course mth 164 6/26 8 query problem 5.6.54 3 cos(2x+`pi) find characteristics and graph using transformations......!!!!!!!!...................................
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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. 22:56:30
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ‘OK’ ------------------------------------------------ Self-critique rating #$&*3 **** query problem 5.6.60 2 cos(2`pi(x-4)) find characteristics and graph using transformations **** explain how you use transformations to construct the graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Determining the amplitude from y=2cos, the amplitude of y=cos is 1, so the max amplitude of the graph is 2 and min amplitude is -2. The period (T) of the graph is determined by 2pi/2pi=1 Because of (x-4), the graph shifts down the x axis by 4 units. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ‘OK’ ------------------------------------------------ Self-critique rating #$&*3 **** Describe the resulting graph by giving its period, its the maximum and minimum y values and its phase shift, and describe how its phase shift affects the graph
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23:20:53 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A= max 2, min -2 Period (T)=1 Phase Shift= -2/pi The graph starts at -2/pi and ends at -2/pi+1. The phase shift is needed to find the end point of 1 cycle, it also helps determine the interval between each time the graph crosses the x axis. confidence rating #$&* 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. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ‘OK’ ------------------------------------------------ Self-critique rating #$&*3 **** query problem 6.1.24 1 - sin^2 x /( 1-cos x) = -cos x
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give the steps in your solution
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1-(1-cos^2 x)/(1-cos x)= 1-(1+cos x)(1-cos x) / (1-cos x)= 1-1-cos x= -cos x confidence rating #$&* 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 #$&*3 **** query problem 6.1.48 sec x / (1 + sec x) = (1-cos x) / sin^2 x give the steps in your solution
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23:47:46 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1/cos x) / (1+1/cos x) = (1-cos x) / (sin^2 x) (1/cos x) / (1+1/cos x)*[(1+1/cos x)(sin^2x)] = (1-cos x) / (sin^2 x )*[(1+1/cos x)(sin^2x)] (1/cos x)(sin^2x) = 1/cos x – cos x = [(1/cos x)(sin^2x)]*(cos x) = (1/cos x – cos x)*(cos x) sin^2 x = 1-cos^2 x = sin^2 x + cos^2 x = 1 confidence rating #$&*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): ‘OK’ ------------------------------------------------ Self-critique rating #$&*3 "