计学(第六版)第七章课后练习答案
管理经济学 原书第六版 课后答案7
Chapter 7: Answers to Questions and Problems1. The four-firm concentration ratio is,4$175,000$150,000$125,000$100,0000.55$1,000,000C +++==.2.a. The HHI is222$200,000$400,000$500,00010,000=3,719$1,100,000$1,100,000$1,100,000HHI ⎡⎤⎛⎞⎛⎞⎛⎞=++⎢⎥⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎢⎥⎣⎦. b. The four-firm concentration ratio is 100 percent.c. If the firms with sales of $200,000 and $400,000 were allowed to merge, the resulting HHI would increase by 1,322 to 5,041. Since the pre-merger HHI exceeds that under the Guidelines (1,800) and the HHI increases by more than that permitted under the Guidelines (100), the merger is likely to be challenged.3. The elasticity of demand for a representative firm in the industry is –1.5, since.5.16.09.09.06.0−=−=⇒−=F F E E .4.a. $100. To see this, solve the Lerner index formula for P to obtain11$35$100110.65P MC L ⎛⎞⎛⎞===⎜⎟⎜⎟−−⎝⎠⎝⎠. b. Since 11P MC L ⎛⎞=⎜⎟−⎝⎠, it follows that the markup factor is 1 2.8610.65⎛⎞=⎜⎟−⎝⎠. That is, the price charged by the firm is 2.86 times the marginal cost of producing the product.c. The above calculations suggest price competition is not very rigorous and that the firm enjoys market power.5.Managers should not specialize in learning to manage a particular type of market structure. Market structure generally evolves over time, and managers must adapt to these changes.6. To the extent that the HHIs are based on too narrow a definition of the product (orgeographic) market or the impact of foreign competition, the merger might be allowed. It might also be allowed if one of the firms is in financial trouble, or if significant economies of scale exist in the industry.7. As shown in the text, the HHI is⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛++⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛++⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛∑=222222112......000,10000,10T n T j T i T T n i T i S S S S S S S S S S S S . (1) When firms i and j merge, the HHI becomes ⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛++⎟⎟⎠⎞⎜⎜⎝⎛+++⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛222221......000,10T n T j i T T S S S S S S S S S . (2) The difference between (2) and (1) is that 22⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛T j T i S S S S becomes 2⎟⎟⎠⎞⎜⎜⎝⎛+T j i S S S .Thus, we can calculate how a merger between firms i and j will change the HHI by knowing only those two firms’ market shares. In general, since()22222⎟⎟⎠⎞⎜⎜⎝⎛++⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛+T j T j i T i T j i S S S S S S S S S S , we know that a merger between firms i and j will cause the HHI to increase by 210,000i j w w ×, where w i and w j are the pre-merger market shares of the two merging firms. Using the information in the problem,()()()800000,102.2.2= represents the increase in the HHI due to the merger.8. No. The conditions for perfect competition include:a. There are many buyers and sellers of products.b. The products are homogenous.c. Consumers and producers have perfect information.d. There is free entry and exit.9. The four-firm concentration ratios in Table 7-2 are likely to overstate the level of concentration in the U.S. Imported beers account for much of the sales in the U.S. It is likely that the brewing industry is much less concentrated than Table 7-2 leads us to believe.10.This industry is most likely monopolistically competitive. Monopolistically competitive industries have concentration measures close to zero, but since eachfirm’s product is slightly differentiated, the Rothschild index will be greater than zero (unlike perfectly competitive markets).11.Monopolistically competitive. In a monopolistically competitive market, there aremany firms, but each firm produces a differentiated product. According to the causal view, the structure of differentiated products causes firms to capitalize on the absence of close substitutes by charging higher prices and earning higher profit. Thus,structure causes conduct resulting in performance. According to the feedback critique, the conduct of firms may determine the market structure. Firms’ products may bedifferentiated because of firms’ conduct in the industry. Examples of such conductinclude advertising and other behavioral tactics that feedback into demand, causingconsumers to view products as differentiated. Thus, it is not at all clear thatdifferentiated products are a structural variable. The willingness of consumers to pay for product variety gives firms an incentive to offer different products (thin-and-crispy pizza, pan pizza, pizza delivery, etc.).12.Merger (a) is the only horizontal merger, and therefore the only merger that would bescrutinized under the Guidelines for horizontal mergers. Merger (b) is a conglomerate merger, while merger (c) is a vertical merger.13.While the pre-merger four-firm concentration ratio is 72 percent, the pre-merger HHIis only 1,535. The merger would increase the HHI by only 100 to 1,635. The merger is unlikely to be blocked based on the merger guidelines.14.If approved, the merger would raise the HHI by ()()()702000,1013.27.2= points (see the solution to problem 7). Since the pre-merger HHI is 3,025, which is greater thanthe Guidelines (1,800), and the HHI increases by 702 (which is greater than the 100points permitted in the Guidelines), it is unlikely that the merger will receiveunconditional approval.15.See Table 7-1.O w n P rice Elasticity of M arket Dem andO w n P rice E lasticityof D em and forR epresentative Firm'sP roduct R othsc hild IndexA griculture-1.8-96.20.019C onstruction-1.0-5.20.192Du rable m anufacturing-1.4-3.50.400No ndu rable m anufacturing-1.3-3.40.382Transportation-1.0-1.90.526Com m unication and utilities-1.2-1.80.667W holesale trade-1.5-1.60.938R etail trade-1.2-1.80.667Finance-0.1-5.50.018S ervices-1.2-26.40.045Table 7-1Based on the Rothschild indices in Table 7-1, wholesale trade most closely resembles a monopoly, while finance most closely resembles perfect competition.16. The Lerner index is $3$0.300.9$3P MC L P −−===, which indicates the firm has considerable market power. This makes sense because the product that the firm sells is currently under patent protection, which essentially makes the firm a legal monopoly.17.Based on the information contained in Table 7-3 of the text, the food and apparel industries are most competitive and therefore probably represent the best match for the expertise of these managers.18. The market for color film in the U.S. is highly concentrated. The five-firmconcentration ratio is 100 percent and Kodak alone accounts for 67 percent of all rolls sold. Market demand for color film is relatively elastic at -1.75; indicating that a 10 