#$&* course Mth 164 2/19 10:31 007. Tangent Function
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Given Solution: `aThe unit-circle points corresponding to the given angles are (1,0), (sqrt(3)/2, 1/2), (sqrt(2)/2, sqrt(2)/2) and (1/2, sqrt(3)/2). So the values of the tangent, each calculated as y / x, are as follows: tan(0) = 0/1 = 0, tan(pi/6) = 1/2 / (sqrt(3)/2) = 1/2 * 2/sqrt(3) = 1/sqrt(3) = sqrt(3)/3, tan(pi/4) = sqrt(2)/2 / (sqrt(2)/2) = 1 and tan(pi/3) = sqrt(3)/2 / (1/2) = sqrt(3). The corresponding graph is shown in Figure 73. The slopes of the graph between 0 and pi/6, between pi/6 and pi/4, and between pi/4 and pi/3 are, respectively: (sqrt(3)/3 - 0)/ (pi/6 - 0) = 6 sqrt(3) / pi = 1.10, (1 - sqrt(3)/3)/(pi/4-pi/6) = 1.62 and (sqrt(3) - 1)/(pi/3 - pi/4) = 2.80. The slopes are increasing, slowly at first, then more quickly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Figure 52 indicates the unit circle positions corresponding to the angles which are multiples of pi/18. The grid shows intervals of .5. Estimate the x coordinates of the first-quadrant points then make a table of tangent (theta) vs. theta for the data from 0 to 8 pi/18. Give your values and use them to extend the graph of tan(theta) through theta = 8 pi/18. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Estimate Pi/18 = .17 Pi/9 = .33 Pi/6 = .58 2pi/9 = .82 5pi/18 = 1.10 pi/3 = 1.68 7pi/18 = 2.71 4pi/9 = 5.80 Real Values Pi/18 = .18 Pi/9 = .36 Pi/6 = .58 2pi/9 = .84 5pi/18 = 1.19 pi/3 = 1.73 7pi/18 = 2.76 4pi/9 = 5.67 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe angles 3 pi/18 = pi/6 and 6 pi/18 = pi/3 are known to be approximately sqrt(3)/3 = .58 and sqrt(3) = 1.73, respectively. The values at the remaining points can be estimated more or less accurately. The actual values, to 2 significant figures, strarting with angle pi/18, are .18, .36, .58, .84, 1.19, 1.7, 2.7 and 5.7. The rapidly increasing slopes as theta exceeds pi/3 are apparent from these numbers. The extended graph shown in Figure 97 depicts only the results for angles through 7 pi/18; the figure would have to be twice as high to include the point (8 pi/18, 5.7), or in reduced for (4 pi/9, 5.6). The colored line segments just below the x axis indicate the multiples of pi/18. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Sketch a series of points on the unit circle approaching the pi/2 position. As we approach closer and closer to pi/2, what happens to the y coordinate? What happens to the x coordinate? Does the y coordinate approach a limiting value? Does the x coordinate approach a limiting value? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As we approach pi/2 the y coordinate grows. As we approach pi/2 the x coordinate shrinks. The y coordinate does approach a limiting value and that value is 1. X also a limiting factor. The factor is 0 on the positive side but if you go to the negative side the limiting factor is -1. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFigure 48 shows a series of points on the unit circle approaching the pi/2 position. It should be clear that the y coordinate approaches 1 and the x coordinate approaches 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. As the angle approaches pi/2 from the first quadrant the y coordinate approaches 1 and the x coordinate approaches 0, as we saw in the last problem. What happens to the ratio 1/x as x take values 0.1, 0.01, 0.001, 0.0001? What happens as x continues to approach 0? Is there a limit to how large 1/x can get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is no limit to what X can be. As we take the values .1, 0.01, 0.001, and 0.0001 we get x values of 10, 100, 1000, and 10000. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf x = 0.1, 0.01, 0.001 and 0.0001 we see that 1/x = 10, 100, 1000 and 10,000. There is no limit to how large 1/x can get; we can make it as large as we wish by choosing x small enough. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. What happens to the tangent of theta as theta approaches pi/2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It continues to get large and approaches infinity. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince the x coordinate approaches zero and the y coordinate approaches 1, it follows that tan(theta) = y / x gets larger and larger, without bound. We say that this quantity approaches infinity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): tan(theta) = y / x gets larger without bound. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q006. As we have seen the value of tangent (theta) exceeds all bounds as theta approaches pi/2 from within the first quadrant. How do we extend the graph of tangent (theta) vs. theta to cover the domain 0 <= theta <= pi/2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We extend it like in the last problem and ise 8pi/18 which is really 4pi/9. This point is a little less than pi/2. Also as the numbers get larger and large they run into of the asymptote of pi/2. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aPreceding solutions and graphs showed how the slopes of the graph increased as we moved from theta = 0 to theta = 8 pi / 18 (which reduces to 4 pi / 9, just a little less than pi/2). The slopes will continue to increase, and the ratio y / x will increase without bound so that the values of the function increase without bound as theta approaches pi / 2. The graph therefore forms a vertical asymptote at theta = pi/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. What are the values of tan(theta) as theta changes from -8 pi/18 to 0 in increments of pi/18? Use your estimates of the coordinates of the first-quadrant points as a basis for your estimates of the values of the tangent function. Extend your graph of the function so that it is graphed for -8 pi / 18 <= theta <= 8 pi / 18, then show what happens as theta approaches -pi/2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Pi/18 = .18 Pi/9 = .36 Pi/6 = .58 2pi/9 = .84 5pi/18 = 1.19 pi/3 = 1.73 7pi/18 = 2.76 4pi/9 = 5.67 -Pi/18 = -.18 -Pi/9 = -.36 -Pi/6 = -.58 -2pi/9 = -.84 -5pi/18 = -1.19 -pi/3 = -1.73 -7pi/18 = -2.76 -4pi/9 = -5.67 As theta approaches your x increases in value and your y values increase in value. The graph will approach the asymptote -pi/2. