#$&* course Mth 164 3:47 pm3/1/15
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Given Solution: The unit-circle points corresponding to the given angles are (1,0), (sqrt(3)/2, 1/2), (sqrt(2)/2, sqrt(2)/2) and (1/2, sqrt(3)/2). So the values of the tangent, each calculated as y / x, are as follows: tan(0) = 0/1 = 0, tan(pi/6) = 1/2 / (sqrt(3)/2) = 1/2 * 2/sqrt(3) = 1/sqrt(3) = sqrt(3)/3, tan(pi/4) = sqrt(2)/2 / (sqrt(2)/2) = 1 and tan(pi/3) = sqrt(3)/2 / (1/2) = sqrt(3). The corresponding graph is shown in Figure 73. The slopes of the graph between 0 and pi/6, between pi/6 and pi/4, and between pi/4 and pi/3 are, respectively: (sqrt(3)/3 - 0)/ (pi/6 - 0) = 6 sqrt(3) / pi = 1.10, (1 - sqrt(3)/3)/(pi/4-pi/6) = 1.62 and (sqrt(3) - 1)/(pi/3 - pi/4) = 2.80. The slopes are increasing, slowly at first, then more quickly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t show my work on here for the slopes of the graph but I did get the correct solutions. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q002. Figure 52 indicates the unit circle positions corresponding to the angles which are multiples of pi/18. The grid shows intervals of .5. Estimate the x coordinates of the first-quadrant points then make a table of tangent (theta) vs. theta for the data from 0 to 8 pi/18. Give your values and use them to extend the graph of tan(theta) through theta = 8 pi/18. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I found the following estimates of x for the given circle with a radius of 5 since there are 10 points of the .5 intervals. Theta cos(theta) tan(theta) Pi/18 4.95 0.17 Pi/9 4.65 0.35 Pi/6 4.3 0.58 2pi/9 3.7 0.86 5pi/18 3.2 1.16 Pi/3 2.5 1.72 7pi/18 1.65 2.8 4pi/9 0.83 5.96 This graph resembles one that increases exponentially. Increasing the most at the last point. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Starting at the original origin we can see that as we approach pi/2 the y values start at 0 on the x axis at the most right part of the circle and increase to the top most point of the circle. The x values start at high values the farthest point of the circle and continuously decrease to the point of 0 on the top most point of the circle on y axis. These x and y values have a mirror effect on the way up the circle towards pi/2 where the y values increase and the x values decrease. Both of these values on increase to the value equivalent to the radius but will not extend any farther. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Figure 48 shows a series of points on the unit circle approaching the pi/2 position. It should be clear that the y coordinate approaches 1 and the x coordinate approaches 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. As the angle approaches pi/2 from the first quadrant the y coordinate approaches 1 and the x coordinate approaches 0, as we saw in the last problem. What happens to the ratio 1/x as x take values 0.1, 0.01, 0.001, 0.0001? What happens as x continues to approach 0? Is there a limit to how large 1/x can get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The x values in 1 /x get smaller the ratio produces a larger value. 1/0.1 = 10 1/0.01 = 100 1/0.001 = 1000 1/0.0001 = 10,000 This number will continue to grow infinitely as x will continue to get smaller infinitely. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. What happens to the tangent of theta as theta approaches pi/2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Tangent = y/x or sin/ cos We see here an increase in the tangent as we approach pi/2. This is because our y values are getting larger divided by x values that are getting smaller. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the x coordinate approaches zero and the y coordinate approaches 1, it follows that tan(theta) = y / x gets larger and larger, without bound. We say that this quantity approaches infinity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. As we have seen the value of tangent (theta) exceeds all bounds as theta approaches pi/2 from within the first quadrant. How do we extend the graph of tangent (theta) vs. theta to cover the domain 0 <= theta <= pi/2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Preceding 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: Compared to the problem pertaining to quadrant I we see the only difference is that the y values for be negative instead of positive. So we will have: Theta sin cos tan -8pi/18 (-4pi/9) -4.95 0.83 -5.96 -7pi/18 -4.65 1.65 -2.8 -6pi/18 (-pi/3) -4.3 2.5 -1.72 -5pi/18 -3.7 3.2 -1.16 -4pi/18 (-2pi/9) -3.2 3.7 -0.86 -3pi/18 (-pi/6) -2.5 4.3 -0.58 -2pi/18 (-pi/9) -1.65 4.65 -0.35 -pi/18 -0.85 4.95 -0.17 0 0 1 0 This graph is a mirror of the graph from the first quadrant stretching out and down from 0 with the largest increase being between -4pi/9 and -7pi/18. We can see here that the vertical asymptote is along the angular position -pi/2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Figure 71 shows the multiples of pi/18 from -8 pi/18 to 0. It is clear that for all values of theta the x coordinates are positive and for all values of theta except 0 the y coordinates negative, which make the values of tan(theta) = y / x negative. Otherwise the values are the same as those obtained for theta = 0 to 8 pi/18. The actual values, to 2 significant figures, starting with angle -pi/18, are -.18, -.36, -.58, -.84, -1.19, -1.7, -2.7 and -5.7. The graph is shown in Figure 8. Note how the graph now approaches the vertical line theta = -pi/2 as an asymptote. The values of y = tan(theta) for -pi/2 < theta < pi/2 combines the graph just obtained with the graph obtained in the preceding exercise, and is shown in Figure 97. