course Mth 164 Self-critique (if necessary):Self-critique Rating: query problem 5.6.54 3 cos(2x+`pi) find characteristics and graph using transformations
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********************************************* Your solution: amplitude = 3, period = pi, phase shift = pi/2 Starting left of the origin, the curve first intercepts the x-axis at (-3pi/4, 0), then curves up to a peak at (-pi/2, 3), then back down across the x-axis again at (-pi/4, 0), then intercepting the y-axis at (, -3) then back up across the x at (pi/4, 0) for one complete cycle. I? not sure why you would use a transformation for this one. I thought you used transformations for the csc and sec functions. Confidence Assessment: 2 ********************************************* 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. 22:56:30
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):had the right graph, just didn? show as many points Self-critique Rating: **** 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. ********************************************* Your solution: I actually just realized this was a separate question when I was typing my answers. I had the description of the graph here and had already typed the self-critique below. Until I read the given solution I was trying to make sense of why I should graph a secant function to graph a cosine function. Now I see that what I should do here is graph y =cos(pi(x-2)) then multiply everything by 2, which gives a vertical stretch factor of 2 and a horizontal compression factor of 2 - the peaks are twice as high and half as far apart.( I constructed my graph as described below as y=2cos(2pi(x-4)), but I didn? understand what was meant by, or how to construct it as, a transformation, until I read the given solution below) Confidence Assessment: 2 ********************************************* 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I think the given solution to this one explains what is meant by ?se transformations to graph?Self-critique Rating: **** 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 ********************************************* Your solution: A= 2, period = 2pi/w = 2pi/2pi = 1, phase shift = -4/2pi = -2/pi Starting to the left of the origin, the curve intercepts the x-axis at (-pi/4, 0), then up to a peak at (-pi/8, 2), then down to intercept the y-axis at (0, -5pi/2), the back up to cross the x-axis at (pi/12, 0) and up to another peak at (pi/6, 2) Confidence Assessment: 2 ********************************************* 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. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: **** query problem 6.1.24 1 - sin^2 x /( 1-cos x) = -cos x
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give the steps in your solution
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********************************************* Your solution: rewrite left side:(1-(1-cos^2))/(1-cos x) = (1-(1-cos x)(1+cos x))/(1-cos x) = 1-1+cos x cos x = -cos x cos x = cos(-x) = -cos x because cos is an even identity - this has to be it because I have tried everything else Confidence Assessment: 2 ********************************************* 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):I got the given solution, after about 12 hours of working on this problem ! (that? why I couldn? finish this one at school.) I didn? open this up as an open query so I didn? know the answers were here. I can? believe I didn? see this one until I had tried everything else. I? afraid that? what will happen on the test. Self-critique Rating: **** 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 ********************************************* Your solution: Rewrite left side : (1/cos x)/(1+(1/cos x))= (1/cos x)/((cos x/cos x)+(1/cos x))=(1/cos x)/((cos x +1)/(cos x)) = (cos x/cos x)/(cos x+1) = 1/cos x +1 Rewrite right side: (1-cos x)/sin^2x = (1-cos x)/(1-cos^2 x) = (1-cos x)/((1+cos x)(1-cos x)) = 1/(1+cos x) Confidence Assessment: 2 ********************************************* 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 - I think my answer is correct but is seems too simple. It? definitely not the first thing I tried.