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Problem 4.12. The infamous crawling tractor sprinkler is located as pictured below, 100 feet South of a \( {10}\mathrm{{ft}} \) . wide sidewalk; notice the hose and sidewalk are not perpendicular. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North along the hose at the rate of \( \frac{1}{2} \) inch/second.
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(a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and find the equation of the line forming the southern boundary of the sidewalk.\n\n(b) After 33 minutes, sketch a picture of the wet portion of the sidewalk; find the length of the wet portion of the Southern edge of the sidewalk.\n\n(c) Find the equation of the line forming the northern boundary of the sidewalk. (Hint: You can use the properties of right triangles.)
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No
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Problem 4.14. Juliet and Mercutio are moving at constant speeds in the \( {xy} \) -plane. They start moving at the same time. Juliet starts at the point \( \left( {0, - 6}\right) \) and heads in a straight line toward the point \( \left( {{10},5}\right) \), reaching it in 10 seconds. Mercutio starts at \( \left( {9, - {14}}\right) \) and moves in a straight line. Mercutio passes through the same point on the \( x \) axis as Juliet, but 2 seconds after she does.
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How long does it take Mercutio to reach the \( y \) -axis?
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No
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Example 5.5.2. You are driving 65 mph from the Kansas state line (mile marker 0) to Salina (mile marker 130) along I-35. Describe a linear function that calculates mile marker after \( \mathrm{t} \) hours. Describe another linear function that will calculate your distance from Salina after \( \mathfrak{t} \) hours.
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Solution. Define a function \( d\left( t\right) \) to be the mile marker after\n\n\n\nFigure 5.11: Distance functions.\n\n\( \mathrm{t} \) hours. Using \
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No
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A software company plans to bring a new product to market. The sales price per unit is $15 and the expense to produce and market \( x \) units is $100(1 + \sqrt{x}). What is the profit potential?
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Two functions control the profit potential of the new software. The first tells us the gross income, in dollars, on the sale of \( x \) units. All of the costs involved in developing, supporting, distributing and marketing \( x \) units are controlled by the expense equation (again in dollars):\n\n\[ g(x) = 15x \]\n(gross income function)\n\n\[ e(x) = 100(1 + \sqrt{x}) \]\n(expense function)\n\nA profit will be realized on the sale of \( x \) units whenever the gross income exceeds expenses; i.e., this occurs when \( g(x) > e(x) \). A loss occurs on the sale of \( x \) units when expenses exceed gross income; i.e., when \( e(x) > g(x) \). Whenever the sale of \( x \) units yields zero profit (and zero loss), we call \( x \) a break-even point; i.e., when \( e(x) = g(x) \).\n\nThe above approach is \
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No
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1. When is the pilot climbing and descending?\n2. When is the pilot at the glider port elevation?\n3. How much time does the pilot spend flying level?
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1. Graphically, we need to determine the portions of the graph that are increasing or decreasing. In this example, it is increasing when \( 0 \leq \mathrm{t} \leq 2 \) and \( 7 \leq \mathrm{t} \leq 9 \) . And, it is decreasing when \( 3 \leq \mathrm{t} \leq 5 \) and \( 9 \leq t \leq {10} \) .\n\n2. Graphically, this question amounts to asking when the elevation is 0 , which is the same as finding when the graph crosses the horizontal axis. We can read off there are four such times: \( t = 0,4,8,{10} \) .\n\n3. Graphically, we need to determine the portions of the graph that are made up of horizontal line segments. This happens when \( 2 \leq \mathrm{t} \leq 3 \) and \( 5 \leq \mathrm{t} \leq 7 \) . So, our pilot flies level for a total of 3 minutes.
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Yes
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Example 6.3.1. Sketch the graph of the multipart function\n\n\[ g\\left( x\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{if }x \\leq - 1 \\\\ 1 + \\sqrt{1 - {x}^{2}} & \\text{if } - 1 \\leq x \\leq 1 \\\\ 1 & \\text{if }x \\geq 1 \\end{array}\\right. \]
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Solution. The graph of \( g\\left( x\\right) \) will consist of three pieces.\n\n\n\nFigure 6.12: Multipart function \( g\\left( x\\right) \) .\n\nThe first case consists of the graph of the function \( y = \) \( g\\left( x\\right) = 1 \) on the domain \( x \\leq - 1 \), this consists of all points on the horizontal line \( y = 1 \) to the left of and including the point \( \\left( {-1,1}\\right) \) . We have \
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Yes
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function \( s = h\left( t\right) \) keeps track of the height of the ball’s center above the floor after \( \mathrm{t} \) seconds. Sketch a reasonable graph of \( s = h\left( t\right) \) .
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Solution. If we take the domain to be \( 0 \leq \mathrm{t} \leq 2 \) (the first 2 seconds), a reasonable graph might look like Figure 6.13. This is a multipart function. Three portions of the graph are decreasing and two portions are increasing. Why doesn't the graph touch the \( t \) axis?
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No
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Describe a sequence of geometric operations leading from the graph of \( y = {x}^{2} \) to the graph of \( y = f\left( x\right) = - 3{\left( x - 1\right) }^{2} + 2 \) .
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To begin with, we can make some initial conclu- sions about the specific shifts, reflections and dilations involved, based on looking at the vertex form of the equation. In addition, by Fact 7.1.1, we know that the vertex of the graph of \( y = f\left( x\right) \) is \( \left( {1,2}\right) \), the line \( x = 1 \) is a vertical axis of symmetry and the parabola opens downward.\n\n- A horizontal shift by \( h = 1 \) yields the graph of \( y = {\left( x - 1\right) }^{2} \) ; this is the fat parabola opening upward with vertex \( \left( {1,0}\right) \) .\n\n- A dilation by \( a = 3 \) yields the graph of \( y = 3(x - 1{)}^{2} \) ; this is the skinny parabola opening upward with vertex \( \left( {1,0}\right) \) .\n\n- A reflection yields the graph of \( y = - 3{\left( x - 1\right) }^{2} \) ; this is the downward opening parabola with vertex \( \left( {1,0}\right) \) .\n\n- A vertical shift by \( k = 2 \) yields the graph of \( y = - 3(x - 1{)}^{2} + 2 \) ; this is the downward opening parabola with vertex \( \left( {1,2}\right) \) .
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Yes
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Find the vertex form of the quadratic function \( y = - 3{x}^{2} + {6x} - 1 \) .
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Since our goal is to put the function in vertex form, we can write down what this means, then try to solve for the unknown constants. Our first step would be to write\n\n\[ - 3{x}^{2} + {6x} - 1 = a{\left( x - h\right) }^{2} + k \]\n\nfor some constants \( a, h, k \) . Now, expand the right hand side of this equation and factor out coefficients of \( x \) and \( {x}^{2} \) :\n\n\[ - 3{x}^{2} + {6x} - 1 = a{\left( x - h\right) }^{2} + k \]\n\n\[ - 3{x}^{2} + {6x} - 1 = a\left( {{x}^{2} - {2xh} + {h}^{2}}\right) + k \]\n\n\[ - 3{x}^{2} + {6x} - 1 = a{x}^{2} - {2xah} + a{h}^{2} + k \]\n\n\[ \left( {-3}\right) {x}^{2} + \left( 6\right) x + \left( {-1}\right) = \left( a\right) {x}^{2} + \left( {-{2ah}}\right) x + \left( {a{h}^{2} + k}\right) . \]\n\nIf this is an equation, then it must be the case that the coefficients of like powers of \( x \) match up on the two sides of the equation in Figure 7.9. Now we have three equations and three unknowns (the \( a, h, k \) ) and we\n\n\n\nFigure 7.9: Balancing the coefficients.\n\ncan proceed to solve for these:\n\n\[ - 3 = a \]\n\n\[ 6 = - {2ah} \]\n\n\[ - 1 = a{h}^{2} + k \]\n\nThe first equation just hands us the value of \( a = - 3 \) . Next, we can plug this value of \( a \) into the second equation, giving us\n\n\[ 6 = - {2ah} \]\n\n\[ = - 2\left( {-3}\right) h \]\n\n\[ = {6h} \]\n\nso \( h = 1 \) . Finally, plug the now known values of \( a \) and \( h \) into the third equation:\n\n\[ - 1 = a{h}^{2} + k \]\n\n\[ = - 3\left( {1}^{2}\right) + \mathrm{k} \]\n\n\[ = - 3 + k \]\n\nso \( k = 2 \) . Our conclusion is then\n\n\[ - 3{x}^{2} + {6x} - 1 = - 3{\left( x - 1\right) }^{2} + 2. \]\n\nNotice, this is the quadratic we studied in Example 7.1.2 on page 90.
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Yes
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Describe the relationship between the graphs of \( y = {x}^{2} \) and \( y = f\left( x\right) = - 4{x}^{2} + {5x} + 2 \) .
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Solution. We will go through the algebra to complete the square, then interpret what this all means in terms of graphical maneuvers. We have\n\n\[ \n- 4{x}^{2} + {5x} + 2 = a{\left( x - h\right) }^{2} + k \]\n\n\[ \n\left( {-4}\right) {x}^{2} + {5x} + 2 = a{x}^{2} + \left( {-{2ah}}\right) x + \left( {a{h}^{2} + k}\right) . \]\n\nThis gives us three equations:\n\n\[ \n- 4 = a \]\n\n\[ \n5 = - {2ah} \]\n\n\[ \n2 = a{h}^{2} + k\text{.} \]\n\nWe conclude that \( a = - 4, h = \frac{5}{8} = {0.625} \) and \( k = \frac{57}{16} = {3.562} \) . So, this tells us that we can obtain the graph of \( y = f\left( x\right) \) from that of \( y = {x}^{2} \) by these steps:\n\n- Horizontally shifting by \( h = {0.625} \) units gives \( y = {\left( x - {0.625}\right) }^{2} \).\n\n- Vertically dilate by the factor \( a = 4 \) gives \( y = 4{\left( x - {0.625}\right) }^{2} \).\n\n- Reflecting across the \( x \) -axis gives \( y = - 4{\left( x - {0.625}\right) }^{2} \).\n\n- Vertically shifting by \( k = {3.562} \) gives \( y = f\left( x\right) = - 4{\left( x - {0.625}\right) }^{2} + {3.562} \).
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Yes
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A drainage canal has a cross-section in the shape of a parabola. Suppose that the canal is 10 feet deep and 20 feet wide at the top. If the water depth in the ditch is 5 feet, how wide is the surface of the water in the ditch?
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Impose an \( {xy} \) -coordinate system so that the parabolic cross-section of the canal is symmetric about the \( y \) -axis and its vertex is the origin. The vertex form of any such parabola is \( y = f\left( x\right) = a{x}^{2} \), for some \( a > 0 \) ; this is because \( \left( {h, k}\right) = \left( {0,0}\right) \) is the vertex and the parabola opens upward! The dimension information given tells us that the points \( \left( {{10},{10}}\right) \) and \( \left( {-{10},{10}}\right) \) are on the graph of \( f\left( x\right) \) . Plugging into the expression for \( f \), we conclude that \( {10} = {100a} \), so \( a = {0.1} \) and \( f\left( x\right) = \left( {0.1}\right) {x}^{2} \) . Finally, if the water is 5 feet deep, we must solve the equation: \( 5 = \left( {0.1}\right) {x}^{2} \), leading to \( x = \pm \sqrt{50} = \pm {7.07} \) . Conclude the surface of the water is 14.14 feet wide when the water is 5 feet deep.
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Yes
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Discuss the graph of the quadratic function \( y = f\\left( x\\right) = \) \( - 2{x}^{2} + {11x} - 4 \) .
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Solution. We need to place the equation \( y = f\\left( x\\right) \) in vertex form. We can simply compute \( a = - 2, h = \\frac{-b}{2a} = \\frac{11}{4} \) and \( k = f\\left( \\frac{11}{4}\\right) = \\frac{89}{8} \), using Fact 7.3.1:\n\n\[ f\\left( x\\right) = - 2{x}^{2} + {11x} - 4 \]\n\n\[ = - 2{\\left( x - \\left( \\frac{11}{4}\\right) \\right) }^{2} + \\frac{89}{8}. \]\n\nThis means that the graph of \( f\\left( x\\right) \) is a parabola opening downward with vertex \( \\left( {\\frac{11}{4},\\frac{89}{8}}\\right) \) and axis \( x = \\frac{11}{4} \) ; see Figure 7.13.
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Yes
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What is the maximum height of the ball and when is this height achieved? When does the ball hit the ground? How high is the cliff?
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Solution. The function \( y\left( t\right) \) is a quadratic function with a negative leading coefficient, so its graph in the ts-coordinate system will be a downward opening parabola. We use a graphing device to get the picture in Figure 7.14(b).\n\nThe vertex is the highest point on the graph, which can be found by writing \( y\left( t\right) \) in standard form using Fact 7.3.1:\n\n\[ y\left( t\right) = - {16}{t}^{2} + {48t} + {50} \]\n\n\[ = - {16}{\left( t - \frac{3}{2}\right) }^{2} + {86} \]\n\nThe vertex of the graph of \( y\left( t\right) \) is \( \left( {\frac{3}{2},{86}}\right) \), so the maximum height of the ball above the level ground is 86 feet, occurring at time \( \mathrm{t} = \frac{3}{2} \).\n\nThe ball hits the ground when its height above the ground is zero; using the quadratic formula:\n\n\[ y\left( t\right) = - {16}{t}^{2} + {48t} + {50} \]\n\n\[ = 0 \]\n\n\[ t = \frac{-{48} \pm \sqrt{{\left( {48}\right) }^{2} - 4\left( {-{16}}\right) \left( {50}\right) }}{2 \cdot {16}} \]\n\n\[ = {3.818}\mathrm{{sec}}\text{or} - {0.818}\mathrm{{sec}} \]\n\nConclude the ball hits the ground after 3.818 seconds. Finally, the height of the cliff is the height of the ball zero seconds after release; i.e., \( y\left( 0\right) = {50} \) feet is the height of the cliff.
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Yes
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A hot air balloon takes off from the edge of a mountain lake. Impose a coordinate system as pictured in Figure 7.15 and assume that the path of the balloon follows the graph of \( y = f\left( x\right) = - \frac{2}{2500}{x}^{2} + \frac{4}{5}x \) . The land rises at a constant incline from the lake at the rate of 2 vertical feet for each 20 horizontal feet. What is the maximum height of the balloon above lake level? What is the maximum height of the balloon above ground level? Where does the balloon land on the ground? Where is the balloon 50 feet above the ground?
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Solution. In the coordinate system indicated, the origin is the takeoff point and the graph of \( y = f\left( x\right) \) is the path of the balloon. Since \( f\left( x\right) \) is a quadratic function with a negative leading coefficient, its graph will be a parabola which opens downward. The difficulty with this problem is that at any instant during the balloon's flight, the \
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No
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Problem 7.8. (a) Suppose \( f\left( x\right) = 3{x}^{2} - 2 \) . Does the point \( \left( {1,2}\right) \) lie on the graph of \( y = f\left( x\right) \) ? Why or why not?
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To determine if the point \( \left( {1,2}\right) \) lies on the graph of \( y = f\left( x\right) \), we need to check if \( f(1) = 2 \). Substituting \( x = 1 \) into the function \( f(x) \), we get \( f(1) = 3(1)^2 - 2 = 3 - 2 = 1 \). Since \( f(1) = 1 \) and not 2, the point \( \left( {1,2}\right) \) does not lie on the graph of \( y = f\left( x\right) \).
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Yes
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A pebble is tossed into a pond. The radius of the first circular ripple is measured to increase at the constant rate of \( {2.3}\mathrm{{ft}}/\mathrm{{sec}} \) . What is the area enclosed by the leading ripple after 6 seconds have elapsed? How much time must elapse so that the area enclosed by the leading ripple is 300 square feet?
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Solution. We know that an object tossed into a pond will generate a series of concentric ripples, which grow steadily larger. We are asked questions that relate the area of the circular region bounded by the leading ripple and time elapsed.\n\nLet \( r \) denote the radius of the leading ripple after \( t \) seconds; units of feet. The area \( A \) of a disc bounded by a leading ripple will be \( A = A\left( r\right) = \pi {r}^{2} \) . This exhibits \( A \) as a function in the variable \( r \) . However, the radius is changing with respect to time:\n\n\[ r = r\left( t\right) = \text{ radius after }t\text{ seconds } = \left( {{2.3}\frac{\text{ feet }}{\text{ sec }}}\right) t\text{ seconds } = {2.3t}\text{ feet. } \]\n\nSo, \( r = r\left( t\right) \) is a function of \( t \) . In the expression \( A = A\left( r\right) \), replace \
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Yes
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Let \( f\left( x\right) = {x}^{2}, g\left( x\right) = x + 1 \) and \( h\left( x\right) = x - 1 \) . Find the formulas for \( f\left( {g\left( x\right) }\right), g\left( {f\left( x\right) }\right), f\left( {h\left( x\right) }\right) \) and \( h\left( {f\left( x\right) }\right) \) . Discuss the relationship between the graphs of these four functions.