percent increase in price leads to a 17.5 percent decline in quantity demand for color film. The Rothschild index indicates that market demand relative to the demand for Kodak color film is 875.0275.1=−−=R , indicating that Kodak’s demand is, roughly, as sensitive to price changes as is the entire market demand. The Lerner index for Kodak is 50.095.6$475.3$95.6$=−=L , indicating that Kodak’s markup factor is 2. For every $1 spent on color film, $0.50 is markup. Taken together, these things suggest that the color film industry in the U.S. closely resembles an oligopoly.19. Note first that a merger with Unilever or Tricor Braun is not a horizontal merger. Moreover, while a horizontal merger with either Dole or Goya is likely to enhance Del Monte’s profitability (profits as a percentage of sales are 8.7 and 7.1, respectively and the Lerner Indices are 0.14 and 0.32, respectively), the market for cannedtomatoes and canned pineapple are highly concentrated. The four-firm concentration ratio and HHI for the canned tomatoes market are, respectively, 86.3 percent and 3,297. Similarly, the four-firm concentration ratio and HHI in the canned pineapple industry are 94.2 percent and 5,457, respectively. This information suggests that potential mergers in these industries are likely to be scrutinized.20.On the surface the industry analysts’ suggestion would represent a merger tomonopoly and the HHI before and after the merger exceeds the threshold for raising antitrust concerns. However, there are several reasons why the merger might beallowed. First, satellite radio may not be its own industry, but rather a smaller part of a larger market that include MP3 players, AM and FM radios, and the like. Their market shares in this more broadly defined market are trivial, and therefore themerger would not impact the HHI in the more broadly defined market. Additionally, the merger might be allowed if the (1) firms could show significant cost savings; (2) rapidly changing technology in the portable music industry would prevent anti-competitive behavior; (3) government viewed the firms as financially unstable; or (4) barriers to entry were low enough to allow competition from new entrants after the merger.21.With number portability, the services of the various providers are now closersubstitutes to each other. One implication is that the cost to consumers of switchingservice providers is not lower, which increases the likelihood that consumers wouldswitch for small price reductions. These factors make the demand for the individualcellular service providers more elastic (increase the price elasticity of demand). Local number portability, however, is unlikely to affect the elasticity of demand for theindustry as a whole. If the elasticity of demand increases for individual firms, butremains constant for the industry, the Rothschild index will decrease.。
(完整版)贾俊平统计学[第六版]思考题答案解析.docx
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第一章：1、什么是统计学？统计学是一门收集、分析、表述、解释数据的科学和艺术。
2、描述统计：研究的是数据收集、汇总、处理、图表描述、概括与分析等统计方法。
推断统计：研究的是如何利用样本数据来推断总体特征。
3、统计学据可以分成哪几种类型，个有什么特点？按照计量尺度不同，分为：分类数据、顺序数据、数值型数据。
分类数据：只能归于某一类别的，非数字型数据。
顺序数据：只能归于某一有序类别的，非数字型数据。
数值型数据：按数字尺度测量的观察值，结果表现为数值。
按收集方法不同。
分为：观测数据、和实验数据观测数据：通过调查或观测而收集到的数据；不控制条件；社会经济领域实验数据：在试验中收集到的数据；控制条件；自然科学领域。
按时间不同，分为：截面数据、时间序列数据截面数据：在相同或近似相同的时间点上收集的数据。
时间序列数据：在不同时间收集的数据。
4、举例说明总体、样本、参数、统计量、变量这几个概念。
总体：是包含全部研究个体的集合，包括有限总体和无限总体（范围、数目判定）样本：从总体中抽取的一部分元素的集合。
参数：用来描述总体特征的概括性数字度量。
（平均数、标准差、比例等）统计量：用来描述样本特征的概括性数字度量。
（平均数、标准差、比例等）变量：是说明样本某种特征的概念，其特点：从一次观察到下一次观察结果会呈现出差别或变化。
（商品销售额、受教育程度、产品质量等级等）（对一千灯泡进行寿命测试，那么这千个灯泡就是总体，从中抽取一百个进行检测，这一百个灯泡的集合就是样本，这一千个灯泡的寿命的平均值和标准差还有合格率等描述特征的数值就是参数，这一百个灯泡的寿命的平均值和标准差还有合格率等描述特征的数值就是统计量，变量就是说明现象某种特征的概念，比如说灯泡的寿命。
）5、变量可以分为哪几类？分类变量：说明事物类别；取值是分类数据。
顺序变量：说明事物有序类别；取值是顺序数据数值型变量：说明事物数字特征；取值是数值型数据。
变量也可以分为：随机变量和非随机变量；经验变量和理论变量6、举例说明离散型变量和连续型变量。
高等数学第六版（上册）第七章课后习题答案
1 高等数学第六版(上册)第七章课后习题答案习题7-11.设 u =a -b +2c ,v =-a +3b -c .试用 a 、b 、c 表示 2u -3v . 解 2u -3v =2(a -b +2c )-3(-a +3b -c )=2a -2b +4c +3a -9b +3c=5a -11b +7c .2.如果平面上一个四边形的对角线互相平分,试用向量证明这是平行四边形.→ → → → → →证 AB =OB -OA ;DC =OC -OD ,→ → → →而OC =-OA ,OD =-OB , → → → → → →所以 DC =-OA +OB =OB -OA =-AB .这说明四边形ABCD 的对边AB =CD 且AB //CD ,从而四边形 ABCD 是平行四边形.3.把∆ABC 的BC 边五等分,设分点依次为D 1、D 2、D 3、D 4,再把→各分点与点A 连接.试以AB =c 、 → → →→BC =a 表示向量D 1A 、D 2 A 、D 3 A 、→ D 4 A .→ → → 解 D 1A =BA -BD 1 =-c -5a , → → → 2 D 2 A =BA -BD 2 =-c -5a , → → → 3 D 3 A =BA -BD 3 =-c -5a , → → → 4 D 4 A =BA -BD 4 =-c -5a .62 +72 +(-6)2 4.已知两点M 1(0, 1, 2)和M 2(1,-1, 0).试用坐标表示式表→ →示向量M 1M 2 及-2M 1M 2 .→解 M 1M 2 =(1, -1, 0)-(0,1, 2) =(1, -2, -2) ,→ -2M 1M 2 =-2(1, -2, -2)=(-2, 4, 4) .5.求平行于向量 a =(6, 7,-6)的单位向量.解 |a |= =11,平行于向量a =(6, 7,-6)的单位向量为 1a =(6, 7, -6) 或-1a =(-6, -7, 6) .|a | 1111 11 |a | 11 11116.在空间直角坐标系中,指出下列各点在哪个卦限？A (1,-2, 3);B (2, 3,-4);C (2,-3,-4);D (-2,-3, 1).解A 在第四卦限,B 在第五卦限,C 在第八卦限,D 在第三卦 限.7.在坐标面上和坐标轴上的点的坐标各有什么特征？指 出下列各点的位置:A (3, 4, 0);B (0, 4, 3);C (3, 0, 0);D (0,-1, 0).解 在 xOy 面上,点的坐标为(x ,y , 0);在 yOz面上,点的坐标为(0,y ,z );在 zOx 面上,点的坐标为(x , 0,z ).在 x 轴上,点的坐标为(x , 0, 0);在 y 轴上,点的坐标为(0,y , 0) ,在z 轴上,点的坐标为(0,0,z ).A 在 xOy 面上,B 在 yOz 面上,C 在 x 轴上,D 在 y 轴上.8.求点(a ,b ,c )关于(1)各坐标面; (2)各坐标轴;(3)坐标原点的对称点的坐标.解 (1)点(a ,b ,c )关于xOy 面的对称点为(a ,b ,-c ),点(a ,b ,c )关于yOz 面的对称点为(-a ,b ,c ),点(a ,b ,c )关于zOx面的对称点为(a ,-b ,c ).(2) 点(a ,b ,c )关于x 轴的对称点为(a ,-b ,-c ),点(a ,b ,c )关于y 轴的对称点为(-a ,b ,-c ),点(a ,b ,c )关于 z 轴的对称点为(-a , -b ,c ).(3) 点(a ,b ,c )关于坐标原点的对称点为(-a ,-b ,-c ).9.自点P 0(x 0,y 0,z 0)分别作各坐标面和各坐标轴的垂线,写出各垂足的坐标.解 在 xOy 面、yOz 面和 zOx 面上,垂足的坐标分别为(x 0,y 0, 0)、(0,y 0,z 0)和(x 0, 0,z 0).在 x 轴、y 轴和 z 轴上,垂足的坐标分别为(x 0, 0, 0),(0,y 0, 0) 和(0, 0,z 0).10.过点P 0(x 0,y 0,z 0)分别作平行于z 轴的直线和平行于xOy 面的平面,问在它们上面的点的坐标各有什么特点？解 在所作的平行于 z 轴的直线上,点的坐标为(x 0,y 0,z );在所作的平行于 xOy 面的平面上,点的坐标为(x ,y ,z 0).11.一边长为a 的立方体放置在xOy面上,其底面的中心在坐标原点,底面的顶点在 x 轴和 y 轴上,求它各顶点的坐标.解因为底面的对角线的长为标分别为2a ,所以立方体各顶点的坐 (,(a ) ,2 , 0, 0), 2, 0, a ), 2 (0,, 0) ,(0, , a ) , 2, 2) . 212.求点 M (4,-3, 5)到各坐标轴的距离.(-3)2 +52 34 42 +52 41 ⎩ 解点M 到x 轴的距离就是点(4,-3, 5)与点(4, 0, 0)之间的距离,即d x = = .点M 到y 轴的距离就是点(4,-3, 5)与点(0,-3, 0)之间的距离,即d y = = .点 M 到 z 轴的距离就是点(4,-3, 5)与点(0, 0, 5)之间的距离, 即d z = =5 .13.在 yOz 面上,求与三点 A (3, 1, 2)、B (4,-2,-2)和 C (0, 5, 1) 等距离的点.解设所求的点为P (0,y ,z )与A 、B 、C 等距离,则→ |PA |2=32 +(y -1)2 +(z -2)2 ,→ |PB |2=42+(y +2)2+(z +2)2,→ | PC |2=(y -5)2 +(z -1)2 .由题意,有→ → → | PA |2=| PB |2=| PC |2 ,⎧32+(y -1)2+(z -2)2=(y -5)2+(z -1)2 ⎨42+(y +2)2+(z +2)2=(y -5)2+(z -1)2 解之得 y =1,z =-2,故所求点为(0, 1,-2).14.试证明以三点 A (4, 1, 9)、B (10,-1, 6)、C (2, 4,3)为顶点的三角形是等腰三角直角三角形.解因为42 +(-3)2 即(2-4)2 +(4-1)2 +(3-9)2 (2-10)2 +(4+1)2 +(3-6)2 2→ |AB |= →|AC |= → |BC |= → → → =7 ,=7 ,=7 ,→ →所以|BC |2=|AB |2+|AC |2,|AB |=|AC |.因此∆ABC 是等腰直角三角形.→ 15.设已知两点M 1(4, 的模、方向余弦和方向角.→2,1) 和M 2(3, 0, 2).计算向量M 1M 2 解 M 1M 2 =(3-4, 0- →|M 1M 2 |= 2, 2-1) =(-1, = 2 ;2,1) ;cos α=-1, 2 cos β 2 cos γ=1; 2α=2π, 3 β=3π, 4 γ=π. 316.