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |. fig 8 fig 97 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. What happens to the value of the tangent function as the angle approaches pi/2 through second-quadrant angles? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The absolute value of these numbers increase but since they are of negative value they actually are decreasing in value as they approach -pi/2. So as the tangent values approach positive pi/2 we will see an increase in the values as they approach from -pi/2 that increase actually continues into the first quadrant. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In 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): I believe I misunderstood the question but I do understand the given solution. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: Since the sine and cosine values in the third quadrant will both be negative this results in the tangent being positive values. These values will create an asymptote along 3 pi/2. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In the third quadrant, as we approach theta = 3 pi/2 on the circle the x coordinates are negative as the approach 0 while the y value approaches -1. Thus tan(theta) = y / x will be positive, with the magnitude of the quantity approaching infinity. We see that tan(theta) approaches +infinity. The graph will be asymptotic to the positive portion of the line theta = 3 pi/2, as in Figure 30. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q010. Sketch a graph of the tangent function from theta = -pi/2 through theta = 3 pi / 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This graph gives 2 separate lines one passing through the coordinates (0,0) and the other running through (pi, 0) along the x axis. These lines create asymptotes at -pi/2, pi/2 and 3pi/2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The 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): This is the same as my graph but is better explained than my solution. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q011. Sketch a graph of y = tan ( 2 x ), showing two complete cycles. What is the length of a cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe a complete cycle would be from one original point all the way around a circle to the same point again. I used -pi/2 >= x >= pi/2 for one cycle since the tangent range is -pi/2 < theta < pi/2 This gave me: Tan(2* -pi/2) = tan (-pi) = 0 Tan(2* -pi/3) = tan (-2pi/3) = 1.73 Tan(2* -pi/6) = tan (-pi/3) = -1.73 Tan(2* 0) = tan (0) = 0 Tan(2* pi/6) = tan (pi/3) = 1.73 Tan(2* pi/3) = tan (2pi/3) = -1.73 Tan(2* pi/2) = tan (pi) = 0 I continued to 2 pi /3 to finish the line Tan(2* 2pi/3) = tan(4 pi /3) = 1.73 This graph gives similar lines to that of tan(theta) with asymptotes between -pi/2, pi/2 and 3pi/2. I’m assuming the 2 lines on this graph are 2 cycles. I may be wrong about that. I’ve extended them and my values for the given equation. There is also the start of another graph to the left of these 2 lines with points at -2pi/3, and -pi. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We begin by sketching a graph of y = tan(theta) for -pi/2 <= theta <= 3 pi/2, as shown in Figure 30. We then let theta = 2x and relabel the graph as indicated in Figure 79, with pi/2 <= theta <= 3 pi/2 replaced by -pi/2 <= 2x <= 3 pi/2 so that we have -pi/4 <= x <= 3 pi/4. This shows us that x goes through a complete cycle between -pi/4 and pi/4 and again between pi/4 and 3 pi/4. Each cycle has length pi/2. Note how the period of the function tan(2x) is 1/2 the period of the tan(x) function, which is pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So -pi/2 <= x <= pi/2 is the length of 2 cycles. This graph is the same as the graph I have drawn except I have the start of another graph to the left of these 2. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q011. Sketch a graph of y = tan ( 2 x + pi/3 ), showing two complete cycles. What is the length of a cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This graph will probably be similar to the last graph except shifted by pi/3 and since the graph tan (2x) is in intervals of pi/3 then this will shift the graph to the right by pi/3. After sketching the graph I see this is true and that means the length of a cycle is also pi/2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We 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. " 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: This graph will probably be similar to the last graph except shifted by pi/3 and since the graph tan (2x) is in intervals of pi/3 then this will shift the graph to the right by pi/3. After sketching the graph I see this is true and that means the length of a cycle is also pi/2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!