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFigure 71 shows the multiples of pi/18 from -8 pi/18 to 0. It is clear that for all values of theta the x coordinates are positive and for all values of theta except 0 the y coordinates negative, which make the values of tan(theta) = y / x negative. Otherwise the values are the same as those obtained for theta = 0 to 8 pi/18. The actual values, to 2 significant figures, starting with angle -pi/18, are -.18, -.36, -.58, -.84, -1.19, -1.7, -2.7 and -5.7. The graph is shown in Figure 8. Note how the graph now approaches the vertical line theta = -pi/2 as an asymptote. The values of y = tan(theta) for -pi/2 < theta < pi/2 combines the graph just obtained with the graph obtained in the preceding exercise, and is shown in Figure 97. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |. fig 8 fig 97 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. What happens to the value of the tangent function as the angle approaches pi/2 through second-quadrant angles? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: These are the values when the second quadrant approaches the first quadrant. The y values are negative and increasing to 0 and the x increases from negative to 0. When you hit the first quadrant the y and x values are still growing to infinity because of the asymptote at pi/2. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIn the second quadrant, as we approach theta = pi/2 on the circle the x coordinates are negative as the approach 0 while the y value approaches +1. Thus tan(theta) = y / x will be negative, with the magnitude of the quantity approaching infinity. We say that tan(theta) approaches -infinity. The graph will be asymptotic to the negative portion of the line theta = pi/2, as indicated in Figure 69, which combines this portion of the graph with the existing graph of the full cycle of the tangent function. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): x coordinates are negative as the approach 0 while the y value approaches 1. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q009. What happens to the value of the tangent function as the angle approaches 3 pi/2 through third-quadrant angles? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As you approach 3pi/2 the x values are increasing from negative to 0 and your y values are decreasing from 0 to -1. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIn the third quadrant, as we approach theta = 3 pi/2 on the circle the x coordinates are negative as the approach 0 while the y value approaches -1. Thus tan(theta) = y / x will be positive, with the magnitude of the quantity approaching infinity. We see that tan(theta) approaches +infinity. The graph will be asymptotic to the positive portion of the line theta = 3 pi/2, as in Figure 30. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The graph approaches 3pi/2 which is another asymptote, which means it, is going to infinity. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q010. Sketch a graph of the tangent function from theta = -pi/2 through theta = 3 pi / 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe graph is shown in Figure 30. Note how the tangent function goes through its complete cycle of values, from -infinity to +infinity, between theta = -pi/2 and theta = pi/2, then again between theta = pi/2 and theta = 3 pi/2. The horizontal 'distance' corresponding to a complete cycle is thus seen to be pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. Sketch a graph of y = tan ( 2 x ), showing two complete cycles. What is the length of a cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe begin by sketching a graph of y = tan(theta) for -pi/2 <= theta <= 3 pi/2, as shown in Figure 30. We then let theta = 2x and relabel the graph as indicated in Figure 79, with pi/2 <= theta <= 3 pi/2 replaced by -pi/2 <= 2x <= 3 pi/2 so that we have -pi/4 <= x <= 3 pi/4. This shows us that x goes through a complete cycle between -pi/4 and pi/4 and again between pi/4 and 3 pi/4. Each cycle has length pi/2. Note how the period of the function tan(2x) is 1/2 the period of the tan(x) function, which is pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. Sketch a graph of y = tan ( 2 x + pi/3 ), showing two complete cycles. What is the length of a cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe begin by sketching a graph of y = tan(theta) for -pi/2 <= theta <= 3 pi/2, as shown in Figure 30. We then let theta = 2x + pi/3 and relabel the graph as indicated in Figure 62, with pi/2 <= theta <= 3 pi/2 replaced by -pi/2 <= 2x + pi/3 <= 3 pi/2 so that we have -5 pi/12 <= x <= 7 pi/12. This graph shows us that x goes through a complete cycle between -5 pi/12 and pi/12 and again between pi/12 and 7 pi/12. Each cycle has length pi/2. Note how the period of the function tan(2x + pi/3) is 1/2 the period of the tan(x) function, which is pi. Complete Assignments 6, 7, including Class Notes, text problems and Web-based problems as specified on the Assts page. When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_. Note that Assignment 7 consists of a test covering Assignments 1-5. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. Sketch a graph of y = tan ( 2 x + pi/3 ), showing two complete cycles. What is the length of a cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe begin by sketching a graph of y = tan(theta) for -pi/2 <= theta <= 3 pi/2, as shown in Figure 30. We then let theta = 2x + pi/3 and relabel the graph as indicated in Figure 62, with pi/2 <= theta <= 3 pi/2 replaced by -pi/2 <= 2x + pi/3 <= 3 pi/2 so that we have -5 pi/12 <= x <= 7 pi/12. This graph shows us that x goes through a complete cycle between -5 pi/12 and pi/12 and again between pi/12 and 7 pi/12. Each cycle has length pi/2. Note how the period of the function tan(2x + pi/3) is 1/2 the period of the tan(x) function, which is pi. Complete Assignments 6, 7, including Class Notes, text problems and Web-based problems as specified on the Assts page. When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_. Note that Assignment 7 consists of a test covering Assignments 1-5. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!