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Solution. If we apply Procedure 8.1.5, we obtain the composition formulas. The four graphs are given on the domain \( - 3 \leq x \leq 3 \), together with the graph of \( f\left( x\right) = {x}^{2} \) .\n\n\[ f\left( {g\left( x\right) }\right) = f\left( {x + 1}\right) = {\left( x + 1\right) }^{2} \]\n\n\[ g\left( {f\left( x\right) }\right) = g\left( {x}^{2}\right) = {x}^{2} + 1 \]\n\n\[ \begin{matrix} f\left( {h\left( x\right) }\right) & = & f\left( {x - 1}\right) = {\left( x - 1\right) }^{2} \end{matrix} \]\n\n\[ h\left( {f\left( x\right) }\right) = h\left( {x}^{2}\right) = {x}^{2} - 1. \]\n\nWe can identify each graph by looking at its vertex:\n\n- \( f\left( x\right) \) has vertex \( \left( {0,0}\right) \)\n\n- \( f\left( {g\left( x\right) }\right) \) has vertex \( \left( {-1,0}\right) \)\n\n- \( g\left( {f\left( x\right) }\right) \) has vertex \( \left( {0,1}\right) \)\n\n- \( f\left( {h\left( x\right) }\right) \) has vertex \( \left( {1,0}\right) \)\n\n\[ \text{-}h\left( {f\left( x\right) }\right) \text{has vertex}\left( {0, - 1}\right) \]\n\nHorizontal or vertical shifting of the graph of \( f\left( x\right) = {x}^{2} \) gives the other four graphs: See Figure 8.4.
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Yes
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Start with the function \( y = f\left( x\right) = {x}^{2} \) on the domain \( - 1 \leq x \leq 1 \) . Find the rule and domain of \( y = f\left( {g\left( x\right) }\right) \), where \( g\left( x\right) = x - 1 \) .
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Solution. We can apply the first statement in Procedure 8.2.1 to find the rule for \( y = f\left( {g\left( x\right) }\right) \) :\n\n\[ y = f\left( {g\left( x\right) }\right) \]\n\n\[ = f\left( {x - 1}\right) = {\left( x - 1\right) }^{2} \]\n\n\[ = {x}^{2} - {2x} + 1\text{. } \]\n\nTo find the domain of \( y = f\left( {g\left( x\right) }\right) \), we apply the second statement in Procedure 8.2.1; this will require that we solve an inequality equation:\n\n\[ - 1 \leq \;g\left( x\right) \; \leq 1 \]\n\n\[ - 1 \leq \left| {x - 1}\right| \leq 1 \]\n\n\[ 0 \leq x \leq 2 \]\n\nThe conclusion is that \( y = f\left( {g\left( x\right) }\right) = {x}^{2} - {2x} + 1 \) on the domain \( 0 \leq x \leq 2 \) .
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Yes
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Let \( y = f\left( z\right) = \sqrt{z}, z = g\left( x\right) = x + 1 \) . What is the largest possible domain so that the composition \( f\left( {g\left( x\right) }\right) \) makes sense?
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Solution. The largest possible domain for \( y = f\left( z\right) \) will consist of all nonnegative real numbers; this is also the range of the function \( f\left( z\right) \) : See Figure 8.6(a).\n\nTo find the largest domain for the composition, we try to find a domain of \( x \) -values so that the range of \( z = g\left( x\right) \) is the domain of \( y = f\left( z\right) \) . So, in this case, we want the range of \( g\left( x\right) \) to be all non-negative real numbers, denoted \( 0 \leq z \) . We graph \( z = g\left( x\right) \) in the \( {xz} \) -plane, mark the desired range \( 0 \leq z \) on the vertical \( z \) -axis, then determine which \( x \) -values would lead to points on the graph with second coordinates in this zone. We find that the domain of all \( x \) -values greater or equal to \( - 1 \) (denoted \( - 1 \leq x \) ) leads to the desired range. In summary, the composition \( y = f\left( {g\left( x\right) }\right) = \sqrt{x + 1} \) is defined on the domain of \( x \) -values \( - 1 \leq x \) .
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Yes
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Verify that the mathematical model for this experiment is given by \( f\left( {g\left( t\right) }\right) \), where \( g\left( t\right) = t - 4 \) .
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Solution. Our expectation is that the plot for this new experiment will have the \
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No
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Write the equations \( y = {8}^{3x} \) and \( y = 7{\left( \frac{1}{2}\right) }^{{2x} - 1} \) in standard exponential form.
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Solution. In both cases, we just use the rules of exponents to maneuver the given equation into standard form:\n\n\[ y = {8}^{3x} \]\n\n\[ = {\left( {8}^{3}\right) }^{x} \]\n\n\[ = {512}^{\mathrm{x}} \]\n\nand\n\n\[ y = 7{\left( \frac{1}{2}\right) }^{{2x} - 1} \]\n\n\[ = 7{\left( \frac{1}{2}\right) }^{2x}{\left( \frac{1}{2}\right) }^{-1} \]\n\n\[ = 7{\left( {\left( \frac{1}{2}\right) }^{2}\right) }^{x}2 \]\n\n\[ = {14}{\left( \frac{1}{4}\right) }^{x} \]
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Yes
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A computer industry spokesperson has predicted that the number of subscribers to geton.com, an internet provider, will grow exponentially for the first 5 years. Assume this person is correct. If geton.com has 100,000 subscribers after 6 months and 750,000 subscribers after 12 months, how many subscribers will there be after 5 years?
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The solution to this problem offers a template for many exponential modeling applications. Since, we are assuming that the number of subscribers \( \mathrm{N}\left( x\right) \), where \( x \) represents years, is a function of exponential type,\n\n\[ N\left( x\right) = {N}_{ \circ }{b}^{x} \]\n\nfor some \( {N}_{ \circ } \) and \( b > 1 \) . We are given two pieces of information about the values of \( \mathrm{N}\left( x\right) \) :\n\n\[ \mathrm{N}\left( {0.5}\right) = {100},{000}\text{; i.e.,}{\mathrm{N}}_{0}{\mathrm{\;b}}^{0.5} = {100},{000}\text{, and} \]\n\n\[ N\left( 1\right) = {750},{000}\text{; i.e.,}\;{N}_{0}b = {750},{000}\text{.} \]\n\nWe can use these two equations to solve for the two unknowns \( {N}_{ \circ } \) and \( b \) as follows: If we divide the second equation by the first, we get\n\n\[ \frac{{\mathrm{b}}^{1}}{{\mathrm{\;b}}^{0.5}} = {7.5} \]\n\n\[ {b}^{0.5} = {b}^{1/2} = \sqrt{b} = {7.5} \]\n\n\[ \therefore \mathrm{b} = {56.25}\text{.} \]\n\nPlugging this value of \( b \) into either equation (say the first one), we can solve for \( {\mathrm{N}}_{ \circ } : {\mathrm{N}}_{ \circ } = \frac{{100},{000}}{{\left( {56.25}\right) }^{0.5}} = {13},{333} \) . We conclude that the number of geton.com subscribers will be predicted by\n\n\[ \mathrm{N}\left( x\right) = {13},{333}{\left( {56.25}\right) }^{x}. \]\n\nIn five years, we obtain \( \mathrm{N}\left( 5\right) = 7,{508},{300},{000},{000} \) subscribers, which exceeds the population of the Earth (which is between 5 and 6 billion)!
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Yes
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Example 11.1.2. At birth, your Uncle Hans secretly purchased a \$5,000 U.S. Savings Bond for \( \$ 2,{500} \) . The conditions of the bond state that the U.S. Government will pay a minimum annual interest rate of \( r = {8.75}\% \) , compounded quarterly. Your Uncle has given you the bond as a gift, subject to the condition that you cash the bond at age 35 and buy a red Porsche.\n\nOn your way to the Dealer, you receive a call from your tax accountant informing you of a \( {28}\% \) tax on the capital gain you realize through cashing in the bond; the capital gain is the selling price of the bond minus the purchase price. Before stepping onto the showroom floor, compute how much cash will you have on hand, after the U.S. Government shares in your profits.
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Solution. The value of your bond after 35 years is computed by the formula in Fact 11.1.1, using \( {P}_{0} = \$ 2,{500}, r = {0.0875}, n = 4 \), and \( t = {35} \) . Plugging this all in, we find that the selling price of the bond is\n\n\[ P\left( {35}\right) = \$ 2,{500}{\left( 1 + \frac{0.0875}{4}\right) }^{4\left( {35}\right) } = \$ {51},{716.42}. \]\n\nThe capital gain will be \$51,716.42 - \$2,500 = \$49,216.42 and the tax due is \( \$ \left( {{49},{216.42}}\right) \left( {0.28}\right) = \$ {13},{780.60} \) . You are left with \( \$ {51},{716.42} \) - \$13,780.60 = \$37,935.82. Better make that a used Porsche!
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Yes
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For any value \( a \), show that the point \( \left( {x, y}\right) = \left( {\cosh \left( a\right) ,\sinh \left( a\right) }\right) \) is on the unit hyperbola.
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(Hint: Verify that \( {\left\lbrack \cosh \left( x\right) \right\rbrack }^{2} - {\left\lbrack \sinh \left( x\right) \right\rbrack }^{2} = 1 \), for all \( x \).)
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No
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If $2,000 is invested in a continuously compounding savings account and we want the value after 12 years to be $130,000, what is the required annual interest rate?
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Solution. In the first scenario,\n\n\[ \n130,000 = 2,000 e^{12r} \n\]\n\n\[ \n65 = e^{12r} \n\]\n\n\[ \n\ln(65) = \ln(e^{12r}) \n\]\n\n\[ \n\ln(65) = 12r \n\]\n\n\[ \nr = \frac{\ln(65)}{12} = 0.3479. \n\]\n\nThis gives an annual interest rate of 34.79%.
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Yes
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Ten years ago, you purchased a house valued at \$80,000. Your plan is to sell the house at some point in the future, when the value is at least \( \$ 1,{000},{000} \) . Assume that the future value of the house can be computed using quarterly compounding and an annual interest rate of \( {4.8}\% \) . How soon can you sell the house?
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Solution. We can use the future value formula to obtain the equation\n\n\[ \n\text{1,000,000} = {80},{000}{\left( 1 + \frac{0.048}{4}\right) }^{4t} \n\]\n\n\[ \n{12.5} = {\left( {1.012}\right) }^{4\mathrm{t}} \n\]\n\nUsing the log base \( b = {1.012} \),\n\n\[ \n{\log }_{1.012}\left( {12.5}\right) = {\log }_{1.012}\left( {\left( {1.012}\right) }^{4t}\right) \n\]\n\n\[ \n{\log }_{1.012}\left( {12.5}\right) = {4t} \n\]\n\n\[ \n\mathrm{t} = \frac{\ln \left( {12.5}\right) }{4\ln \left( {1.012}\right) } = {52.934}. \n\]\n\nSince you have already owned the house for 10 years, you would need to wait nearly 43 years to sell at the desired price.
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Yes
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A loudspeaker manufacturer advertises that their model no. 801 speaker produces a sound pressure level of 87 db when a reference test tone is applied. A competing speaker company advertises that their model X-1 speaker produces a sound pressure level of 93 db when fed the same test signal. What is the ratio of the two sound intensities produced by these speakers?
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Solution. If we let \( {I}_{1} \) and \( {I}_{2} \) refer to the sound intensities of the two speakers reproducing the test signal, then we have two equations:\n\n\[ {87} = {10}{\log }_{10}\left( \frac{{\mathrm{I}}_{1}}{{\mathrm{I}}_{0}}\right) \]\n\n\[ {93} = {10}{\log }_{10}\left( \frac{{\mathrm{I}}_{2}}{{\mathrm{I}}_{0}}\right) \]\n\nUsing log properties, we can solve the first equation for \( {I}_{1} \) :\n\n\[ {87} = {10}{\log }_{10}\left( \frac{{\mathrm{I}}_{1}}{{\mathrm{I}}_{0}}\right) = {10}{\log }_{10}\left( {\mathrm{I}}_{1}\right) - {10}{\log }_{10}\left( {\mathrm{I}}_{0}\right) \]\n\n\[ {\log }_{10}\left( {\mathrm{I}}_{1}\right) = {8.7} + {\log }_{10}\left( {\mathrm{I}}_{0}\right) \]\n\n\[ {10}^{{\log }_{10}\left( {\mathrm{I}}_{1}\right) } = {10}^{{8.7} + {\log }_{10}\left( {\mathrm{I}}_{0}\right) } \]\n\n\[ {\mathrm{I}}_{1} = {10}^{8.7}{10}^{{\log }_{10}\left( {\mathrm{I}}_{0}\right) } = {10}^{8.7}{\mathrm{I}}_{0}. \]\n\nSimilarly, we find that \( {\mathrm{I}}_{2} = {10}^{9.3}{\mathrm{I}}_{0} \) . This means that the ratio of the intensities will be\n\n\[ \frac{{\mathrm{I}}_{2}}{{\mathrm{I}}_{1}} = \frac{{10}^{9.3}{\mathrm{I}}_{0}}{{10}^{8.7}{\mathrm{I}}_{0}} = {10}^{0.6} = {3.98}. \]\n\nThis means that the test signal on the \( \mathrm{X} - 1 \) speaker produces a sound pressure level nearly 4 times that of the same test signal on the no. 801 speaker.
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Yes
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Problem 12.3. Rewrite each function in the form \( y = {A}_{ \circ }{e}^{at} \), for appropriate constants \( {A}_{ \circ } \) and \( a \) .
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(a) \( y = {13}\left( {3}^{\mathrm{t}}\right) \)\n(b) \( y = 2{\left( \frac{1}{8}\right) }^{\mathrm{t}} \)\n(c) \( y = - 7{\left( {1.567}\right) }^{\mathrm{t} - 3} \)\n(d) \( y = - {17}{\left( {2.005}\right) }^{-t} \)\n(e) \( y = 3{\left( {14.24}\right) }^{4t} \)
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Yes
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Consider the parallelogram-shaped region \( \mathcal{R} \) with vertices \( \left( {0,2}\right) ,\left( {0, - 2}\right) ,\left( {1,0}\right) \), and \( \left( {-1,0}\right) \) . Use the reflection principle to find functions whose graphs bound \( \mathcal{R} \) .
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Solution. Here is a picture of the region \( \mathcal{R} \) : First off, using the two point formula for the equation of a line, we find that the line \( {\ell }_{1} \) passing through the points \( P = \left( {0,2}\right) \) and \( Q = \left( {1,0}\right) \) is the graph of the function \( y = {f}_{1}\left( x\right) = \) \( - {2x} + 2 \) . By Fact 13.2.1 (i), \( {\ell }_{2} \) is the graph of the equation \( - y = - {2x} + 2 \), which we can write as the function \( y = {f}_{2}\left( x\right) = {2x} - 2 \) . By Fact 13.2.1 (ii) applied to \( {\ell }_{2} \), the line \( {\ell }_{3} \) is the graph of the function \( y = {f}_{3}\left( x\right) = - {2x} - 2 \) . Finally, by Fact 13.2.1 (i) applied to \( {\ell }_{3} \), the line \( {\ell }_{4} \) is the graph of the equation \( - y = - {2x} - 2 \), which we can write as the function \( y = {f}_{4}\left( x\right) = {2x} + 2 \) .
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Yes
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Describe the relationship between the graphs of\n\n\\[ y = f\\left( x\\right) \\; = \\sqrt{1 - {\\left( x + 1\\right) }^{2}}, \\]\n\n\\[ y = - f\\left( x\\right) \\text{ 的面积 } = - \\sqrt{1 - {\\left( x + 1\\right) }^{2}}\\text{, and } \\]\n\n\\[ y = - {4f}\\left( x\\right) = - 4\\sqrt{1 - {\\left( x + 1\\right) }^{2}}. \\]
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Solution. The graph of \\( y = f\\left( x\\right) \\) is an upper semicircle of radius 1 centered at the point \\( \\left( {-1,0}\\right) \\) . To obtain the picture of the graph of \\( y = - {4f}\\left( x\\right) \\), we first reflect \\( y = f\\left( x\\right) \\) across the \\( x \\) -axis; this gives us the graph of \\( y = - f\\left( x\\right) \\) . Then, we vertically dilate this picture by a factor of \\( c = 4 \\) to get the graph of \\( \\frac{y}{4} = - f\\left( x\\right) \\), which is the same as the graph of the equation \\( y = - {4f}\\left( x\\right) \\) . See Figure 13.12.
