1.\u00a0<\/strong>[latex]s(t)=2t^3-3t^2-12t+8[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 2.\u00a0<\/strong>[latex]s(t)=2t^3-15t^2+36t-10[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739303945\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739303945\"]\r\na. [latex]v(t)=6t^2-30t+36, \\, a(t)=12t-30[\/latex]\r\nb. Speeds up: [latex](2,2.5)\\cup (3,\\infty)[\/latex]; Slows down: [latex](0,2)\\cup (2.5,3)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 3.\u00a0<\/strong>[latex]s(t)=\\dfrac{t}{1+t^2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 4.\u00a0<\/strong>A rocket is fired vertically upward from the ground. The distance [latex]s[\/latex] in feet that the rocket travels from the ground after [latex]t[\/latex] seconds is given by [latex]s(t)=-16t^2+560t[\/latex].<\/p>\r\n\r\n a. [latex]464 \\, \\text{ft\/s}^2[\/latex]\r\nb. [latex]-32 \\, \\text{ft\/s}^2[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 5.\u00a0<\/strong>A ball is thrown downward with a speed of 8 ft\/s from the top of a 64-foot-tall building. After [latex]t[\/latex] seconds, its height above the ground is given by [latex]s(t)=-16t^2-8t+64[\/latex].<\/p>\r\n\r\n 6.\u00a0<\/strong>The position function [latex]s(t)=t^2-3t-4[\/latex] represents the position of the back of a car backing out of a driveway and then driving in a straight line, where [latex]s[\/latex] is in feet and [latex]t[\/latex] is in seconds. In this case, [latex]s(t)=0[\/latex] represents the time at which the back of the car is at the garage door, so [latex]s(0)=-4[\/latex] is the starting position of the car, 4 feet inside the garage.<\/p>\r\n\r\n 7.\u00a0<\/strong>The position of a hummingbird flying along a straight line in [latex]t[\/latex] seconds is given by [latex]s(t)=3t^3-7t[\/latex] meters.<\/p>\r\n\r\n 8.\u00a0<\/strong>A potato is launched vertically upward with an initial velocity of 100 ft\/s from a potato gun at the top of an 85-foot-tall building. The distance in feet that the potato travels from the ground after [latex]t[\/latex] seconds is given by [latex]s(t)=-16t^2+100t+85[\/latex].<\/p>\r\n\r\n a. 84 ft\/s, \u221284 ft\/s\r\nb. 84 ft\/s\r\nc. [latex]\\frac{25}{8}[\/latex] sec\r\nd. [latex]-32 \\, \\text{ft\/s}^2[\/latex] in both cases\r\ne. [latex]\\frac{1}{8}(25+\\sqrt{965})[\/latex] sec\r\nf. [latex]-4\\sqrt{965}[\/latex] ft\/s<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 9.\u00a0<\/strong>The position function [latex]s(t)=t^3-8t[\/latex] gives the position in miles of a freight train where east is the positive direction and [latex]t[\/latex] is measured in hours.<\/p>\r\n\r\n 10.\u00a0<\/strong>The following graph shows the position [latex]y=s(t)[\/latex] of an object moving along a straight line.<\/p>\r\n a. Velocity is positive on [latex](0,1.5)\\cup (6,7)[\/latex], negative on [latex](1.5,2)\\cup (5,6)[\/latex], and zero on [latex](2,5)[\/latex].\r\nb.<\/p>\r\n 11.\u00a0<\/strong>The cost function, in dollars, of a company that manufactures food processors is given by [latex]C(x)=200+\\dfrac{7}{x}+\\dfrac{x^2}{7}[\/latex], where [latex]x[\/latex] is the number of food processors manufactured.<\/p>\r\n\r\n 12.\u00a0<\/strong>The price [latex]p[\/latex] (in dollars) and the demand [latex]x[\/latex] for a certain digital clock radio is given by the price-demand function [latex]p=10-0.001x[\/latex].<\/p>\r\n\r\n a. [latex]R(x)=10x-0.001x^2[\/latex]\r\nb. [latex]R^{\\prime}(x)=10-0.002x[\/latex]\r\nc. $6 per item, $0 per item<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 13. [T]<\/strong> A profit is earned when revenue exceeds cost. Suppose the profit function for a skateboard manufacturer is given by [latex]P(x)=30x-0.3x^2-250[\/latex], where [latex]x[\/latex] is the number of skateboards sold.<\/p>\r\n\r\n 14. [T]<\/strong> In general, the profit function is the difference between the revenue and cost functions: [latex]P(x)=R(x)-C(x)[\/latex].