course MTH 163 The q_a_ and query exercises for assignment 9. Įq^d Ų~}assignment #009
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22:03:55 `q001. Note that this assignment has 2 questions For the function y = 1.1 x + .8, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2? What therefore is the rise between these two points, and what is the run? What is the average slope of the graph between these two points? Be sure to simplify your result.
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RESPONSE --> For the coordinates of the point x=x1 the x-value will be x1 and the y-value will be f(x1) or 1.1*x1+.8). The coordinates are therefore (x1 , 1.1*x1+.8). For the coordinates of the point x=x2 the x-value will be x2 and the y-value will be f(x2) or (1.1*x2+.8). The coordinates are therefore (x2 , 1.1*x2+.8). The rise between these two points will be (y2-y1), or (1.1*x2+.8) - (1.1*x1+.8) The run between these two points will be (x2-x1). The average slope will be: [(1.1*x2+.8) - (1.1*x1+.8)] / (x2-x1) = [(1.1*x2) - (1.1*x1)] / (x2-x1) = 1.1(x2-x1) / (x2-x1) = 1.1 / 1 = 1.1 confidence assessment: 2
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22:05:09 In terms of the symbol x1 the coordinates of the x = x1 point are ( x1, 1.1 x1 + .8) and the coordinates of the x = x2 point are ( x2, 1.1 x2 + .8). The rise between the two points is therefore rise = (1.1 x2 + .8) - (1.1 x1 + .8) = 1.1 x2 + .8 - 1.1 x1 - .8 = 1.1 x2 - 1.1 x1. The run is run = x2 - x1. The slope is therefore (1.1 x2 - 1.1 x1) / (x2 - x1) = 1.1 (x2 - x1) / (x2 - x1) = 1.1.
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RESPONSE --> I didn't subtract out my b-value (.8) until the process of calculating the slope, but otherwise did everything correctly. I understand how all this works. self critique assessment: 2
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22:09:23 `q002. For the function y = 3.4 x + 7, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2? What therefore is the rise between these two points, and what is the run? What is the average slope of the graph between these two points? Be sure to simplify your result.
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RESPONSE --> For the coordinates of the point x=x1 the x-value will be x1 and the y-value will be f(x1) or (3.4*x1+7). The coordinates are therefore (x1 , 3.4*x1+7). For the coordinates of the point x=x2 the x-value will be x2 and the y-value will be f(x2) or (3.4*x2+7). The coordinates are therefore (x2 , 3.4*x2+7). The rise between these two points will be (y2-y1), or (3.4*x2+7) - (3.4*x1+7) = 3.4*x1 + 7 - 3.4*x2 + 7 = 3.4*x1 - 3.4*x2 The run between these two points will be (x2-x1). The average slope will be: 3.4*x2 - 3.4*x1 / x2-x1 = 3.4(x2-x1) / x2-x1 = 3.4 / 1 = 3.4 confidence assessment: 2
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22:09:39 In terms of the symbol x1 the coordinates of the x = x1 point are ( x1, 3.4 x1 + 7) and the coordinates of the x = x2 point are ( x2, 3.4 x2 + 7). The rise between the two points is therefore rise = (3.4 x2 + 7) - (3.4 x1 + 7) = 3.4 x2 + 7 - 3.4 x1 - 7 = 3.4 x2 - 3.4 x1. The run is run = x2 - x1. The slope is therefore (3.4 x2 - 3.4 x1) / (x2 - x1) = 3.4 (x2 - x1) / (x2 - x1) = 3.4.
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RESPONSE --> I understand. self critique assessment: 2
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_Ɖ؛ZӴݪ˽D assignment #009 009. `query 9 Precalculus I 10-10-2007
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23:36:10 Symbolic calculation of slope, preliminary exercise What was the function, between which two points were you to calculate the average slope and how did you get this slope?