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Yes
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The problem is to describe a sequence of geometric maneuvers that transform the graph of \( y = {x}^{2} \) into the graph of \( y = - 3{\left( x - 1\right) }^{2} + 2 \) .
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The idea is to rewrite \( y = - 3{\left( x - 1\right) }^{2} + 2 \) as a composition of \( y = {x}^{2} \) with four other functions, each of which corresponds to a horizontal shift, vertical shift, reflection or dilation. Once we have done this, we can read off the order of geometric operations using the order of composition. Along the way, pay special attention to the exact order in which we will be composing our functions; this will make a big difference.\n\nTo begin with, we can isolate four key numbers in the equation:\n\n<table><thead><tr><th colspan=\
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No
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Problem 13.7. Describe how each graph differs from that of \( y = {x}^{2} \) .
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(a) \( y = 2{x}^{2} \)\n(b) \( y = {x}^{2} - 5 \)\n(c) \( y = {\left( x - 4\right) }^{2} \)\n(d) \( y = {\left( 3x - {12}\right) }^{2} \)\n(e) \( y = 2{\left( 3x - {12}\right) }^{2} - 5 \)
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Yes
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Sketch the graph of the function \( f\left( x\right) = \frac{{3x} - 1}{{2x} + 7} \) .
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Solution. We begin by finding the asymptotes of \( f \) .\n\nThe denominator is equal to zero when \( {2x} + 7 = 0 \), i.e., when \( x = - 7/2 \) . As a result, the vertical asymptote for this function is the vertical line \( x = - 7/2 \) .\n\nBy taking the ratio of the coefficients of \( x \) in the numerator and denominator, we can find that the horizontal asymptote is the horizontal line\n\n\[ y = \frac{3}{2} \]\n\nWe then sketch these two asymptotes. The last thing we need is a single point. For instance, we may evaluate \( f\left( 0\right) \) :\n\n\[ f\left( 0\right) = \frac{-1}{7} \]\nand so the point \( \left( {0, - 1/7}\right) \) is on the graph. With this information, we know that the curve lies below the horizontal asymptote to the right of the vertical asymptote, and consequently the curve lies above the horizontal asymptote to the left of the vertical asymptote.\n\nWe graph the result in Figure 14.2.
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Yes
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Let \( f\left( x\right) = \frac{{2x} + 3}{{5x} - 7} \) .
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\[ f\left( x\right) = \frac{{2x} + 3}{{5x} - 7} \cdot \frac{\frac{1}{5}}{\frac{1}{5}} = \frac{\frac{2}{5}x + \frac{3}{5}}{x - \frac{7}{5}} \]
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Yes
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Find the linear-to-linear rational function \( f\left( x\right) \) such that \( f\left( {10}\right) = {20}, f\left( {20}\right) = {32} \) and \( f\left( {25}\right) = {36} \).
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Solution. Since \( f\left( x\right) \) is a linear-to-linear rational function, we know\n\n\[ f\left( x\right) = \frac{{ax} + b}{x + c} \]\n\nfor constants \( a, b \), and \( c \) . We need to find \( a, b \) and \( c \) .\n\nWe know three things.\n\nFirst, \( f\left( {10}\right) = {20} \) . So\n\n\[ f\left( {10}\right) = \frac{{10a} + b}{{10} + c} = {20} \]\n\nwhich we can rewrite as\n\n\[ {10a} + b = {200} + {20c}. \]\n\n(14.1)\n\nSecond, \( f\left( {20}\right) = {32} \) . So\n\n\[ f\left( {20}\right) = \frac{{20a} + b}{{20} + c} = {32} \]\n\nwhich we can rewrite as\n\n\[ {20a} + b = {640} + {32c}. \]\n\n(14.2)\n\nThird, \( f\left( {25}\right) = {36} \) . So\n\n\[ f\left( {25}\right) = \frac{{25a} + b}{{25} + c} = {36} \]\nwhich we can rewrite as\n\n\[ {25a} + b = {900} + {36c}. \]\n\n(14.3)\n\nThese three numbered equations are enough algebraic material to solve for \( a, b \), and \( c \) . Here is one way to do that.\n\nSubtract equation 14.1 from equation 14.2 to get\n\n\[ {10a} = {440} + {12c} \]\n\n(14.4)\n\nand subtract equation 14.2 from equation 14.3 to get\n\n\[ {5a} = {260} + {4c} \]\n\n(14.5)\n\nNote that we've eliminated \( b \) . Now multiply this last equation by 2 to get\n\n\[ {10a} = {520} + {8c} \]\n\nSubtract equation 14.4 from this to get\n\n\[ 0 = {80} - {4c} \]\n\nwhich easily give us \( \mathrm{c} = {20} \).\n\nPlugging this value into equation 14.4, we can find \( a = {68} \), and then we can find \( b = - {80} \).\n\nThus,\n\n\[ f\left( x\right) = \frac{{68x} - {80}}{x + {20}} \]\n\nWe can check that we have done the algebra correctly by evaluating \( f\left( x\right) \) at \( x = {10}, x = {20} \) and \( x = {25} \) . If we get \( f\left( {10}\right) = {20}, f\left( {20}\right) = {32} \) and \( f\left( {25}\right) = {36} \), then we’ll know our work is correct.
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Yes
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Clyde makes extra money selling tickets in front of the Safeco Field. The amount he charges for a ticket depends on how many he has. If he only has one ticket, he charges \$100 for it. If he has 10 tickets, he charges \$80 a piece. But if he has a large number of tickets, he will sell them for \$50 each. How much will he charge for a ticket if he holds 20 tickets?
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Solution. We want to give a linear-to-linear rational function relating the price of a ticket \( y \) to the number of tickets \( x \) that Clyde is holding. As we saw above, we can assume the function is of the form\n\n\[ y = \frac{ax + b}{x + c} \]\n\nwhere \( a, b \) and \( c \) are numbers. Note that \( y = a \) is the horizontal asymptote. When \( x \) is very large, \( y \) is close to 50 . This means the line \( y = 50 \) is a horizontal asymptote. Thus \( a = 50 \) and\n\n\[ y = \frac{50x + b}{x + c} \]\n\nNext we plug in the point \( \left( 1,100 \right) \) to get a linear equation in \( b \) and \( c \).\n\n\[ 100 = \frac{50 \cdot 1 + b}{1 + c} \]\n\n\[ 100 \cdot \left( 1 + c \right) = 50 + b \]\n\n\[ 50 = b - 100c \]\n\nSimilarly, plugging in (10,80) and doing a little algebra (do it now!) gives another linear equation \( 300 = b - 80c \). Solving these two linear equations simultaneously gives \( c = 12.5 \) and \( b = 1300 \). Thus our function is\n\n\[ y = \frac{50x + 1300}{x + 12.5} \]\n\nand, if Clyde holds 20 tickets, he will charge\n\n\[ y = \frac{50 \cdot 20 + 1300}{20 + 12.5} = \$ 70.77 \]\n\nper ticket.
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Yes
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If a 16 inch pizza is cut into 12 equal slices, what is the area of a single slice?
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This can be solved using a general principle:\n\n(Area of a part) \( = \) (area of the whole) \( \times \) (fraction of the part)\n\nSo, for our pizza:\n\n\[ \text{(area one slice)} = \text{(area whole pie)} \times \text{(fraction of pie)} \]\n\n\[ = \left( {{8}^{2}\pi }\right) \left( \frac{1}{12}\right) \]\n\n\[ = \frac{16\pi }{3}\text{. } \]
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Yes
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If the rear wheel is 28 inches in diameter, determine the angular speed of a location on the rear tire. A pebble becomes stuck to the tread of the rear tire. Describe the location of the pebble after 1 second and 0.1 second.
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Solution. The tires will be rotating in a counterclockwise direction and the radius \( r = \frac{1}{2}{28} = {14} \) inches. The other given quantity, \
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No
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What is the angular speed (in RPM) of a CD if the laser is at the beginning, located \( \frac{3}{4} \) inches from the center of the disc ? What is the angular speed (in RPM) of a CD if the laser is at the end, located 2 inches from the center of the disc ? Find the location of the laser if the angular speed is 350 RPM.
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Solution. This is an application of Fact 16.2.1. Let \( {\omega }_{3/4} \) be the angular speed at the start and \( {\omega }_{2} \) the angular speed at the end of the \( {CD} \) ; the subscript is keeping track of the laser distance from the \( {CD} \) center.\n\n\[{\omega }_{2} = \frac{\left( {2835}\text{ inches }/\mathrm{{min}}\right) }{\left( 2\left( 2\right) \pi \text{ inches }/\mathrm{{rev}}\right) } = {225.6}\mathrm{{RPM}}\]\n\n\[{\omega }_{3/4} = \frac{{2835}\mathrm{{inches}}/\mathrm{{min}})}{\left( 2\left( {0.75}\right) \pi \mathrm{{inches}}/\mathrm{{rev}}\right) } = {601.6}\mathrm{{RPM}}\]\n\nTo answer the remaining question, let \( r \) be the radial distance from the center of the \( {CD} \) to the laser location on the \( {CD} \) . If the angular speed \( {\omega }_{r} \) at this location is \( {350}\mathrm{{RPM}} \), we have the equation\n\n\[{350}\mathrm{{RPM}} = {\omega }_{\mathrm{r}}\]\n\n\[= \;\frac{\left( 2,{835}\;\frac{\text{inches}}{\text{minute}}\right) }{\left( 2r\pi \;\frac{\text{inches}}{\text{revolution}}\right) }\]\n\n1.289 inches \( = r \) .\n\nSo, when the laser is 1.289 inches from the center, the \( {CD} \) is moving 350 RPM.
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Yes
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Problem 16.4. Lee is running around the perimeter of a circular track at a rate of 10 \( \mathrm{{ft}}/\mathrm{{sec}} \) . The track has a radius of 100 yards. After 10 seconds, Lee turns and runs along a radial line to the center of the circle. Once he reaches the center, he turns and runs along a radial line to his starting point on the perimeter. Assume Lee does not slow down when he makes these two turns.
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(a) Sketch a picture of the situation.\n\n(b) How far has Lee traveled once he returns to his starting position?\n\n(c) How much time will elapse during Lee's circuit?\n\n(d) Find the area of the pie shaped sector enclosed by Lee's path.
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No
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Example 17.1.1. You are preparing to make your final shot at the British Pocket Billiard World Championships. The position of your ball is as in Figure 17.2, and you must play the ball off the left cushion into the lower-right corner pocket, as indicated by the dotted path. For the big money, where should you aim to hit the cushion?
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Solution. This problem depends on two basic facts. First, the angles of entry and exit between the path the cushion will be equal. Secondly, the two obvious right triangles in this picture are similar triangles. Let \( x \) represent the distance from the bottom left corner to the impact point of the ball's path: Properties of similar triangles tell us that the ratios of common sides are equal: \( \frac{4}{5 - x} = \frac{12}{x} \) . If we solve this equation for \( x \), we obtain \( x = \frac{15}{4} = {3.75} \) feet.
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Yes
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A plane is flying 2000 feet above sea level toward a mountain. The pilot observes the top of the mountain to be \( {18}^{ \circ } \) above the horizontal, then immediately flies the plane at an angle of \( {20}^{ \circ } \) above horizontal. The airspeed of the plane is \( {100}\mathrm{{mph}} \). After 5 minutes, the plane is directly above the top of the mountain. How high is the plane above the top of the mountain (when it passes over)? What is the height of the mountain?
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Solution. We can compute the hypotenuse of \( \bigtriangleup \) LPT by using the speed and time information about the plane:\n\n\[ \left| \overline{\mathrm{{PT}}}\right| = \left( {{100}\mathrm{{mph}}}\right) \left( {5\text{ minutes }}\right) \left( {1\text{ hour }/{60}\text{ minutes }}\right) = \frac{25}{3}\text{ miles. } \]\n\nThe definitions of the trigonometric ratios show:\n\n\[ \left| \overline{\mathrm{{TL}}}\right| = \frac{25}{3}\sin \left( {20}^{ \circ }\right) = {2.850}\text{ miles, and } \]\n\n\[ \left| \overline{\mathrm{{PL}}}\right| = \frac{25}{3}\cos \left( {20}^{ \circ }\right) = {7.831}\text{ miles. } \]\n\nWith this data, we can now find \( \left| \overline{EL}\right| \) :\n\n\[ \left| \overline{\mathrm{{EL}}}\right| = \left| \overline{\mathrm{{PL}}}\right| \tan \left( {18}^{ \circ }\right) = {2.544}\text{miles.} \]\n\nThe height of the plane above the peak is \( \left| \overline{\mathrm{{TE}}}\right| = \left| \overline{\mathrm{{TL}}}\right| - \left| \overline{\mathrm{{EL}}}\right| = {2.850} - {2.544} = \) \( {0.306} \) miles \( = 1,{616} \) feet. The elevation of the peak above sea level is given by: Peak elevation \( = \) plane altitude \( + \left| \overrightarrow{EL}\right| = \left| \overrightarrow{SP}\right| + \left| \overrightarrow{EL}\right| = 2,{000} + \) \( \left( {2.544}\right) \left( {5,{280}}\right) = {15},{432} \) feet.
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Yes
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A Forest Service helicopter needs to determine the width of a deep canyon. While hovering, they measure the angle \( \gamma = {48}^{ \circ } \) at position B (see picture), then descend 400 feet to position A and make two measurements of \( \alpha = {13}^{ \circ } \) (the measure of \( \angle \mathrm{{EAD}} \) ), \( \beta = {53}^{ \circ } \) (the measure of \( \angle \mathrm{{CAD}} \) ). Determine the width of the canyon to the nearest foot.
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Solution. We will need to exploit three right triangles in the picture: \( \bigtriangleup {BCD},\bigtriangleup {ACD} \), and \( \bigtriangleup {ACE} \). Our goal is to compute \( \left| \overline{ED}\right| = \) \( \left| \overline{CD}\right| - \left| \overline{CE}\right| \), which suggests more than one right triangle will come into play.\n\nThe first step is to use \( \bigtriangleup {BCD} \) and \( \bigtriangleup {ACD} \) to obtain a system of two equations and two unknowns involving some of the side lengths; we will then solve the system. From the definitions of the trigonometric ratios,\n\n\[ \left| \overline{CD}\right| = \left( {{400} + \left| \overline{AC}\right| }\right) \tan \left( {48}^{ \circ }\right) \]\n\n\[ \left| \overline{\mathrm{{CD}}}\right| = \left| \overline{\mathrm{{AC}}}\right| \tan \left( {53}^{ \circ }\right) \text{.} \]\n\nPlugging the second equation into the first and rearranging we get\n\n\[ \left| \overline{AC}\right| = \frac{{400}\tan \left( {48}^{ \circ }\right) }{\tan \left( {53}^{ \circ }\right) - \tan \left( {48}^{ \circ }\right) } = 2,{053}\text{ feet. } \]\n\nPlugging this back into the second equation of the system gives\n\n\[ \left| \overline{CD}\right| = \left( {2053}\right) \tan \left( {53}^{ \circ }\right) = {2724}\text{ feet. } \]\n\nThe next step is to relate \( \bigtriangleup {ACD} \) and \( \bigtriangleup {ACE} \), which can now be done in an effective way using the calculations above. Notice that the measure of \( \angle {CAE} \) is \( \beta - \alpha = {40}^{ \circ } \). We have\n\n\[ \left| \overline{CE}\right| = \left| \overline{AC}\right| \tan \left( {40}^{ \circ }\right) = \left( {2053}\right) \tan \left( {40}^{ \circ }\right) = 1,{723}\text{ feet. } \]\n\nAs noted above, \( \left| \overline{ED}\right| = \left| \overline{CD}\right| - \left| \overline{CE}\right| = 2,{724} - 1,{723} = 1,{001} \) feet is the width of the canyon.
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Yes
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Michael is test driving a vehicle counter-clockwise around a desert test track which is circular of radius 1 kilometer. He starts at the location pictured, traveling \( {0.025}\frac{\text{rad}}{\text{sec}} \) . Impose coordinates as pictured. Where is Michael located (in xy-coordinates) after 18 seconds?