<\/p>\r\n Suppose the price-demand and cost functions for the production of cordless drills is given respectively by [latex]p=143-0.03x[\/latex] and [latex]C(x)=75,000+65x[\/latex], where [latex]x[\/latex] is the number of cordless drills that are sold at a price of [latex]p[\/latex] dollars per drill and [latex]C(x)[\/latex] is the cost of producing [latex]x[\/latex] cordless drills.<\/p>\r\n\r\n a. [latex]C^{\\prime}(x)=65[\/latex]\r\nb. [latex]R(x)=143x-0.03x^2, \\, R^{\\prime}(x)=143-0.06x[\/latex]\r\nc. [latex]R^{\\prime}(1000)=83, \\, R^{\\prime}(4000)=-97[\/latex]. At a production level of 1000 cordless drills, revenue is increasing at a rate of $83 per drill; at a production level of 4000 cordless drills, revenue is decreasing at a rate of $97 per drill.\r\nd. [latex]P(x)=-0.03x^2+78x-75000, \\, P^{\\prime}(x)=-0.06x+78[\/latex]\r\ne. [latex]P^{\\prime}(1000)=18, \\, P^{\\prime}(4000)=-162[\/latex]. At a production level of 1000 cordless drills, profit is increasing at a rate of $18 per drill; at a production level of 4000 cordless drills, profit is decreasing at a rate of $162 per drill.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 15.\u00a0<\/strong>A small town in Ohio commissioned an actuarial firm to conduct a study that modeled the rate of change of the town\u2019s population. The study found that the town\u2019s population (measured in thousands of people) can be modeled by the function [latex]P(t)=-\\frac{1}{3}t^3+64t+3000[\/latex], where [latex]t[\/latex] is measured in years.<\/p>\r\n\r\n 16. [T]<\/strong> A culture of bacteria grows in number according to the function [latex]N(t)=3000\\left(1+\\dfrac{4t}{t^2+100}\\right)[\/latex], where [latex]t[\/latex] is measured in hours.<\/p>\r\n\r\n a. [latex]N^{\\prime}(t)=3000(\\frac{-4t^2+400}{(t^2+100)^2})[\/latex]\r\nb. [latex]N^{\\prime}(0)=120, \\, N^{\\prime}(10)=0, \\, N^{\\prime}(20)=-14.4, \\, N^{\\prime}(30)=-9.6[\/latex]\r\nc. The bacteria population increases from time 0 to 10 hours; afterwards, the bacteria population decreases.\r\nd. [latex]N''(0)=0, \\, N''(10)=-6, \\, N''(20)=0.384, \\, N''(30)=0.432[\/latex]. The rate at which the bacteria is increasing is decreasing during the first 10 hours. Afterwards, the bacteria population is decreasing at a decreasing rate.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 17.\u00a0<\/strong>The centripetal force of an object of mass [latex]m[\/latex] is given by [latex]F(r)=\\dfrac{mv^2}{r}[\/latex], where [latex]v[\/latex] is the speed of rotation and [latex]r[\/latex] is the distance from the center of rotation.<\/p>\r\n\r\n The following questions (18-19) concern the population (in millions) of London by decade in the 19th century, which is listed in the following table.<\/p>\r\n\r\n 18. [T]<\/strong><\/p>\r\n\r\n a. [latex]P(t)=0.03983+0.4280[\/latex]\r\nb. [latex]P^{\\prime}(t)=0.03983[\/latex]. The population is increasing.\r\nc. [latex]P''(t)=0[\/latex]. The rate at which the population is increasing is constant.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 19. [T]<\/strong><\/p>\r\n\r\n For the following exercises (20-21), consider an astronaut on a large planet in another galaxy. To learn more about the composition of this planet, the astronaut drops an electronic sensor into a deep trench. The sensor transmits its vertical position every second in relation to the astronaut\u2019s position. The summary of the falling sensor data is displayed in the following table.<\/p>\r\n\r\n 20. [T]<\/strong><\/p>\r\n\r\n a. [latex]p(t)=-0.6071x^2+0.4357x-0.3571[\/latex]\r\nb. [latex]p^{\\prime}(t)=-1.214x+0.4357[\/latex]. This is the velocity of the sensor.\r\nc. [latex]p''(t)=-1.214[\/latex]. This is the acceleration of the sensor; it is a constant acceleration downward.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 21. [T]<\/strong><\/p>\r\n\r\n The following problems (22-24) deal with the Holling type I, II, and III equations. These equations describe the ecological event of growth of a predator population given the amount of prey available for consumption.