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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 r... | (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)... | No |
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, - ... | How long does it take Mercutio to reach the \( y \) -axis? | No |
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. | 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 \ | No |
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? | 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(gr... | No |
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? | 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 qu... | Yes |
Example 6.3.1. Sketch the graph of the multipart function\n\n\[ g\\left( x\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{if }x \\leq - 1 \\\\ 1 + \\sqrt{1 - {x}^{2}} & \\text{if } - 1 \\leq x \\leq 1 \\\\ 1 & \\text{if }x \\geq 1 \\end{array}\\right. \] | 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\\... | Yes |
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 \( s = h\left( t\right) \) . | 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? | No |
Describe a sequence of geometric operations leading from the graph of \( y = {x}^{2} \) to the graph of \( y = f\left( x\right) = - 3{\left( x - 1\right) }^{2} + 2 \) . | 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... | Yes |
Find the vertex form of the quadratic function \( y = - 3{x}^{2} + {6x} - 1 \) . | 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 a... | Yes |
Describe the relationship between the graphs of \( y = {x}^{2} \) and \( y = f\left( x\right) = - 4{x}^{2} + {5x} + 2 \) . | 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}\righ... | Yes |
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? | 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) ... | Yes |
Discuss the graph of the quadratic function \( y = f\\left( x\\right) = \) \( - 2{x}^{2} + {11x} - 4 \) . | 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... | Yes |
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? | 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 w... | Yes |
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 verti... | 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 prob... | No |
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? | 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 ... | Yes |
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... | 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; u... | Yes |
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 g... | 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\lef... | Yes |
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 \( g\left( x\right) = x - 1 \) . | 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( ... | Yes |
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? | 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 = ... | Yes |
Verify that the mathematical model for this experiment is given by \( f\left( {g\left( t\right) }\right) \), where \( g\left( t\right) = t - 4 \) . | Solution. Our expectation is that the plot for this new experiment will have the \ | No |
Write the equations \( y = {8}^{3x} \) and \( y = 7{\left( \frac{1}{2}\right) }^{{2x} - 1} \) in standard exponential form. | 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{... | Yes |
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 ... | 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... | Yes |
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 cond... | 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( ... | Yes |
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. | (Hint: Verify that \( {\left\lbrack \cosh \left( x\right) \right\rbrack }^{2} - {\left\lbrack \sinh \left( x\right) \right\rbrack }^{2} = 1 \), for all \( x \).) | No |
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? | 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%. | Yes |
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 s... | 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}\lef... | Yes |
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 ... | 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}}... | Yes |
Problem 12.3. Rewrite each function in the form \( y = {A}_{ \circ }{e}^{at} \), for appropriate constants \( {A}_{ \circ } \) and \( a \) . | (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) \( y = 3{\left( {14.24}\right) }^{4t} \) | Yes |
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 \( \mathcal{R} \) . | 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) =... | Yes |
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}}.... | 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 o... | Yes |
The problem is to describe a sequence of geometric maneuvers that transform the graph of \( y = {x}^{2} \) into the graph of \( y = - 3{\left( x - 1\right) }^{2} + 2 \) . | 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 compos... | No |
Problem 13.7. Describe how each graph differs from that of \( y = {x}^{2} \) . | (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) \( y = 2{\left( 3x - {12}\right) }^{2} - 5 \) | Yes |
Sketch the graph of the function \( f\left( x\right) = \frac{{3x} - 1}{{2x} + 7} \) . | 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 deno... | Yes |
Let \( f\left( x\right) = \frac{{2x} + 3}{{5x} - 7} \) . | \[ f\left( x\right) = \frac{{2x} + 3}{{5x} - 7} \cdot \frac{\frac{1}{5}}{\frac{1}{5}} = \frac{\frac{2}{5}x + \frac{3}{5}}{x - \frac{7}{5}} \] | Yes |
Find the linear-to-linear rational function \( f\left( x\right) \) such that \( f\left( {10}\right) = {20}, f\left( {20}\right) = {32} \) and \( f\left( {25}\right) = {36} \). | 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... | Yes |
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 ... | 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... | Yes |
If a 16 inch pizza is cut into 12 equal slices, what is the area of a single slice? | 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\[ = \fr... | Yes |
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. | Solution. The tires will be rotating in a counterclockwise direction and the radius \( r = \frac{1}{2}{28} = {14} \) inches. The other given quantity, \ | No |
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... | 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 }/\... | Yes |
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 s... | (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. | No |
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 hi... | 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: Prope... | Yes |
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 p... | 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... | Yes |
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 \( \angle \mathrm{{EAD}} \) ), \( \b... | 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... | Yes |
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? | 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.} \... | Yes |
Where are the drivers located (in xy-coordinates) after 18 seconds? | 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 \... | Yes |
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. | 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... | Yes |
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 fli... | We impose a coordinate system in Fig-\n\nure 17.21(a), where \ | No |
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... | Impose a coordinate system, and give the \( x \) - and \( y \) -coordinates of Charlie after one minute of running. | No |
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 Jan... | 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) =... | Yes |
Problem 18.4. These graphs represent periodic functions. Describe the period in each case. |  | No |
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 o... | 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... | No |
