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Unformatted text preview: Section 4.6
34. Step 1: s shadow length sun's angle of elevation Step 2: At the instant in question, s 60 ft and
d dt 185 Step 6:
d dt (20)( 2) (10)(1) 102 202 0.1 radian/sec 5.73 degrees/sec To the nearest degree, the angle is changing at the rate 0.0015 radian/min. of 6 degrees per second. 0.27 /min Step 3: We want to find Step 4: tan Step 5:
ds dt 80 or s s ds . dt 36. Step 1:
A c a 80 cot 120 O b B 80 csc 2 d dt 80 and 60 5 . 4 Step 6: Note that, at the moment in question, since tan 0
2 a b c
4 and so csc 5 distance from O to A distance from O to B distance from A to B Step 2: At the instant in question, a b 3 nautical miles,
da dt , we have sin
ds dt 5 2 80 (0.0015 ) 4 12 in ft 0.1875 1 ft min 5 nautical miles,
db dt 14 knots, and 21 knots. Step 3: We want to find Step 4: Law of Cosines:
dc . dt 2.25 in./min 7.1 in./min
ds Since dt 0, the rate at which the shadow length is c2 c2 a2 a2 b2 b2 2ab cos 120 ab decreasing is positive. The shadow length is decreasing at the rate of approximately 7.1 in./min. 35. Step 1: a distance from origin to A b distance from origin to B angle shown in problem statement Step 2: At the instant in question, a 10 m, and b 20 m.
da dt Step 5: 2c
dc dt 2a da dt 2b db dt a db dt b da dt Step 6: Note that, at the instant in question, c a2 2(7)
dc dt dc 14 dt dc dt b2 ab (5)2 2(3)(21) (3)2 (5)(3) (3)(14) 49 7 2 m/sec, db dt 1 m/sec, 2(5)(14) 413 29.5 knots (5)(21) Step 3: We want to find Step 4: tan Step 5:
d dt 1 1
a 2 b d . dt The ships are moving apart at a rate of 29.5 knots. 37.
dy dx dx 10(1 x 2) 2(2x) dx dt dt dx Since 3 cm/sec, we have dt dy 60x cm/sec. dt (1 x2)2 dy dt dx 20x x 2 ) 2 dt a or b a tan 1 b (1 b da dt a b2 db dt b da dt a db dt a2 b2 (a) (b) (c) dy dt dy dt dy dt [1 60( 2) ( 2)2]2 60(0) 0 2)2 120 52 24 cm/sec 5 (1 0 cm/sec 0.00746 cm/sec 60(20) (1 202)2 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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