Section 3.1 Homework Solutions. 1. y = 5, so dy dx = y = 3x, so dy dx = y = x 12, so dy. dx = 12x11. dx = 12x 13
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1 Math y = 5, so dx = 0 2. y = 3x, so dx = 3 3. y = x 12, so dx = 12x11 4. y = x 12, so dx = 12x y = x 4/3, so dx = 4 3 x1/3 6. y = 8t 3, so = 24t2 7. y = 3t 4 2t 2, so = 12t3 4t 8. y = 5x + 13, so dx = 5 Section 3.1 Homework Solutions 9. fx) = 1 x 4 can be rewritten as fx) = x 4, so f x) = 4x fq) = q , so f q) = 3q y = x 2 + 5x + 9, so dx = 2x y = 6x 3 + 4x 2 2x, so dx = 18x2 + 8x y = 3x 2 + 7x 9, so dx = 6x y = 8t 3 4t t 3, so = 24t2 8t y = 4.2q 2 0.5q , so dq = 8.4q y = 3x 4 4x 3 6x + 2, so dx = 12x3 12x gt) = 1 t 5 can be rewritten as gt) = t 5, so g t) = 5t fz) = 1 z 6.1 can be rewritten as fz) = z 6.1, so f z) = 6.1z y = 1 r 7/2 can be rewritten as y = r 7/2, so dr = 7 2 r 9/2
2 20. y = x can be rewritten as y = x 1/2, so dx = 1 2 x 1/2 21. hθ) = 1 3 θ can be rewritten as hθ) = θ 1/3, so h θ) = 1 3 θ 4/3 22. fx) = 1 x 3 can be rewritten as fx) = x 3/2, so f x) = 3 2 x 5/2 23. y = 3t 5 5 t + 7 t can be rewritten as y = 3t5 5t 1/2 + 7t 1, so = 15t4 5 2 t 1/2 7t y = z z can be rewritten as y = z z 1, so = 2z 1 2 z y = 3t t 1 t 2 can be rewritten as y = 3t2 + 12t 1/2 t 2, so = 6t 6t 3/2 + 2t ht) = 3 t + 4 t 2 can be rewritten as ht) = 3t 1 + 4t 2, so h t) = 3t 2 8t y = xx + 1) can be rewritten as y = x 3/2 + x 1/2, so dx = 3 2 x1/ x 1/2 28. hθ) = θθ 1/2 θ 2 ) can be rewritten as hθ) = θ 1/2 θ 1, so h θ) = 1 2 θ 1/2 + θ fx) = kx 2, so f x) = 2kx 30. y = ax 2 + bx + c, so dx = 2ax + b 31. Q = ap 2 + bp 3, so dq dp = 2aP + 3bP v = at 2 + b t 2 can be rewritten as v = at2 + bt 2, so dv = 2at 2bt P = a + b t can be rewritten as P = a + bt 1/2, so dp = 1 2 bt 1/2 34. V = 4 3 πr2 b, so dv dr = 8 3 πrb 35. w = 3ab 2 q, so dw dq = 3ab2 36. hx) = ax + b c can be rewritten as hx) = a c x + b c, so h x) = a c 37. ft) = t 2 4t + 5, so f t) = 2t 4 which gives f 1) = 2 and f 2) = 0. You can also graph y = x 2 4x + 5 and use the built-in /dx button at x = 1 and x = fx) = x 2 + 1, so f x) = 2x. Thus f 0) = 0, f 1) = 2, f 2) = 4, and f 1) = 2. You can also graph y = x and use the built-in /dx button at x = 0, x = 1, x = 2, and x = 1.