设向量的方向余弦分别满足(1)cos α=0; (2)cos β=1;(3) cos α=cos β=0,问这些向量与坐标轴或坐标面的关系如何？解(1)当cos α=0时,向量垂直于x 轴,或者说是平行于yOz面.(2) 当cos β=1时,向量的方向与y 轴的正向一致,垂直于 zOx 面.(3) 当cos α=cos β=0时,向量垂直于x 轴和y 轴,平行于z轴,垂直于 xOy 面.17.设向量r 的模是4,它与轴u 的夹角是60︒,求r 在轴u 上的投影.(10-4)2 +(-1-1)2 +(6-9)2 (-1)2 +( 2)2 +12解 Pr j r =|r |⋅cos π=4⋅1=2 .u 3 218.一向量的终点在点 B (2,-1, 7),它在 x 轴、y 轴和 z轴上的投影依次为 4,-4, 7.求这向量的起点 A 的坐标.解设点A 的坐标为(x ,y ,z ).由已知得⎧⎪2-x =4 ⎨-1-y =-4 ,⎪⎩7-z =7解得 x =-2,y =3,z =0.点 A 的坐标为 A (-2, 3, 0).19.设m =3i +5j +8k ,n =2i -4j -7k 和p =5i +j -4k .求向量a =4m +3n -p 在x 轴上的投影及在y 轴上的分向量.解因为a =4m +3n -p=4(3i +5j +8k )+3(2i -4j -7k )-(5i +j -4k )=13i +7j +15k ,所以 a =4m +3n -p 在 x 轴上的投影为 13,在 y 轴上的分向量 7j .17 习题 7-21.设 a =3i -j -2k ,b =i +2j -k ,求(1)a ⋅b 及 a ⨯b ; (2)(-2a )⋅3b 及 a ⨯2b ; (3)a 、b 夹角的余弦.解(1)a ⋅b =3⨯1+(-1)⨯2+(-2)⨯(-1)=3,i j k a ⨯b = 3-1 - 2 =5i +j + 7k .1 2 -1(2)(-2a )⋅3b =-6a ⋅b =-6⨯3=-18,a ⨯2b =2(a ⨯b )=2(5i +j +7k )=10i +2j +14k .(3)cos(a ,^b )=|a ⋅b |= 3 =3 .|a ||b |14 6 2212.设 a 、b 、c 为单位向量,且满足 a +b +c =0,求 a ⋅b +b ⋅c +c ⋅a .解因为a +b +c =0,所以(a +b +c )⋅(a +b +c )=0,即a ⋅a +b ⋅b +c ⋅c +2a ⋅b +2a ⋅c +2c ⋅a =0,于是 a ⋅b +b ⋅c +c ⋅a =-1(a ⋅a +b ⋅b +c ⋅c )=-1(1+1+1)=-3.2 2 2→ → 3.已知M 1(1,-1, 2)、M 2(3, 3, 1)和M 3(3, 1, 3).求与M 1M 2 、M 2M 3 同时垂直的单位向 量.→ 解 M 1M 2=(3-1,3+1,1-2)=(2,4,-1),→M 2M 3 =(3-3,1-3, 3-1) =(0, - 2, 2) . → → i j k n =M 1M 2⨯M 2M 3 = 2 4 0 -2 -1 =6i - 4 j - 4k ,2|n |=36+16+16=2,e =±1(6i - 4 j -4k )=±1(3i - 2 j - 2k ) 为所求向量.2 17 174.设质量为100kg 的物体从点 M 1(3, 1, 8)沿直线称动到点 M 2(1, 4,2),计算重力所作的功(长度单位为 m ,重力方向为 z 轴负方向).解 F =(0, 0,-100⨯9. 8)=(0, 0,-980), →S =M 1M 2 =(1-3, 4-1, 2-8) =(-2, 3, -6) .W =F ⋅S =(0, 0,-980)⋅(-2, 3,-6)=5880(焦耳).→ 5.在杠杆上支点O 的一侧与点O 的距离为x 1的点P 1处,有一与O P 1 成角θ1的力F 1→作用着;在O 的另一侧与点O 的距离为x 2的点P 2处,有一与OP 2 成角θ1的力F 1作用着.问 θ1、θ2、x 1、x 2、|F 1|、|F 2|符合怎样的条件才能使杠杆保持平衡？解因为有固定转轴的物体的平衡条件是力矩的代数和为零,再注意到对力矩正负的22 + 22 +12OA OB 规定可得,使杠杆保持平衡的条件为x 1|F 1|⋅sin θ1-x 2|F 2|⋅sin θ2=0,即 x 1|F 1|⋅sin θ1=x 2|F 2|⋅sin θ2.6.求向量 a =(4,-3, 4)在向量 b =(2, 2, 1)上的投影.解Pr j b a =a ⋅e b =a ⋅b =1a ⋅b = 1 (4,-3,4)⋅(2,2,1)=1(4⨯2-3⨯2+4⨯1)=2. |b ||b | 3 7.设 a =(3, 5,-2),b =(2, 1, 4),问 λ与 μ有怎样的关系,能使得 λa +μb 与 z 轴垂直? 解 λa +μb =(3λ+2μ, 5λ+μ,-2λ+4μ),λa +μb 与 z 轴垂 ⇔λa +μb ⊥k⇔(3λ+2μ, 5λ+μ,-2λ+4μ)⋅(0, 0, 1)=0,即-2λ+4μ=0,所以 λ=2μ.当 λ=2μ时,λa +μb 与 z 轴垂直.8.试用向量证明直径所对的圆周角是直角.→ → 证明 设 AB 是圆 O 的直径,C 点在圆周上,则OB =-OA ,→ →|OC |=|OA | .→→→ → → → → → → → →→ 因为 AC ⋅BC =(OC -OA )⋅(OC -OB ) =(OC -OA )⋅(OC +OA ) =|OC |2 -|OA |2 =0 ,→ →所以 AC ⊥BC ,∠C =90︒.9.设已知向量 a =2i -3j +k ,b =i -j +3k 和 c =i -2j ,计算: (1)(a ⋅b )c -(a ⋅c )b ; (2)(a +b )⨯(b +c );(3)(a ⨯b )⋅c .解 (1)a ⋅b =2⨯1+(-3)⨯(-1)+1⨯3=8,a ⋅c =2⨯1+(-3)⨯(-2)=8,(a ⋅b )c -(a ⋅c )b =8c -8b =8(c -b )=8[(i -2j )-(i -j +3k )]=-8j -24k .(2)a +b =3i -4j +4k ,b +c =2i -3j +3k ,i j (a +b )⨯(b +c )=3 -4 2 -3 k 4 =-j -k . 3i (3) a ⨯b = 2 1 j k -3 1 =-8i -5 j +k , -1 3(a ⨯b )⋅c =-8⨯1+(-5)⨯(-2)+1⨯0=2.→ 10.已知OA =i +3 j , →OB =j +3k ,求 ∆OAB 的面积.→→ → →解 根据向量积的几何意义,于是∆OAB 的面积为S =1| →⨯→| . 2|OA ⨯OB | 表示以OA 和 OB 为邻边的平行四边形的面积,19 →→ i jk →→因为OA ⨯OB = 1 0 3 =-3i -3 j +k , |OA ⨯OB |=0 1 3 (-3)3 +(-3)2 +12= ,所以三角形∆OAB 的面积为 S =1|→⨯→|=1 19 . 2 OA OB 2 12.试用向量证明不等式:a 2 +a 2 +a 2b 2 +b 2 +b 2 ≥|a b +a b +a b | , 1 2 3 1 2 3 11 2 2 33其中a 1、a 2、a 3、b 1、b 2、b 3为任意实数,并指出等号成立的条件.解设a =(a 1,a 2,a 3),b =(b 1,b 2,b 3),则有a ⋅b =|a |⋅|b |cos(a ,^b )≤|a |⋅|b | ,于是 a 2 +a 2 +a 2b 2 +b 2 +b 2 ≥|a b +a b +a b | , 1 2 3 1 2 3 11 2 2 3 3其中当cos(a ,^b ) =1时,即 a 与 b 平行是等号成立.14 6 x 2 +y 2 +z 2(x - 2)2 +(y -3)2 +(z - 4)2 29 y 2 +z 2 x 2 +y 2 习题 7-31.一动点与两定点(2, 3, 1)和(4, 5, 6)等距离,求这动点的轨迹方程.解设动点为M (x ,y ,z ),依题意有(x -2)2+(y -3)2+(z -1)2=(x -4)2+(y -5)2+(z -6)2,即4x +4y +10z -63=0. 2.建立以点(1, 3,-2)为球心,且通过坐标原点的球面方程.解 球的半径R =12+32+(-2)2= ,球面方程为(x -1)2+(y -3)2+(z +2)2=14, 即 x 2+y 2+z 2-2x -6y +4z =0.3.方程x 2+y 2+z 2-2x +4y +2z =0 表示什么曲面?解由已知方程得(x 2-2x +1)+(y 2+4y +4)+(z 2+2z +1)=1+4+1,即(x -1)2+(y +2)2 +(z +1)2=(6)2 ,所以此方程表示以(1,-2,-1)为球心,以为半径的球面.4.求与坐标原点O 及点(2, 3, 4)的距离之比为1:2 的点的全体所组成的曲面的方程,它表示怎样曲面？解设点(x ,y ,z )满足题意,依题意有=1 , 2 化简整理得(x +2)2 +(y +1)2 +(z +4)2 =116,3 3 9它表示以(-2,-1,-4)为球心,以2 为半径的球面.3 3 35.将zOx 坐标面上的抛物线z 2=5x 绕x 轴旋转一周,求所生成的旋转曲面的方程.解将方程中的z 换成± 得旋转曲面的方程y 2+z 2=5x . 6.将zOx 坐标面上的圆x 2+z 2=9绕z 轴旋转一周,求所生成的旋转曲面的方程.解将方程中的x 换成± 得旋转曲面的方程x 2+y 2+z 2=9.7.将 xOy 坐标面上的双曲线 4x 2-9y 2=36 分别绕 x 轴及 y轴旋转一周,求所生成的旋转曲面的方程.解双曲线绕x 轴旋转而得的旋转曲面的方程为4x2-9y2-9z2=36.双曲线绕y 轴旋转而得的旋转曲面的方程为4x2+4z2-9y2=36.8.画出下列方程所表示的曲面:(1) (x -a)2 +y2 =(a)2 ;2 2(2) -x2 +y2=1 ;4 9(3) x2 +z2 =1 ;9 4(4)y 2-z =0;(5)z =2-x 2.9.指出下列方程在平面解析几何中和在空间解析几何中分别表示什么图形:(1)x =2;解在平面解析几何中,x =2 表示平行于y 轴的一条直线;在空间解析几何中,x =2 表示一张平行于 yOz 面的平面.(2)y =x +1;解 在平面解析几何中,y =x +1 表示一条斜率是 1,在 y 轴上的截距也是 1的直线;在空间解析几何中，y =x +1 表示一张平行于 z 轴的平面.(3)x 2+y 2=4;解 在平面解析几何中,x 2+y 2=4 表示中心在原点,半径是 4 的圆;在空间解析几何中, x 2+y 2=4 表示母线平行于 z 轴,准线为 x 2+y 2=4 的圆柱面.(4)x 2-y 2=1.解 在平面解析几何中,x 2-y 2=1 表示双曲线;在空间解析几何中,x 2-y 2=1表示母线平行于 z 轴的双曲面.10.说明下列旋转曲面是怎样形成的:(1) x 2 +y 2 +z 2 =1 ; 4 9 9解这是xOy 面上的椭圆x 2 +y 2 =1绕x 轴旋转一周而形成的,或是zOx 面上的椭圆 4 9x2 +z2 =1 绕 x 轴旋转一周而形成的.4 9(2) x2 -y2+z2 =1 ; 4解这是xOy 面上的双曲线x2 -y2=1 y 轴旋转一周而形成的,或是yOz 面上的双曲4线-y2+z2 =1绕 y 轴旋转一周而形成的.4(3)x2-y2-z2=1;解这是xOy 面上的双曲线x2-y2=1 绕x 轴旋转一周而形成的,或是zOx 面上的双曲线x2-z2=1 绕 x 轴旋转一周而形成的.(4)(z-a)2=x2+y2.解这是zOx 面上的曲线(z-a)2=x2绕z轴旋转一周而形成的,或是yOz面上的曲线(z-a)2=y2绕 z 轴旋转一周而形成的.11.画出下列方程所表示的曲面:(1)4x2+y2-z2=4;(2)x2-y2-4z2=4;(3)z=x2 +y2 .3 4 9⎨y =2 4-x 2-y 2 ⎨x -y =0 ⎨x 2+2 2z 习题 7-41.画出下列曲线在第一卦限内的图形:(1) ⎧x =1 ; ⎩(2)⎧z = ; ⎩(3) ⎧x 2 +y 2 =a 2 . ⎩⎨y =2x -3 ⎩ ⎩⎨x 2+2 2z -y =0 ⎨x 2+2 2z -y =0 ⎨x 2+2 2z -y =0 ⎨z =0 ⎨y =x 2.指出下方程组在平面解析几何中与在空间解析几何中分别表示什么图形:(1) ⎧y = 5x +1 ; ⎩ 解 在平面解析几何中,⎧y = 5x +1 表示直线y =5x +1 与y =2x -3 的交点(-4, -17) ;在空 ⎨y =2x -3 3 3 间解析几何中,⎧y = 5x +1 表示平面y =5x +1 与y =2x -3 的交线,它表示过点(-4, -17, 0) ,并且行于 z 轴.⎨y =2x -3 3 3 ⎧⎪x 2+y 2 =1(2)⎨4 9 .⎪⎩y =3 ⎧⎪x 2+y 2 =1x 2 y 2 解 在平面解析几何中, ⎨4 9 ⎪⎩y =3 表示椭圆 4 9 =1 与其切线y =3 的交点(0,3);在 ⎧⎪x 2+y 2 =1 x 2 y 2 空间解析几何中, ⎨4 9 ⎪⎩y =3 表示椭圆柱面 4 +9 =1 与其切平面 y =3 的交线. 3.分别求母线平行于 x 轴及 y 轴而且通过曲线⎧2x 2 +y 2 +z 2 =16 的柱面方程. ⎩解把方程组中的x 消去得方程3y 2-z 2=16,这就是母线平行于x 轴且通过曲线⎧2x 2 +y 2 +z 2 =16 的柱面方程. ⎩把方程组中的y 消去得方程3x 2+2z 2=16,这就是母线平行于y 轴且通过曲线⎧2x 2 +y 2 +z 2 =16 的柱面方程. ⎩ 4.求球面 x 2+y 2+z 2=9 与平面 x +z =1 的交线在 xOy 面上的投影的方程.解由x +z =1 得z =1-x 代入x 2+y 2+z 2=9 得方程2x 2-2x +y 2=8,这是母线平行于z 轴,准线为球面x 2+y 2+z 2=9 与平面x +z =1 的交线的柱面方程,于是所求的投影方程为⎧2x 2 - 2x +y 2 =8 . ⎩5.将下列曲线的一般方程化为参数方程:(1) ⎧x 2 +y 2 +z 2 =9 ; ⎩⎨z =0 ⎨z =0解将y =x 代入x 2+y 2+z 2=9 得2x 2+z 2=9,即令 x =3cos t ,则 z =3sin t .2故所求参数方程为x 2 3 2 +z 2 =1 . 32 x = 3cos t , 2 y = 3cos t ,z =3sin t .2(2) ⎧(x -1)2 +y 2 +(z +1)2 = 4 . ⎩ 解 将 z =0 代入(x -1)2+y 2+(z +1)2=4 得(x -1)2+y 2=3.令 x =1+ 3cos t ,则 y =3 sin t ,于是所求参数方程为x =1+ 3 cos t , y = 3 sin t ,z =0.⎧⎪x =a c os θ 6.求螺旋线⎨y =a sin θ在三个坐标面上的投影曲线的直角坐标方程.⎪⎩z =b θ解由前两个方程得x 2+y 2=a 2,于是螺旋线在xOy 面上的投影曲线的直角坐标方程为⎧x 2 +y 2 =a 2 . ⎩由第三个方程得θ=z 代入第一个方程得 bx =cos z ,即 z =b arccos x ,a b a于是螺旋线在zOx 面上的投影曲线的直角坐标方程为⎧⎪z =b arccos x . ⎨ a ⎪⎩y =0由第三个方程得θ=z 代入第二个方程得 by =sin z ,即 z =b arcsin y ,a b a于是螺旋线在yOz 面上的投影曲线的直角坐标方程为⎧⎪x =0 ⎨z =b arcsin y ⎩⎪ a .a 2 -x 2 -y 2 a 2 -x 2 -y 27.