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Sample Final Exam Fall 2011

# Sample Final Exam Fall 2011 - Math 2123.2 — Fall 2011 ~...

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Unformatted text preview: Math 2123.2 — Fall 2011 ~ Sample Final Exam Solution To receive credit, you must show your work. 1. Let L be the line through the points (—1, 3) and (2,1). (a) Find the slope of L. (b) Write an equation for L. (c) Find the :c—intercept of L. f . f (d) Find the y—intercept of L. , 1 — 3 2 "A __ 1 E f (a) The slope is 2 _ -1 = —3. ’ . . . 2 . 2 4 (b) We use the pomt—slope. An equation is y — 1 : —§(x — 2). It may be rewritten as y— 1 = —§a:+ g 2 7 or asyz —§m+§. (c) To ﬁnd the x—intercept we solve 7&3: + g = 0. It gives in: = g and so :3 = :1 - g = g. The 7 m—intercept is (5, 0). 2 7 7 (d) From the slope—intercept form equation y = —§a: + g, the y—intercept is (0, §)' —f/r 2 WJ‘\ .1 ’ ‘2‘ I R y—eﬂtb *1“1‘- 1113 ¢ 2. Use appropriate derivative rules to calculate the derivative of 71" (275333 The function can be rewritten as y : x(x2 + 1)1/2. To ﬁnd the derivative, we use the product rule and then the generalized power rule: ,. 7 ( .,.. -' 1 X J" d_y ‘\ 2 1 2 1/2d d 1 p“ ((93, +1) / )+ (:02 +1) 56(1) \._r .1, (1—33 :‘jdé ﬂ , , /_..d -__ . V :1 X 2 __ —1/2 1/2 (x 1) 61—93(902 +1)+(12+1) 1 0’01: ZX (12--1)—1/2.(2\$)+(12+ 1W2 m : I 2(1'2 + 1)— 1/2 +(122 + 1)1/2 :~(ac2 1—2)1/2(x +9162 +1) =(cc2— 1) 1”(2.71:2 +1) _ 23:2 +1 x2+1' if 1, 2 —:c 1 2 :3: [The simpliﬁcations from the third line on are optional.] ”9 ‘ t/27. 1‘ a} I an} 1-1131, ‘ #51 (341+?) 2': 1—11/5Céw £7 . "I 1/! ‘{ Tye/1:; 11/“ng E w' 2 ; ref] 7 Z I ’ refit-“1% a /1) ;~ 1 W? 7 \$2 4y2 [/K ‘/,A_../~ — — -—- : 1 .1 ’f k m 9 9 C LI fl .uL&L\‘ 1 33.2. _ 4-2 = 1. 0‘ ‘ j: 3 ‘ L i C} (a) The vertices are (:3, 0). (b) The asymptotes are y = ::¥:c = ::%:c. (G) Since 32 + (3/2)2 = 9 + (9/4) 2 45/4, the foci are (::\/45/4, 0) m (:34, 0). (d) The sketch is shown to the left. 1 X ‘* 2 4 ‘f 4 1 1,1 _ “a“ 1: T? -. C. — i. I Os L K! ( LL”. ) Z 1/“ F ‘ 2‘ —« ﬁ ‘3 j :C / 6. Evaluate /1\/_ -—3— 2da:. .1 . \ —~1 “C 1/2 {E 4/ * w j» \ ,— / 4 cs 1 A _,_‘ ..._. f "'C —3—d\$=1\/—x_3dcc 1"2 \ ‘gﬁt C C :26» [‘1‘ ‘/ ‘/1‘/2\/§ \$3 /2M___,,_,L m.‘\ / 2,13, .Y CHIC/A- 2 1‘ ) ‘ X 1 _2 “7:1 \ 3; .51 : 2~ —a: i / ‘ _2 9” 1/2 .1“— m (r: 1 ’ =_L1\$:1 git... J‘ V ., \‘ ﬂ 3327;: 1/2 ,a’/ i V * g :71, C ELL 1 «f9 z 1 1 - «i (1)? ﬂ (1/2? ‘3; , 1 + (/4. N '51:": in? LEFT" ’— . l V 2 _: 2 “x 1’1 3M" 1 f if .3 1. Jr /\ i. i ﬂ : i :57 \ Z.» 1 K i b 1. fﬂf'jﬂj ‘ {1‘1 ﬂ fyﬂFv / o ”“5 ' ’61 s I 5. Consider the hyperbola with equation '\ J \L L13 _ "1.. \ m2_4y2:9 7*“ 13W at (a) Find the vertices of the hyperbole. 1 ‘ (b) Find the asymptotes of the hyperbola. ”X,“ (c) Find the foci of the hyperbola. ‘33 (d) Sketch the auxiliary rectangle, the asymptotes, the foci, and the hyperbola on the Cartesian coordinate system shown below. We start by rewriting the equation in standard form: [V \/ g x \ 4. w it \:~ , ”7"“: l w» J 9. Lety=m4~4m3. H "at 1‘ l (a) Find the critical values of a3. “ ’ J I (b) Make a Sign chart for the derivative. {/1 V J (a) We have y’ 2 4w3 — 12a:2 2 4902(35 A 3) and this is zero when ct = 0 or :c = 3. The critical sic-values arew=0and\$=3. (b) ‘ a: l test value