Chapter 3 112

# Chapter 3 112 - 262 CHAPTER 3 PROBLEMS PLUS 29 5 By similar...

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Unformatted text preview: 262 CHAPTER 3 PROBLEMS PLUS 29. _ 5 By similar triangles. % = 116 i 7" = 51—2. The volume of the cone is 2 _1 2_1 5h _257r3‘dVﬁ257r db 4 . ‘_ \/ V—E‘IT'I" h—57r<1—6> h—ﬁh.soE—ﬁh2E.Nowtherateot ‘6 l \[281 h , change of the volume is also equal to the difference of what is being added i Z (2 cma/min) and what is oozing out (knrl. where 7rrl is the area of the cone and k: . . . dV IS a proportionality constant). Thus. a? 2 2 — kn’rl. dV dh 5 10 25 l 10 Equating the two expressions for :1? and substituting h # 10. dt = 0.3. 7‘ ' i182 ' 8 and V587 : E 125k \/ 281 750 . 4:) l2; 281. we get 3—:%(10)2(—0.3) : 2 7 k17r%% 281 4:} ———:;—4—— : 2 + 3%. Solvmg for k gives us k : M. To maintain a certain height. the rate of oozing. kTrrl. must equal the rate of the liquid 25077 \/ 281 _ dV 256 + 375w 25 5 \/ 281 256 + 375% 3 . b' d’;tht. :O.k: [A t . . ————z11.204cm mln. elng poure in a IS dt 7r?" 2507r \/2—8T 7r 8 8 128 / ...
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