求上半球0≤z ≤的投影. 与圆柱体x 2+y 2≤ax (a >0)的公共部分在xOy 面和zOx 面上 解圆柱体x 2+y 2≤ax 在xOy 面上的投影为x 2+y 2≤ax ,它含在半球0≤z ≤在 xOy 面上的投影 x 2+y 2≤a 2内,所以半球与圆柱体的公共部分在 xOy 面上的投影为 x 2+y 2≤ax .为求半球与圆柱体的公共部分在 zOx 面上的投影,由圆柱面方程 x 2+y 2=ax 得 y 2=ax -x 2,代入半球面方程z = ,得z =zOx 面上的投影为(0≤x ≤a ),于是半球与圆柱体的公共部分在0≤z ≤a 2 -ax (0≤x ≤a ),即 z 2+ax ≤a 2, 0≤x ≤a ,z ≥0.8.求旋转抛物面 z =x 2+y 2(0≤z ≤4)在三坐标面上的投影.解 令 z =4 得 x 2+y 2=4,于是旋转抛物面 z =x 2+y 2(0≤z ≤4)在 xOy 面上的投影为 x 2+y 2≤4. 令 x =0得 z =y 2,于是旋转抛物面 z =x 2+y 2(0≤z ≤4)在 yOz 面上的投影为 y 2≤z ≤4.令 y =0得 z =x 2,于是旋转抛物面 z =x 2+y 2(0≤z ≤4)在 zOx 面上的投影为 x 2≤z ≤4.a 2 -x 2 -y 2 a 2 -ax习题 7-51.求过点(3, 0,-1)且与平面3x-7y+5z-12=0 平行的平面方程.解所求平面的法线向量为n=(3,-7, 5),所求平面的方程为3(x-3)-7(y-0)+5(z+1)=0,即 3x-7y+5z-4=0.2.求过点M0(2, 9,-6)且与连接坐标原点及点M0的线段OM0垂直的平面方程.解所求平面的法线向量为n=(2, 9,-6),所求平面的方程为2(x-2)+9(y-9)-6(z-6)=0,即 2x+9y-6z-121=0.3.求过(1, 1,-1)、(-2,-2, 2)、(1,-1, 2)三点的平面方程.解 n1=(1,-1, 2)-(1, 1,-1)=(0,-2,3),n1=(1,-1, 2)-(-2,-2, 2)=(3, 1,0),所求平面的法线向量为i jn=n1⨯n2= 0 -23 1所求平面的方程为k3 =-3i +9 j +6k , 0-3(x-1)+9(y-1)+6(z+1)=0,即 x-3y-2z=0.4.指出下列各平面的特殊位置,并画出各平面:(1)x=0;解 x=0 是 yOz 平面.(2)3y-1=0;解3y-1=0 是垂直于y 轴的平面,它通过y 轴上的点(0, 1, 0).322 +(-2)2 +11(3)2x -3y -6=0;解 2x -3y -6=0 是平行于 z 轴的平面,它在 x 轴、y轴上的截距分别是 3 和-2.(4)x - 3y =0;解x - 3y =0是通过z 轴的平面,它在xOy 面上的投影的斜3(5)y +z =1;解y +z =1 是平行于x 轴的平面,它在y 轴、z 轴上的截距均 为 1.(6)x -2z =0;解 x -2z =0 是通过 y 轴的平面.(7)6x +5-z =0.解 6x +5-z =0 是通过原点的平面.5.求平面 2x -2y +z +5=0与各坐标面的夹角的余弦.解此平面的法线向量为 n =(2,-2, 1).此平面与yOz 面的夹角的余弦为cos α=cos(n ^,i )=n ⋅i = 2 =2;|n |⋅|i | 22+(-2)2+11 3此平面与zOx 面的夹角的余弦为cos β=cos(n ,^j )=n ⋅j = -2 =-2;|n |⋅|j | 3此平面与xOy 面的夹角的余弦为cos γ=cos(n ,^ k )=n ⋅k = 1 =1.|n |⋅|k | 22+(-2)2+11 36.一平面过点(1, 0,-1)且平行于向量 a=(2, 1, 1)和 b=(1,-1, 0),试求这平面方程.解所求平面的法线向量可取为i jn=a⨯b=2 11-1 所求平面的方程为k1 =i +j -3k , 0(x-1)+(y-0)-3(z+1)=0,即 x+y-3z-4=0.7.求三平面x+3y+z=1, 2x-y-z=0,-x+2y+2z=3 的交点.解解线性方程组⎧⎪x+3y+z=1⎨2x-y-z=0⎪⎩-x+2y+2z=3得 x=1,y=-1,z=3.三个平面的交点的坐标为(1,-1, 3).8.分别按下列条件求平面方程:(1)平行于zOx面且经过点(2,-5,3);解所求平面的法线向量为j =(0, 1, 0),于是所求的平面为0⋅(x-2)-5(y+5)+0⋅(z-3)=0,即 y=-5.(2)通过 z 轴和点(-3, 1,-2);解 所求平面可设为 Ax+By=0.因为点(-3, 1,-2)在此平面上,所以-3A+B=0,将B=3A 代入所设方程得Ax+3Ay=0,所以所求的平面的方程为x+3y=0,(3)平行于 x 轴且经过两点(4, 0,-2)和(5, 1, 7).解所求平面的法线向量可设为n=(0,b,c).因为点(4,0,-2) 和(5, 1, 7)都在所求平面上,所以向量n1=(5, 1, 7)-(4, 0,-2)=(1, 1, 9)与n 是垂直的,即b+9c=0,b=-9c ,于 是 n=(0,-9c,c)=-c(0, 9,-1).所求平面的方程为9(y-0)-(z+2)=0,即 9y-z-2=0.9.求点(1, 2, 1)到平面x+2y+2z-10=0 的距离.解点(1, 2, 1)到平面x+2y+2z-10=0 的距离为d =|1+2⨯2+2⨯1-10|=1.12 + 22 +22⎨2x +y +z =4 ⎨2x +y +z =4 ⎨2x +z =4习题 7-6 1.求过点(4,-1, 3)且平行于直线 x -3=y =z -1的直线方程.2 1 5解所求直线的方向向量为s =(2, 1, 5),所求的直线方程为x -4=y +1=z -3.2 1 52.求过两点 M 1(3,-2, 1)和 M 2(-1, 0, 2)的直线方程.解 所求直线的方向向量为 s =(-1, 0, 2)-(3,-2, 1)=(-4, 2, 1), 所求的直线方程为x -3=y +2=x -1.-4 2 13.用对称式方程及参数方程表示直线⎧x -y +z =1 . ⎩解 平面 x -y +z =1 和 2x +y +z =4 的法线向量为 n 1=(1,-1, 1), n 2=(2, 1, 1),所求直线的方向向量为i s =n 1⨯n 2 = 1 2 j k-1 1 =-2i +j +3k . 1 1在方程组⎧x -y +z =1 ⎩ 中,令 y =0,得⎧x +z =1 ⎩,解 得 x =3, z =-2.于是点(3, 0,-2)为所求直线上的点.所求直线的对称式方程为x -3=y =z +2;-2 1 3参数方程为x =3-2t ,y =t ,z =-2+3t .⎨3x +5y -2z +1=0 ⎩ ⎩⎨-2x +y +z =7 ⎨2x -y -z =0 4.求过点(2, 0,-3)且与直线⎧x -2y +4z -7=0 ⎩垂直的平面 方程.解所求平面的法线向量n 可取为已知直线的方向向量,即i j n =(1, -2, 4)⨯(3, 5, -2)= 1- 2 3 5 k 4=-16i +14 j +11k . -2所平面的方程为-16(x -2)+14(y -0)+11(z +3)=0,即 16x -14y -11z -65=0.5.求直线⎧5x -3y +3z -9=0与直线⎧2x +2y -z +23=0的夹角 的余弦.⎨3x -2y +z =0 ⎨3x +8y +z -18=0 解两直线的方向向量分别为i s 1 = 5 3 i j k-33 =3i + 4 j -k , -2 1j ks 2 = 22 38 -1 =10i -5 j +10k . 1两直线之间的夹角的余弦为cos(s ,^s )= s 1⨯s 2 1 2 | s |⋅|s | 1 2= 3⨯10+4⨯(-5)+(-1)⨯10 =0. 32 +42 +(-1)2 102 +(-5)2 +1026.证明直线⎧x +2y -z =7 ⎩ 与直线⎧3x +6y -3z =8平行. ⎩解两直线的方向向量分别为i j s 1=1 2 -21 i j k-1 =3i +j +5k , 1ks 2 = 3 2 6 -3=-9i -3j -15k .-1-1因为 s 2=-3s 1,所以这两个直线是平行的.7.求过点(0, 2,4)且与两平面 x +2z =1 和 y -3z =2平行的直线方程. 解因为两平面的法线向量n 1=(1, 0, 2)与n 2=(0, 1,-3)不平行,所以两平面相交于一直线,此直线的方向向量可作为所求直线的方向向量s ,即i j s = 10 01 k 2=-2i +3j +k . -3所求直线的方程为 x =y -2=z -4.-2 3 1 8.求过点(3, 1,-2)且通过直线 x -4=y +3=z 的平面方程.5 2 1 解所求平面的法线向量与直线x -4=y +3=z 的方向向量5 2 1s 1=(5, 2, 1)垂直.因为点(3, 1,-2)和(4,-3, 0)都在所求的平面上,所以所求平面的法线向量与向量s 2=(4,-3,0)-(3,1,-2)=(1,-4,2) 也是垂直的.因此所求平面的法线向量可取为⎨x -y -z =0 ⎨x -y -z =0i j n =s 1 ⨯s 2 =5 2 1-4 k 1 =8i -9 j -22k . 2所求平面的方程为8(x -3)-9(y -1)-22(z +2)=0,即 8x -9y -22z -59=0.9.求直线⎧x +y +3z =0 与平面 x -y -z +1=0 的夹角. ⎩解已知直线的方向向量为i s =(1,1, 3)⨯(1, -1, -1)=1 1 j k 1 3 -1 -1=2i + 4 j -2k =2(i + 2 j -k ) , 已知平面的法线向量为 n =(1,-1,-1).因为s ⋅n =2⨯1+4⨯(-1)+(-2)⨯(-1)=0,所以 s ⊥n ,从而直线⎧x +y +3z =0 与平面 x -y -z +1=0 的夹角为 0. ⎩10.试确定下列各组中的直线和平面间的关系: (1) x +3=y +4=z 和 4x -2y -2z =3;-2 -7 3解 所给直线的方向向量为 s =(-2,-7,3),所给平面的法线向量为 n =(4,-2,-2).因为 s ⋅n =(-2)⨯4+(-7)⨯(-2)+3⨯(-2)=0,所以 s ⊥n ,从而所给直线与所给平面平行.又因为直线上的点(-3,-4, 0)不满足平面方程 4x -2y -2z =3,所以所给直线不在所给平面上. (2) x =y =z 和 3x -2y +7z =8;3 -2 7⎩ ⎩解 所给直线的方向向量为 s =(3,-2,7),所给平面的法线向量为 n =(3,-2, 7).因为 s =n ,所以所给直线与所给平面是垂直的. (3) x -2=y +2=z -3和 x +y +z =3.3 1 -4解 所给直线的方向向量为 s =(3,1,-4),所给平面的法线向量为 n =(1, 1, 1).因为 s ⋅n =3⨯1+1⨯1+(-4)⨯1=0,所以 s ⊥n ,从而所给直线与所给平面平行.又因为直线上的点(2,-2,3)满足平面方程 x +y +z =3,所以所给直线在所给平面上. 11.求过点(1,2,1)而与两直线⎧x +2y -z +1=0 和⎧2x -y +z =0平行的平面的方程.⎨x -y +z -1=0 ⎨x -y +z =0 解已知直线的方向向量分别为i s 1=(1,2,-1)⨯(1,-1,1)=1 1 i s 1 =(2, -1,1)⨯(1, -1,1)=2 1 j k 2 -1 =i -2 j -3k ,-1 1j k -1 1 =-j -k . -1 1所求平面的法线向量可取为i n =s 1⨯s 2 =1 0 所求平面的方程为j k -2 -3 =-i +j -k ,-1 -1-(x -1)+(y -2)-(z -1)=0,即 x -y +z =0.12.求点(-1, 2, 0)在平面 x +2y -z +1=0 上的投影.⎨2x -y +z -4=0 解平面的法线向量为n =(1, 2,-1).过点(-1, 2, 0)并且垂直于已知平面的直线方程为x +1=y -2=z .1 2 -1将此方程化为参数方程x =-1+t ,y =2+2t ,z =-t ,代入平面方程 x +2y -z +1=0 中,得(-1+t )+2(2+2t )-(-t )+1=0,解得t =-2.再将t =-2代入直线的参数方程,得x =-5,y =2, 3 3 3 3z =2.于是点(-1, 2, 0) 在平面x +2y -z +1=0上的投影为点 3 (-5, 2, 2) . 2 3 313.求点 P (3,-1,2)到直线⎧x +y -z +1=0 ⎩ 解已知直线的方向向量为的距离.i s =(1,1,-1)⨯(2,-1,1)=1 2 j k 1 -1=-3j -3k . -1 1过点P 且与已知直线垂直的平面的方程为-3(y +1)-3(z -2)=0,即 y +z -1=0.解线性方程组⎧⎪x +y -z +1=0⎨2x -y +z -4=0 ,⎪⎩y +z -1=0得 x =1, y =-1, 2 z =3.22 ⎨2x -y +z -4=0 ⎨3x -y -2z -9=0 点P (3,-1, 2)到直线⎧x +y -z +1=0 ⎩与点(1, -1, 3) 间的距离,即的距离就是点P (3,-1, 2) 2 2d =. 214.设M 0是直线L 外一点,M 是直线L 上任意一点,且直线的方向向量为s ,试证:点M 0到直线L 的距离→ d =|M 0M ⨯s |. |s |→解设点M 0到直线L 的距离为d ,L 的方向向量s =MN ,根→ → 据向量积的几何意义,以M 0M 和MN 为邻边的平行四边形的面积为→ → →|M 0M ⨯MN |=|M 0M ⨯s | ,→ → → 又以 M 0M 和 MN 为邻边的平行四边形的面积为 d ⋅|MN |=d ⋅| s | . 因此→d ⋅|s |=| → M 0M ⨯s |, d =|M 0M ⨯s |. | s | 15.求直线⎧2x -4y +z =0 ⎩在平面4x -y +z =1 上的投影直线 的方程.解过已知直线的平面束方程为(2+3λ)x +(-4-λ)y +(1-2λ)z -9λ=0.为在平面束中找出与已知平面垂直的平面,令(4 -1, 1)⋅(2+3λ,-4-λ, 1-2λ)=0,⎩ 即4⋅(2+3λ)+(-1)⋅(-4-λ)+1⋅(1-2λ)=0.解之得λ=-13.将λ=-13代入平面束方程中,得11 1117x +31y -37z -117=0.故投影直线的方程为⎧4x -y +z =1 ⎨17x +31y -37z -117 =0 16.画出下列各曲面所围成的立体图形:(1)x =0,y =0,z =0,x =2,y =1, 3x +4y +2z -12=0;(2)x =0,z =0,x =1,y =2, z =y ;4.(3)z=0,z=3,x-y=0,x-3y=0,x2+y2=1(在第一卦限内);(4)x=0,y=0,z=0,x2+y2=R2,y2+z2=R2(在第一卦限内).总习题七1.填空(1)设在坐标系[O;i,j,k]中点A和点M的坐标依次为(x0,y0,z0)和(x,y,z),则在[A;i,j,k]→坐标系中,点M 的坐标为,向量OM的坐标为.解 M(x-x0,y-y0,z-z0),→OM =(x, y, z) .提示:自由向量与起点无关,它在某一向量上的投影不会因起点的位置的不同而改变.