#$&* course Mth 164 February 22 11:36 AM query problem 5.6.54 3 cos(2x+`pi) find characteristics and graph using transformations
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The amplititude is 3 The period is 2pi/B which is 2pi/2 and that would be pi Asin(Bx-C)+D Phase shift is C/B and this case would be pi/2 The graph will stretch verticaly from 3 to -3 and horizontally from -pi to pi. The graph would peak at 3 for pi/2 then be at its lowest at -3 for 3pi/2. It would cross the x axis at pi and 2pi. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: STUDENT SOLUTION: ** Here are two solutions provided by students from previous years: The amplitude is 3 The Period = T= (2`pi)/`omega = 2`pi/(2) = `pi The phase shift = `phi/(`omega) = `pi/2 The graph of y = 3 cos (2x + `pi) will lie between -3 and 3 on the y axis. One cycle will begin at x = `phi/(`omega) = `pi/2 and will end at 2`pi/(`omega) + `phi/(`omega) = (2`pi)/2 + (`pi)/2 = (3`pi)/2. We then divide the interval of [`pi/2, 3`pi/2] into (4) subintervals each of length `pi divided by 4 = `pi/4: [`pi/2, 3`pi/4], [3`pi/4, `pi], [`pi, 5`pi/4], [5`pi/4, 3`pi/2]. The five key points for the graph are: (`pi/2, 3), (3`pi/4, 0), (`pi, -3), (5`pi/4, 0), and (3`pi/2, 3). ANOTHER STUDENT SOLUTION (consistent with preceding but with different details provided): the graph of this function has a maximum point of y=+3 and a minimum point of y=-3. at the origin the graph touches the point y=-3. and whenever x= pi, 2pi and 3pi y=-3. and at the points x= (-pi),-2pi,-3pi y= -3. when x=pi/2 and 3pi/2 and -pi/2 and -3pi/2 y= 3. to solve for the amplitude and the period and the phase shift we use the equation y= Acos((omega)(x)-phi). so the amplitude of the equation is the absolute value of A which is 3. so A=3. the period is 2pi/2 which is pi so there is a period at pi. and the phase shift is phi/omega. which in this case is pi/2. so the phase shift is pi/2. STUDENT QUESTION Was my given answer correct? I want to make sure that I’m understanding this correctly. INSTRUCTOR RESPONSE your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv 3cos tells me the amplitude will be 3. The value 2x tells me that the cycle will run 2 times. The addition of pi to 2x indicates that this wave will shift to the right by the value pi. Your first task would be to check your answers against the given answers, which were as follows: The amplitude is 3 The Period = T= (2`pi)/`omega = 2`pi/(2) = `pi The phase shift = `phi/(`omega) = `pi/2 Your amplitude is correct. Your other two answers don't quite get to the necessary information. A careful self-critique would have noted this and attempted to address the discrepancy in terms of the details of the given solution. The period is the change in x necessary for the argument of the function to change by 2 pi. The expression 2 x will change by 2 pi when x changes by pi, so the period is pi. In the expression 2 x + pi, you can't immediately see the phase shift. If you factor the expression into 2 ( x + pi/2 ) then you can see that the phase shift is -pi/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 **** query problem 5.6.60 2 cos(2`pi(x-4)) find characteristics and graph using transformations **** explain how you use transformations to construct the graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Your amplitude is 2 The period is 2pi/2pi the phase shift is c/b or -4/2pi = -2/pi 2<=0<=-2 The graph will peak at 4 between 2 and -2. Peaks are at intervals of one so peaks are at 4,5,6,etc.. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: ** Starting with y = cos(x), which has amplitude 1 and period 2 `pi and which peaks on the y axis (i.e., at x = 0) and at every interval of 2 pi on the x axis, we apply the appropriate transformations as follows: We first multiply the function by 2, which doubles all the y coordinates, stretching the graph vertically by factor 2. This doubles the amplitude from 1 to 2. Next we multiply x by 2 `pi, which compresses the graph in the horizontal direction by factor 2 `pi. So the period of the function is changed from 2 `pi to 2 `pi / (2 `pi) = 1. We then replace x by x - 4, which shifts the graph 4 units to the right. Our graph now has a peak at x = 4. It oscillates between max value y = +2 and min value y = -2, peaking at x = 4 and at regular intervals of 1 so that peaks occur at x = 4, 5, 6, . . . as well as 3, 2, 1, . . . . ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: By using transformations to construct this graph, I would start with y = cos x graph. Then vertically stretch this graph by factor of 2 for y = 2 cos x. Then I would horizontally stretch this graph by a factor of 2`pi for y = 2 cos (2`pi x), ** this is a horizontal compression by factor 2--the graph is compressed in the x direction, from period 2 pi to period pi ** then I would horizontally shift this graph by a factor of 4 (to the right) ** you shift it 4 units to the right; a factor is something you multiply by ** STUDENT QUESTIONS Oops ! Please advise on this …. When I saw (2`pi(x-4)) I immediately multiplied to get 2piX - 8pi. This was not right. Are you saying that this simplifies to x-4? Please explain. Also, why is it that I always shift to the right? In this equation I wanted to say shift left (-4< 0). Sorry for all the questions, I just want to understand this before I take the