3 39. fx) = x 2 + 3x 5, so f x) = 2x + 3 which gives f 0) = 3, f 3) = 9, and f 2) = ft) = 700 3t 2, so f t) = 6t. Thus f5) = 625 and f 5) = 30. Five years later in 2010, the height of the sand dune is 625 centimeters and decreasing at a rate of 30 centimeters per year. 41. P t) = t 3 + 4t + 1, so P t) = 3t Thus P 2) = ft) = 2t 3 4t 2 + 3t 1, so f t) = 6t 2 8t + 3 and f t) = 12t ft) = t 4 3t 2 + 5t, so f t) = 4t 3 6t + 5 and f t) = 12t Zt) = 300t 2, so Z t) = 600t. Thus Z4) = 4800 and Z 4) = After 4 months, there are 4800 mussels in the bay and they are growing at a rate of 2400 mussels per month. 45. fx) = x 3 and f x) = 3x 2 so that f2) = 8 and f 2) = 12. Thus we need the equation of a line through the point 2, 8) with a slope of 12. From the point-slope formula we get y 8 = 12x 2) which simplifies to our answer y = 12x 16. From the graphs we see that the estimates will be underestimates. 46. ft) = 6t t 2 and f t) = 6 2t so that f4) = 8 and f 4) = 2. Thus we need the equation of a line through the point 4, 8) with a slope of 2. From the point-slope formula we get y 8 = 2t 4) which simplifies to our answer y = 2t fx) = 2x 3 5x 2 + 3x 5 and f x) = 6x 2 10x + 3 so that f1) = 5 and f 1) = 1. Thus we need the equation of a line through the point 1, 5) with a slope of 1. From the point-slope formula we get y + 5 = 1x 1) which simplifies to our answer y = x Cq) = q 2, so C q) = 4q. Thus C 25) = 100 dollars per item, so the marginal cost of producing the 25th item is $ fx) = x 10x 2, so f x) = x. Thus f5) = 770 and f 5) = 40. If 5 pounds of fertilizer are used on each acre, then the yield on each acre is 770 bushels and increasing at a rate of 40 bushels per pound of fertilizer. More fertilizer should be used. 59. y = t 2, so velocity = = 32t. The height of the ball is decreasing as it falls so the sign of the velocity, as expected, is negative. The height of the ball is 0 when it hits the ground. Now solving 0 = t 2, we get that t ± Only the positive value makes sense here. Plugging this into the velocity equation gives feet per second. This converts to about miles per hour.
4 Math 122 Section 3.2 Homework Solutions 1. fx) = 2e x + x 2, so f x) = 2e x + 2x 2. P = 3t 3 + 2e t, so dp = 9t2 + 2e t 3. y = 5t 2 + 4e t, so = 10t + 4et 4. fx) = x x, so f x) = 3x 2 + ln 3)3 x 5. y = 2 x + 2 x 3 can be rewritten as y = 2x + 2x 3, so dx = ln 2)2x 6x 4 6. y = 5 5 t t, so = 5ln 5)5t + 6ln 6)6 t 7. fx) = 2 x x, so f x) = ln 2)2 x + 2ln 3)3 x 8. y = 4 10 x x 3, so dx = 4ln 10)10x 3x 2 9. y = 3x 2 4 x, so dx = 3 2ln 4)4x 10. y = 5 2 x 5x + 4, so dx = 5ln 2)2x P t) = ) t, so P t) = 3000ln 1.02)1.02 t 12. P t) = ) t, so P t) = 12.41ln 0.94)0.94 t 13. P t) = Ce t, so P t) = Ce t 14. y = B + Ae t, so = Aet 15. fx) = Ae x Bx 2 + C, so f x) = Ae x 2Bx 16. y = 10 x + 10 x can be rewritten as y = 10x + 10x 1, so dx = ln 10)10x 10x R = 3 ln q, so dr dq = 3 q 18. D = 10 ln p, so dd dp = 1 p 19. y = t ln t, so = 2t + 5 t 20. Rq) = q 2 2 ln q, so R q) = 2q 2 q 21. y = x 2 + 4x 3 ln x, so dx = 2x x
5 22. ft) = Ae t + B ln t, so f t) = Ae t + B t 24. y = 3 x, so dx = ln 3)3x. At x = 1, we get y = 3 and = 3 ln 3. So our point is 1, 3) and the dx slope is 3 ln 3. Now the point-slope formula gives y 3 = 3 ln 3x 1) which simplifies to our answer y = 3 ln 3) x ln 3. Thus y 3.3x The graph of fx) = 1 e x crosses the x-axis at x = 0. f x) = e x so that f 0) = 1. Thus we need the equation of a line through the point 0, 0) with a slope of 1. From the point-slope formula we get y 0 = 1x 0) which simplifies to our answer y = x. 26. ft) = ) t, so f t) = 1040ln 1.3)1.3 t. We obtain f0) = 1040 and f 0) so in the year 2000, the worldwide production of solar power is 1040 megawatts and increasing at megawatts per year. We also obtain f15) and f 15) so in the year 2015, the worldwide production of solar power is expected to be megawatts and increasing at megawatts per year. 27. P = ) t, so dp = 10.8ln 0.994)0.994)t. Thus at t = 20, we have P and dp So in the year 2010, the population is predicted to be 9,575,279 and decreasing by 57,625 people per year. 28. V = ) t, so dv = 75ln 1.35)1.35)t dollars per year. 29. Since P 0 = 1 we get P = 1.05) t and P = ln 1.05)1.05) t. At t = 10, P 0.08 so prices are rising by 8 cents per year. 31. P = ) t so dp = 4.1ln dp 1.02)1.02)t. At t = 0, increasing by billion people per year. At t = 25, dp increasing by billion people per year so in 1975, population is so in 2000, population is 33. y = ln x and y = 1 x so that y1) = 0 and y 1) = 1. Thus we need the equation of a line through the point 1, 0) with a slope of 1. From the point-slope formula we get y 0 = 1x 1) which simplifies to our answer y = x 1. Thus ln x x 1 for values of x near 1. We get ln , ln 2 1, ln , and ln From the graphs of y = ln x and y = x 1, we see that these approximate values are larger than the true values. 34. fx) = e x and f x) = e x so that f0) = 1 and f 0) = 1. Thus we need the equation of a line through the point 0, 1) with a slope of 1. From the point-slope formula we get y 1 = 1x 0) which simplifies to our answer y = x + 1. From the graphs of y = e x and y = x + 1 it is clear that e x 1 + x for all values of x.