(2)设数λ1、λ2、λ3不全为0,使λ1a+λ2b+λ3c=0,则a、b、c三个向量是的.解共面.(3)设a=(2,1,2),b=(4,-1,10),c=b-λa,且a⊥c,则λ= .解 3.提示:因为 a⊥c,所以 a⋅c=0.又因为由a⋅c=a⋅b-λa⋅a=2⨯4+1⨯(-1)+2⨯10-λ(22+12+22)=27-9λ,所以λ=3.(4)设a、b、c都是单位向量,且满足a+b+c=0,则a⋅b+b⋅c+c⋅a= .解 -3.2提示:因为a+b+c=0,所以(a+b+c)⋅(a+b+c)=0,即a⋅a+b⋅b+c⋅c+2a⋅b+2a⋅c+2c⋅a=0,于是a⋅b+b⋅c+c⋅a=-1(a⋅a+b⋅b+c⋅c)=-1(1+1+1)=-3.2 2 2(5)设|a|=3,|b|=4,|c|=5,且满足a+b+c=0,则|a⨯b+b⨯c+c⨯a|= .解36.提示:c=-(a+b),a⨯b+b⨯c+c⨯a=a⨯b-b⨯(a+b)-(a+b)⨯a=a⨯b-b⨯a-b⨯a=3a⨯b,|a⨯b+b⨯c+c⨯a|=3|a⨯b|=3|a|⋅|b|=3⋅3⋅4=36.2.在y轴上求与点 A(1,-3, 7)和点 B(5, 7,-5)等距离的点.解设所求点为M(0,y, 0),则有12+(y+3)2+72=52+(y-7)2+(-5)2,即(y+3)2=(y-7)2,解得 y=2,所求的点为 M(0, 2, 0).3.已知 ∆ABC 的顶点为 A(3,2,-1)、B(5,-4,7)和C(-1,1,2),求从顶点 C 所引中线的长度.解线段AB 的中点的坐标为(3+5, 2-4, -1+7) =(4, -1, 3) .所求中线的长度为2 2 2.→→→4.设∆ABC 的三边BC =a 、CA=b 、AB =c ,三边中点依次为D、E、F,试用向量a、d = (4+1)2 +(-1-1)2 +(3- 2)2 = 30→→→b 、c 表示AD 、BE 、CF ,并证明→→→AD +BE +CF =0 . 解 →=→+→=c +1a , ADABBD 2→=→+→=a +1b , BEBCCE 2→=→+→=b +1c . CFCAAF 2→→→3 3AD +BE +CF =2 (a +b +c )=2(-c +c )=05.试用向量证明三角形两边中点的连线平行于第三边,且其长度等于第三边长度的一半.证明设D ,E 分别为AB ,AC 的中点,则有 →=→-→=1( →-→) , DE AE AD 2AC AB→ →→→→BC =BA +AC =AC -AB ,所以→1→DE =2BC ,从而 DE //BC ,且|DE |=1| BC | .26.设|a +b |=|a -b |,a =(3,-5, 8),b =(-1, 1,z ),求 z .解a +b =(2,-4, 8+z ),a -b =(4,-6, 8-z ).因为|a +b |=|a -b |,所以,解得 z =1.7.设|a |=, |b |=1,(a ,^b ) =π,求向量 a +b 与 a -b 的夹角.6解 |a +b |2=(a +b )⋅(a +b )=|a |2+|b |2+2a ⋅b =|a |2+|b |2+2|a |⋅|b |cos(a ,^b ) =3+1+2|a -b |2=(a -b )⋅(a -b )=|a |2+|b |2-2a ⋅b =|a |2+|b |2-2|a |⋅|b |cos(a ,^b )=3+1-2 设向量a +b 与a -b 的夹角为θ,则3 cos π=7 ,63 cos π=1 .6cos θ=(a +b )⋅(a -b )= |a |2 -|b |2 |a +b |⋅|a -b | θ=arccos 2.7|a +b |⋅|a -b | 22 +(-4)2 +(8+z )2 = 42 +(-6)2 +(8-z )2 38.设 a +3b ⊥7a -5b ,a -4b ⊥7a -2b ,求(a ,^b ) . 解 因为 a +3b ⊥7a -5b ,a -4b ⊥7a -2b , 所以 (a +3b )⋅(7a -5b )=0,(a -4b )⋅(7a -2b )=0, 即 7|a |2+16a ⋅b -15|b |2=0, 7|a |2-30a ⋅b +8|b |2=0, 又以上两式可得|a |=|b |= 2 a ⋅b ,于是cos(a ,^b ) =a ⋅b =1,(a ,^b ) =π.|a |⋅|b | 239.设 a =(2,-1,-2),b =(1, 1,z ),问 z 为何值时(a ,^b ) 最小？并求出此最小值. 解 cos(a ,^b ) =a ⋅b =1-2z . |a |⋅|b | 3 2+z 2因为当 0<(a ,^b )<π时,cos(a ,^b ) 为单调减函数.求(a ,^b ) 的最小值也就是求 f (z )=1-2z2的最大值.3 2+z 2令 f '(z ) =1⋅-4-z =0 ,得 z =-4.3 (2+z 2)3/ 2当z =-4时,cos(a ,^b )=2,所以(a ,^b )=2min2 410.设|a |=4, |b |=3, (a ,^b ) =π,求以 a +2b 和 a -3b 为边的平行四边形的面积.6解 (a +2b )⨯(a -3b )=-3a ⨯b +2b ⨯a =5b ⨯a .以a +2b 和a -3b 为边的平行四边形的面积为|(a + 2b )⨯(a -3b )|=5|b ⨯a |=5|b |⋅|a |sin(a ,^b ) =5⋅3⋅4⋅1=30 .211.设 a =(2,-3, 1),b =(1,-2, 3),c =(2, 1, 2),向量 r 满足 r ⊥a ,r ⊥b , Prj c r =14,求 r . 解 设 r =(x ,y ,z ).因为r ⊥a ,r ⊥b ,所以r ⋅a =0,r ⋅b =0,即 2x -3y +z =0,x -2y +3z =0.又因为Prj cr =14,所以r ⋅1c =14 ,即|c |2x +y +2z =42. 解线性方程组⎪(x -1)2+(y +1)2+(z -2)2⎧⎪2x -3y +z =0 ⎨x -2y +3z =0 , ⎪⎩2x +y +2z =42得 x =14,y =10,z =2,所以 r =(14, 10, 2).i j k另解 因为 r ⊥a ,r ⊥b ,所以 r 与a ⨯b =2-3 1 -2 1 =-7i -5 j -k 平行,故可设 r =λ(7, 5, 1).3又因为Prj c r =14,所以r ⋅1c =14 ,r ⋅c =42,即|c |λ(7⨯2+5⨯1+1⨯2)=42,λ=2,所以 r =(14, 10, 2).12.设a =(-1, 3, 2),b =(2,-3,-4),c =(-3, 12, 6),证明三向量a 、b 、c 共面,并用a 和b 表示 c .证明向量a 、b 、c 共面的充要条件是(a ⨯b )⋅c =0.因为i j a ⨯b =-1 3 2 -3 k2 =-6i -3k ,- 4 (a ⨯b )⋅c =(-6)⨯(-3)+0⨯12+(-3)⨯6=0,所以向量a 、b 、c 共面.设c =λa +μb ,则有(-λ+2μ, 3λ-3μ, 2λ-4μ)=(-3, 12, 6), 即有方程组⎧-λ+ 2μ=-3 ⎨3λ-3μ=12 , ⎪⎩2λ-4μ=6解之得λ=5,μ=1,所以 c =5a +b .13.已知动点 M (x ,y ,z )到 xOy 平面的距离与点 M 到点(1, -1, 2)的距离相等,求点 M 的轨迹方程.解根据题意,有|z |= ,或 z 2=(x -1)2+(y +1)2+(z -2)2,化简得(x -1)2+(y +1)2=4(z -1),这就是点 M 的轨迹方程.14.指出下列旋转曲面的一条母线和旋转轴:(1)z =2(x 2+y 2);解 旋转曲面的一条母线为 zOx 面上的曲线 z =2x 2,旋转轴为 z 轴.2⎨x =0 ⎨x =0⎨x =0 (2) x 2 +y 2 +z 2=1 ; 36 9 36解 旋转曲面的一条母线为 xOy 面上的曲线 x 2 +y 2=1,旋转轴为 y 轴.(3)z 2=3(x 2+y 2);36 9解 旋转曲面的一条母线为 yOz 面上的曲线 z = 3 y ,旋转轴为 z 轴.(4) x 2 -y 2 -z 2=1. 4 4解 旋转曲面的一条母线为 xOy 面上的曲线 x 2-y 2=1 ,旋转轴为 x 轴.415.求通过点 A (3, 0, 0)和 B (0, 0, 1)且与 xOy 面成 π角的平面的方程.3 解 设所求平面的法线向量为 n =(a ,b ,c ).→BA =(3, 0, -1) ,xOy 面的法线向量为 k =(0, 0, 1).→ 按要求有n ⋅ =0, n ⋅k =cos π,BA⎧⎪3a -c =0 |n |⋅|k | 3即解之得 c =3a , b =± 26a .于是所求的平面的方程为(x -3) ±26 y +3z =0 ,即x + 26y +3z =3,或 x -26 y +3z =3 .16.设一平面垂直于平面z =0,并通过从点(1, -1, 1)到直线⎧y -z +1= 0的垂线,求此平 ⎩面方程.解 直线⎧y -z +1= 0的方向向量为 s =(0, 1,-1)⨯(1, 0, 0)=(0,-1,-1). ⎩设点(1, -1, 1)到直线⎧y -z +1= 0的垂线交于点(x 0,y 0,z 0).因为点(x 0,y 0,z 0)在直线⎩⎨x =0⎧y -z +1= 0上,所以(x 0,y 0,z 0)=(0,y 0,y 0+1).于是,垂线的方向向量为 ⎩s 1=(-1,y 0+1,y 0).显然有s ⋅s 1=0,即-y 0-1-y 0=0, y =-1. 2 从而 s 1 =(-1, y 0 +1, y 0) =(-1, 1, -1) .2 2所求平面的法线向量可取为n =k ⨯s 1 =k ⨯(-i +1j -1k ) =-1i -j ,所求平面的方程为2 2 2 -1(x -1) -(y +1) =0 ,即 x +2y +1=0 217.求过点(-1, 0, 4),且平行于平面3x -4y +z -10=0,又与直线x +1=y -3=z相交的直1 12 线的方程.解过点(-1, 0, 4),且平行于平面3x -4y +z -10=0 的平面的方程为3(x +1)-4(y -0)+(z -4)=0,即 3x -4y +z -1=0. 将直线x +1=y -3=z化为参数方程 x =-1+t ,y =3+t ,z =2t ,代入平面方程 3x -4y +z -1=0,1 12 得3(-1+t )-4(3+t )+2t -1=0,解得t =16.于是平面3x -4y +z -1=0 与直线x +1=y -3=z的交点的坐标为(15, 19,32),这也1 12 是所求直线与已知直线的交点的坐标.所求直线的方向向量为s =(15, 19, 32)-(-1, 0, 4)=(16, 19, 28), 所求直线的方程为x +1=y =z -4. 16 19 2818.已知点 A (1, 0, 0)及点 B (0, 2, 1),试在 z 轴上求一点 C ,使∆ABC 的面积最小.→ 解 设所求的点为 C (0, 0,z ),则 AC =(-1, 0, z ) ,→BC =(0, - 2, z -1) .→→i j k因为 AC ⨯BC =-10 0 - 2 z z -1= 2z i +(z -1) j + 2k ,所以∆ABC 的面积为x 2 +y 2 x 2 +y 2 ⎨z = 22(x -1)+(y -1) ⎩ ⎩ ⎩⎨y =0 ⎩ ⎨x =0 ⎩ ⎩⎨z =0⎩ ⎩S =1|→⨯→|=14z 2 +(z -1)2 + 4 . 2AC BC 2令dS =1⋅ 8z +2(z -1)=0,得z =1,所求点为C (0,0,1). dz 4 4z 2 +(z -1)2 +4 5519.求曲线⎧z = 2-x 2 -y 2 ⎩ 在三个坐标面上的投影曲线的方程. 解在xOy 面上的投影曲线方程为⎧(x -1)2 +(y -1)2 = 2-x 2 -y 2 ,即⎧x 2 +y 2 =x +y .⎨z =0 ⎨z =0 在zOx 面上的投影曲线方程为⎧z =(x -1)2+(± ⎨y =0 2-x 2 -z -1)2 ,即⎧2x 2+ 2xz +z 2- 4x -3z + 2=0 .⎩ 在yOz 面上的投影曲线方程为⎧z =(±⎨x =0 2-y 2 -z -1)2 +(y -1)2 ,即⎧2y 2+ 2yz +z 2- 4y -3z + 2=0 .⎩20.求锥面z =与柱面z 2=2x 所围立体在三个坐标面上的投影.解锥面与柱面交线在xOy 面上的投影为⎧2x =x 2 +y 2 ,即⎧(x -1)2 +y 2 =1 ,⎨z =0 ⎨z =0 所以,立体在 xOy 面上的投影为⎧(x -1)2 +y 2 ≤1.⎩ 锥面与柱面交线在yOz 面上的投影为 ⎧⎪z =⎧⎪(z 2-2)2+y 2=1⎨ ,即⎨2 , ⎪⎩x ⎪⎩x =0⎧⎪(z 2-2)2+y 2≤1 所以,立体在yOz 面上的投影为⎨2 .⎪⎩x =0锥面z = 与柱面z 2=2x 与平面y =0 的交线为⎧z =|x |和⎧z = 2x ,⎨y =0 ⎨y =0所以,立体在zOx 面上的投影为⎩x 2 +y 2⎧x ≤z≤ ⎨y =021.画出下列各曲面所围立体的图形:(1)抛物柱面 2y 2=x ,平面 z =0 及x =y=z =1 ;4 2 2(2)抛物柱面 x 2=1-z ,平面 y =0,z =0 及 x +y =1;(3) 圆锥面z = 及旋转抛物面z =2-x 2-y 2;(4) 旋转抛物面x 2+y 2=z ,柱面y 2=x ,平面z =0及x =1.2x.。
伍德里奇计量经济学导论第6版笔记和课后习题答案
第1章计量经济学的性质与经济数据1.1复习笔记考点一:计量经济学★1计量经济学的含义计量经济学,又称经济计量学,是由经济理论、统计学和数学结合而成的一门经济学的分支学科,其研究内容是分析经济现象中客观存在的数量关系。
2计量经济学模型(1)模型分类模型是对现实生活现象的描述和模拟。
根据描述和模拟办法的不同,对模型进行分类,如表1-1所示。
(2)数理经济模型和计量经济学模型的区别①研究内容不同数理经济模型的研究内容是经济现象各因素之间的理论关系,计量经济学模型的研究内容是经济现象各因素之间的定量关系。
②描述和模拟办法不同数理经济模型的描述和模拟办法主要是确定性的数学形式,计量经济学模型的描述和模拟办法主要是随机性的数学形式。
③位置和作用不同数理经济模型可用于对研究对象的初步研究,计量经济学模型可用于对研究对象的深入研究。
考点二:经济数据★★★1经济数据的结构(见表1-3)2面板数据与混合横截面数据的比较(见表1-4)考点三:因果关系和其他条件不变★★1因果关系因果关系是指一个变量的变动将引起另一个变量的变动,这是经济分析中的重要目标之计量分析虽然能发现变量之间的相关关系,但是如果想要解释因果关系,还要排除模型本身存在因果互逆的可能,否则很难让人信服。
2其他条件不变其他条件不变是指在经济分析中,保持所有的其他变量不变。
“其他条件不变”这一假设在因果分析中具有重要作用。
1.2课后习题详解一、习题1.假设让你指挥一项研究，以确定较小的班级规模是否会提高四年级学生的成绩。
(i)如果你能指挥你想做的任何实验，你想做些什么?请具体说明。
(ii)更现实地，假设你能搜集到某个州几千名四年级学生的观测数据。
你能得到它们四年级班级规模和四年级末的标准化考试分数。
你为什么预计班级规模与考试成绩成负相关关系?(iii)负相关关系一定意味着较小的班级规模会导致更好的成绩吗?请解释。
答:(i)假定能够随机的分配学生们去不同规模的班级，也就是说，在不考虑学生诸如能力和家庭背景等特征的前提下，每个学生被随机的分配到不同的班级。