I y’ | sign 2 > -, . ’ ‘ ‘5‘- (~oo,0) —1 716 7 a L1 l ,, 0“ [forum / u“ l' (073) 1 —8 4 / ”iv 3‘;- tg. \ f ‘ (3, oo) 4 64 + (c) Since the derivative is negative both before and after, the critical point at (0,0) is neither a maximum point nor a minimum point. Since the derivative is negative before and positive after, the critical point at (3, —27) is a relative minimum point. //"/7\\‘ 10. Find the volume of the solid obtained by revolving the region below the parabola y é\l— m2 and above the m—axis about the y-axis. [Use the shell method] “V We start by sketching the region with a typical element of area displayed. The corresponding element of volume is dV 2 277\$(l—x2) dab : 27T(:L’ — 3:3) dx./ Thus the total volume / ls , ~~~W ~7~A~ 7.,7.._»«r~-~~ m“, r,__/ r» 7 f 1, > V 23/ (.327T(.71 — \$3)d.’E 0f _ _ 1 2 1 4 (11:1 —27r(2a: — 4x ) x20 _ 1 2 1 4 1 2 _ 1 4 a 277(2(1) 4(1) ) 27r(2(0) 4(0) ) 1 _ 27f - Z — 27T ' 0 2 ~ 0 “1.4..“ .\‘/» ‘ 1 . 1., i. 1 ﬂ 1 / r1 \ 1 l .1 1 1 " _ 1 7. Evaluate 7 ,1 + dx. V. . = 1 90 \i--/ . 1 d . We use the General Power Form with u— — 1 +\$ - :2 11—504 and n — 1/2 Then du- - 31—1- dsc = —.1‘"2 (.156. M» , m . a: Note that this 1{ off by a negative Sign (a factor of —1)\frorn the integral, so we have to correct this ﬁrst. mm _. 1-1 /' . \ - . V, .x 1.. 1 F7 1/2 “1‘ b1111“\1_,/—2— 1+— dxzﬁé/ (1+ —> -——2dx 1 -11 A“ / 11‘ l ‘ 1u1/2+1 S ’1 : ’) ,l,« + C ((1/2) + 1 (\ M \$15. # '3 [The last line is optional.] 8. Evaluate/ (2§.—:1)6‘2dm. M ' C1“: 11,11 0 . \\ We use the General Power Form with u— — 2m — 1 and n— — 6. Then du: 2—: dm— - 2 dz“. When ac- # 0, u— — 2(0) — 1: ——1 and when a:- — 1, u — 2(1)— 1: 1. Note that du is ofl from the given integral by a factor of 2 so we have to correct this a U1)» 3‘1-1 g 3 (a 1 1 1 1 ’3 f0 (21:1 1)6d:c = 5/0 (2:10 - 1)62 dye». . c1 11, " \A “xvi“: LJA“ 21/1 U6 du (1 1 2 ”1 \15 {3‘64 _1 1 “711:1 \ &\\1\.¥ &\\ —§ 7“ 11:71 1. _L 7 L 7 — 1411) T 141*1 ) . 1 -1"; [/1 ., l1 _ 14 14 . 1 1 1, : ZE' \C1 1 1*» \ 1 1:1... 1 g” \1 ‘\ . 1 1 . 1 1 1, <1 1 , 1=~ 1 -' 3. Consider the circle with equation x2+y2—2x+4y+4:0. (a) Find the center of the circle. (b) Find the radius of the circle. 'r \ (c) Sketch the circle on the Cartesian coordinate system shown below. T )L )' Ail l l’ 45 , ~. .3; ( 7) (L‘ 1)2 + (y +7??? This is the equation of the circle in standard form. v-«mwwwwmwmw (a) The center is (1, *2). (b) The radius is 1. (c) The circle is shown on the graph. l f ‘7/ 1 ‘ 7:“‘dﬂ ’f";§F f - 3 7 . . . . w,» i» I " 4. The function y = 1: + ﬂ + EEK/has a critical pomt at (1, 3.5). y 3» .a l 3 { m 7: \L A \(L‘h’ 2;: 1. a x (a) Calculate y’. . :~ ; . i‘a ' ~ , , “ e 4 (b) Calculate y”i /_ ’ {Fiﬁ 6 .u 3;? 3 ’ ,Ni '1 a We haveyzx+ac1/2+ —:c_1 and so .3 “x. \ 2 ” ’4 :5 > i i J i 7* W» l .3 “‘ y121+1\$71/2+é.(_1)m72 ‘ 2 2 1 3 :1 _ 71/2w_ —2- + 2:1: , 2w 1 1 3 1 b h // _ 0 . —3/2 . 2 j—3 __ —3/2 —3. ( ) We avey + 2 23: 2 a — 435 +336 1 3 (c) We have y”|\$=1 : —Z(1)—3/2 + 3(1)_3 : ZZ > 0. Since this is positive, the second derivative test says that the critical point (1, 3.5) is a relative minimum point ...
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Sample Final Exam Fall 2011 - Math 2123.2 — Fall 2011 ~...

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