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}\l... | 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 ... | Yes |
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 \( \mathr... | 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{... | Yes |
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 ... | (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 val... | No |
Find all values of \( \theta \) (an angle) that make this equation true: \( \sin \left( \theta \right) = \frac{1}{2} \) . | 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![7bc6... | No |
If two sides of a right triangle have lengths 1 and \( \sqrt{3} \) as pictured below, what are the acute angles \( \alpha \) and \( \beta \) ? | 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 fun... | Yes |
Find two acute angles \( \theta \) so that the following equation is satisfied:\n\n\[ \frac{9}{4{\cos }^{2}\left( \theta \right) } = \frac{{25}^{2}}{16}\left( {1 - {\cos }^{2}\left( \theta \right) }\right) . \] | 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... | Yes |
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} \... | 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 \(... | Yes |
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 befo... | Solution. A picture of the situation is shown in Fig-\n\n (a) The physical layout. \n\n(b) Modeling the problem.... | Yes |
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? | 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- ... | Yes |
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 ... | 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 l... | No |
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. | 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 t... | Yes |
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 ... | 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 co... | Yes |
Suppose \( z = 3 - {4i} \) and \( w = 2 + {7i} \). | 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\[ = {34} + {13i}\text{.} \] | Yes |
Example 2.2.2 Exploring the polar form. | On the left side of the following diagram, we plot the points \( z = 2{e}^{{i\pi }/4}, w = \) \( 3{e}^{{i\pi }/2}, v = - 2{e}^{{i\pi }/6}, u = 3{e}^{-{i\pi }/3} \) .   {e}^{i\left( {\theta + \beta }\right) }.\] | 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( {\c... | Yes |
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? | \[ r{e}^{i\theta } = - \left| r\right| {e}^{i\theta } \] \[ = \left( {e}^{i\pi }\right) \cdot \left| r\right| {e}^{i\theta }\text{ since } - 1 = {e}^{i\pi } \] \[ = \left| r\right| {e}^{i\left( {\theta + \pi }\right) }\text{, by Theorem 2.2.3.} \] Thus, by adding \( \pi \) to the angle if necessary, we may always assum... | Yes |
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\\] | \\[ \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\\] | Yes |
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 \( \beta \) form... | 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 }\r... | No |
Let \( \\theta \) be an angle, and define \( {R}_{\\theta } : \\mathbb{C} \\rightarrow \\mathbb{C} \) by \( {R}_{\\theta }\\left( z\\right) = {e}^{i\\theta }z \). | 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\... | Yes |
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 \( A \) . | 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... | Yes |
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 \( k \) . | 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 c... | Yes |
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 \( \mathbb{C} \). | 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 comple... | Yes |
Theorem 3.1.9 Suppose \( T \) is a general linear transformation.\na. \( T \) maps lines to lines. | 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 g... | Yes |
Theorem 3.1.12 General linear transformations preserve angles. | 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... | Yes |
Some Euclidean isometries of \( \mathbb{C} \). | 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\... | Yes |
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 \( {z}_{0} \) . | 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 \( {r}_{i} \) deno... | Yes |
Inversion in the unit circle. | 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| {{... | Yes |
Theorem 3.2.4 There exists a unique cline through any three distinct points in \( \mathbb{C} \) . | 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 ... | Yes |
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. | 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 eq... | Yes |
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}. ... | 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}... | No |
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 \( C \) . | 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 \( z \) p... | Yes |
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 \( {i}_{C}\left( D\right) \) . | 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.... | Yes |
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. | 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}^... | Yes |
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. | 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 \( {L}_{1} \) . Ne... | Yes |
Theorem 3.3.1 Any general linear transformation extended to the domain \( {\mathbb{C}}^{ + } \) fixes \( \infty \) . | 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, \( T\left( \infty \right) = \infty \) . | Yes |
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 \( \theta \) . | 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 intersec... | No |
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 \( {ad} - {bc} \neq 0 \) . | 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} = \f... | Yes |
Theorem 3.4.2 The Möbius transformation\n\n\[ T\left( z\right) = \frac{{az} + b}{{cz} + d} \] has the inverse transformation | \[ {T}^{-1}\left( z\right) = \frac{-{dz} + b}{{cz} - a}. \] | Yes |
Theorem 3.4.6 Any Möbius transformation \( T : {\mathbb{C}}^{ + } \rightarrow {\mathbb{C}}^{ + } \) fixes 1,2, or all points of \( {\mathbb{C}}^{ + } \) . | 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 Exa... | Yes |
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( ... | 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 \(... | Yes |
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, \( T\left( {C}_{1}\right) = {C}_{2} \) . | 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 se... | Yes |
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 \( p + q = {z}_{\infty } + {w}_{\infty } \). | 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} ... | Yes |
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 | 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\... | No |
Example 3.5.9 Fix -1 and 1 and send \( i \mapsto \infty \) . | 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 | Yes |
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. | \[ z = T\left( z\right) \] \[ z\left( {{3z} - 5}\right) = {7z} - {12} \] \[ 3{z}^{2} - {12z} + {12} = 0 \] \[ {z}^{2} - {4z} + 4 = 0 \] \[ {\left( z - 2\right) }^{2} = 0 \] \[ z = 2\text{.} \] 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 ... | Yes |
Show that \( \mathcal{T} \) is a group of transformations. | 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 closu... | Yes |
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. | 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 transfo... | No |
Lemma 4.1.16 For any points \( z, w, v \) in \( \mathbb{C} \) , \n\n\[ \left| {z - w}\right| \leq \left| {z - v}\right| + \left| {v - w}\right| \] | 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 im... | Yes |
Theorem 5.1.3 Any Möbius transformation in \( \mathcal{H} \) is the composition of two reflections of the hyperbolic plane. | 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 un... | Yes |
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