统计学(第六版)贾俊平-课后习题及答案
目录第一章P10 (1)第二章P34 (2)第三章P66 (3)第四章P94 (8)第七章P176 (11)第八章Ｐ212 (15)第10 章Ｐ258 (17)第11 章Ｐ291 (21)第13 章P348 (26)第14 章P376 (30)第一章P10一、思考题1.1什么是统计学？1.2解释描述统计和推断统计。
1.3统计数据可分为哪几种类型？不同类型的数据各有什么特点？1.4解释分类数据、顺序数据和数值型数据的含义。
1.5举例说明总体、样本、参数、统计量、变量这几个概念。
1.6变量可分为哪几类？1.7举例说明离散型变量和连续型变量。
1.8请举出统计应用的几个例子。
1.9请举出应用统计的几个领域。
1.1 指出下面变量的类型：（1）年龄（2）性别（3）汽车产量（4）员工对企业某项改革措施的态度（赞成、中立、反对）（5）购买商品时的支付方式（现金、信用卡、支票）（1）数值型变量。
（2）分类变量。
（3）离散型变量。
（4）顺序变量。
（5）分类变量。
1.2 某研究部门准备抽取 2000 个职工家庭推断该城市所有职工家庭的年人均收入。
要求：（1）描述总体和样本。
（2）指出参数和统计量。
（1）总体是该市所有职工家庭的集合；样本是抽中的 2000 个职工家庭的集合。
（2）参数是该市所有职工家庭的年人均收入；统计量是抽中的 2000 个职工家庭的年人均收入。
1.3 一家研究机构从 IT 从业者中随机抽取 1000 人作为样本进行调查，其中 60%的人回答他们的月收入在5000 元以上，50%的人回答他们的消费支付方式是用信用卡。
回答下列问题：（1）这一研究的总体是什么？（2）月收入是分类变量、顺序变量还是数值型变量？（3）消费支付方式是分类变量、顺序变量还是数值型变量？（4）这一研究涉及截面数据还是时间序列数据？（1）总体是所有 IT 从业者的集合。
（2）数值型变量。
（3）分类变量。
（4）截面数据。
1.4 一项调查表明，消费者每月在网上购物的平均花费是 200 元，他们选择在网上购物的主要原因是“价格便宜”。
C++ Primer Plus 第六版 第七章编程题答案
//7.13.10#include<iostream>double add(double x,double y);double subtraction(double x,double y);double calculate(double x,double y ,double (*ps)(double a ,double b));int main(){using namespace std;cout << "Please enter two number: \n";double x,y;cin >> x >>y ;if(cin){double result_add,result_subtraction;result_add = calculate(x,y,add);result_subtraction = calculate(x,y,subtraction);cout << "The sum of "<< x << " and "<< y << "is " << result_add<<endl;cout << "The subtraction of "<< x << " and "<< "is "<< result_subtraction << endl;cout << "Enter a number: ";cin >> x;cout << "Enter another number: ";cin >> y;}return 0;}double add(double x,double y){return x+y;}double subtraction(double x,double y){return x-y;}double calculate(double x,double y,double (*ps)(double,double)){double a;a = (*ps)(x,y);return a;}/*//7.13.9#include<iostream>using namespace std ;const int SLEN = 30 ;struct student {char fullname[SLEN];char hobby[SLEN];int ooplevel;};// getinfo() has two arguments: a pointer to the first element of // an array of student structures and an int representing the// number of elements of the array. The function solicits and// stores data about students. It terminates input upon filling // the array or upon encoutering a blank line for the student// name. The function returns the actual number of array elements // filled.int getinfo(student pa[], int n);// dispay1() takes a student structure as an argument// and displays its contentsvoid display1(student st);// display2() takes the address of student structure as an// argument and displays the structure's contentsvoid display2(const student * ps);// display3() takes the address of the first element of an array // of student structures and the number of array elements as// arguments and displays the contents of the structuresvoid display3(const student pa[], int n);int main(){cout << "Enter class size: ";int class_size;cin >> class_size ;while (cin.get() != '\n')continue ;student * ptr_stu = new student[class_size];int entered = getinfo(ptr_stu,class_size);for (int i = 0 ; i< entered; i++){display1(ptr_stu[i]);display2(&ptr_stu[i]);}display3(ptr_stu,entered);delete [] ptr_stu ;return 0 ;}int getinfo(student pa[],int n){int i = 0 ;while ( i < n){cout << "Enter your name: ";cin.get( pa[i].fullname,SLEN) ;char stop;stop =cin.get();if (stop!= '\0')break;cout << "Enter your hobby: ";cin.getline(pa[i].hobby ,SLEN);cout << "Enter the ooplevel: ";cin >> pa[i].ooplevel ;cin.get();i++;}return i ;}void display1(student st){using namespace std ;cout << st.fullname << endl<< st.hobby <<endl<< st.ooplevel << endl ;}void display2(const student * ps){using namespace std ;cout << ps->fullname << endl<< ps->hobby << endl<< ps->ooplevel << endl ;}void display3(const student pa[],int n){using namespace std ;int i = 0 ;while (i < n){cout << pa[i].fullname << endl<< pa[i].hobby << endl<< pa[i].ooplevel << endl ;i++;}}*//*//7.13.8(b)#include<iostream>struct expense{double ex[4];};const int Seasons = 4;const char arr[4][10] ={"Spring","Summer","Fall","Winter"};void fill(expense * ps);void show(const expense * ps);int main(){using namespace std ;expense esp;fill(&esp);show(&esp);return 0 ;}void fill(expense * ps){using namespace std ;for (int i = 0 ;i < Seasons; i++){cout << "Enter " << arr[i] <<" expenses: ";cin >> ps->ex[i];}}void show(const expense * ps){using namespace std ;double total = 0.0;cout << "\nEXPENSES\n";for (int i = 0 ;i< Seasons;i++){cout << arr[i] << ": $" << ps->ex[i] << endl ;total += ps->ex[i];}cout << "Total Expenses: $" << total << endl ;}*//*//7.13.8 ?#include<iostream>const int Seasons = 4 ;const char arr[4][10] ={"Spring","Summer","Fall","Winter"};void fill(double arr[]);void show(const double arr[]);int main(){using namespace std ;double expenses[4];fill(expenses);show(expenses);return 0 ;}void fill(double ar[]){using namespace std ;for (int i = 0 ;i < Seasons; i++){cout << "Enter " << arr[i] <<" expenses: ";cin >> ar[i];}}void show(const double ar[]){using namespace std ;double total = 0.0;cout << "\nEXPENSES\n";for (int i = 0 ;i< Seasons;i++){cout << arr[i] << ": $" << ar[i] << endl ;total += ar[i];}cout << "Total Expenses: $" << total << endl ; }*//*//7.13.7#include<iostream>const int Max = 5;double * fill_array(double arr[], int limit);void show_array(double arr[],double * ps);void revalue(double r , double arr[], double * ps);int main(){using namespace std ;double properties[Max];double * ps ;ps = fill_array(properties, Max);show_array(properties,ps);if (ps != properties){cout << "Enter revaluation factor: ";double factor ;while (!(cin>> factor)){cin.clear();while (cin.get() != '\n')continue;cout << "Bad input; please enter a number: ";}revalue(factor,properties,ps);show_array(properties,ps);}cout << "Done.\n";cin.get();cin.get();return 0;}double * fill_array(double arr[], int limit){using namespace std ;double temp ;int i ;for (i = 0 ;i< limit; i++){cout << "Enter value #" << (i+1) << ": ";cin >> temp ;if (!cin){cin.clear();while (cin.get() != '\n')continue ;cout << "Bad input; input process terminated.\n";break;}else if (temp < 0)break;arr[i] = temp ;}return &arr[i-1];}void show_array(double arr[],double * ps){using namespace std ;for (int i = 0; ps != &arr[i];i++){cout << "Property #"<< (i+1) << ": &";cout << arr[i] << endl ;}}void revalue(double r ,double arr[], double * ps) {for (int i = 0 ;ps != &arr[i]; i++)arr[i] *= r ;}*//*//7.13.6#include<iostream>const int Max = 100 ;int Fill_array(double arr[],int length);void Show_array(const double arr[],int length); void Reverse_array(double arr[],int length);int main(){using namespace std ;double arr[Max];int Size;Size = Fill_array(arr,Max);Show_array(arr,Size);cout << endl ;Reverse_array(arr,Size);Show_array(arr,Size);return 0 ;}int Fill_array(double arr[],int Max){using namespace std ;int i = 0 ;cout << "Please enter number!!!