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Solution. Let \( M\left( t\right) \) be the point on the circle of motion representing Michael’s location after \( t \) seconds and \( \theta \left( t\right) \) the angle swept out the by Michael after \( t \) seconds. Since we are given the angular speed, we get\n\n\[ \theta \left( t\right) = {0.025t}\text{radians.} \]\n\nSince the angle \( \theta \left( \mathrm{t}\right) \) is in central standard position, we get\n\n\[ M\left( t\right) = \left( {\cos \left( {\theta \left( t\right) }\right) ,\sin \left( {\theta \left( t\right) }\right) }\right) = \left( {\cos \left( {0.025t}\right) ,\sin \left( {0.025t}\right) }\right) . \]\n\nSo, after 18 seconds Michael’s location will be \( M\left( {18}\right) = \left( {{0.9004},{0.4350}}\right) \).
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Yes
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Where are the drivers located (in xy-coordinates) after 18 seconds?
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Solution. Let \( M\left( t\right) \) be the point on the circle of motion rep-\n\nresenting Michael’s location after \( t \) seconds. Likewise, let A(t) be the point on the circle of motion representing Angela’s location after \( t \) seconds. Let \( \theta \left( t\right) \) be the angle swept out the by Michael and \( \alpha \left( \mathrm{t}\right) \) the angle swept out by Angela after \( t \) seconds.\n\nSince we are given the angular speeds, we get\n\n\[ \theta \left( \mathrm{t}\right) = {0.025}\mathrm{t}\text{radians, and} \]\n\n\[ \alpha \left( t\right) = {0.03t}\text{radians.} \]\n\nFrom the previous Example 17.4.2,\n\n\[ M\left( t\right) = \left( {\cos \left( {0.025t}\right) ,\sin \left( {0.025t}\right) }\right) \text{, and} \]\n\n\[ M\left( {18}\right) = \left( {{0.9004},{0.4350}}\right) \text{.} \]\n\nAngela’s angle \( \alpha \left( t\right) \) is NOT in central standard position, so we must observe that \( \alpha \left( t\right) + \pi = \beta \left( t\right) \), where \( \beta \left( t\right) \) is in central standard position: See Figure 17.14(b). We conclude that\n\n\[ A\left( t\right) = \left( {\cos \left( {\beta \left( t\right) }\right) ,\sin \left( {\beta \left( t\right) }\right) }\right) \]\n\n\[ = \left( {\cos \left( {\pi + {0.03t}}\right) ,\sin \left( {\pi + {0.03t}}\right) }\right) \text{.} \]\n\nSo, after 18 seconds Angela’s location will be \( A\left( {18}\right) = \) \( \left( {-{0.8577}, - {0.5141}}\right) \) .
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Yes
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Suppose Cosmo begins at the position \( \mathrm{R} \) in the figure, walking around the circle of radius 20 feet with an angular speed of \( \frac{4}{5} \) RPM counterclockwise. After 3 minutes have elapsed, describe Cosmo’s precise location.
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Solution. Cosmo has traveled \( 3\frac{4}{5} = \frac{12}{5} \) revolutions. If \( \theta \) is the angle traveled after 3 minutes, \( \theta = \left( {\frac{12}{5}\mathrm{{rev}}}\right) \left( {{2\pi }\frac{\mathrm{{radians}}}{\mathrm{{rev}}}}\right) = \) \( \frac{24\pi }{5} \) radians \( = {15.08} \) radians. By (15.5.1), we have \( x = \) \( {20}\cos \left( {\frac{24\pi }{5}\mathrm{{rad}}}\right) \; = \; - {16.18}\;\mathrm{{feet}}\;\mathrm{{and}}\;y\; = \;{20}\sin \left( {\frac{24\pi }{5}\;\mathrm{{rad}}}\right) \; = \) 11.76feet. Conclude that Cosmo is located at the point \( S = \left( {-{16.18},{11.76}}\right) \) . Using (15.1), \( \theta = {864}^{ \circ } = 2\left( {360}^{ \circ }\right) + {144}^{ \circ } \) ; this means that Cosmo walks counterclockwise around the circle two complete revolutions, plus \( {144}^{ \circ } \) .
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Yes
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Three airplanes depart SeaTac Airport. A NorthWest flight is heading in a direction \( {50}^{ \circ } \) counterclockwise from East, an Alaska flight is heading \( {115}^{ \circ } \) counterclockwise from East and a Delta flight is heading \( {20}^{ \circ } \) clockwise from East. Find the location of the Northwest flight when it is 20 miles North of SeaTac. Find the location of the Alaska flight when it is 50 miles West of SeaTac. Find the location of the Delta flight when it is 30 miles East of SeaTac.
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We impose a coordinate system in Fig-\n\nure 17.21(a), where \
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No
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Problem 17.9. Charlie and Alexandra are running around a circular track with radius 60 meters. Charlie started at the westernmost point of the track, and, at the same time, Alexandra started at the northernmost point. They both run counterclockwise. Alexandra runs at 4 meters per second, and will take exactly 2 minutes to catch up to Charlie.
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Impose a coordinate system, and give the \( x \) - and \( y \) -coordinates of Charlie after one minute of running.
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No
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Assume that the number of hours of daylight in Seattle during 1994 is given by the function \( \mathrm{d}\left( \mathrm{t}\right) = {3.7}\sin \left( {\frac{2\pi }{366}\left( {\mathrm{t} - {80.5}}\right) }\right) + {12} \), where \( \mathrm{t} \) represents the day of the year and \( \mathrm{t} = 0 \) corresponds to January 1. How many hours of daylight will there be on May 11?
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Solution. To solve the problem, you need to consult a calendar, finding every month has 31 days, except: February has 28 days and April, June, September and November have 30 days. May 11 is the \( {31} + {28} + {31} + {30} + {11} = \) \( {131}^{\text{st }} \) day of the year. So, there will be \( d\left( {131}\right) = {3.7}\sin \left( {2\left( {50.5}\right) \pi /{366}}\right) + {12} = \) 14.82 hours of daylight on May 11.
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Yes
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Problem 18.4. These graphs represent periodic functions. Describe the period in each case.
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No
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The temperature (in \( {}^{ \circ }\mathrm{C} \) ) of Adri-N’s dorm room varies during the day according to the sinusoidal function \( \mathrm{d}\left( \mathrm{t}\right) = 6\sin \left( {\frac{\pi }{12}\left( {\mathrm{t} - {11}}\right) }\right) + \) 19, where t represents hours after midnight. Roughly sketch the graph of \( \mathrm{d}\left( \mathrm{t}\right) \) over a 24 hour period.. What is the temperature of the room at 2:00 pm? What is the maximum and minimum temperature of the room?
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Solution. We begin with the rough sketch. Start by taking an inventory of the constants in this sinusoidal function:\n\n\[ d\left( t\right) = 6\sin \left( {\frac{\pi }{12}\left( {t - {11}}\right) }\right) + {19} = A\sin \left( {\frac{2\pi }{B}\left( {t - C}\right) }\right) + D. \]\n\nConclude that \( A = 6, B = {24}, C = {11}, D = {19} \) . Following the first four steps of the procedure outlined, we can sketch the lines \( y = D = {19} \) , \( y = D \pm A = {19} \pm 6 \) and three points where the graph crosses the mean line (see Figure 19.8).\n\n\n\nFigure 19.8: Sketching the mean D and amplitude A.\n\nAccording to the fifth step in the sketching procedure, we can plot the maxima \( \left( {C + \frac{1}{4}B, D + A}\right) = \left( {{17},{25}}\right) \) and the minima \( \left( {C + \frac{3}{4}B, D - A}\right) = \left( {{29},{13}}\right) \) . We then \
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No
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Assume that the number of hours of daylight in Seattle is given by a sinusoidal function \( \mathrm{d}\left( \mathrm{t}\right) \) of time. During 1994, assume the longest day of the year is June 21 with 15.7 hours of daylight and the shortest day is December 21 with 8.3 hours of daylight. Find a formula \( \mathrm{d}\left( \mathrm{t}\right) \) for the number of hours of daylight on the \( {\mathrm{t}}^{\text{th }} \) day of the year.
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Solution. Because the function \( d\left( t\right) \) is assumed to be sinusoidal, it has the form \( y = A\sin \left( {\frac{2\pi }{B}\left( {t - C}\right) }\right) + D \), for constants \( A, B, C \), and \( D \) . We simply need to use the given information to find these constants. The largest value of the function is 15.7 and the smallest value is 8.3. Knowing this, from the above discussion we can read off : \[ D = \frac{{15.7} + {8.3}}{2} = {12}\;A = \frac{{15.7} - {8.3}}{2} = {3.7}. \] To find the period, we need to compute the time between two successive maximum values of \( d\left( t\right) \) . To find this, we can simply double the time length of one-half period, which would be the length of time between successive maximum and minimum values of \( d\left( t\right) \) . This gives us the equation \[ B = 2\text{(days between June 21 and December 21)} = 2\left( {183}\right) = {366}\text{.} \] Locating the final constant \( \mathrm{C} \) requires the most thought. Recall, the longest day of the year is June 21, which is day 172 of the year, so \[ \mathrm{C} = \text{(day with max daylight)} - \frac{\mathrm{B}}{4} = {172} - \frac{366}{4} = {80.5}\text{.} \] In summary, this shows that \[ d\left( t\right) = {3.7}\sin \left( {\frac{2\pi }{366}\left( {t - {80.5}}\right) }\right) + {12}. \]
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Yes
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The depth of a migrating salmon below the water surface changes according to a sinusoidal function of time. The fish varies between 1 and 5 feet below the surface of the water. It takes the fish 1.571 minutes to move from its minimum depth to its successive maximum depth. It is located at a maximum depth when \( \mathrm{t} = {4.285} \) minutes. What is the formula for the function \( \mathrm{d}\left( \mathrm{t}\right) \) that predicts the depth of the fish after t minutes? What was the depth of the salmon when it was first spotted? During the first 10 minutes, how many times will the salmon be exactly 4 feet below the surface of the water?
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Solution. We know that \( d\left( t\right) = A\sin \left( {\frac{2\pi }{B}\left( {x - C}\right) }\right) + D \), for appropriate constants \( A, B, C \), and \( D \) . We need to use the given information to extract these four constants. The amplitude and mean are easily found using the above formulas:\n\n\[ A = \frac{\text{ max depth } - \text{ min depth }}{2} = \frac{5 - 1}{2} = 2 \]\n\n\[ D = \frac{\max \text{ depth } + \min \text{ depth }}{2} = \frac{5 + 1}{2} = 3. \]\n\nThe period can be found by noting that the information about the time between a successive minimum and maximum depth will be half of a period (look at the picture in Figure 19.13):\n\n\[ B = 2\left( {1.571}\right) = {3.142} \]\n\nFinally, to find C we\n\n\[ \mathrm{C} = \left( \text{time of maximum depth}\right) - \frac{\mathrm{B}}{4} = {4.285} - \frac{3.142}{4} = {3.50}. \]\n\nThe formula is now\n\n\[ d\left( t\right) = 2\sin \left( {\frac{2\pi }{3.142}\left( {t - {3.5}}\right) }\right) + 3 = 2\sin \left( {{2t} - 7}\right) + 3 \]\n\nThe depth of the salmon when it was first spotted is just\n\n\[ d\left( 0\right) = 2\sin \left( {-7}\right) + 3 = {1.686}\text{ feet. } \]\n\nFinally, graphically, the last question amounts to determining how many times the graph of \( d\left( t\right) \) crosses the line \( y = 4 \) on the domain \( \left\lbrack {0,{10}}\right\rbrack \) . This can be done using Figure 19.13. A simultaneous picture of the two graphs is given, from which we can see the salmon is exactly 4 feet below the surface of the water six times during the first 10 minutes.
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Yes
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Problem 19.2. A weight is attached to a spring suspended from a beam. At time \( t = 0 \), it is pulled down to a point \( {10}\mathrm{\;{cm}} \) above the ground and released. After that, it bounces up and down between its minimum height of 10 \( \mathrm{{cm}} \) and a maximum height of \( {26}\mathrm{\;{cm}} \), and its height \( h\left( t\right) \) is a sinusoidal function of time \( t \) . It first reaches a maximum height 0.6 seconds after starting.
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(a) Follow the procedure outlined in this section to sketch a rough graph of \( h\left( t\right) \) . Draw at least two complete cycles of the oscillation, indicating where the maxima and minima occur.\n\n(b) What are the mean, amplitude, phase shift and period for this function?\n\n(c) Give four different possible values for the phase shift.\n\n(d) Write down a formula for the function \( h\left( t\right) \) in standard sinusoidal form; i.e. as in 19.1.1 on Page 254.\n\n(e) What is the height of the weight after 0.18 seconds?\n\n(f) During the first 10 seconds, how many times will the weight be exactly \( {22}\mathrm{\;{cm}} \) above the floor? (Note: This problem does not require inverse trigonometry.)
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No
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Find all values of \( \theta \) (an angle) that make this equation true: \( \sin \left( \theta \right) = \frac{1}{2} \) .
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Solution. We begin with a graphical reinterpretation: the solutions correspond to the places where the graphs of \( z = \sin \left( \theta \right) \) and \( z = \frac{1}{2} \) intersect in the \( {\theta z} \) -coordinate system. Recalling Figure 18.11, we can picture these two graphs simultaneously as below:\n\n\n\nFigure 20.2: Where does \( \sin \left( \theta \right) \) cross \( z = \frac{1}{2} \) ?\n\nThe first thing to notice is that these two graphs will cross an infinite number of times, so there are infinitely many solutions to Example 20.1.1! However, notice there is a predictable spacing of the crossing points, which is just a manifestation of the periodicity of the sine function. In fact, if we can find the two crossing points labeled \
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No
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If two sides of a right triangle have lengths 1 and \( \sqrt{3} \) as pictured below, what are the acute angles \( \alpha \) and \( \beta \) ?
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Solution. By the Pythagorean Theorem the remaining side has length\n\n\[ \sqrt{1 + {\left( \sqrt{3}\right) }^{2}} = 2 \]\n\nSince \( \tan \left( \alpha \right) = \sqrt{3} \), we need to solve this equation for \( \alpha \) . Graphically, we need to determine where \( z = \sqrt{3} \) crosses the graph of the tangent function:\n\nFrom Fact 20.1.2, there will be infinitely many solutions to our equation, but notice that there is exactly one solution in the interval \( \left\lbrack {-\frac{\pi }{2},\frac{\pi }{2}}\right\rbrack \) and we can find it using Table 17.1:\n\n\[ \tan \left( \frac{\pi }{3}\right) = \tan \left( {60}^{ \circ }\right) = \frac{\sin \left( {60}^{ \circ }\right) }{\cos \left( {60}^{ \circ }\right) } = \frac{\left( \frac{\sqrt{3}}{2}\right) }{\left( \frac{1}{2}\right) } = \sqrt{3}. \]\n\nSo, \( \alpha = \frac{\pi }{3} \) radians \( = {60}^{ \circ } \) is the only acute angle solution and \( \beta = {180}^{ \circ } - \) \( {60}^{ \circ } - {90}^{ \circ } = {30}^{ \circ } \) .