6 Math d 4x 2 + 1) 7) = 74x 2 + 1) 6 8x) = 56x4x 2 + 1) 6 dx 2. fx) = x + 1) 99, so f x) = 99x + 1) 98 1) = 99x + 1) R = q 2 + 1) 4, so dr dq = 4q2 + 1) 3 2q) = 8qq 2 + 1) 3 4. w = t 2 + 1) 100, so dw = 100t2 + 1) 99 2t) = 200tt 2 + 1) w = t 3 + 1) 100, so dw = 100t3 + 1) 99 3t 2 ) = 300t 2 t 3 + 1) w = 5r 6) 3, so dw dr = 35r 6)2 5) = 155r 6) 2 Section 3.3 Homework Solutions 7. y = s can be rewritten as y = s 3 + 1) 1/2, so ds = 1 2 s3 + 1) 1/2 3s 2 ) = 8. ft) = e 3t, so f t) = e 3t 3) = 3e 3t 9. y = e 0.7t, so = e0.7t 0.7) = 0.7e 0.7t 10. y = e 4t, so = e 4t 4) = 4e 4t 11. P = e 0.2t, so dp = e 0.2t 0.2) = 0.2e 0.2t 12. P = 50e 0.6t, so dp = 50e 0.6t 0.6) = 30e 0.6t 13. P = 200e 0.12t, so dp = 200e0.12t 0.12) = 24e 0.12t 14. y = 12 3x 2 + 2e 3x, so dx = 6x + 2e3x 3) = 6x + 6e 3x 15. C = 123q 2 5) 3, so dc dq = 363q2 5) 2 6q) = 216q3q 2 5) fx) = 6e 5x + e x2, so f x) = 6e 5x 5) + e x2 2x) = 30e 5x 2xe x2 3s 2 2 s y = 5e 5t+1, so = 5e5t+1 5) = 25e 5t w = e 3t2, so dw = 6t) = 6te e 3t2 3t2 19. w = e s, so dw ) 1 ds = e s 2 s 1/2 = e s 2 s We used that dds s) = 12 s 1/2 )
7 20. y = ln 5t + 1), so = 1 5t + 1 5) = 5 5t fx) = ln 1 x), so f x) = 1 1 1) = 1 x 1 x 22. ft) = ln t 2 + 1), so f t) = 1 t t) = 23. fx) = ln 1 e x ), so f x) = 24. fx) = ln e x + 1), so f x) = 1 e x + 1 ex ) = ft) = 5 ln 5t + 1), so f t) = 5 5t + 1 2t t e x e x 1)) = e x 1 e x = 1 e x 1 ) ex e x gt) = ln 4t + 9), so g t) = 1 4t + 9 4) = 4 4t y = 5 + ln 3t + 2), so = 1 3t + 2 3) = 3 3t + 2 5) = 25 5t Q = 100t 2 + 5) 0.5, so dq = 50t2 + 5) 0.5 2t) = 100tt 2 + 5) y = 5x + ln x + 2), so dx = x + 2 1) = x y = 5 + e x ) 2, so dx = 25 + ex ) e x ) = 2e x 5 + e x ) 31. P = 1 + ln x) 0.5, so dp dx = ln x) ) 1 x d ex + 1) = d e x + 1) 1/2) = 1 dx dx 2 ex + 1) 1/2 e x ) = e x 2 e x fx) = 1 x 2 can be rewritten as fx) = 1 x 2 ) 1/2, so f x) = x2 ) 1/2 2x) = x 1 x fθ) = e θ + e θ ) 1, so f θ) = e θ + e θ ) 2 e θ e θ ) 35. fx) = x 1) 3 and f x) = 3x 1) 2 so that f2) = 1 and f 2) = 3. Thus we need the equation of a line through the point 2, 1) with a slope of 3. From the point-slope formula we get y 1 = 3x 2) which simplifies to our answer y = 3x y = e 2t, so = e 2t 2). At t = 0, we get y = 1 and = 2. So our point is 0, 1) and the slope is 2. Now the point-slope formula gives y 1 = 2t 0) which simplifies to our answer y = 2t + 1.
