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## ENGINEERING MATH I

by: Evert Christiansen

21

0

2

# ENGINEERING MATH I MATH 151

Evert Christiansen
Texas A&M
GPA 3.92

Amy Austin

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COURSE
PROF.
Amy Austin
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 151 at Texas A&M University taught by Amy Austin in Fall. Since its upload, it has received 21 views. For similar materials see /class/226011/math-151-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
FaH2007DJa 1151 Final Exam Practice Answers courtesy Amy Austin Final Exam Practice Sections 11 65 1 a lt 23 7 gt b lt 1 3 gt c 153 d Vector projection lt 2565 gt scalar projection 10 2 Avector equation lt 1 21 2 10t gt parametric equations 1 2 1 21 y 2 2 10t 3 above force22588 N below force23660 N 4 Formula to use f Ur 2 Iim 39 h gtU h a I39m 7 39 quot mnx 1 b 2 7 1 m x 3y 5 a QC 1 b 7 11 1 c 1 d The limit does not eXist because Iirgl at 2 17 17 h and In at 2 5 17 h e 3 1 f 6 a 3 b 2 c 5 d The limit does not eXist e Not continuous at 1 2 3 not in domain not con tinuous at 1 2 1 1 2 5 and 1 2 7 the limit does not eXist Not differentiable at 1 2 1 1 2 3 1 2 5 and 1 2 7 not continuous implies not differentiable Also not differentiable at 1 2 1 and 1 2 6 because of sharp comers gt1 9 0 39 a Not continuous at 1 2 0 because fur does not eXist Continuous for all other values of 1 b Not continuous at 1 2 1 because at 2 1 yet f1 2 1 c a 2 6 b 2 3 xy120 1133 31225 horizontal asymptote y 2 0 vertical asymptote 1 2 1 4 m 2 8 equation 3 2 8r 1 gt In 1 f 1r 2 In 1 a 2 32 b y i311 I1 I a for 1212 11134 161 1 1212 b f t 312 cos1 12 214 sin1 t2 c G m 12tm 12 1x 1 002013 1 56 y In 27 2 3 In 27 at In 3 Att 2 1 y 17 2 633 1 horizontal tangent y 2 61 verical tangent 1 2 O I r w LU 2 In2 12r 2 Qt 2 I112 12r 2 18r 22 The linear and quadratic approximations are useful in approximat ing the function for 1 suf ciently close to a dh r19 cmmrn 53 feet per second 05 feet per second Jr 2 2 1 2 Since this is not in the domain there is no solution I 26 27 28 2 D 30 31 3 33 3 4 3 U 36 46 47 Equot 90 NEW E 39 90313 11 125 Jr 2 t y 2 626t f r 2 tzmr ln1x3x3 2 2y 9 3 z 1 x3131 2 1 Where 1 1 313 72811 minutes t 2 944 hours Inc 309 Dec 003 Local Min 3 33 Local MaX None Concave up QC 0 and 2 0c concave down 0 2 points ofin ection O 6 and 2 22 0 QC Absolute Max 1 Absolute min 5 critical values 1 2 1 1 2 1 1 2 5 f inc 11 509 fdec QC 1 15 local min 13 2 1r 2 5 local maX 1 2 1 fcu 000 and 5 0c fcd 0 r1 in ection points 1 2 O 1 2 5 a 63 b 0 c 0 2515 21311 2 2112 1448745691 5 29 7L 3 3 1 3931 W lt f 5 feet by 10 feet long 21 1 5 0 4 DO U 5 ln t2 1 C 1 7T 1103 11111 13n cos 3x C 23 111 5 211 1 C

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