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Yes
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Find two acute angles \( \theta \) so that the following equation is satisfied:\n\n\[ \frac{9}{4{\cos }^{2}\left( \theta \right) } = \frac{{25}^{2}}{16}\left( {1 - {\cos }^{2}\left( \theta \right) }\right) . \]
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Solution. Begin by multiplying each side of the equation by \( {\cos }^{2}\left( \theta \right) \) and rearranging terms:\n\n\[ \frac{9}{4} = \frac{{25}^{2}}{16}\left( {1 - {\cos }^{2}\left( \theta \right) }\right) {\cos }^{2}\left( \theta \right) \]\n\n\[ 0 = \frac{{25}^{2}}{16}{\cos }^{4}\left( \theta \right) - \frac{{25}^{2}}{16}{\cos }^{2}\left( \theta \right) + \frac{9}{4}. \]\n\nTo solve this equation for \( \theta \), we use what is called the technique of substitution. The central idea is to bring the quadratic formula into the picture by making the substitution \( z = {\cos }^{2}\left( \theta \right) \):\n\n\[ 0 = \frac{{25}^{2}}{16}{z}^{2} - \frac{{25}^{2}}{16}z + \frac{9}{4} \]\n\nApplying the quadratic formula, we obtain\n\n\[ z = \frac{\frac{{25}^{2}}{16} \pm \sqrt{{\left( \frac{{25}^{2}}{16}\right) }^{2} - 4\left( \frac{{25}^{2}}{16}\right) \left( \frac{9}{4}\right) }}{\frac{2\left( {25}^{2}\right) }{16}} = {0.9386}\text{ or }{0.06136}. \]\n\nWe now use the fact that \( z = {\cos }^{2}\left( \theta \right) \) and note the cosine of an acute angle is non-negative to conclude that\n\n\[ {\cos }^{2}\left( \theta \right) = {0.9386}\;\text{ or }\;{\cos }^{2}\left( \theta \right) = {0.06136} \]\n\n\[ \cos \left( \theta \right) = {0.9688}\;\text{ or }\;\cos \left( \theta \right) = {0.2477}. \]\n\nFinally, we use the inverse cosine function to arrive at our two acute angle solutions:\n\n\[ \cos \left( \theta \right) = {0.9688}\; \Rightarrow \;\theta = {\cos }^{-1}\left( {0.9688}\right) = {14.35}^{ \circ } \]\n\n\[ \cos \left( \theta \right) = {0.2477}\; \Rightarrow \;\theta = {\cos }^{-1}\left( {0.2477}\right) = {75.66}^{ \circ } \]
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Yes
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A 32 ft ladder leans against a building (as shown below) making an angle \( \alpha \) with the wall. OSHA (Occupational Safety and Health Administration) specifies a “safety range” for the angle \( \alpha \) to be \( {15}^{ \circ } \leq \alpha \leq {44}^{ \circ } \) . If the base of the ladder is \( \mathrm{d} = {10} \) feet from the house, is this a safe placement? Find the highest and lowest points safely accessible.
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Solution. If \( d = {10} \), then \( \sin \left( \alpha \right) = \frac{10}{32} \), so the principal solution is \( \alpha = {\sin }^{-1}\left( \frac{10}{32}\right) = {18.21}^{ \circ } \) ; this lies within the safety zone. From the picture, it is clear that the highest point safely reached will occur precisely when \( \alpha = {15}^{ \circ } \) and as this angle increases, the height \( h \) decreases until we reach the lowest safe height when \( \alpha = {44}^{ \circ } \) . We need to solve two right triangles. If \( \alpha = {15}^{ \circ } \), then \( h = {32}\cos \left( {15}^{ \circ }\right) = {30.9}\mathrm{{ft}} \) . If \( \alpha = {44}^{ \circ } \), then \( h = {32}\cos \left( {44}^{ \circ }\right) = {23.02}\mathrm{{ft}} \) .
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Yes
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A Coast Guard jet pilot makes contact with a small unidentified propeller plane 15 miles away at the same altitude in a direction 0.5 radians counterclockwise from East. The prop plane flies in the direction 1.0 radians counterclockwise from East. The jet has been instructed to allow the prop plane to fly 10 miles before intercepting. In what direction should the jet fly to intercept the prop plane? If the prop plane is flying 200 mph, how fast should the jet be flying to intercept?
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Solution. A picture of the situation is shown in Fig-\n\n (a) The physical layout. \n\n(b) Modeling the problem.\n\nFigure 20.9: Visualizing the Coast Guard problem.\n\nure 20.9(a). After a look at the picture, three right triangles pop out and beg to be exploited. We highlight these in the Figure 20.9(b), by imposing a coordinate system and labeling the various sides of our triangles. We will work in radian units and label \( \theta \) to be the required intercept heading.\n\nWe will first determine the sides \( x + y \) and \( u + w \) of the large right triangle. To do this, we have\n\n\[ x = {15}\cos \left( {0.5}\right) = {13.164}\text{ miles,}\]\n\n\[ y = {10}\cos \left( {1.0}\right) = {5.403}\text{miles,}\]\n\n\[ w = {15}\sin \left( {0.5}\right) = {7.191}\text{miles, and}\]\n\n\[ u = {10}\sin \left( {1.0}\right) = {8.415}\text{miles.}\]\n\nWe now have \( \tan \left( \theta \right) = \frac{w + u}{x + u} = {0.8405} \), so the principal solution is \( \theta = {0.699} \) radians, which is about \( {40.05}^{ \circ } \) . This is the only acute angle solution, so we have found the required intercept heading.\n\nTo find the intercept speed, first compute your distance to the intercept point, which is the length of the hypotenuse of the big right triangle: \( d = \) \( \sqrt{{\left( {18.567}\right) }^{2} + {\left( {15.606}\right) }^{2}} = {24.254} \) miles. You need to travel this distance in the same amount of time \( T \) it takes the prop plane to travel 10 miles at 200 mph; i.e. T \( = \frac{10}{200} = \) 0.05 hours. Thus, the intercept speed \( s \) is\n\n\[ s = \frac{\text{ distance traveled }}{\text{ time T elapsed }} = \frac{24.254}{0.05} = {485}\mathrm{{mph}}. \]
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Yes
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Example 20.3.4. A rigid 14 ft pole is used to vault. The vaulter leaves and returns to the ground when the tip is 6 feet high, as indicated. What are the angles of the pole with the ground on takeoff and landing?
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Solution. From the obvious right triangles in the picture, we are interested in finding angles \( \theta \) where \( \sin \left( \theta \right) = \frac{6}{14} = \frac{3}{7} \) . The idea is to proceed in three steps:\n\n- Find the principal solution of the equation \( \sin \left( \theta \right) = \frac{3}{7} \) ;\n\n- Find all solutions of the equation \( \sin \left( \theta \right) = \frac{3}{7} \) ;\n\n- Use the constraints of the problem to find \( \alpha \) and \( \beta \) among the set of all solutions.\n\nSolving the equation \( \sin \left( \theta \right) = \frac{3}{7} \) involves finding the points on the unit circle with \( y \) -coordinate equal to \( \frac{3}{7} \) . From the picture, we see there are two such points, labeled P and Q.\n\nThe coordinates of these points will be \( P = \left( {\cos \left( \alpha \right) ,\frac{3}{7}}\right) \) and \( Q = \left( {\cos \left( \beta \right) ,\frac{3}{7}}\right) \) . Notice, \( \alpha \) is the principal solution of our equation \( \sin \left( \theta \right) = \frac{3}{7} \), since \( 0 \leq \alpha \leq {90}^{ \circ } \) ; so \( \alpha = {\sin }^{-1}\left( \frac{3}{7}\right) = {25.38}^{ \circ } \) . In general, the solutions come in two basic flavors:\n\n\[ \theta = \alpha + 2\mathrm{k}\left( {180}^{ \circ }\right) \]\n\n\[ = {25.38}^{ \circ } + {2k}\left( {180}^{ \circ }\right) \]\n\n\nor\n\n\[ \theta = \beta + {2k}\left( {180}^{ \circ }\right) \]\n\nwhere \( k = 0, \pm 1, \pm 2, \pm 3,\ldots \) To find the angle \( \beta \), we can use basic properties of the circular functions:\n\n\[ \sin \left( \beta \right) = \sin \left( \alpha \right) = - \sin \left( {-\alpha }\right) = \sin \left( {{180}^{ \circ } - \alpha }\right) = \sin \left( {154.62}^{ \circ }\right) .\n\nThis tells us \( \beta = {154.62}^{ \circ } \) .
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Yes
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Assume that the number of hours of daylight in your hometown during 1994 is given by the function \( \mathrm{d}\left( \mathrm{t}\right) = {3.7}\sin \left( {\frac{2\pi }{366}\left( {\mathrm{t} - {80.5}}\right) }\right) + 12 \), where t represents the day of the year. Find the days of the year during which there will be approximately 14 hours of daylight?
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Solution. To begin, we want to roughly sketch the graph of \( y = d\left( t\right) \) on the domain \( 0 \leq \mathrm{t} \leq {366} \). If you apply the graphing procedure discussed in Chapter 19, you obtain the sinusoidal graph below on the larger domain \( - {366} \leq \mathrm{t} \leq {732} \). (The reason we use a larger domain is so that the \
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No
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Theorem 1.2.1 The Pythagorean Theorem. In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.
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Proof. Suppose we have right triangle \( {ABC} \) as in Figure 1.2.2 with right angle at \( C \), and side lengths \( a, b \), and \( c \), opposite corners \( A, B \) and \( C \), respectively. In the figure we have extended squares from each leg of the triangle, and labeled various corners. We have also constructed the line through \( C \) parallel to \( {AD} \) , and let \( L \) and \( M \) denote the points of intersection of this line with \( {AB} \) and \( {DE} \) , respectively. One can check that \( \bigtriangleup {KAB} \) is congruent to \( \bigtriangleup {CAD} \) . Moreover, the area of \( \bigtriangleup {KAB} \) is one half the area of the square \( {AH} \) . This is the case because they have equal base (segment \( {KA} \) ) and equal altitude (segment \( {AC} \) ). By a similar argument, the area of \( \bigtriangleup {DAC} \) is one half the area of the parallelogram \( {AM} \) . This means that square \( {AH} \) and parallelogram \( {AM} \) have equal areas, the value of which is \( {b}^{2} \) . One may proceed as above to argue that the areas of square \( {BG} \) and parallelogram \( {BM} \) are also equal, with value \( {a}^{2} \) . Since the area of square \( {BD} \), which equals \( {c}^{2} \), is the sum of the two parallelogram areas, it follows that \( {a}^{2} + {b}^{2} = {c}^{2} \).
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Yes
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Consider the surface obtained by identifying the edges of the hexagon as indicated in Figure 1.3.8. In particular, the edges are matched according to their labels and arrow orientation. So, if a ship flies off the hexagonal screen at a spot on the edge marked \( a \), say, then it reappears at the matching spot on the other edge marked \( a \) .
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However, the angle of each corner is \( {120}^{ \circ } \), and gluing them together will create a cone point, as pictured below. Similarly, she would find that the other corners of the hexagon meet in groups of two, creating two additional cone points. As with the Coneland Example 1.3.5, the pilot can distinguish a corner point from an interior point here. She can look at triangles: a triangle containing one of the cone points will have angle sum greater than \( {180}^{ \circ } \) ; any other triangle will have angle sum equal to \( {180}^{ \circ } \) .\n\nSo the surface is not homogeneous, if it is drawn in the plane. However, the surface does admit a homogeneous geometry. We can get rid of the cone points if we can increase each corner angle of the hexagon to \( {180}^{ \circ } \) . Then, two corners would come together to form a perfect \( {360}^{ \circ } \) patch about the point.\n\nBut how can we increase the corner angles? Put the hexagon on the sphere! Imagine stretching the hexagon onto the northern hemisphere of a sphere (see Figure 1.3.10). In this case we can think of the 6 points of our hexagon as lying on the equator. Then each corner angle is \( {180}^{ \circ } \) , each edge is still a line (geodesic), and when we glue the edges, each pair of corner angles adds up to exactly \( {360}^{ \circ } \), so the surface is homogeneous. The homogeneous geometry of this surface is the geometry of the sphere (elliptic geometry), not the geometry of the plane (Euclidean geometry).
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Yes
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Suppose \( z = 3 - {4i} \) and \( w = 2 + {7i} \).
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Then \( z + w = 5 + {3i} \), and\n\n\[ z \cdot w = \left( {3 - {4i}}\right) \left( {2 + {7i}}\right) \]\n\n\[ = 6 + {28} - {8i} + {21i} \]\n\n\[ = {34} + {13i}\text{.} \]
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Yes
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Example 2.2.2 Exploring the polar form.
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On the left side of the following diagram, we plot the points \( z = 2{e}^{{i\pi }/4}, w = \) \( 3{e}^{{i\pi }/2}, v = - 2{e}^{{i\pi }/6}, u = 3{e}^{-{i\pi }/3} \) .   To convert \( z = - 3 + {4i} \) to polar form, refer to the right side of the diagram. We note that \( r = \sqrt{9 + {16}} = 5 \), and \( \tan \left( \alpha \right) = 4/3 \), so \( \theta = \pi - {\tan }^{-1}\left( {4/3}\right) \approx {2.21} \) radians. Thus, \[ - 3 + {4i} = 5{e}^{i\left( {\pi - {\tan }^{-1}\left( {4/3}\right) }\right) } \approx 5{e}^{2.21i}. \]
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Yes
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Theorem 2.2.3 The product of two complex numbers in polar form is given by\n\n\[ r{e}^{i\theta } \cdot s{e}^{i\beta } = \left( {rs}\right) {e}^{i\left( {\theta + \beta }\right) }.\]
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Proof. We use the definition of the complex exponential and some trigonometric identities.\n\n\[ r{e}^{i\theta } \cdot s{e}^{i\beta } = r\left( {\cos \theta + i\sin \theta }\right) \cdot s\left( {\cos \beta + i\sin \beta }\right) \]\n\n\[ = \left( {rs}\right) \left( {\cos \theta + i\sin \theta }\right) \cdot \left( {\cos \beta + i\sin \beta }\right) \]\n\n\[ = {rs}\left\lbrack {\cos \theta \cos \beta - \sin \theta \sin \beta + \left( {\cos \theta \sin \beta + \sin \theta \cos \beta }\right) i}\right\rbrack \]\n\n\[ = {rs}\left\lbrack {\cos \left( {\theta + \beta }\right) + \sin \left( {\theta + \beta }\right) i}\right\rbrack \]\n\n\[ = {rs}\left\lbrack {e}^{i\left( {\theta + \beta }\right) }\right\rbrack \]
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Yes
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When representing a complex number \( z \) in polar form as \( z = r{e}^{i\theta } \), we may assume that \( r \) is non-negative. If \( r < 0 \), then what happens?
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\[ r{e}^{i\theta } = - \left| r\right| {e}^{i\theta } \] \[ = \left( {e}^{i\pi }\right) \cdot \left| r\right| {e}^{i\theta }\text{ since } - 1 = {e}^{i\pi } \] \[ = \left| r\right| {e}^{i\left( {\theta + \pi }\right) }\text{, by Theorem 2.2.3.} \] Thus, by adding \( \pi \) to the angle if necessary, we may always assume that \( z = r{e}^{i\theta } \) where \( r \) is non-negative.
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Yes
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We convert the following quotient to Cartesian form:\n\n\\[ \n\\frac{2 + i}{3 + {2i}} = \\frac{2 + i}{3 + {2i}} \\cdot \\frac{3 - {2i}}{3 - {2i}} \n\\]
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\\[ \n= \\frac{\\left( {6 + 2}\\right) + \\left( {-4 + 3}\\right) i}{9 + 4} \n\\]\n\n\\[ \n= \\frac{8 - i}{13} \n\\]\n\n\\[ \n= \\frac{8}{13} - \\frac{1}{13}i \n\\]
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Yes
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Theorem 2.4.2 Any line in \( \mathbb{C} \) can be expressed by the equation \( \left| {z - \gamma }\right| = \left| {z - \beta }\right| \) for suitably chosen points \( \gamma \) and \( \beta \) in \( \mathbb{C} \), and the set of all points (Euclidean) equidistant from distinct points \( \gamma \) and \( \beta \) forms a line.
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Proof. Given two points \( \gamma \) and \( \beta \) in \( \mathbb{C}, z \) is equidistant from both if and only if \( {\left| z - \gamma \right| }^{2} = {\left| z - \beta \right| }^{2} \) . Expanding this equation, we obtain\n\n\[ \left( {z - \gamma }\right) \left( \overline{z - \gamma }\right) = \left( {z - \beta }\right) \left( \overline{z - \beta }\right) \]\n\n\[ {\left| z\right| }^{2} - \overline{\gamma }z - \gamma \overline{z} + {\left| \gamma \right| }^{2} = {\left| z\right| }^{2} - \overline{\beta }z - \beta \overline{z} + {\left| \beta \right| }^{2} \]\n\n\[ \overline{\left( \beta - \gamma \right) }z + \left( {\beta - \gamma }\right) \bar{z} + \left( {{\left| \gamma \right| }^{2} - {\left| \beta \right| }^{2}}\right) = 0. \]\n\nThis last equation has the form of a line, letting \( \alpha = \overline{\left( \beta - \gamma \right) } \) and \( d = {\left| \gamma \right| }^{2} - {\left| \beta \right| }^{2} \) .\n\nConversely, starting with a line we can find complex numbers \( \gamma \) and \( \beta \) that do the trick. In particular, if the given line is the perpendicular bisector of the segment \( {\gamma \beta } \), then \( \left| {z - \gamma }\right| = \left| {z - \beta }\right| \) describes the line. We leave the details to the reader.
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No
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Let \( \\theta \) be an angle, and define \( {R}_{\\theta } : \\mathbb{C} \\rightarrow \\mathbb{C} \) by \( {R}_{\\theta }\\left( z\\right) = {e}^{i\\theta }z \).