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RESPONSE --> The function was y = .1 x^2 - 3 We were to calculate the average slope between the points x = -2 and x = 7. For the first point I substituted the first x-value given into the function: y = .1 -2^2 - 3 y = .1 * 4 - 3 y = .4 - 3 y = -2.6 Giving a coordinate of (-2 , -2.6) For the second point I substituted the second x-value given into the function: y = .1 7^2 - 3 y = .1 * 49 - 3 y = 4.9 - 3 y = 1.9 Giving a coordinate of (7 , 1.9) Next I calculate the rise and the run between the points: Run = x2 - x1 = 7 - -2 = 9 Rise = y2 - y1 = 1.9 - -2.6 = 4.5 Slope = (y2 - y1) / (x2 - x1) = 4.5 / 9 = .5 confidence assessment: 2
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23:40:01 ** For the function y = .1 x^2 - 3 between x = -2 and x = 7 we get: slope = (y2 - y1) / (x2 - x1). For x1 = 2 and x2 = 7 we have y2 = .1 * 7^2 - 3 = 1.9 and y1 = .1 * 2^2 - 3 = -2.6, so slope = (1.9 - (-2.6) ) / ( 7 - 2) = 4.5 / 5 = .9. **
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RESPONSE --> I don't understand how the x1 value changed from a -2 as stated in the given problem to a 2 as stated in the solution. self critique assessment: 1
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23:42:43 problem 2 symbolic expression for slope, fn depth(t). What is the expression for the slope between the two specified t values?
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RESPONSE --> The slope equals the rise over the run: (y2 - y1) / (x2 - x1) = (depth(30) - depth(10)) / (30 - 10) = (depth(30) - depth(10)) / 20 confidence assessment: 2
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23:42:55 ** The function is given a name: depth(t). t values are 10 and 30. So rise = depth(30) - depth(10) and run = 30 - 10 = 20. Thus slope = [ depth(30) - depth(10) ] / 20 . **
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RESPONSE --> I understand. self critique assessment: 2
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23:44:24 What is the rise between the two specified t values?
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RESPONSE --> The rise between the two t-values: (y2 - y1) = [depth(30) - depth(10)] confidence assessment: 2
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23:44:32 ** The rise is the change in depth. The two depths are depth(10) and depth(30). The change in depth is final depth - initial depth, which gives us the expression depth(30)-depth(10) **
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RESPONSE --> self critique assessment:
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23:45:19 What is the run between the two specified t values?
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RESPONSE --> The run between the two specified t-values: (x2 - x1) = 30 - 10 = 20 confidence assessment: 2
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23:45:30 ** run = 30 - 10 = 20 **
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RESPONSE --> I understand. self critique assessment: 2
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23:46:55 What therefore is the slope and what does it mean?
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RESPONSE --> The slope, being the rise divided by the run, is therefore: [depth(30) - depth(10)] / 20 There is not much more we can do with this without knowing a function to apply it to. confidence assessment: 2
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23:47:54 ** rise = depth(30)-(depth(10) indicates change in depth. run = 30 - 10 = 20 = change in clock time. Slope = rise / run = (depth(30) - depth(10) ) / 20, which is the average rate at which depth changes with respect to clock time between t = 10 and t = 30. **
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RESPONSE --> I should've specified how the x- and y-values represented depth and clock time, respectively. self critique assessment: 1
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23:50:05 problem 5 graph points corresponding to load1 and load2 What are the coordinates of the requested graph points?
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RESPONSE --> The graph points corresponding to load = load1 are (3 , springLength(3)) The graph points corresponding to load = load2 are (10 , springLength(10)) confidence assessment: 2
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23:57:44 ** The horizontal axis is the 'load' asix, the vertical axis is the springLength axis. The load axis coordinates are load1 and load2. The corresponding spring lengths are springLength(load1) and springLength(load2). The springLength axis coordinates are springLength(load1) and springLength(load2). The graph points are thereofore (load1, springLength(load1) ) and (load2, springLength(load2) ). **
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RESPONSE --> I jumped the gun a little and had already substituted in the given load-values, but I understand how the general expressions work. self critique assessment: 1
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23:59:48 What is your expression for the average slope of the graph between load1 and load2?