8 37. fx) = 10e 0.2x and f x) = 2e 0.2x so that f4) = 10e and f 4) = 2e Thus we need the equation of a line through the point 4, 4.493) with a slope of From the point-slope formula we get y = 0.899x 4) which simplifies to our answer y = 0.899x Cq) = e 0.05q, so C q) = 30e 0.05q 0.05). We let q = 50 to get the cost of $ and marginal cost of $18.27 per item. 44. H = e 0.02t, so dh = 16e 0.02t 0.02) = 0.32e 0.02t. At t = 0, the temperature is decreasing by 0.32 degrees per minute. At t = 10, the temperature is decreasing by 0.26 degrees per minute.
9 Math 122 Section 3.4 Homework Solutions 1. fx) = 2x + 1)3x 2) f x) = 2x + 1) 3x 2) + 2x + 1)3x 2) f x) = 2)3x 2) + 2x + 1)3) f x) = 12x 1 fx) = 6x 2 x 2 f x) = 12x 1 2. fx) = x 2 x ) f x) = x 2) x ) + x 2) x ) f x) = 2x) x ) + x 2) 3x 2) f x) = 5x x fx) = x 5 + 5x 2 f x) = 5x x 3. fx) = xe x f x) = x) e x ) + x) e x ) f x) = 1) e x ) + x) e x ) f x) = e x + xe x 4. ft) = te 2t f t) = t) e 2t) + t) e 2t) f t) = 1) e 2t) + t) e 2t 2 ) f t) = e 2t 2te 2t 5. y = 5xe x2 ) dx = 5x) e x2 + 5x) e x2) dx = 5) e x2) + 5x) ) e x2 2x
10 dx = 5ex2 + 10x 2 e x2 6. y = t 2 3t + 1) 3 = t 2) ) 3t + 1) 3 + t 2) 3t + 1) 3) = 2t) 3t + 1) 3) + t 2) 33t + 1) 2 3 ) = 2t3t + 1)3 + 9t 2 3t + 1) 2 7. y = x ln x dx = x) ln x) + x) ln x) dx = 1) ln x) + x) dx = ln x + 1 ) 1 x 8. y = t 2 + 3)e t = t ) e t ) + t ) e t) = 2t) e t) + t ) e t) = et t 2 + 2t + 3 ) 9. z = 3t + 1)5t + 2) = 3t + 1) 5t + 2) + 3t + 1) 5t + 2) = 3) 5t + 2) + 3t + 1) 5) = 30t y = t 3 7t 2 + 1)e t = t 3 7t ) e t ) + t 3 7t ) e t) = 3t 2 14t ) e t) + t 3 7t ) e t)
11 = et t 3 4t 2 14t + 1 ) 11. P = t 2 ln t dp = t 2) ln t) + t 2 ) ln t) dp = 2t) ln t) + t 2) dp = 2t ln t + t ) 1 t 12. ft) = 5 t + 6 t 2 ft) = 5t 1 + 6t 2 f t) = 5t 2 12t fx) = x2 + 3 x fx) = x + 3x 1 f x) = 1 3x R = 3qe q dr dq = 3q) e q) + 3q) e q) dr dq = 3) e q) + 3q) e q 1 ) dr dq = 3e q 3qe q 15. y = te t2 e t2) ) = t) e t2 + t) = 1) e t2) ) + t) e t2 2t = 2t 2 e e t2 t2 16. fz) = ze z