test. INSTRUCTOR RESPONSE No problem. I'm glad to clarify. As you hopefully learned in first-semester precalculus, replacing x by x - h in the function y = f(x) gives you the function y = f (x - h). The graph of f(x - h) is shifted h units in the x direction, relative to the graph of y = f(x). For the present function we could say that f(x) = 2 cos(2 pi x), so that f(x - 4) = 2 cos(2 pi ( x - 4 ) ). f(x - 4) is the function given in this question. The graph of f(x - 4) is shifted 4 units to the right of the graph of f(x). The reason replacing x by x - 4 shifts the graph to the right is that if x takes a certain value, then in order to give x - 4 the same value, the quantity x has to be 4 units greater. So the graph of f(x-4) lies 4 units 'greater', or 4 units to the right, of the graph of f(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 **** Describe the resulting graph by giving its period, its the maximum and minimum y values and its phase shift, and describe how its phase shift affects the graph
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23:20:53 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It's period is 2pi/2pi which is 1 and max and min y values are 2 and -2. The phase shift would be -4/2pi which is a -2/pi. The max and min value of y are 2 and -2. The phase shift will shift by -2/pi units to the right. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: STUDENT SOLUTION: The amplitude is 2 The period is T = 2`pi/(`omega) = 2`pi/(2`pi) = 1. The phase shift = `phi/(`omega) = -4 / (2`pi) = -2/`pi. The graph will lie between 2 and -2 on the y-axis. One cycle will begin at x = `phi/(`omega) = -4/2`pi = -2/`pi and will end at 2`pi/(`omega) + `phi/(`omega) = 2`pi/2`pi + (-4)/2`pi = 1 - 2/`pi. Divide the interval of [-2/`pi, 1-2/`pi] into four subintervals each of length 1 divided by 4 = 1/4 ------And here's where I get lost in the math. INSTRUCTOR COMMENT: ** If you just show the interval from -2 / `pi to -2 / `pi + 1 as containing the entire cycle you won't be far wrong. However you can easily enough add increments of 1/4 to the starting point - 2 / `pi to get -2 / `pi + 1/4, -2 / pi + 1/2, -2 / `pi + 3/4 and -2/`pi + 1. These numbers would have to be approximated. -2/`pi for example is about -.64 or so. ** STUDENT QUESTION Sorry still trying to understand this ……. Where did sine go? INSTRUCTOR RESPONSE Good question, and no need to apologize. sin^2(x) + cos^2(x) = 1, a consequence of the unit-circle definition of sine and cosine, and the Pythagorean Theorem. So sin^2(x) = 1 - cos^2(x). In the above, sin^2(x) was simply replaced by 1 - cos^2(x).
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23:20:53
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand everything but how to calculate 2/pi. ???I know what 2pi is but what would be 2/pi??? ------------------------------------------------ Self-critique Rating:
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give the steps in your solution
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 -( sin^2(x)/(1-cos x) = 1- (1-cos^2(x))/(1-cos x) = sin^2(x) + cos^2(x) = 1 so sin^2(x)=1-cos^2(x) 1 - 1 - cos x = - cos x confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: ** 1 -( sin^2(x)/(1-cos x) = 1- (1-cos^2(x))/(1-cos x) = 1- [(1-cos x)(1+cos x)/(1-cos x) = 1- (1+cos x) = 1 - 1 - cos x = - cos x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 **** query problem 6.1.48 sec x / (1 + sec x) = (1-cos x) / sin^2 x give the steps in your solution
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23:47:46 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first thing we need to do is to change everything to sines and cosines. To co that we need to remember that sec(x) = 1/cos(x) so we need to multiply that by both sides. [ 1 / cos(x) ] / ( 1 + 1 / cos(x) ] = (1 - cos(x) ) / sin^2(x) We can now see a common denominator of (1+1/cos(x)) * sin^2(x) so we need to multiply that be both sides to have [ 1 / cos(x) ] * sin^2(x) = (1 - cos(x) ) ( 1 + 1 / cos(x) ) sin^2(x) / cos(x) = 1 + (1 / cos(x)) - cos(x) - 1 or sin^2 / cos(x) = [ 1 / cos(x)] - cos(x) Multiply both sides by the denominator cos(x) sin^2(x) = 1 - cos^2(x) sin^2 x + cos^2 x = 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: ** There are many ways to rearrange this equation to prove the identity. Here we will start by changing everything to sines and cosines using sec(x) = 1 / cos(x). We get [ 1 / cos(x) ] / ( 1 + 1 / cos(x) ] = (1 - cos(x) ) / sin^2(x). Multiplying both sides by the common denominator (1 + 1 / cos(x) ) * sin^2(x) we get [ 1 / cos(x) ] * sin^2(x) = (1 - cos(x) ) ( 1 + 1 / cos(x) ). Multiplying out the right-hand side and simplifying the left we have sin^2(x) / cos(x) = 1 + (1 / cos(x)) - cos(x) - 1 or since 1 - 1 = 0 just sin^2 / cos(x) = [ 1 / cos(x)] - cos(x). Multiplying both sides by the only remaining denominator cos(x) we have sin^2(x) = 1 - cos^2(x), which we rearrange into the basic Pythagorean identity sin^2 x + cos^2 x = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 2 I admit that I had to look at the given solution on this one to work out the problem. I thought that I started with the sin^2x woudl be the same as 1-cos^2x but I was really getting confused as to what to do with sec. ------------------------------------------------ Self-critique Rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!