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This transformation causes points in the plane to rotate about the origin by the angle \( \\theta \) . (If \( \\theta > 0 \) the rotation is counterclockwise, and if \( \\theta < 0 \) the rotation is clockwise.) To see this is the case, suppose \( z = r{e}^{i\\beta } \), and notice that\n\n\[ \n{R}_{\\theta }\\left( z\\right) = {e}^{i\\theta }r{e}^{i\\beta } = r{e}^{i\\left( {\\theta + \\beta }\\right) }.\n\]
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Yes
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Theorem 3.1.5 If \( T \) and \( S \) are two transformations of the set \( A \), then the composition \( S \circ T \) is also a transformation of the set \( A \) .
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Proof. We must prove that \( S \circ T : A \rightarrow A \) is 1-1 and onto.\n\nThat \( S \circ T \) is onto:\n\nSuppose \( c \) is in \( A \) . We must find an element \( a \) in \( A \) such that \( S \circ T\left( a\right) = c \) .\n\nSince \( S \) is onto, there exists some element \( b \) in \( A \) such that \( S\left( b\right) = c \) .\n\nSince \( T \) is onto, there exists some element \( a \) in \( A \) such that \( T\left( a\right) = b \) .\n\nThen \( S \circ T\left( a\right) = S\left( b\right) = c \), and we have demonstrated that \( S \circ T \) is onto.\n\nThat \( S \circ T \) is 1-1:\n\nAgain, we prove the contrapositive. In particular, we show that if \( S \circ T\left( {a}_{1}\right) = \n\n\( S \circ T\left( {a}_{2}\right) \) then \( {a}_{1} = {a}_{2} \) .\n\nIf \( S\left( {T\left( {a}_{1}\right) }\right) = S\left( {T\left( {a}_{2}\right) }\right) \) then \( T\left( {a}_{1}\right) = T\left( {a}_{2}\right) \) since \( S \) is 1-1.\n\nAnd \( T\left( {a}_{1}\right) = T\left( {a}_{2}\right) \) implies that \( {a}_{1} = {a}_{2} \) since \( T \) is 1-1 .\n\nTherefore, \( S \circ T \) is 1-1 .
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Yes
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Suppose \( k > 0 \) is a real number. The transformation \( T\left( z\right) = {kz} \) is called a dilation; such a map either stretches or shrinks points in the plane along rays emanating from the origin, depending on the value of \( k \) .
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Indeed, if \( z = x + {yi} \), then \( T\left( z\right) = {kx} + {kyi} \), and \( z \) and \( T\left( z\right) \) are on the same line through the origin. If \( k > 1 \) then \( T \) stretches points away from the origin. If \( 0 < k < 1 \), then \( T \) shrinks points toward the origin. In either case, such a map is called a dilation.
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Yes
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Consider the general linear transformation \( T\left( z\right) = {az} + b \), where \( a, b \) are in \( \mathbb{C} \) and \( a \neq 0 \). We show \( T \) is a transformation of \( \mathbb{C} \).
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That \( T \) is onto:\n\nLet \( w \) denote an arbitrary element of \( \mathbb{C} \). We must find a complex number \( z \) such that \( T\left( z\right) = w \). To find this \( z \), we solve \( w = {az} + b \) for \( z \). So, \( z = \frac{1}{a}\left( {w - b}\right) \) should work (since \( a \neq 0, z \) is a complex number). Indeed, \( T\left( {\frac{1}{a}\left( {w - b}\right) }\right) = a \cdot \left\lbrack {\frac{1}{a}\left( {w - b}\right) }\right\rbrack + b = w \). Thus, \( T \) is onto.\n\nThat \( T \) is one-to-one:\n\nTo show that \( T \) is 1-1 we show that if \( T\left( {z}_{1}\right) = T\left( {z}_{2}\right) \), then \( {z}_{1} = {z}_{2} \).\n\nIf \( {z}_{1} \) and \( {z}_{2} \) are two complex numbers such that \( T\left( {z}_{1}\right) = T\left( {z}_{2}\right) \), then \( a{z}_{1} + b = a{z}_{2} + b \). By subtracting \( b \) from both sides we see that \( a{z}_{1} = a{z}_{2} \), and then dividing both sides by \( a \) (which we can do since \( a \neq 0 \) ), we see that \( {z}_{1} = {z}_{2} \). Thus, \( T \) is \( 1 - 1 \) as well as onto, and we have proved \( T \) is a transformation.
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Yes
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Theorem 3.1.9 Suppose \( T \) is a general linear transformation.\na. \( T \) maps lines to lines.
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Proof. a. We prove that if \( L \) is a line in \( \mathbb{C} \) then so is \( T\left( L\right) \) . A line \( L \) is described by the line equation\n\n\[ \n{\alpha z} + \overline{\alpha z} + d = 0 \n\]\n\nfor some complex constant \( \alpha \) and real number \( d \) . Suppose \( T\left( z\right) = {az} + b \) is a general linear transformation (so \( a \neq 0 \) ). All the points in \( T\left( L\right) \) have the form \( w = {az} + b \) where \( z \) satisfies the preceeding line equation. It follows that \( z = \frac{1}{a}\left( {w - b}\right) \) and when we plug this into the line equation we see that\n\n\[ \n\alpha \frac{w - b}{a} + \bar{\alpha }\frac{\overline{w - b}}{\bar{a}} + d = 0 \n\]\n\nwhich can be rewritten\n\n\[ \n\frac{\alpha }{a}w + \frac{\bar{\alpha }}{\bar{a}}\bar{w} + d - \frac{\alpha b}{a} - \frac{\overline{\alpha b}}{\bar{a}} = 0. \n\]\n\nNow, for any complex number \( \beta \) the sum \( \beta + \bar{\beta } \) is a real number, so in the above expression, \( d - \left( {\frac{\alpha b}{a} + \frac{\overline{\alpha b}}{\bar{a}}}\right) \) is a real number. Therefore, all \( w \) in \( T\left( L\right) \) satisfy a line equation. That is, \( T\left( L\right) \) is a line.
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Yes
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Theorem 3.1.12 General linear transformations preserve angles.
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Proof. Suppose \( T\left( z\right) = {az} + b \) where \( a \neq 0 \) . Since the angle between curves is defined to be the angle between their tangent lines, it is sufficient to check that the angle between two lines is preserved. Suppose \( {L}_{1} \) and \( {L}_{2} \) intersect at \( {z}_{0} \), and \( {z}_{i} \) is on \( {L}_{i} \) for \( i = 1,2 \), as in the following diagram.\n\n\n\nThen,\n\n\[ \angle \left( {{L}_{1},{L}_{2}}\right) = \arg \left( \frac{{z}_{2} - {z}_{0}}{{z}_{1} - {z}_{0}}\right) .\n\]\n\nSince general linear transformations preserve lines, \( T\left( {L}_{i}\right) \) is the line through \( T\left( {z}_{0}\right) \) and \( T\left( {z}_{i}\right) \) for \( i = 1,2 \) and it follows that\n\n\[ \angle \left( {T\left( {L}_{1}\right), T\left( {L}_{2}\right) }\right) = \arg \left( \frac{T\left( {z}_{2}\right) - T\left( {z}_{0}\right) }{T\left( {z}_{1}\right) - T\left( {z}_{0}\right) }\right) \n\]\n\n\[ = \arg \left( \frac{a{z}_{2} + b - a{z}_{0} - b}{a{z}_{1} + b - a{z}_{0} - b}\right) \n\]\n\n\[ = \arg \left( \frac{{z}_{2} - {z}_{0}}{{z}_{1} - {z}_{0}}\right) \n\]\n\n\[ = \angle \left( {{L}_{1},{L}_{2}}\right) \text{.} \n\]\n\nThus \( T \) preserves angles.
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Yes
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Some Euclidean isometries of \( \mathbb{C} \).
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It is perhaps clear that translations, which move each point in the plane by the same amount in the same direction, ought to be isometries. Rotations are also isometries. In fact, the general linear transformation \( T\left( z\right) = {az} + b \) will be a Euclidean isometry so long as \( \left| a\right| = 1 \) :\n\n\[ \left| {T\left( z\right) - T\left( w\right) }\right| = \left| {{az} + b - \left( {{aw} + b}\right) }\right| \]\n\n\[ = \left| {a\left( {z - w}\right) }\right| \]\n\n\[ = \left| a\right| \left| {z - w}\right| \]\n\nSo, \( \left| {T\left( z\right) - T\left( w\right) }\right| = \left| {z - w}\right| \Leftrightarrow \left| a\right| = 1 \) . Translations and rotations about a point in \( \mathbb{C} \) are general linear transformations of this type, so they are also Euclidean isometries.
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Yes
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Theorem 3.1.17 A translation of \( \mathbb{C} \) is the composition of reflections about two parallel lines. A rotation of \( \mathbb{C} \) about a point \( {z}_{0} \) is the composition of reflections about two lines that intersect at \( {z}_{0} \) .
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Proof. Given the translation \( {T}_{b}\left( z\right) = z + b \) let \( {L}_{1} \) be the line through the origin that is perpendicular to segment \( {0b} \) as pictured in Figure 3.1.18(a). Let \( {L}_{2} \) be the line parallel to \( {L}_{1} \) through the midpoint of segment \( {0b} \) . Also let \( {r}_{i} \) denote reflection about line \( {L}_{i} \) for \( i = 1,2 \) .\n\nNow, given any \( z \) in \( \mathbb{C} \), let \( L \) be the line through \( z \) that is parallel to vector \( b \) (and hence perpendicular to \( {L}_{1} \) and \( {L}_{2} \) ). The image of \( z \) under the composition \( {r}_{2} \circ {r}_{1} \) will be on this line. To find the exact location, let \( {z}_{1} \) be the intersection of \( {L}_{1} \) and \( L \), and \( {z}_{2} \) the intersection of \( {L}_{2} \) and \( L \) ,(see the figure). To reflect \( z \) about \( {L}_{1} \) we need to translate it along \( L \) twice by the vector \( {z}_{1} - z \) . Thus \( {r}_{1}\left( z\right) = z + 2\left( {{z}_{1} - z}\right) = 2{z}_{1} - z. \)\n\nNext, to reflect \( {r}_{1}\left( z\right) \) about \( {L}_{2} \), we need to translate it along \( L \) twice by the vector \( {z}_{2} - {r}_{1}\left( z\right) \) . Thus,\n\n\[ \n{r}_{2}\left( {{r}_{1}\left( z\right) }\right) = {r}_{1}\left( z\right) + 2\left( {{z}_{2} - {r}_{1}\left( z\right) }\right) = 2{z}_{2} - {r}_{1}\left( z\right) = 2{z}_{2} - 2{z}_{1} + z. \n\]\n\nNotice from Figure 3.1.18(a) that \( {z}_{2} - {z}_{1} \) is equal to \( b/2 \) . Thus \( {r}_{2}\left( {{r}_{1}\left( z\right) }\right) = z + b \) is translation by \( b \) .
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Yes
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Inversion in the unit circle.
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The unit circle in \( \mathbb{C} \), denoted \( {\mathbb{S}}^{1} \), is the circle with center \( {z}_{0} = 0 \) and radius \( r = 1 \) . The equation for the point \( {z}^{ * } \) symmetric to a point \( z \neq 0 \) with repsect to \( {\mathbb{S}}^{1} \) thus reduces from \( \left| {z - {z}_{0}}\right| \cdot \left| {{z}^{ * } - {z}_{0}}\right| = {r}^{2} \) to \[ \left| z\right| \cdot \left| {z}^{ * }\right| = 1 \] Moreover, \( {z}^{ * } \) is just a scaled version of \( z \) since they are on the same ray through the origin. That is, \( {z}^{ * } = {kz} \) for some positive real number \( k \) . Plug this description of \( {z}^{ * } \) into the symmetry point equation to see \( \left| z\right| \cdot \left| {kz}\right| = 1 \), which implies \( k = 1/{\left| z\right| }^{2} \) . Thus, \( {z}^{ * } = \left( {1/{\left| z\right| }^{2}}\right) z \) . Moreover, \( {\left| z\right| }^{2} = z \cdot \bar{z} \), so inversion in the unit circle \( {\mathbb{S}}^{1} \) may be written as \[ {i}_{{\mathbb{S}}^{1}}\left( z\right) = 1/\bar{z} \]
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Yes
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Theorem 3.2.4 There exists a unique cline through any three distinct points in \( \mathbb{C} \) .
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Proof. Suppose \( u, v \), and \( w \) are distinct complex numbers. If \( v \) is on the line through \( u \) and \( w \) then this line is the unique cline through the three points. Otherwise, the three points do not lie on a single line, and we may build a circle through these three points as demonstrated in Figure 3.2.5. Construct the perpendicular bisector to segment \( {uv} \), and the perpendicular bisector to segment \( {vw} \) . These bisectors will intersect because the three points are not collinear. If we call the point of intersection \( {z}_{0} \), then the circle centered at \( {z}_{0} \) through \( w \) is the unique cline through the three points.
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Yes
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Theorem 3.2.6 Inversion in a circle maps clines to clines. In particular, if a cline goes through the center of the circle of inversion, its image will be a line; otherwise the image of a cline will be a circle.
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Proof. We prove the result in the case of inversion in the unit circle. The general proof will then follow, since any inversion is the composition of this particular inversion together with translations and dilations, which also preserve clines by Theorem 3.1.9.\n\nSuppose the cline \( C \) is described by the cline equation\n\n\[ \n{cz}\bar{z} + {\alpha z} + \overline{\alpha z} + d = 0, \]\n\nwhere \( c, d \in \mathbb{R},\alpha \in \mathbb{C} \) .\n\nWe want to show that the image of this cline under inversion in the unit circle, \( {i}_{{\mathbb{S}}^{1}}\left( C\right) \), is also a cline. Well, \( {i}_{{\mathbb{S}}^{1}}\left( C\right) \) consists of all points \( w = 1/\bar{z} \), where \( z \) satisfies the cline equation for \( C \) . We show that all such \( w \) live on a cline.\n\nIf \( z \neq 0 \) then we may multiply each side of the cline equation by \( 1/\left( {z \cdot \bar{z}}\right) \) to\n\nobtain\n\[ \nc + \alpha \frac{1}{\bar{z}} + \bar{\alpha }\frac{1}{z} + d\frac{1}{z}\frac{1}{\bar{z}} = 0. \]\n\nBut since \( w = 1/\bar{z} \) and \( \bar{w} = 1/z \), this equation reduces to\n\n\[ \nc + \alpha \cdot w + \bar{\alpha } \cdot \bar{w} + {dw}\bar{w} = 0, \]\nor\n\n\[ \n{dw}\bar{w} + \alpha \cdot w + \bar{\alpha } \cdot \bar{w} + c = 0. \]\n\nThus, the image points \( w \) form a cline equation. If \( d = 0 \) then the original cline \( C \) passed through the origin, and the image cline is a line. If \( d \neq 0 \) then \( C \) did not pass through the origin, and the image cline is a circle. (In fact, we must also check that \( {\left| \alpha \right| }^{2} > {dc} \) . This is the case because the original cline equation ensures \( {\left| \alpha \right| }^{2} > {cd} \) .)
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Yes
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Lemma 3.2.7 Suppose \( C \) is the circle with radius \( r \) centered at \( o \), and \( p \) is a point outside \( C \) . Let \( s = \left| {p - o}\right| \) . If a line through \( p \) intersects \( C \) at points \( m \) and \( n \), then\n\n\[ \left| {p - m}\right| \cdot \left| {p - n}\right| = {s}^{2} - {r}^{2}. \]
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Proof. Suppose the line through \( p \) does not pass through the center of \( C \), as in the diagram below. Let \( q \) be the midpoint of segment \( {mn} \), and let \( d = \left| {q - o}\right| \) as in the diagram. Note also that the line through \( q \) and \( o \) is the perpendicular bisector of segment \( {mn} \) . In particular, \( \left| {m - q}\right| = \left| {q - n}\right| \) .\n\n\n\nThe Pythagorean theorem applied to \( {\Delta pqo} \) gives\n\n\[ {\left| p - q\right| }^{2} + {d}^{2} = {s}^{2} \]\n\n(1)\n\nand the Pythagorean theorem applied to \( {\Delta nqo} \) gives\n\n\[ {\left| q - n\right| }^{2} + {d}^{2} = {r}^{2} \]\n\n(2)\n\nBy subtracting equation (2) from (1), we have\n\n\[ {\left| p - q\right| }^{2} - {\left| q - n\right| }^{2} = {s}^{2} - {r}^{2}, \]\n\nwhich factors as\n\n\[ \left( {\left| {p - q}\right| - \left| {q - n}\right| }\right) \left( {\left| {p - q}\right| + \left| {q - n}\right| }\right) = {s}^{2} - {r}^{2}. \]\n\nSince \( \left| {p - q}\right| - \left| {q - n}\right| = \left| {p - m}\right| \) and \( \left| {p - q}\right| + \left| {q - n}\right| = \left| {p - n}\right| \), the result follows.\n\nThe case that the line through \( p \) goes through the center of \( C \) is left as an exercise.