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RESPONSE --> The average slope of the graph between load1 and load2: (y2-y1) / (x2-x1) = (springLength(10) - springLength(3)) / (10-3) (springLength(10) - springLength(3)) / 7 confidence assessment: 2
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00:00:43 ** rise = springLength(load2) - springLength(load1) run = load2 - load1 so slope = [ springLength(load2) - springLength(load1)] / (load2 - load1). **
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RESPONSE --> Again I mesread the question and substituted in the given load-values, but I understand how it works. self critique assessment: 1
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00:02:29 problem 6 symbolic expression for slope of depth function
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RESPONSE --> The symbolic expression for slope of the function: (y2-y1) / (x2-x1) = [depth(x2) - depth(x1)] / (x2-x1) confidence assessment: 2
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00:03:42 ** the name of the function is depth(t). We need the slope between t = t1 and t = t2. The depths are depth(t1) and depth(t2). Thus rise is depth(t2) - depth(t1) and run is t2 - t1. Slope is [ depth(t2) - depth(t1) ] / (t2 - t1). **
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RESPONSE --> I should've explained how I reached my rise and run. Also, I used the symbol ""x"" where I should've used ""t"". I understand how the function works though. self critique assessment: 1
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00:14:17 problem 8 average rate from formula f(t) = 40 (2^(-.3 t) ) + 25 intervals of partition (10,20,30,40) What average rate do you get from the formula? Show your steps.
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RESPONSE --> To estimate the average rate between the time intervals 10 and 20 I substitute the individual points of time into the function: f(10) = 40 (2^(-.3*10) ) + 25 f(10) = 40 2^(-3) + 25 f(10) = 40 (.125) + 25 f(10) = 5 + 25 f(10) = 30 f(20) = 40 (2^(-.3*20) ) + 25 f(20) = 40 2^(-6) + 25 f(20) = 40 (.0156) + 25 f(20) = .625 + 25 f(20) = 25.625 To find the average rate between the time intervals 10 and 20: (y2-y1) / (x2-x1) = (25.625 - 30) / (20 - 10) = (-4.375) / (10) = -.4375 To estimate the average rate between the time intervals 20 and 30 I substitute the individual points of time into the function: f(30) = 40 (2^(-.3*30) ) + 25 f(30) = 40 2^(-9) + 25 f(30) = 40 (.00195) + 25 f(30) = .0781 + 25 f(30) = 25.0781 To find the average rate between the time intervals 20 and 30: (y2-y1) / (x2-x1) = (25.0781 - 25.625) / (30 - 20) = (-.5469) / (10) = -.05469 To estimate the average rate between the time intervals 30 and 40 I substitute the individual points of time into the function: f(40) = 40 (2^(-.3*40) ) + 25 f(40) = 40 2^(-12) + 25 f(40) = 40 (.000244) + 25 f(40) = .00977 + 25 f(40) = 25.00977 To find the average rate between the time intervals 30 and 40: (y2-y1) / (x2-x1) = ( 25.00977- 25.0781) / (30 - 20) = (-.06833) / (10) = -.006833 confidence assessment: 2
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00:14:38 ** ave rate = change in depth / change in t. For the three intervals we get (f(20)-f(10))/(20-10) = (25.625 - 30 )/(20 - 10) = -4.375 / 10 = -.4375 (f(30)-f(20))/(30-20) = (25.07813 - 25.625)/(30 - 20) = -.5469 / 10 = -.05469. (f(40)-f(30))/(40-30) = (25.00977 - 25.07813)/(40 - 30) = -.0684 / 10 = -.00684. **
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RESPONSE --> I understand. self critique assessment: 2
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00:17:06 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Not exactly part of the assignment, but one of the very last steps in the class notes struck me as interesting. In explaining the basic form of the slope=slope equation; how it's possible to be rearranged, but is more efficient to remember and associate with a picture when in it's most basic and simplest form. self critique assessment: 2
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00:17:39 STUDENT RESPONSE: Ummm I know the slope formula is (y2-y1)/(x2-x1), but I always just put the number into the expression in the order I see them, but that is ok because I keep the order and get the correct answere because the y2,y1,x2,x1 or all relative. I am correct in doing this? INSTRUCTOR COMMENT: In other words you use (y1 - y2) / (x1 - x2) instead of (y2 - y1) / (x2 - x1). It's more conventional to regard, say, 10 as x1 and 20 as x2, so f(20) is y2 and f(10) is y1. If you start from the lower x number and change to the higher the difference is higher - lower, and this is the way we usually think about changes. According to this convention we calculate change in y as y2 - y1 and change in x as x2 - x1. You are doing (y1 - y2) / (x1 - x2) and you get a negative change in x, a negative denominator, and if you are thinking about change from the first quantity to the second this is backwards. However as you say both numerator and denominator follow the same order so you still get the right answer, since (y1-y2)/(x1-x2)= (y2-y1) / (x2-x1). **
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RESPONSE --> self critique assessment:
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