Chapter 3 77 - 34 35 36 37 SECTION 3.10 RELATED HATES 227...

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Unformatted text preview: 34. 35. 36. 37. SECTION 3.10 RELATED HATES 227 We are given that 2—: = 4(27r) = 8n rad/min. cc 2 3tan19 => 3 :ll—acz3sec2 Bd—f When$:1,tan0=%,2sosec6=1+(%)2 2130and t dm 80” N p x 2% — 3(30)(87r)— T N 83. 8km/min We are given that d_m = 300 km/ h. By the Law of Cosines, dt X y2 = m2 + 12 ~ 2(1)(m)cos 120° :32 +1 *234—5) 2 (c2 +m+1,so . dy dm dm dy_ 2m+1dm ) — — — .Afterl minute 1 2ydt: 29” dt+d—t : dt 2y dt' m:3—6000:5km :yzx/52+5+1:\/31km ;> 1 @ : 2(5) +1(300) = £9 2 296 km/h. dt 21/33 Vi dac dy . . We are given that a — 3 mi/h and— dt — —2 m1/h. By the Law of Cosmes, y Z 222m2+y2~2$ycos45°:m2+y2—\/§my => dz dcc dy dy d2: , 1 . 1 t =—h, x 22dt 29: dt+2y dt fimdt fly dt After 5m1nues[ 4 ] v13—6 2 wehavex:%andy=§=% :> 22:(%)2+(%)2—\/§(%)(§) :> z:T\/_and 2 13— 26 d2— 2 amen wen—fie —E— f 4‘13 6 Wm“ m 2.125 mi/h. )3] _ \/13—6\/_ Let the distance between the runner and the friend he Z. Then by the Law A‘ of Cosines, 200—61 22 : 2002 + 1002 — 2 ~ 200 4 100 ~ c030 : 50,000 — 40,000 0050 (1). Differentiating implicitly with respect to t, we obtain 22 g 2 ~40 000(‘ sin 0) CL: .Now if D IS the distance run when the angle is 19 radians, then by the formula for the length of an an: on a circle, 3 : r0, we have D = 1000, so d0 1 dB 7 d6 — 100 —D => E — m d—t 1—00 To substitute into the expression for a we must know sin 19 at the time when Z : 200, which we find from (*): 2002 = 50,000 — 40,000 c050 (1) cos 0 = :1 => sin19 : 1 — (if 2 ¢T_ Substituting, we get 2(200) 2%: 40,000\/TB (W70) : dZ/dt : 134E % 6.78 m/s. Whether the distance between them is increasing or decreasing depends on the direction in which the runner is running. ...
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