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No
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Theorem 3.2.8 Suppose \( C \) is a circle in \( \mathbb{C} \) centered at \( {z}_{0} \), and \( z \neq {z}_{0} \) is not on \( C \) . A cline through \( z \) is orthogonal to \( C \) if and only if it goes through \( {z}^{ * } \), the point symmetric to \( z \) with respect to \( C \) .
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Proof. Assume \( C \) is the circle of radius \( r \) centered at \( {z}_{0} \), and \( D \) is a cline through a point \( z \neq {z}_{0} \) not on \( C \) . Let \( {z}^{ * } \) denote the point symmetric to \( z \) with respect to \( C \) .\n\nFirst, suppose \( D \) is a line through \( z \) . A line through \( z \) passes through \( {z}^{ * } \) if and only if it passes through the center of \( C \), which is true if and only if the line is orthogonal to \( C \) . Thus, the line \( D \) through \( z \) contains \( {z}^{ * } \) if and only if it is orthogonal to \( C \), and the theorem is proved in this case.\n\nNow assume \( D \) is a circle through \( z \) . Let \( o \) and \( k \) denote the center and radius of \( D \), respectively. Set \( s = \left| {{z}_{o} - o}\right| \), and let \( t \) denote a point of intersection of \( C \) and \( D \) as pictured below.\n\n\n\nWe must argue that \( C \) and \( D \) are orthogonal if and only if \( {z}^{ * } \) is on \( D \) . Now, \( C \) and \( D \) are orthogonal if and only if \( \angle {ot}{z}_{0} \) is right, which is the case if and only if \( {r}^{2} = {s}^{2} - {k}^{2} \) by the Pythagorean theorem. Applying Lemma 3.2.7 to the point \( {z}_{0} \) (which is outside \( D \) ) and the line through \( {z}_{0} \) and \( z \), we see that\n\n\[ \left| {{z}_{0} - z}\right| \cdot \left| {{z}_{0} - w}\right| = {s}^{2} - {k}^{2} \]\n\n(1)\n\nwhere \( w \) is the second point of intersection of the line with circle \( D \) .\n\nNote also that as symmetric points, \( z \) and \( {z}^{ * } \) satisfy the equation\n\n\[ \left| {{z}_{0} - z}\right| \cdot \left| {{z}_{0} - {z}^{ * }}\right| = {r}^{2}. \]\n\n(2)\n\nThus, if we assume \( {z}^{ * } \) is on \( D \), then it must be equal to the point \( w \), in which case equations (1) and (2) above tell us \( {s}^{2} - {k}^{2} = {r}^{2} \) . It follows that \( D \) is orthogonal to \( C \) . Conversely, if \( D \) is orthogonal to \( C \), then \( {s}^{2} - {k}^{2} = {r}^{2} \), so \( \left| {{z}_{0} - w}\right| = \left| {{z}_{0} - {z}^{ * }}\right| \) . Since \( {z}^{ * } \) and \( w \) are both on the ray \( \overrightarrow{{z}_{0}z} \) it must be that \( {z}^{ * } = w \) . In other words, \( {z}^{ * } \) is on \( D \) .
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Yes
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Theorem 3.2.12 Inversion preserves symmetry points. Let \( {i}_{C} \) denote inversion in a cline \( C \) . If \( p \) and \( q \) are symmetric with respect to a cline \( D \), then \( {i}_{C}\left( p\right) \) and \( {i}_{C}\left( q\right) \) are symmetric with respect to the cline \( {i}_{C}\left( D\right) \) .
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Proof. Assume \( C \) is the cline of inversion, and assume \( p \) and \( q \) are symmetric with respect to a cline \( D \) as in Figure 3.2.13 (where \( C \) and \( D \) are represented as circles).\n\n\n\nFigure 3.2.13 Inversion preserves symmetry points: If \( p \) and \( q \) are symmetric with respect to \( D \) and we invert about cline \( C \) then the image points are symmetric with respect to the image of \( D \) .\n\nWe may construct two clines \( E \) and \( F \) that go through \( p \) and \( q \) . In the figure, cline \( E \) is a circle and cline \( F \) is a line. These clines intersect \( D \) at right angles (Theorem 3.2.8). Since inversion preserves clines and angle magnitudes, we know that \( {E}^{ * } = {i}_{C}\left( E\right) \) and \( {F}^{ * } = {i}_{C}\left( F\right) \) are clines intersecting the cline \( {D}^{ * } = {i}_{C}\left( D\right) \) at right angles. Both \( {E}^{ * } \) and \( {F}^{ * } \) contain \( {p}^{ * } = {i}_{C}\left( p\right) \), so they both contain the point symmetric to \( {p}^{ * } \) with respect to \( {D}^{ * } \) (Theorem 3.2.8), but the only other point common to both \( {E}^{ * } \) and \( {F}^{ * } \) is \( {q}^{ * } = {i}_{C}\left( q\right) \) . Thus, \( {p}^{ * } \) and \( {q}^{ * } \) are symmetric with respect to \( {D}^{ * } \) .
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Yes
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Theorem 3.2.14 Apollonian Circles Theorem. Let \( p, q \) be distinct points in \( \mathbb{C} \), and \( k > 0 \) a positive real number. Let \( D \) consist of all points \( z \) in \( \mathbb{C} \) such that \( \left| {z - p}\right| = k\left| {z - q}\right| \) . Then \( D \) is a cline.
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Proof. If \( k = 1 \), the set \( D \) is a Euclidean line, according to Theorem 2.4.2, so we assume \( k \neq 1 \) . Let \( C \) be the circle centered at \( p \) with radius 1 . Suppose \( z \) is an arbitrary point in the set \( D \) . Inverting about \( C \), let \( {z}^{ * } = {i}_{C}\left( z\right) \) and \( {q}^{ * } = {i}_{C}\left( q\right) \) as in the following diagram.\n\n\n\nObserve first that \( {\Delta p}{z}^{ * }{q}^{ * } \) and \( {\Delta pqz} \) are similar.\n\nIndeed, \( \left| {p - z}\right| \cdot \left| {p - {z}^{ * }}\right| = 1 = \left| {p - q}\right| \cdot \left| {p - {q}^{ * }}\right| \) by the definition of the inversion transformation, so we have equal side-length ratios\n\n\[ \frac{\left| p - {z}^{ * }\right| }{\left| p - q\right| } = \frac{\left| p - {q}^{ * }\right| }{\left| p - z\right| } \]\n\nand the included angles are equal, \( \angle {q}^{ * }p{z}^{ * } = \angle {qpz} \) .\n\nIt follows that\n\n\[ \frac{\left| z - q\right| }{\left| p - q\right| } = \frac{\left| {z}^{ * } - {q}^{ * }\right| }{\left| {z}^{ * } - p\right| } \]\n\nfrom which we derive\n\n\[ \left| {{z}^{ * } - {q}^{ * }}\right| = \left| {{z}^{ * } - p}\right| \cdot \frac{\left| z - q\right| }{\left| p - q\right| } \]\n\n\[ = \left\lbrack {\left| {{z}^{ * } - p}\right| \cdot \left| {z - p}\right| }\right\rbrack \cdot \frac{\left| z - q\right| }{\left| z - p\right| } \cdot \frac{1}{\left| p - q\right| } \]\n\n\[ = 1 \cdot \frac{1}{k} \cdot \frac{1}{\left| p - q\right| } \]\n\nThus, the set \( D \) of all points \( z \) satisfying \( \left| {z - p}\right| = k\left| {z - q}\right| \) has image \( {i}_{C}\left( D\right) \) under this inversion consisting of all points \( {z}^{ * } \) on a circle centered at \( {q}^{ * } \) with radius \( {\left( k\left| p - q\right| \right) }^{-1} \) . Since inversion preserves clines and \( p \) is not on \( {i}_{C}\left( D\right) \), it follows that \( D \) itself is a circle.
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Yes
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Theorem 3.2.16 Suppose we have two clines that do not intersect, and at least one of them is a circle. Then there exist two points, \( p \) and \( q \), that are symmetric with respect to both clines.
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Proof. First, assume one cline is a line \( L \), and the other is a circle \( C \) centered at the point \( {z}_{0} \) as pictured in Figure 3.2.15. Let \( {L}_{1} \) be the line through \( {z}_{0} \) that is perpendicular to \( L \), and let \( {z}_{1} \) be the point of intersection of \( L \) and \( {L}_{1} \) . Next, construct the circle \( {C}_{1} \) having the diameter \( {z}_{0}{z}_{1} \) . Circle \( {C}_{1} \) intersects circle \( C \) at some point, which we call \( t \) . Notice that \( \angle {z}_{0}t{z}_{1} \) is right, and so the circle \( {C}_{2} \) centered at \( {z}_{1} \) through \( t \) is orthogonal to \( C \) . Furthermore, the center of \( {C}_{2} \) , \( {z}_{1} \), lies on line \( L \), so \( {C}_{2} \) is orthogonal to \( L \) . Let \( p \) and \( q \) be the two points at which \( {C}_{2} \) intersects \( {L}_{1} \) . By construction, and by using Theorem 3.2.8, \( p \) and \( q \) are symmetric to both \( C \) and \( L \) .
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Yes
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Theorem 3.3.1 Any general linear transformation extended to the domain \( {\mathbb{C}}^{ + } \) fixes \( \infty \) .
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Proof. If \( T\left( z\right) = {az} + b \) where \( a \) and \( b \) are complex constants with \( a \neq 0 \) , then by limit methods from calculus, as \( \left| {z}_{n}\right| \rightarrow \infty ,\left| {a{z}_{n} + b}\right| \rightarrow \infty \) as well. Thus, \( T\left( \infty \right) = \infty \) .
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Yes
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Theorem 3.3.5 Stereographic projection preserves angles. That is, if two curves on the surface of the sphere intersect at angle \( \theta \), then their image curves in \( {\mathbb{C}}^{ + } \) also intersect at angle \( \theta \) .
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Thus, if two curves in \( {\mathbb{C}}^{ + } \) intersect at \( \infty \) we may define the angle at which they intersect to equal the angle at which their pre-image curves under stereographic projection intersect. The angle at which two parallel lines intersect at \( \infty \) is 0 . Furthermore, if two lines intersect at a finite point \( p \) as well as at \( \infty \), the angle at which they intersect at \( \infty \) equals the negative of the angle at which they intersect at \( p \) . As a consequence, we may say that inversion about a circle preserves angle magnitudes at all points in \( {\mathbb{C}}^{ + } \) .
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No
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Theorem 3.4.1 The function\n\n\[ T\left( z\right) = \frac{{az} + b}{{cz} + d} \]\n\nis a transformation of \( {\mathbb{C}}^{ + } \) if and only if \( {ad} - {bc} \neq 0 \) .
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Proof. First, suppose \( T\left( z\right) = \left( {{az} + b}\right) /\left( {{cz} + d}\right) \) and \( {ad} - {bc} \neq 0 \) . We must show that \( T \) is a transformation. To show \( T \) is one-to-one, assume \( T\left( {z}_{1}\right) = T\left( {z}_{2}\right) \) .\n\nThen\n\[ \frac{a{z}_{1} + b}{c{z}_{1} + d} = \frac{a{z}_{2} + b}{c{z}_{2} + d} \]\n\nCross multiply this expression and simplify to obtain\n\n\[ \left( {{ad} - {bc}}\right) {z}_{1} = \left( {{ad} - {bc}}\right) {z}_{2}. \]\nSince \( {ad} - {bc} \neq 0 \) we may divide this term out of the expression to see \( {z}_{1} = {z}_{2} \), so \( T \) is 1-1 and it remains to show that \( T \) is onto.\n\nSuppose \( w \) in \( {\mathbb{C}}^{ + } \) is given. We must find \( z \in {\mathbb{C}}^{ + } \) such that \( T\left( z\right) = w \) . If \( w = \infty \), then \( z = - d/c \) (which is \( \infty \) if \( c = 0 \) ) does the trick, so assume \( w \neq \infty \) . To find \( z \) such that \( T\left( z\right) = w \) we solve the equation\n\n\[ \frac{{az} + b}{{cz} + d} = w \]\n\nfor \( z \), which is possible so long as \( a \) and \( c \) are not both 0 (causing the \( z \) terms to vanish). Since \( {ad} - {bc} \neq 0 \), we can be assured that this is the case, and solving\n\nfor \( z \) we obtain\n\[ z = \frac{-{dw} + b}{{cw} - a}. \]\n\nThus \( T \) is onto, and \( T \) is a transformation.\n\nTo prove the converse we show the contrapositive. We suppose \( {ad} - {bc} = 0 \) and show \( T\left( z\right) = \left( {{az} + b}\right) /\left( {{cz} + d}\right) \) is not a transformation by tackling two cases.\n\nCase 1: \( {ad} = 0 \) . In this case, \( {bc} = 0 \) as well, so \( a \) or \( d \) is zero, and \( b \) or \( c \) is zero. In all four scenarios, one can check immediately that \( T \) is not a transformation of \( {\mathbb{C}}^{ + } \) . For instance, if \( a = c = 0 \) then \( T\left( z\right) = b/d \) is neither 1-1 nor onto \( {\mathbb{C}}^{ + } \) .\n\nCase 2: \( {ad} \neq 0 \) . In this case, all four constants are non-zero, and \( a/c = b/d \) . Since \( T\left( 0\right) = b/d \) and \( T\left( \infty \right) = a/c, T \) is not 1-1, and hence not a transformation of \( {\mathbb{C}}^{ + } \) .
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Yes
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Theorem 3.4.2 The Möbius transformation\n\n\[ T\left( z\right) = \frac{{az} + b}{{cz} + d} \] has the inverse transformation
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\[ {T}^{-1}\left( z\right) = \frac{-{dz} + b}{{cz} - a}. \]
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Yes
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Theorem 3.4.6 Any Möbius transformation \( T : {\mathbb{C}}^{ + } \rightarrow {\mathbb{C}}^{ + } \) fixes 1,2, or all points of \( {\mathbb{C}}^{ + } \) .
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Proof. To find fixed points of \( T\left( z\right) = \left( {{az} + b}\right) /\left( {{cz} + d}\right) \) we want to solve\n\n\[ \frac{{az} + b}{{cz} + d} = z \]\n\nfor \( z \), which gives the quadratic equation\n\n\[ c{z}^{2} + \left( {d - a}\right) z - b = 0. \]\n\n(1)\n\nIf \( c \neq 0 \) then, as discussed in Example 2.4.3, equation (1) must have 1 or 2 solutions, and there are 1 or 2 fixed points in this case.\n\nIf \( c = 0 \) and \( a \neq d \), then the transformation has the form \( T\left( z\right) = \left( {{az} + b}\right) /d \) , which fixes \( \infty \) . From equation (1), \( z = b/\left( {d - a}\right) \neq \infty \) is a fixed point as well. So we have 2 fixed points in this case.\n\nIf \( c = 0 \) and \( a = d \), then equation (1) reduces to \( 0 = - b \), so \( b = 0 \) too, and the transformation is the identity transformation \( T\left( z\right) = \left( {{az} + 0}\right) /\left( {{0z} + a}\right) = z \) . This transformation fixes every point.