\n";for (i;i < Max;i++){cout << "Enter number to the arr[" << i << "]: ";cin >> arr[i];if (!cin){cin.clear();while (cin.get() != '\n')continue;cout << "Not a number!!! the process terminated";break;}}return i ;}void Show_array(const double arr[],int length){using namespace std ;cout << endl;int i = 0 ;while (i < length){cout << arr[i] << " ";i++;}}void Reverse_array(double arr[],int length){int j ;j = length/2 ;int k = 1;while (k <=j){double arr1 ;arr1 = arr[k];arr[k]= arr[length-1-k];arr[length-1-k]=arr1;k++;}}*//*//7.13.5#include<iostream>long double function(unsigned number);int main(){using namespace std ;long double sj;unsigned number ;cout << "Enter a number: ";cin >> number;sj = function(number);cout << number << "! = " << sj << endl ;return 0 ;}long double function(unsigned number){long double sj ;if (number == 0)sj = 1;elsesj = number * function(number -1 );return sj ;}*//*//7.13.4#include<iostream>long double probability(unsigned numbers, unsigned picks);int main(){using namespace std ;double ch[4];cout << "ch[4]数簓组哩?中D，?ch[0]:total1,ch[1]:choices1;ch[2]:total2,ch[3]:choices2\n";for (int i = 0; i<4;i++){cout << "the " << i+1 << " numbers: ";cin >> ch[i];if (!cin){cin.clear();while(cin.get() != '\n')continue;cout << "bad input";break;}else if (ch[i] < 0)break;}long double p1;p1 = probability(ch[0],ch[1]);long double p2;p2 = probability(ch[2],ch[3]);long double p3 ;p3 = p1 * p2 ;cout << "you have one chance in " << p3 << " of win.\n";return 0;}long double probability(unsigned numbers, unsigned picks){long double result = 1.0 ;long double n ;unsigned p ;for (n = numbers, p = picks; p>0;n--,p--)result = result * n/p;return result;}*//*//7.13.3#include<iostream>struct box{char maker[40];float height;float width ;float length ;float volume ;};void display(box b1);void function(box * ps);int main(){box b1 ={"Jack H",3.2,2.3,6.5};function(&b1);display(b1);return 0;}void display(box b1){using namespace std;cout <<"Maker: " << b1.maker <<endl;cout << "Height: " << b1.height << endl;cout <<"Width: " << b1.width << endl ;cout << "Length: " << b1.length << endl ;cout << "Volume: " << b1.volume << endl ;}void function(box * ps){ps->volume = ps->height * ps->length * ps ->width ; }*//*//7.13.2#include<iostream>const int Max = 10;int input(double golf[] , int max);void show( const double golf[], int length);double mean(const double golf[],int length);int main(){double golf[Max];int size;size =input(golf ,Max);//std::cout<<size<<std::endl;show(golf,size);double mean_golf ;mean_golf = mean(golf,size);std::cout << std::endl ;std::cout << mean_golf <<std::endl;return 0;}int input(double golf[], int max){using namespace std;int i = 0;double gol_f;for (i;i<max;i++){cout << "Enter your points: ";cin >> gol_f ;if (!cin){cin.clear();while (cin.get() != '\n')continue;cout << "Bad input; input process terminated.\n";break;}else if (gol_f < 0)break;golf[i] = gol_f;}return i;}void show(const double golf[],int length){using namespace std;for (int i = 0 ;i <length ;i++){cout << golf[i] << " ";}}double mean(const double golf[],int length) {int i = 0;double sum =0.0 ;while (i<length){sum = sum + golf[i] ;i++;}//std::cout << sum << std::endl;double mean_g ;mean_g = sum / length;//std::cout << mean_g<<std::endl;return mean_g ;}*//*//7.13.1#include<iostream>double mean(double a,double b);int main(){using namespace std;double a;double b;cin >> a >> b;while(a!=0&&b!=0){double c;c=mean(a,b);cout << "the mean of "<<a <<" and "<< b << " is: "<< c <<endl;cin >> a;cin >> b;}return 0 ;}double mean(double a,double b){double mean_c;mean_c= 2.0*a*b/(a+b);return mean_c;}*/。
投资学第六版第7章课后答案
1. Because very few securities will exhibit perfectly positive correlation,diversification will tend to reduce portfolio risk. Thus, for any given level of expected return, one would expect that portfolios will exhibit lower risk (lie further to the west in the feasible set) than individual portfolios (which will therefore lie to the east in the feasible set).2. Diversified portfolios are more efficient than individual securities. That is,diversified portfolios provide the investor with higher expected returns for given levels of risk and/or lower risk for given levels of expected return when compared with individual securities.Diagrammatically, individual securities will lie in the eastern portion of the feasible set. Hence they are dominated by diversified portfolios, which lie in the northwestern portion of the feasible set, including those on the efficient set.3. The macroeconomic forces that impact the U.S. economy tend to have a strongeffect on the earnings (and, hence, stock prices) of all domestic corporations, although the magnitude of this effect will vary among industries and specific firms.For example, a recession causes most companies to experience a downturn in earnings. While some companies may be more severely affected than others, nevertheless, the broad influence of a recession on general economic activity likely results in most companies' stocks performing poorly.Companies whose stocks would be expected to have a high positive covariance are auto and steel companies. When auto sales are strong (weak), the demand for steel generally rises (falls). The earnings of companies in both industries would rise and fall at roughly the same time and this movement would likely be anticipated by the earlier rise and fall of their stocks' prices.Companies whose stocks would be expected to have a low covariance are banks and gold mining firms. Rising interest rates and poor business conditions generally produce declining bank earnings. At the same time, a pessimistic economic outlook often causes investors to increase their demand for gold, which increases the price of gold and, therefore, the earnings of gold mining firms. The result is that the stock prices of banks and gold mining firms will not likely move in the same direction.4. It is the fact that all stocks do not have high positive covariances that causesdiversification to benefit the investor. That is, by diversifying, investors can reduce portfolio risk and thereby create more efficient portfolios. If stocks did have high positive covariances, then holding a well-diversified portfolio would not result in meaningful reductions in risk relative to holding individual securities.5. If the security in question had significant negative correlation with the rest of thesecurities in the portfolio, Mule might consider purchasing it even though it had anegative expected return. The diversifying nature of the security might reduce the risk of the portfolio sufficiently to make it attractive despite its inferior return potential.6. Given the expected returns and variance-covariance estimates for