12 fz) = z 1/2 e z f z) = z 1/2) ) e z + z 1/2) e z) ) 1 e f z) = ) 2 z 1/2 z + z 1/2) e z 1 ) f z) = 1 2 z 1/2 e z z 1/2 e z 17. gp) = p ln 2p + 1) g p) = p) ln 2p + 1)) + p) ln 2p + 1)) ) 1 g p) = 1) ln 2p + 1)) + p) 2p g p) = ln 2p + 1) + 2p 2p ft) = te 5 2t f t) = t) e 5 2t) + t) e 5 2t) f t) = 1) e 5 2t) + t) e 5 2t 2 ) f t) = e 5 2t 2te 5 2t 19. fw) = 5w 2 + 3)e w2 f w) = 5w ) e w2) + 5w ) e w2) f w) = 10w) f w) = e w2 10w w ) e w2) + 5w ) ) e w2 2w 20. y = x 2 x dx = x) 2 x ) + x) 2 x ) dx = 1) 2x ) + x) ln 2) 2 x ) dx = 2x + ln 2)x2 x 21. w = t 3 + 5t)t 2 7t + 2)
13 dw = t 3 + 5t ) t 2 7t + 2 ) + t 3 + 5t ) t 2 7t + 2 ) dw = 3t ) t 2 7t + 2 ) + t 3 + 5t ) 2t 7) 22. z = te 3t + e 5t) 9 = 9 te 3t + e 5t) 8 te 3t + e 5t) = 9 te 3t + e 5t) 8 1) e 3t ) + t) e 3t 3 ) + e 5t 5 ) = 9 te 3t + e 5t) 8 e 3t + 3te 3t + 5e 5t) 23. fx) = x e x f x) = x) e x ) x) e x ) e x ) 2 f x) = 1) ex ) x) e x ) e x ) 2 f x) = 1 x e x 24. w = 3z 1 + 2z dw = 3z) 1 + 2z) 3z) 1 + 2z) 1 + 2z) 2 dw = 3) 1 + 2z) 3z) 2) 1 + 2z) 2 dw = z) z = 1 t 1 + t = 1 t) 1 + t) 1 t) 1 + t) 1 + t) 2 = = 1) 1 + t) 1 t) 1) 1 + t) t) 2
14 26. y = ex 1 + e x dx = ex ) 1 + e x ) e x ) 1 + e x ) 1 + e x ) 2 dx = ex ) 1 + e x ) e x ) e x ) 1 + e x ) 2 dx = e x 1 + e x ) w = 3y + y2 5 + y dw = 3y + y2 ) 5 + y) 3y + y 2 ) 5 + y) 5 + y) 2 dw = 3 + 2y) 5 + y) 3y + y2 ) 1) 5 + y) 2 dw = y2 + 10y y) y = 1 + z ln z = 1 + z) ln z) 1 + z) ln z) ln z) 2 1) ln z) 1 + z) 1 ) z = ln z) 2 = ln z 1 1 z ln z) ft) = ae bt f t) = ae bt b f t) = abe bt 30. fx) = ax 2 + b ) 3 f x) = 3 ax 2 + b ) 2 2ax f x) = 6ax ax 2 + b ) 2
15 31. fx) = axe bx f x) = ax) e bx) + ax) e bx) f x) = a) e bx) + ax) e bx b ) f x) = ae bx abxe bx 32. fx) = ax + b cx + k f x) = ax + b) cx + k) ax + b) cx + k) cx + k) 2 f x) = f x) = a) cx + k) ax + b) c) cx + k) 2 ak bc cx + k) gα) = e αe 2α g α) = e αe 2α αe 2α) g α) = e αe 2α 1) e 2α) + α) e 2α 2 )) g α) = e αe 2α e 2α 2αe 2α) 34. fx) = 3x + 8)2x 5) fx) = 6x 2 + x 40 f x) = 12x + 1 f x) = fx) = x 2 e x and f x) = 2xe x x 2 e x so that f0) = 0 and f 0) = 0. Thus we need the equation of a line through the point 0, 0) with a slope of 0. From the point-slope formula we get y 0 = 0x 0) which simplifies to our answer y = fx) = 2x 5 x + 1 and f 2)x + 1) 2x 5)1) 7 x) = = x + 1) 2 x + 1) so that f0) = 5 and f 0) = 2 7. Thus we need the equation of a line through the point 0, 5) with a slope of 7. From the point-slope formula we get y + 5 = 7x 0) which simplifies to our answer y = 7x 5.
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