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Yes
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Theorem 3.4.12 Invariance of Cross Ratio. Suppose \( {z}_{0},{z}_{1},{z}_{2} \), and \( {z}_{3} \) are four distinct points in \( {\mathbb{C}}^{ + } \), and \( T \) is any Möbius transformation. Then\n\n\[ \left( {{z}_{0},{z}_{1};{z}_{2},{z}_{3}}\right) = \left( {T\left( {z}_{0}\right), T\left( {z}_{1}\right) ;T\left( {z}_{2}\right), T\left( {z}_{3}\right) }\right) . \]\n
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Proof. Let \( T \) be an arbitrary Möbius transformation, and define \( S\left( z\right) = \) \( \left( {z,{z}_{1};{z}_{2},{z}_{3}}\right) \), which sends \( {z}_{1} \mapsto 1,{z}_{2} \mapsto 0 \), and \( {z}_{3} \mapsto \infty \) . Notice, the composition \( S \circ {T}^{-1} \) is a Möbius transformation that sends \( T\left( {z}_{1}\right) \mapsto 1, T\left( {z}_{2}\right) \mapsto 0 \), and \( T\left( {z}_{3}\right) \mapsto \infty \) . So this map can be expressed as a cross ratio:\n\n\[ S \circ {T}^{-1}\left( z\right) = \left( {z, T\left( {z}_{1}\right) ;T\left( {z}_{2}\right), T\left( {z}_{3}\right) }\right) . \]\n\nPlugging \( T\left( {z}_{0}\right) \) into this transformation, we see\n\n\[ S \circ {T}^{-1}\left( {T\left( {z}_{0}\right) }\right) = \left( {T\left( {z}_{0}\right), T\left( {z}_{1}\right) ;T\left( {z}_{2}\right), T\left( {z}_{3}\right) }\right) . \]\n\nOn the other hand, \( S \circ {T}^{-1}\left( {T\left( {z}_{0}\right) }\right) = S\left( {z}_{0}\right) \), which equals \( \left( {{z}_{0},{z}_{1};{z}_{2},{z}_{3}}\right) \) . So we have proved that\n\n\[ \left( {{z}_{0},{z}_{1};{z}_{2},{z}_{3}}\right) = \left( {T\left( {z}_{0}\right), T\left( {z}_{1}\right) ;T\left( {z}_{2}\right), T\left( {z}_{3}\right) }\right) . \]
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Yes
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Theorem 3.4.15 Given any two clines \( {C}_{1} \) and \( {C}_{2} \), there exists a Möbius transformation \( T \) that maps \( {C}_{1} \) onto \( {C}_{2} \) . That is, \( T\left( {C}_{1}\right) = {C}_{2} \) .
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Proof. Let \( {p}_{1} \) be a point on \( {C}_{1} \) and \( {q}_{1} \) and \( {q}_{1}^{ * } \) symmetric with respect to \( {C}_{1} \) . Similarly, let \( {p}_{2} \) be a point on \( {C}_{2} \) and \( {q}_{2} \) and \( {q}_{2}^{ * } \) be symmetric with respect to \( {C}_{2} \) . Build the Möbius transformation that sends \( {p}_{1} \mapsto {p}_{2},{q}_{1} \mapsto {q}_{2} \) and \( {q}_{1}^{ * } \mapsto {q}_{2}^{ * } \) . Then \( T\left( {C}_{1}\right) = {C}_{2} \
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Yes
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Lemma 3.5.7 Suppose \( T \) is a Möbius transformation that fixes distinct finite points \( p \) and \( q \), sends \( {z}_{\infty } \) to \( \infty \), and sends \( \infty \) to \( {w}_{\infty } \). Then \( p + q = {z}_{\infty } + {w}_{\infty } \).
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Proof. Suppose \( T \) satisfies the conditions of the lemma. Then \( T \) has normal form \[ \frac{z - p}{z - q} = \lambda \frac{T\left( z\right) - p}{T\left( z\right) - q} \] where \( \lambda = r{e}^{i\theta } \). Plug \( z = {z}_{\infty } \) into the normal form to see \[ \frac{{z}_{\infty } - p}{{z}_{\infty } - q} = \lambda \cdot 1 \] Plug \( z = \infty \) into the normal form to see \[ 1 = \lambda \frac{{w}_{\infty } - p}{{w}_{\infty } - q} \] Next solve each equation for \( \lambda \), set them equal, cross multiply, and simplify as follows to get the result: \[ \frac{{z}_{\infty } - p}{{z}_{\infty } - q} = \frac{{w}_{\infty } - q}{{w}_{\infty } - p} \] \[ \left( {{z}_{\infty } - p}\right) \left( {{w}_{\infty } - p}\right) = \left( {{w}_{\infty } - q}\right) \left( {{z}_{\infty } - q}\right) \] \[ {p}^{2} - p{w}_{\infty } - p{z}_{\infty } = {q}^{2} - q{w}_{\infty } - q{z}_{\infty } \] \[ {p}^{2} - {q}^{2} = p\left( {{z}_{\infty } + {w}_{\infty }}\right) - q\left( {{z}_{\infty } + {w}_{\infty }}\right) \] \[ \left( {p - q}\right) \left( {p + q}\right) = \left( {p - q}\right) \left( {{z}_{\infty } + {w}_{\infty }}\right) \] \[ p + q = {z}_{\infty } + {w}_{\infty } \] This completes the proof.
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Yes
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Theorem 3.5.8 If \( T \) is a Möbius transformation that fixes two distinct, finite points \( p \) and \( q \), sends \( {z}_{\infty } \) to \( \infty \), and sends \( \infty \) to \( {w}_{\infty } \), then\n\n\[ T\left( z\right) = \frac{{w}_{\infty }z - {pq}}{z - {z}_{\infty }}. \]\n
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Proof. In the proof of Lemma 3.5.7, we found that the constant \( \lambda \) in the normal form of \( T \) is\n\n\[ \lambda = \frac{{z}_{\infty } - p}{{z}_{\infty } - q} \]\n\nIt follows that \( T \) has the normal form\n\n\[ \frac{z - p}{z - q} = \left( \frac{{z}_{\infty } - p}{{z}_{\infty } - q}\right) \cdot \frac{T\left( z\right) - p}{T\left( z\right) - q}. \]\n\nSolve this expression for \( T\left( z\right) \) and reduce it using the fact that \( p + q = {z}_{\infty } + {w}_{\infty } \) to get the expression for \( T \) that appears in the statement of the theorem. The details are left to the reader.
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No
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Example 3.5.9 Fix -1 and 1 and send \( i \mapsto \infty \) .
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By Lemma 3.5.7, the Möbius transformation sends \( \infty \) to \( - i \), so by Theorem 3.5.8,\n\n\[ T\left( z\right) = \frac{-{iz} - \left( 1\right) \left( {-1}\right) }{z - i} = \frac{-{iz} + 1}{z - i}. \]\n
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Yes
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Consider \( T\left( z\right) = \left( {{7z} - {12}}\right) /\left( {{3z} - 5}\right) \). To find its normal form we start by finding its fixed points.
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\[ z = T\left( z\right) \] \[ z\left( {{3z} - 5}\right) = {7z} - {12} \] \[ 3{z}^{2} - {12z} + {12} = 0 \] \[ {z}^{2} - {4z} + 4 = 0 \] \[ {\left( z - 2\right) }^{2} = 0 \] \[ z = 2\text{.} \] So \( T \) is parabolic and has normal form \[ \frac{1}{T\left( z\right) - 2} = \frac{1}{z - 2} + d \] To find \( d \) plug in the image of another point. Using the original description of the map, we know \( T\left( 0\right) = {2.4} \), so \[ \frac{1}{0.4} = \frac{1}{-2} + d \] so that \( d = 3 \). The normal form is then \[ \frac{1}{T\left( z\right) - 2} = \frac{1}{z - 2} + 3 \]
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Yes
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Show that \( \mathcal{T} \) is a group of transformations.
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Solution. To verify \( \mathcal{T} \) forms a group, we must check the three properties.\n\n1. \( \mathcal{T} \) contains the identity: Since \( 0 \in \mathbb{C},\mathcal{T} \) contains \( {T}_{0}\left( z\right) = z + 0 = z \) , which is the identity transformation of \( \mathbb{C} \).\n\n2. \( \mathcal{T} \) has closure: Suppose \( {T}_{b} \) and \( {T}_{c} \) are in \( \mathcal{T} \) . Then \( {T}_{b} \circ {T}_{c}\left( z\right) = \) \( {T}_{b}\left( {z + c}\right) = \left( {z + c}\right) + b = z + \left( {b + c}\right) \) . But this map is exactly the translation \( {T}_{b + c} \), which is in \( \mathcal{T} \) since \( b + c \in \mathbb{C} \) . Thus, the composition of two translations is again a translation. Notationally, we have shown that \( {T}_{b} \circ {T}_{c}\left( z\right) = {T}_{b + c}\left( z\right) \).\n\n3. \( \mathcal{T} \) contains inverses: Suppose \( {T}_{b} \) is in \( \mathcal{T} \), and consider \( {T}_{-b} \), which is in \( \mathcal{T} \) since \( - b \in \mathbb{C} \) . Note that \( {T}_{b} \circ {T}_{-b}\left( z\right) = {T}_{b}\left( {z - b}\right) = \left( {z - b}\right) + b = z \) , and \( {T}_{-b} \circ {T}_{b}\left( z\right) = {T}_{-b}\left( {z + b}\right) = \left( {z + b}\right) - b = z \) . Thus, \( {T}_{-b} \) is the inverse of \( {T}_{b} \), and this inverse is in \( \mathcal{T} \) . Notationally, \( {T}_{b}^{-1} = {T}_{-b} \) : the inverse of translation by \( b \) is translation back by \( - b \) .
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Yes
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Theorem 4.1.10 An invariant set \( \mathcal{D} \) of figures in a geometry \( \left( {S, G}\right) \) is minimally invariant if and only if any two figures in \( \mathcal{D} \) are congruent.
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Proof. First assume \( \mathcal{D} \) is a minimally invariant set in the geometry \( \left( {S, G}\right) \) , and suppose \( A \) and \( B \) are arbitrary figures in \( \mathcal{D} \) . We must show that \( A \cong B \) .\n\nWe begin by constructing a new set of figures, the one consisting of \( A \) and all transformations of \( A \) . In particular, define\n\n\[ \mathcal{A} = \{ T\left( A\right) \mid T \in G\} .\n\]\n\nNotice that for any \( T \in G, T\left( A\right) \) is in the set \( \mathcal{D} \) since \( \mathcal{D} \) is invariant. This means that \( \mathcal{A} \) is a subset of \( \mathcal{D} \).\n\nFurthermore, \( \mathcal{A} \) itself is an invariant set, thanks to the group nature of \( G \) . In particular, if \( C \) is any member of \( \mathcal{A} \), then \( C = {T}_{0}\left( A\right) \) for some particular \( {T}_{0} \) in \( G \) . Thus, applying any transformation \( T \) to \( C \) ,\n\n\[ T\left( C\right) = T\left( {{T}_{0}\left( A\right) }\right) = T \circ {T}_{0}\left( A\right) \]\n\nand since \( T \circ {T}_{0} \) is again a transformation in \( G, T \circ {T}_{0}\left( A\right) \) lives in \( \mathcal{A} \).\n\nSo we’ve established two facts: (1) \( \mathcal{A} \) is a subset of \( \mathcal{D} \), and (2) \( \mathcal{A} \) is an invariant set. Since \( \mathcal{D} \) is a minimally invariant set it follows by definition that \( \mathcal{A} = \mathcal{D} \). This means that the given set \( B \), which is in \( \mathcal{D} \), is also in \( \mathcal{A} \). That is, \( A \cong B \).\n\nThe proof of the other direction is left as an exercise for the reader.
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No
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Lemma 4.1.16 For any points \( z, w, v \) in \( \mathbb{C} \) , \n\n\[ \left| {z - w}\right| \leq \left| {z - v}\right| + \left| {v - w}\right| \]
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Proof. If \( v = w \) then the the result holds, so we assume \( v \neq w \) . Since \( d \) is invariant under Euclidean transformations, we may assume that \( v = 0 \) and \( w = r > 0 \) is a point on the positive real axis. (Translate the plane by \( - v \) to send \( v \) to 0, and then rotate about 0 until the image of \( w \) under the translation lands on the positive real axis.) Thus, it's enough to show that for any complex number \( z \) and any positive real number \( r \) , \n\n\[ \left| {z - r}\right| \leq \left| z\right| + r \] \n\nNotice that \n\n\[ \left| {z - r}\right| \leq \left| z\right| + r \Leftrightarrow {\left| z - r\right| }^{2} \leq {\left( \left| z\right| + r\right) }^{2} \] \n\n\[ \Leftrightarrow \left( {z - r}\right) \left( {\bar{z} - r}\right) \leq {\left| z\right| }^{2} + {2r}\left| z\right| + {r}^{2} \] \n\n\[ \Leftrightarrow {\left| z\right| }^{2} - r\left( {z + \bar{z}}\right) + {r}^{2} \leq {\left| z\right| }^{2} + {2r}\left| z\right| + {r}^{2} \] \n\n\[ \Leftrightarrow - \left( {z + \bar{z}}\right) \leq 2\left| z\right| \] \n\n\[ \Leftrightarrow - 2\operatorname{Re}\left( z\right) \leq 2\left| z\right| \] \n\n\[ \Leftrightarrow - \operatorname{Re}\left( z\right) \leq \left| z\right| \] \n\nBy letting \( z = a + {bi} \), we may restate the last inequality as \n\n\[ - a \leq \sqrt{{a}^{2} + {b}^{2}} \] \n\nwhich is true since \( - a \leq \left| a\right| = \sqrt{{a}^{2}} \leq \sqrt{{a}^{2} + {b}^{2}} \) .
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Yes
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Theorem 5.1.3 Any Möbius transformation in \( \mathcal{H} \) is the composition of two reflections of the hyperbolic plane.
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Proof. Suppose \( T \) is a Möbius transformation that sends \( \mathbb{D} \) to itself. This means some point in \( \mathbb{D} \), say \( {z}_{0} \), gets sent to the origin,0 . Let \( {z}_{0}^{ * } \) be the point symmetric to \( {z}_{0} \) with respect to \( {\mathbb{S}}_{\infty }^{1} \) . Since \( T \) sends the unit circle to itself, and Möbius transformations preserve symmetry points, it follows that \( T \) sends \( {z}_{0}^{ * } \) to \( \infty \) . Furthermore, some point \( {z}_{1} \) on \( {\mathbb{S}}_{\infty }^{1} \) gets sent to the point 1 .\n\nIf \( {z}_{0} = 0 \), then \( {z}_{0}^{ * } = \infty \), and \( T \) fixes 0 and \( \infty \) . Then by Example 3.5.3, \( T\left( z\right) = r{e}^{i\theta }z \) is a dilation followed by a rotation. However, since \( T \) also sends \( \mathbb{D} \) onto \( \mathbb{D} \), the dilation factor must be \( r = 1 \) . So \( T \) is simply a rotation about the origin, which is the composition of two hyperbolic reflections about Euclidean lines through the origin.\n\nNow assume \( {z}_{0} \neq 0 \) . In this case, by using \( {z}_{0},{z}_{0}^{ * } \), and \( {z}_{1} \) as anchors, we may achieve \( T \) via two hyperbolic reflections, as follows:\n\nFirst, invert about a circle \( C \) orthogonal to \( {\mathbb{S}}_{\infty }^{1} \) that sends \( {z}_{0} \) to the origin.\n\nSuch a circle does indeed exist, and we will construct it now. As in Figure 5.1.4, draw circle \( {C}_{1} \) with diameter \( 0{z}_{0}^{ * } \) . Let \( p \) be a point of intersection of \( {C}_{1} \) and the unit circle \( {\mathbb{S}}_{\infty }^{1} \) . Construct the circle \( C \) through \( p \) centered at \( {z}_{0}^{ * } \) . Since \( \angle {0p}{z}_{0}^{ * } \) is right, \( {\mathbb{S}}_{\infty }^{1} \) is orthogonal to \( C \), so inversion about \( C \) sends \( {\mathbb{S}}_{\infty }^{1} \) to itself. Furthermore, since \( {z}_{0}^{ * } \) gets sent to \( \infty \) and symmetry points must be preserved, inversion in \( C \) sends \( {z}_{0} \) to 0 .\n\nThus, the first inversion takes \( {z}_{0} \) to 0 and \( {z}_{0}^{ * } to \( \infty \) . To build \( T \) we must also send \( {z}_{1} \) to 1 . Note that inversion in circle \( C \) will have sent \( {z}_{1} \) to some point \( {z}_{1}^{\prime } \) on the unit circle (since \( {\mathbb{S}}_{\infty }^{1} \) is sent to itself). Now reflect across the line through the origin that bisects angle \( \angle {10}{z}_{1}^{\prime } \) . This sends \( {z}_{1}^{\prime } \) to 1, sends \( \mathbb{D} \) to \( \mathbb{D} \), and leaves 0 and \( \infty \) fixed. Composing these two inversions yields a Möbius transformation that sends \( {z}_{0} \) to \( 0,{z}_{0}^{ * } \) to \( \infty \) and \( {z}_{1} \) to 1 . Since a Möbius transformation is uniquely determined by the image of three points, this Möbius transformation is \( T \) .
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