9mm i m; m 9m; FLAGL/ ﬁ+§é+cﬁzﬁ (b) Do the following lines intersect? ﬁt} 2 —2} + (1,4, 1H (ﬁt) 2 1, —3} + (2, 1,2}t Mm; 1m; couw HAVPEN 1H omeowr TIMES) -- We 3(a) = 2W) (Nor chzfm) 4+1: = 3+2; 544,} = {1‘5 : —3435 ft : 95'1 ‘Hl/t = s >6ma t5 ‘14‘95 “4(95‘7) ‘ 5 BAclo ‘H <25 -‘/ = 5 \$5 : 5 S = O l: = -1 ,', VB) 1145/ ImERSEcT fr 3. (10 points} In this exercise, you’ll demonstrate the Cauchy-Schwarz inequality in R” using a technique which is more general than the proof provided in lecture. For any two vectors 5.15 E R”, I'E- 'tﬁl S ll'ﬁll ll'tﬁll. with equality iff '5 or 15 is a scalar multiple of the other. _. Show the statement is true if '5 or 15 is 0. Explain why the function = + t15||2 is non-negative. ﬁﬁF-W :1 0‘93 ‘-.—/‘-.—/‘-.—/ Show that the function is a quadratic in t by n1ultiplying-out and collecting like terms until it is in the form a t2 + b t + c. (d) Use the quadratic discrin1inant ()2 — 411 (3 together with the previous two parts to conclude the inequality. The rest of the statement . . with equality iff . . . } will not be graded, but you should try to determine why it is true (but not on this sheet of paper}. @ 7,0 g.» 9.3 =0 75//7H=o =l> [0/5 O)E1c. / L9 IT’S ~0sz GWMZED‘ @ MW MM 212W“ WWI“ b @ M1330 =1> bm‘I/ac o 9» (mwf— 7/IWI“l/WII“< 0 => (Mrs Hiya/WI" ‘75 IV-wI</!71//IWI/ 4. (10 points) (a) In lecture we saw that. the diagonals of a parallelogram bisect. each other. Show that. the diagonals of a parallel- ogram are orthogonal to each other iff the parallelogram is a rhombus. (b) Determine the matrix A which would map the indicated parallelogram onto the unit. square. 3‘} cl 5353 H r~1 p A -___) X). H r—ﬂ A I» p * ‘xJ 6&9 THERE :5 ANOTHER szaﬂ PEI/E13923 WIENHEW. balm CS 1:1? 5. (10 points) (a) Let R9 : R2 —> R2 denote the linear transformation which rotates a vector counterclockwise around the origin by I9. Recall that the matrix of R9 is given by: “as [233% m] Therefore Rwﬂb) = Hg 0 R9, and since composition of linear operators is given by matrix multiplication, it must be true that [quHRg] = [Rmﬂm]. Use this to fact to establish the following trigonometric identities. cos 9+qD')=cosl9 cosqt'v —sinl9 sings sin(l9 + qD') : sin(l9) cos(¢'v) + cos(l9) sin(qD') 0M) ‘MWW illelll3 l : [Rm = . wﬂ/Wf/ V W) WWW) WWW U 9% wow WWW) .MW) 004W) _ . Wm) WM?) m(9)mwx-M(93M(v) -'[amovwltﬂ[email protected](goﬂ lww W lle wwwwwmmmw (b) Let 5,56 W. Show: “as”? HEM?" — (a 552 = (det.[a,5])2 ll (mantras «of “a M)“ 94., 21,949. 39; 5194.29. 99, 9,3 afab‘f’fa‘f L9 'l’ 03 L1 fagl’g ‘l’ Qa1ﬁﬂl ‘1 Q, “a La : La -QG1QQL1LQIJ— U <0©@ (D 6. (10 points) Suppose ‘511 (13 Show that. |det(A)| = ’5 - (E x 5)’ 459 Lemme Nora}. 7. (10 points} Additional questions which won’t. be graded, but. that. you should know how to solve. 1. The product. in R is cancellable: 11 7Q f} and a3: : try] :> 3: = 3;. Show that. the dot. product. in R”, n. > 1 and cross product. in R3 are not. cancellable. _. 2. Showthat. 6%6andE-fzﬁ-ﬁandd'xfzﬁx'ﬁ]:>:i:':y. 3. The product. in R is associative because (3: y} z = 3: (37.2} = 3: y .3. Because it. does not. matter where the parentheses are inserted, there is no chance of ambiguity and we usually drop the parentheses. Show that the cross product is not associative by ﬁnding some vectors in R3 such that. (if X X 2? 7Q :17? X X 2?}. R3; = 3'; :0 15M m @ aha-g -I> 3492—9) =0 =l> ELM-1?) axa =ax§ a» anal—g): 3 -> a” (ﬁg) 4> 3.06%?) 3:» 5(Z-g).(2-g)=o 4, ,2_g=3 Jefo ...
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