all securities, aninvestor can construct the efficient set. This information, combined with the unique risk-return preferences of the investor, allows the investor to determine his or her optimal portfolio. Diagrammatically, this optimal portfolio lies at the point of tangency between the investor's indifference curves and the efficient set.7. The standard deviation of a two-security portfolio is given by:[]σσσρσσp A A B B A B AB A B X X X X =++22122/In Dode's case:= [(.35)²(20)² + (.65)²(25)² + 2(.35)(.65)(20)(25)12]½= [49 + 264 + 22812]½The portfolio's standard deviation will be at a minimum when the correlation between securities A and B is -1.0. That is:= [49 + 264 - 228]½= 9.2%The portfolio's standard deviation will be at a maximum when the correlation between securities A and B is +1.0. That is: = [49 + 264 + 228]½= 23.3%17. The beta of a portfolio is defined as the weighted average of the componentsecurities' betas. In the case of Siggy's portfolio:ββP i i i X ==∑13= (.30 ⨯ 1.20) + (.50 ⨯ 1.05) + (.20 ⨯ 0.90)= 1.07Further, the standard deviation of a portfolio can be expressed as:()σβσσεp P I p =+22212/= [(1.07)²(18)² + (.30)²(5.0)² +(.50)²(8.0)²+ (.20)²(2.0)²]½= [370.9 + 2.3 + 16.0 + 0.2]½= [389.4]½ = 19.7%18.The total risk of a portfolio can be expressed as: σβσσεp p I p =+222Further, the unique risk (σεp 2) is the weighted average of the unique risks of the portfolio's individual securities. In the case of the first portfolio with four equal-weighted securities:()()σεp i 2221430025=⨯=∑..= 56.25 ⨯ 4 = 225.0Therefore the total risk of the first portfolio is:0.225)20()00.1(2221+⨯=σ= 625.01 = 25.0%In the case of the second portfolio with ten equal-weighted securities:()()σεp i 22211030010=⨯=∑..= 9.0 ⨯ 10 = 90.0Therefore the total risk of the second portfolio is:σ222210020900=⨯+(.)().= 490.02 = 22.1%。
计学(第六版)第七章课后练习答案
第七章 课后练习答案7.1（1）已知：96.1%,951,25,40,52/05.0==-===z x n ασ。
样本均值的抽样标准差79.0405===n x σσ （2）边际误差55.140596.12/=⨯==n z E σα7.2 （1）已知：96.1%,951,120,49,152/05.0==-===z x n ασ。
样本均值的抽样标准差14.24915===n x σσ（2）边际误差20.4491596.12/=⨯==n z E σα（3）由于总体标准差已知，所以总体均值μ的95%的置信区间为20.4120491596.11202/±=⨯±=±n z x σα即()2.124,8.1157.3 已知：96.1%,951,104560,100,854142/05.0==-===z x n ασ。
由于总体标准差已知，所以总体均值μ的95%的置信区间为144.167411045601008541496.11045602/±=⨯±=±n z x σα即)144.121301,856.87818(7.4（1）已知：645.1%,901,12,81,1002/1.0==-===z s x n α。
由于100=n 为大样本，所以总体均值μ的90%的置信区间为：974.18110012645.1812/±=⨯±=±n s z x α即)974.82,026.79(（2）已知：96.1%,951,12,81,1002/05.0==-===z s x n α。
由于100=n 为大样本，所以总体均值μ的95%的置信区间为：352.2811001296.1812/±=⨯±=±n s z x α即)352.83,648.78(（3）已知：58.2%,991,12,81,1002/05.0==-===z s x n α。
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第七章 课后练习答案7.1（1）已知：96.1%,951,25,40,52/05.0==-===z x n ασ。
样本均值的抽样标准差79.0405===n x σσ （2）边际误差55.140596.12/=⨯==n z E σα7.2 （1）已知：96.1%,951,120,49,152/05.0==-===z x n ασ。
样本均值的抽样标准差14.24915===n x σσ（2）边际误差20.4491596.12/=⨯==n z E σα（3）由于总体标准差已知，所以总体均值μ的95%的置信区间为20.4120491596.11202/±=⨯±=±n z x σα即()2.124,8.1157.3 已知：96.1%,951,104560,100,854142/05.0==-===z x n ασ。
由于总体标准差已知，所以总体均值μ的95%的置信区间为144.167411045601008541496.11045602/±=⨯±=±n z x σα即)144.121301,856.87818(7.4（1）已知：645.1%,901,12,81,1002/1.0==-===z s x n α。
由于100=n 为大样本，所以总体均值μ的90%的置信区间为：974.18110012645.1812/±=⨯±=±n s z x α即)974.82,026.79(（2）已知：96.1%,951,12,81,1002/05.0==-===z s x n α。
由于100=n 为大样本，所以总体均值μ的95%的置信区间为：352.2811001296.1812/±=⨯±=±n s z x α即)352.83,648.78(（3）已知：58.2%,991,12,81,1002/05.0==-===z s x n α。
由于100=n 为大样本，所以总体均值μ的99%的置信区间为：096.3811001258.2812/±=⨯±=±n s z x α即)096.84,940.77(7.5（1）已知：96.1%,951,5.3,25,602/05.0==-===z x n ασ。
由于总体标准差已知，所以总体均值μ的95%的置信区间为：89.025605.396.1252/±=⨯±=±n z x σα即)89.25,11.24(（2）已知：33.2%,981,89.23,6.119,752/02.0==-===z s x n α。
由于75=n 为大样本，所以总体均值μ的98%的置信区间为：43.66.1197589.2333.26.1192/±=⨯±=±n s z x α即)03.126,17.113(（3）已知：645.1%,901,974.0,419.3,322/1.0==-===z s x n α。
由于32=n 为大样本，所以总体均值μ的90%的置信区间为：283.0419.332974.0645.1419.32/±=⨯±=±n s z x α即)702.3,136.3(7.6（1）已知：总体服从正态分布，96.1%,951,500,8900,152/05.0==-===z x n ασ。
由于总体服从正态分布，所以总体均值μ的95%的置信区间为：03.25389001550096.189002/±=⨯±=±n z x σα即)03.9153,97.8646(（2）已知：总体不服从正态分布，96.1%,951,500,8900,352/05.0==-===z x n ασ。
虽然总体不服从正态分布，但由于35=n 为大样本，所以总体均值μ的95%的置信区间为：65.16589003550096.189002/±=⨯±=±n z x σα即)65.9065,35.8734(（3）已知：总体不服从正态分布，σ未知，645.1%,901,500,8900,352/1.0==-===z s x n α。
虽然总体不服从正态分布，但由于35=n 为大样本，所以总体均值μ的95%的置信区间为：03.139890035500645.189002/±=⨯±=±n s z x α即)03.9039,97.8760(（4）已知：总体不服从正态分布，σ未知，58.2%,991,500,8900,352/01.0==-===z s x n α。
虽然总体不服从正态分布，但由于35=n 为大样本，所以总体均值μ的95%的置信区间为：05.21889003550058.289002/±=⨯±=±n s z x α即)05.9118,95.8681(7.7已知：36=n ，当α为0.1，0.05，0.01时，相应的58.2,96.1,645.12/01.02/05.02/1.0===z z z 。
根据样本数据计算得：61.1,32.3==s x 。
由于36=n 为大样本，所以平均上网时间的90%的置信区间为：44.032.33661.1645.132.32/±=⨯±=±n s z x α即)76.3,88.2(平均上网时间的95%的置信区间为：53.032.33661.196.132.32/±=⨯±=±n s z x α即)85.3,79.2(平均上网时间的99%的置信区间为：69.032.33661.158.232.32/±=⨯±=±n s z x α即)01.4,63.2(7.8已知：总体服从正态分布，但σ未知，8=n 为小样本，365.2)18(,05.02/05.0=-=t α。
根据样本数据计算得：46.3,10==s x 。
总体均值μ的95%的置信区间为：89.210846.3365.2102/±=⨯±=±n s t x α即)89.12,11.7(7.9已知：总体服从正态分布，但σ未知，16=n 为小样本，131.2)116(,05.02/05.0=-=t α。
根据样本数据计算得：113.4,375.9==s x 。
从家里到单位平均距离的95%的置信区间为：191.2375.914113.4131.2375.92/±=⨯±=±n s t x α即)57.11,18.7(7.10已知： 96.1%,951,93.1,5.149,362/05.0==-===z s x n α。
由于35=n 为大样本，所以总体均值μ的95%的置信区间为：63.05.1493693.196.15.1492/±=⨯±=±n s z x α即)13.150,87.148(（2）使用了中心极限定理。
7.11（1）已知：总体服从正态分布，但σ未知，50=n 为大样本，96.1,05.02/05.0==z α。
根据样本数据计算得：63.11)(,32.101121=--===∑∑==n f x M s nf M x k i i i k i i i 。
该种食品平均重量的95%的置信区间为： 45.032.1015063.196.132.1012/±=⨯±=±n s z x α即)77.101,87.100(（2）根据样本数据可知，样本合格率为9.05045==p ，该种食品合格率的95%的置信区间为： 08.09.050)9.01(9.096.19.0)1(2/±=-⨯±=-±n p p z p α 即)98.0,82.0(7.12已知：总体服从正态分布，但σ未知，25=n 为小样本，797.2)125(,01.02/01.0=-=t α。
根据样本数据计算得：871.0,128.16==s x 。
总体均值μ的95%的置信区间为：487.0128.1625871.0797.2128.162/±=⨯±=±n s t x α即)62.16,64.15(7.13已知：总体服从正态分布，但σ未知，18=n 为小样本，740.1)118(,1.02/1.0=-=t α。
根据样本数据计算得：80.7,56.13==s x 。
平均每周加班时间的95%的置信区间为：20.356.131880.7740.156.132/±=⨯±=±n s t x α即)76.16,36.10(7.14（1）已知：58.2,01.0,51.0,442/01.0====z p n α。
总体比例π的99%的置信区间为：19.051.044)51.01(51.058.251.0)1(2/±=-⨯±=-±n p p z p α 即)70.0,32.0(（2）已知：96.1,05.0,82.0,3002/05.0====z p n α。
总体比例π的95%的置信区间为：04.082.0300)82.01(82.096.182.0)1(2/±=-⨯±=-±n p p z p α 即)86.0,78.0(（3）已知：645.1,1.0,48.0,11502/1.0====z p n α。
总体比例π的90%的置信区间为：02.048.01150)48.01(48.0645.148.0)1(2/±=-⨯±=-±n p p z p α 即)50.0,46.0(7.15已知：96.1,645.1,05.0;1.0,23.0,2002/05.02/1.0=====z z p n α。
总体比例π的90%的置信区间为：05.023.0200)23.01(23.0645.123.0)1(2/±=-⨯±=-±n p p z p α 即)28.0,18.0(总体比例π的95%的置信区间为：06.023.0200)23.01(23.096.123.0)1(2/±=-⨯±=-±n p p z p α即)29.0,17.0(7.16已知：58.2,01.0,200,10002/01.0====z E ασ。
应抽取的样本容量为：167200100058.2)(2222222/=⨯==E z n σα 7.17（1）已知：05.2,04.0,40.0,02.02/04.0====z E απ。