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Chapter 3 73

Chapter 3 73 - SECTION3.10 RELATEDRATES D 223 d 14 We are...

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Unformatted text preview: SECTION3.10 RELATEDRATES D 223 d . 14. We are given that —\$ = 24 tt/s. dt dy da: (a) 23 y2 =(90—m)2+902 => 2y— d y=2(90—\$)<—E>. When 30 2 45. y : V452 + 902 = 45 x/g so @_90—m [email protected] H 45 24:) _a \/9/0—x dt_ y dt 45f“ \/§’ H x so the distance from second base is decreasing at a rate of % m 10.7 ft/s. (b) Due to the symmetric nature of the problem in part (a), we expect to get the same answer—and we do. dz dm dz 45 24 2: 2 2 —_2 —Wh —45 :45 5 —= 24 —~10-7ft8~ z :1: +90 :> 22 dt :vdt enx z \/_ SO dt 457( )= x/g / — — 2 cmz/min. . d 15. A = ébh, where b is the base and h is the altitude. We are glven that %— — 1 cm/min and d1:— dA_ 2(1 dh db Using the Product Rule we have— dt — b— dt + h (17>. When h — 10 and A — 100 we have 100=%b(10) :, §b=10 : b:20.so2—;<20-1:10::)> a 4_20+10% : g) : 41—020 : —1.6cm/min. 15. GivenZ—i/=~1m/s ﬁnd\$when\$=8mzy 2352 +1 \$ yj—f _2m%:3- :> %:%%:—%When\$:8,y=\/ﬁ—5,SO (fl—:6 : *VT‘j—E’. Thus, the boat approaches the dock at g m 1.01 m/s. 17. We are given that % = 35 km/h and 3—: : 25 km/h. 22 2 (cc + 1;)2 +1002 dz dy => 22E:2(\$+y)<d—t+ +d—t.> At420..0PM x=4(35):140and y — 4(25) — 100 —> z — «(140 : 100)2 1 1002 _ «67,60 = 260. so dz _ _m+y dy _140+100 720 dt‘ z (d‘tdt +— — T(35+ 25) — —3 ~ 55.4 km/h. 18. Let D denote the distance from the origin (0, 0) to the point on the curve y = ﬂ D: \/(m—0>2+<y—0)2= \/x2+<\/5)2= vw2+w :» dB 1 2 71/2 dm 2334—1 dm dm —=—m+a: 2m+1—:K— W _: — dt 2( ) ( )dt 2 m2+mdt' ith dt 3whenx— 4 dB 9 27 _=h3 2—m302 t 2 20() 4 5 cm/s ...
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