# n e n en1 e exercise set 113 1 a s1 2 s2 125

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Unformatted text preview: town A (m) (m) (m) (m) f (x) of cut (m2 ) 21,000 22,000 23,000 24,000 25,000 26,000 27,000 28,000 29,000 30,000 31,000 133 120 106 108 106 98 100 102 96 91 88 108 106 104 102 100 98 96 94 92 90 88 25 14 2 6 6 0 4 8 4 1 0 875 336 24 96 96 0 56 144 56 11 0 31,000 The total volume of dirt to be excavated, in cubic meters, is f (x) dx. By Simpson’s 21,000 Rule this is approximately 31,000 − 21,000 [875 + 4 · 336 + 2 · 24 + . . . + 4 · 11 + 0] = 1,229,000 m3 . 3 · 10 The total cost of the railroad from N to B is about 4 · 1,229,000 ≈ 4,916,000 dollars. 3. The total cost if trenches are used everywhere is about 17,984,000 + 5,670,853 + 4,916,000 = 28,570,853 dollars. 12 π 5 ≈ 119.27 m2 . The length of the 2 tunnel is 1000 m, so the volume of dirt to be removed is about 1000AT ≈ 1,119,269.91 m3 , and the drilling and dirt-piling costs are 8 · 1000AT ≈ 954,159 dollars. (b) To extend the tunnel from a length of x meters to a length of x + dx meters, we must move a volume of AT dx cubic meters of dirt a distance of about x meters. So the cost of this extension is about 0.06 × AT dx dollars. The cost of moving all of the dirt in the tunnel is therefore 4. (a) The cross-sectional area of a tunnel is AT = 80 + 1000 0.06 × AT dx = 0.06AT 0 x2 2 1000 = 30,000AT ≈ 3,578,097 dollars. 0 (c) The total cost of the tunnel is about 954,159 + 3,578,097 ≈ 4,532,257 dollars. 5. The total cost of the railroad, using a tunnel, is 17,894,000 + 4,532,257 + 4,916,000 + 27,432,257 dollars, which is smaller than the cost found in Exercise 3. It will be cheaper to build the railroad if a tunnel is used. CHAPTER 10 Mathematical Modeling with Differential Equations EXERCISE SET 10.1 3 1. y = 2x2 ex /3 = x2 y and y (0) = 2 by inspection. 2. y = x3 − 2 sin x, y (0) = 3 by inspection. dy dy = c; (1 + x) = (1 + x)c = y dx dx (b) second order; y = c1 cos t − c2 sin t, y + y = −c1 sin t − c2 cos t + (c1 sin t + c2 cos t) = 0 3. (a) ﬁrst order; c dy + y = 2 − e−x/2 + 1 + ce−x/2 + x − 3 = x − 1 dx 2 (b) second order; y = c1 et − c2 e−t , y − y = c1 et + c2 e−t − c1 et + c2 e−t = 0 4. (a) ﬁrst order; 2 5. dy dy y2 dy 1 dy =x + y, (1 − xy ) = y 2 , = y dx dx dx dx 1 − xy 6. 