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Unformatted text preview: 1. (16 pts) The position of an asteroid at time t > 0 is given by \ 77(75)— — e i+2€ sintj +26 cost/c :75, t<2/2 (7272/2; 257727 2,}
(a) (3 pts) Verify that the asteroid’s Speed, as a function of t, is 36.
Uzi/K2222 622/2U U222 V2272: 32<2 2/2/2227 2272727> ,2 ._//‘ /2 2H),) 7 1/ 2222/
(72’1“ 2 ///I/H 2/7717 250.42, 27/75/2732; jam/227 22172251? 252722 2722227/712 / ((2727 2/2 (202/747 ’ ,:7///I 2 “MM A.,;,,W,nﬂ,.. r ,,,.,,,7,“:, ‘,,,,L “.777, , I /yﬁjr_w (l / f . I j "I
7;; (I: 2 21227 7 __7 </  I / ,1 222,277, I, 2 24/727 2, “2722 «2/72,, I 2 272/2” 2 1. 212/ 2 r 7257 7 775, a, ”97”,” 2 4/w / A 4/1
2/} 22M 2 2 2 (2 1:: ’3‘; 2 1/
(b) (2 pts) Find the arclength, as a functiOn of time t, of the asteroid’s
path from t— — —oo to time t. 7 / ﬂ
2’2? {7 j) , I71 7" / I; 2 27'
7222 :1“ 2 22 $222222” ~79; f 72259“ (/2272 )22: / 3 1726 2 A/ 2 2{
L272 / V07 2 <2 I I (4:; 22; 221’ "A7 , (1.72 (c ) (4 pts) Find the unit tangent vector 21—2 and the unit normal vector
N as functions of t. 2’) j: , 1/71/02 77222124072“ 2/0727 27/2/22) 22 H ,, , ”22
1“, Jr, 7: / \ l :2, 2 7 (25' j t
1. )I , C» /2 2/ I
5/ 3 ML 2 j (272/ ,7 7 / /
:7,» ~\/ ( ) 2/ ,2, r '2: /\ J [1/ I \) 7)
2 <2)/ J (22772 f, 12/2I7/ 3 272/227 7 [2 2 ,7 \(22) (1262 {2/} </ ///2( god
. _ A, ,, ‘ j " ’ ’ ' 7)
3y , , 7 z ,7 f7, MAJ/2 277/2 2 1' 71/) 2 7222272 . 7’
222 22 7» 222/2 (1M2, 722222 2(5’222 (2252/2 " ’7 2/” 2' / 2 ’e I MI W 2
T: 21 L
(3) x2 27 (d) (3 pts) Find the curvature of the asteroid’s path as a function of t. )4 M W— , ”if/l &O ﬂ" J’s/i, (e) (4 pts) Find the tangential and normal components of the asteroid’s
acceleration/(1T and cm, as functions of 25:7 , (f_( 2. (8 pts, 2 pts each) Match each of the following four functions With the
graphs on the following page.
. Function: . . Graph: 2/ \/ 1 .
(/M/ W: 0/ ﬂ” ,, > f($, y) : g Slﬂ<$2 —— 1/2) ___C_‘.__..__ (/II ’75 / a (KW/W31 7 ﬂ A
W WW» ‘ III/<75! ﬂaw) = m ___.___ 4 f($,y) = '3Sln(4x +y)
I/ ”WI (WW/Ci; (II/‘6? 4
I”, It] {2/ \/// (Wm,6 cw; 5mg, MW 3. (10 pts) For each of the following limits, compute it or prove that it does
not exist. For full credit you must justify your answers! 1 . ac—l—y
a 5 ts 11m 1 293+ 8111 M
( ) < p ) (””11711r2>( 31) (Im— 1 + ly+2l) 15/ 1011111671, 17011‘71111 1 (1/14 1 11 (/7717 161/} (1% 11A 1; 7/1 /V1ol\/f/ 61/6/151111’7/1 My: 1911/11“ 11 1/ 1/ (,1? 12% 14/1 171/1( 11/ 11/2/111/ 17/1 1 I ,,
17/1/7114, (7W11/I/11 1 1‘7 [111 113W 1/ 1 1 1131,1111 [/,/‘/11:2%1711i/})11(1111/1211//Q 1 b 5 It 1 my 42,5121
()( ps) (931/)1’If100) 1132+y4 1,. 1 ””1111 W1 7’1 ”11 A415; @’//7/“‘7(l,«z)/ 411 m1: 1/ {/1/1/‘1 17/71/11 4. (14 pts) You have designed a rollercoaster track so that at the point
(—10, 5, —24), the center of the osculating circle is at (30, —3, 20). (All
positions are in meters) Suppose that When the rollercoaster car gets
to that point, its acceleration is (16 —5, 5)m 73, and it is slowing down. (a) (3 pts) Find the unit normal vector N to the track at that point. (Hint: Remember. when computing a magnitude factor out anything that you can ) Vain 1775le WI JIM/II I70 (III?! I777 I15” «III/I71 </V (7 (77/37 7777/ 2777 (7777/77) 77 <LW A (/(I/ L/ <M/ 7/ ((>
ARK7777777 II/i717I3/I ((7 J (974/217 I”? 7" L’( 7W7 7/77///71;/// I; .\ /II {M ~’,' I‘ 1 // ,: '1)! (b) (3 pts) Find the normal component Of the car; s acceleration (LN. I / ;_I 1'”. 57/ \
é” ﬂl A} <75, 57/ J > \ .37 19, I52“; /
:fI/ / 3) U) 1/ J (M ‘ / (c ) (5 pts) Find the rate at Which the car is slowing down, and the
direction 1n Which it is moving (as a unit vector). n77 In I“ [7/ 77 4,! I I II, III II,< I/ I55 'I‘ \ ’7' 7 , “7' " r g!
{7’9 / <_ / 1/1 / > /<' 1;} g/ f > / \, N V \[éi’Wf/y (Ii/Iw/I (437 ( pts) HOW fast IS the car geing at that point? 7 (HM/w 7,777 M 6,7177, 7) 7777\4/77Iug (II77 {9:75. ALIII’AII III 7‘ H 7(< {Ii/“Q! 7,7 I 77/17/57 t: éﬂ m ‘l I, I II III ~ 1 I II I III / I III TII I ...
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