<\/p>\r\n\r\n 22. [T]<\/strong> The Holling type I equation<\/span> is described by [latex]f(x)=ax[\/latex], where [latex]x[\/latex] is the amount of prey available and [latex]a>0[\/latex] is the rate at which the predator meets the prey for consumption.<\/p>\r\n\r\n a.<\/p>\r\n 23. [T]<\/strong> The Holling type II equation is described by [latex]f(x)=\\dfrac{ax}{n+x}[\/latex], where [latex]x[\/latex] is the amount of prey available and [latex]a>0[\/latex] is the maximum consumption rate of the predator.<\/p>\r\n\r\n 24. [T]<\/strong> The Holling type III equation is described by [latex]f(x)=\\dfrac{ax^2}{n^2+x^2}[\/latex], where [latex]x[\/latex] is the amount of prey available and [latex]a>0[\/latex] is the maximum consumption rate of the predator.<\/p>\r\n\r\n a.<\/p>\r\n 25. [T]<\/strong> The populations of the snowshoe hare (in thousands) and the lynx (in hundreds) collected over 7 years from 1937 to 1943 are shown in the following table. The snowshoe hare is the primary prey of the lynx.<\/p>\r\n\r\n For the following exercises (1-3), the given functions represent the position of a particle traveling along a horizontal line.<\/p>\n 1.\u00a0<\/strong>[latex]s(t)=2t^3-3t^2-12t+8[\/latex]<\/p>\n<\/div>\n<\/div>\n 2.\u00a0<\/strong>[latex]s(t)=2t^3-15t^2+36t-10[\/latex]<\/p>\n 3.\u00a0<\/strong>[latex]s(t)=\\dfrac{t}{1+t^2}[\/latex]<\/p>\n<\/div>\n<\/div>\n 4.\u00a0<\/strong>A rocket is fired vertically upward from the ground. The distance [latex]s[\/latex] in feet that the rocket travels from the ground after [latex]t[\/latex] seconds is given by [latex]s(t)=-16t^2+560t[\/latex].<\/p>\n a. [latex]464 \\, \\text{ft\/s}^2[\/latex] 5.\u00a0<\/strong>A ball is thrown downward with a speed of 8 ft\/s from the top of a 64-foot-tall building. After [latex]t[\/latex] seconds, its height above the ground is given by [latex]s(t)=-16t^2-8t+64[\/latex].<\/p>\n 6.\u00a0<\/strong>The position function [latex]s(t)=t^2-3t-4[\/latex] represents the position of the back of a car backing out of a driveway and then driving in a straight line, where [latex]s[\/latex] is in feet and [latex]t[\/latex] is in seconds. In this case, [latex]s(t)=0[\/latex] represents the time at which the back of the car is at the garage door, so [latex]s(0)=-4[\/latex] is the starting position of the car, 4 feet inside the garage.<\/p>\n 7.\u00a0<\/strong>The position of a hummingbird flying along a straight line in [latex]t[\/latex] seconds is given by [latex]s(t)=3t^3-7t[\/latex] meters.<\/p>\n 8.\u00a0<\/strong>A potato is launched vertically upward with an initial velocity of 100 ft\/s from a potato gun at the top of an 85-foot-tall building. The distance in feet that the potato travels from the ground after [latex]t[\/latex] seconds is given by [latex]s(t)=-16t^2+100t+85[\/latex].<\/p>\n a. 84 ft\/s, \u221284 ft\/s 9.\u00a0<\/strong>The position function [latex]s(t)=t^3-8t[\/latex] gives the position in miles of a freight train where east is the positive direction and [latex]t[\/latex] is measured in hours.<\/p>\n 10.\u00a0<\/strong>The following graph shows the position [latex]y=s(t)[\/latex] of an object moving along a straight line.<\/p>\n a. Velocity is positive on [latex](0,1.5)\\cup (6,7)[\/latex], negative on [latex](1.5,2)\\cup (5,6)[\/latex], and zero on [latex](2,5)[\/latex]. 11.\u00a0<\/strong>The cost function, in dollars, of a company that manufactures food processors is given by [latex]C(x)=200+\\dfrac{7}{x}+\\dfrac{x^2}{7}[\/latex], where [latex]x[\/latex] is the number of food processors manufactured.<\/p>\n 12.\u00a0<\/strong>The price [latex]p[\/latex] (in dollars) and the demand [latex]x[\/latex] for a certain digital clock radio is given by the price-demand function [latex]p=10-0.001x[\/latex].<\/p>\n\r\n \t