2x + y 2 + 2xy dy = 0, by inspection. dx d y e3x = 0, ye3x = C, y = Ce−3x dx dy = −3dx, ln |y | = −3x + C1 , y = ±e−3x eC1 = Ce−3x separation of variables: y d (b) IF: µ = e−2 dt = e−2t , [ye−2t ] = 0, ye−2t = C, y = Ce2t dt dy = 2dt, ln |y | = 2t + C1 , y = ±eC1 e2t = Ce2t separation of variables: y dx 7. (a) IF: µ = e3 8. (a) IF: µ = e−4 = e3x , x dx = e−2x , separation of variables: 2 2 2 d y e−2x = 0, y = Ce2x dx 2 2 dy = 4x dx, ln |y | = 2x2 + C1 , y = ±eC1 e2x = Ce2x y d y et = 0, y = Ce−t dt dy = −dt, ln |y | = −t + C1 , y = ±eC1 e−t = Ce−t separation of variables: y (b) IF: µ = e 9. dt = et , 1 1 y y dy = dx, ln |y | = ln |x| + C1 , ln = C1 , = ±eC1 = C , y = Cx y x x x 10. dy 1 = x2 dx, tan−1 y = x3 + C, y = tan 2 1+y 3 11. √ √ x dy 2 2 = −√ dx, ln |1 + y | = − 1 + x2 + C1 , 1 + y = ±e− 1+x eC1 = Ce− 1+x , 2 1+y 1+x √ 1+x2 y = Ce− 13 x +C 3 −1 343 Exercise Set 10.1 1 x3 dx y 2 = ln(1 + x4 ) + C1 , 2y 2 = ln(1 + x4 ) + C, y = ± [ln(1 + x4 ) + C ]/2 , 42 1+x 4 12. y dy = 13. 14. 1 +y y 17. 18. dy = ex dx, ln |y | + y 2 /2 = ex + C ; by inspection, y = 0 2 2 dy = −x dx, ln |y | = −x2 /2 + C1 , y = ±eC1 e−x /2 = Ce−x /2 y 15. ey dy = 16. 344 sin x dx = sec x tan x dx, ey = sec x + C , y = ln(sec x + C ) cos2 x dy x2 + C, y = tan(x + x2 /2 + C ) = (1 + x) dx, tan−1 y = x + 2 1+y 2 dy dx 1 1 y−1 = , −+ dy = csc x dx, ln = ln | csc x − cot x| + C1 , y2 − y sin x y y−1 y 1 y−1 = ±eC1 (csc x − cot x) = C (csc x − cot x), y = ; y 1 − C (csc x − cot x) by inspection, y = 0 is also a solution. 1 3 cos y dy = dx, dy = 3 cos x dx, ln | sin y | = 3 sin x + C1 , tan y sec x sin y sin y = ±e3 sin x+C1 = ±eC1 e3 sin x = Ce3 sin x 19. µ = e 20. µ = e2 21. µ = e 22. 23. 24. 25. 3dx x dx dx ex dx = ex + C , y = e−2x + Ce−3x = e3x , e3x y = 2 = ex , 2 2 2 2 d 12 1 y ex = xex , yex = ex + C, y = + Ce−x dx 2 2 = ex , ex y = 1 dy + 2y = , µ = e dx 2 2dx ex cos(ex )dx = sin(ex ) + C , y = e−x sin(ex ) + Ce−x = e2x , e2x y = 1 1 1 2x e dx = e2x + C , y = + Ce−2x 2 4 4 2 2 1 x dy +2 y = 0, µ = e (x/(x +1))dx = e 2 ln(x +1) = dx x + 1 C d y x2 + 1 = 0, y x2 + 1 = C, y = √ dx x2 + 1 1 dy +y = ,µ=e dx 1 + ex dx = ex , ex y = x2 + 1, ex dx = ln(1 + ex ) + C , y = e−x ln(1 + ex ) + Ce−x 1 + ex 1 1 dy d + y = 1, µ = e (1/x)dx = eln x = x, [xy ] = x, xy = x2 + C, y = x/2 + C/x dx x dx 2 3 1 (a) 2 = y (1) = + C, C = , y = x/2 + 3/(2x) 2 2 (b) 2 = y (−1) = −1/2 − C, C = −5/2, y = x/2 − 5/(2x) 345 26. Chapter 10 2 2 x2 dy = x dx, ln |y | = + C1 , y = ±eC1 ex /2 = Cex /2 y 2 2 (a) 1 = y (0) = C so C = 1, y = ex 12 1 = y (0) = C , so y = ex /2 (b) 2 2 27. µ = e− x dx = e−x 2 2 y = −1 + Cex 28. µ = e dt /2 /2 , e−x 30. /2 y= xe−x 2 /2 dx =...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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