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Chapter 4 110 - 372 CI CHAPTEIM APPLICATIONS OF...

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Unformatted text preview: 372 CI CHAPTEIM APPLICATIONS OF DIFFERENTIATION 61. (a) If k = energy/km over land. then energy/km over water : 1.4k. So the total energy is E: 1.4.Im/25+ac2 +k(13—m).0 g a: g 13. andso£:—fl——k. dw (25 + 52w Set fl = 0: 1.4/61: 2 k(25 + :52) d2: “2 => 1.96m2 : x2 + 25 => 0.96902 = 25 2» a: : 0596 m 5.1. Testing against the value ofE at the endpoints: E(O) = 1.4k(5) + 13k 2 20k. E(5.1) z 17.91:. E(13) % 19.5k. Thus. to minimize energy, the bird should fly to a point about 5.1 km from B. (b) If W/ L is large. the bird would fly to a point C that is closer to B than to D to minimize the energy used flying over water. If W/ L is small. the bird would fly to a point C that is closer to D than to B to minimize the . . dE ch dlstanceofthefli ht. E:W\/25+m2+L 13—51; :> —— : —— —L:0when g ( I den 25 + x2 /2 2 g 2 # By the same sort of argument as in part (a). this ratio will give the minimal expenditure of energy if the bird heads for the point an km from B. (C) For flight direct to D. a: : 13. so from part (b). W/L : @ m 1.07. There is no value of W/L for which the bird should fly directly to B. But note that lim+(W/ L) : 00. so if the point at which E is a minimum is zaO close to B. then W/L is large. (d) Assuming that the birds instinctively choose the path that minimizes the energy expenditure. we can use the equation for dE/dx = 0 from part (a) with 1.4k : 6. ac : 4. and k 2 1: (c)(4) : 1 - (25 + 42)1/2 c : \/41/4 m 1.6. strength of source 62. (a) I (ac) o< 2 . Adding the intensities from the left and right lightbulbs, (distance from source) k k k k: : + 1(m):$2+d2+(10—x)2+d2 acQ-I—d2 $2—20x+100+d2. (b) The magnitude of the constant k won’t affect the location of the point of maximum intensity, so for convenience 2w 2 (a: — 10) (2:2 + d2)2 _ (m2 A 2018 + 100 + d2)” Substituting d : 5 into the equations for I (:c)and I ’(m) we get we take k : 1. I'(m) = — 1 1 22‘, 20;: , 10) 1 : — d 1’ : __ — —————- 5(1’) 1:2 + 25 T x2 — 203: + 125 an 5(95) (:32 + 25)2 (x2 i 205 + 125)2 0.06 0-005 From the graphs. it appears that I 5 (m) k ...
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