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<\/span>\r\nc. Acceleration is positive on [latex](5,7)[\/latex], negative on [latex](0,2)[\/latex], and zero on [latex](2,5)[\/latex].\r\nd. The object is speeding up on [latex](6,7)\\cup (1.5,2)[\/latex] and slowing down on [latex](0,1.5)\\cup (5,6)[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n\r\n \t
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\r\n \r\nYears since 1800<\/th>\r\n Population (millions)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n 1<\/td>\r\n 0.8795<\/td>\r\n<\/tr>\r\n \r\n 11<\/td>\r\n 1.040<\/td>\r\n<\/tr>\r\n \r\n 21<\/td>\r\n 1.264<\/td>\r\n<\/tr>\r\n \r\n 31<\/td>\r\n 1.516<\/td>\r\n<\/tr>\r\n \r\n 41<\/td>\r\n 1.661<\/td>\r\n<\/tr>\r\n \r\n 51<\/td>\r\n 2.000<\/td>\r\n<\/tr>\r\n \r\n 61<\/td>\r\n 2.634<\/td>\r\n<\/tr>\r\n \r\n 71<\/td>\r\n 3.272<\/td>\r\n<\/tr>\r\n \r\n 81<\/td>\r\n 3.911<\/td>\r\n<\/tr>\r\n \r\n 91<\/td>\r\n 4.422<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n \t
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\r\n \r\nTime after dropping (s)<\/th>\r\n Position (m)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n 0<\/td>\r\n 0<\/td>\r\n<\/tr>\r\n \r\n 1<\/td>\r\n \u22121<\/td>\r\n<\/tr>\r\n \r\n 2<\/td>\r\n \u22122<\/td>\r\n<\/tr>\r\n \r\n 3<\/td>\r\n \u22125<\/td>\r\n<\/tr>\r\n \r\n 4<\/td>\r\n \u22127<\/td>\r\n<\/tr>\r\n \r\n 5<\/td>\r\n \u221214<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n \t
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<\/span>\r\nb. [latex]f^{\\prime}(x)=a[\/latex]. The more increase in prey, the more growth for predators.\r\nc. [latex]f''(x)=0[\/latex]. As the amount of prey increases, the rate at which the predator population growth increases is constant.\r\nd. This equation assumes that if there is more prey, the predator is able to increase consumption linearly. This assumption is unrealistic because we would expect there to be some saturation point at which there is too much prey for the predator to consume adequately.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n\r\n \t
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<\/span>\r\nb. [latex]f^{\\prime}(x)=\\frac{2axn^2}{(n^2+x^2)^2}[\/latex]. When the amount of prey increases, the predator growth increases.\r\nc. [latex]f''(x)=\\frac{2an^2(n^2-3x^2)}{(n^2+x^2)^3}[\/latex]. When the amount of prey is extremely small, the rate at which predator growth is increasing is increasing, but when the amount of prey reaches above a certain threshold, the rate at which predator growth is increasing begins to decrease.\r\nd. At lower levels of prey, the prey is more easily able to avoid detection by the predator, so fewer prey individuals are consumed, resulting in less predator growth.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n\r\n \r\nPopulation of snowshoe hare (thousands)<\/th>\r\n Population of lynx (hundreds)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n 20<\/td>\r\n 10<\/td>\r\n<\/tr>\r\n \r\n 55<\/td>\r\n 15<\/td>\r\n<\/tr>\r\n \r\n 65<\/td>\r\n 55<\/td>\r\n<\/tr>\r\n \r\n 95<\/td>\r\n 60<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n \t
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\nb. Speeds up: [latex](2,2.5)\\cup (3,\\infty)[\/latex]; Slows down: [latex](0,2)\\cup (2.5,3)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n
\nb. [latex]-32 \\, \\text{ft\/s}^2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n
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\nb. 84 ft\/s
\nc. [latex]\\frac{25}{8}[\/latex] sec
\nd. [latex]-32 \\, \\text{ft\/s}^2[\/latex] in both cases
\ne. [latex]\\frac{1}{8}(25+\\sqrt{965})[\/latex] sec
\nf. [latex]-4\\sqrt{965}[\/latex] ft\/s<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n
<\/span><\/p>\n\n
\nb.<\/p>\n<\/span>
\nc. Acceleration is positive on [latex](5,7)[\/latex], negative on [latex](0,2)[\/latex], and zero on [latex](2,5)[\/latex].
\nd. The object is speeding up on [latex](6,7)\\cup (1.5,2)[\/latex] and slowing down on [latex](0,1.5)\\cup (5,6)[\/latex].\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n
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