#$&* course Mth 163 2/16/13 around 11p.m. 010.
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Given Solution: The graph of y = x consists of a straight line through the origin, having slope 1. This line has basic points (0,0) and (1,1). The points of the graph of y = .5 x all lie twice as close to the x-axis as the points of the graph of y = x. The point (0,0) of the y = x graph is already on the x-axis, so the corresponding point on the graph of y = .5 x is also (0,0). The point (1,1) of the y = x graph lies 1 unit above the x-axis, so the corresponding point on the graph of y = .5 x will lie twice as close, or .5 units above the x-axis, so that the corresponding point is (1, .5). The graph of y = .5 x Thus passes through the points (0,0) and (1,.5). Of course this result could have been found by simply plugging 0 and 1 into the function y = .5 x, but the point here is to see that we can get the same result if we think of moving all points twice as close. This order thinking will be useful when dealing with more complex functions. Thinking along similar lines we expect the points of the graph of y = 2 x to all lie twice as far from the x-axis as the points of the function y = x. Thus the two basic points (0,0) and (1,1) of the y = x graph will correspond to the points (0,0) and (1,2) on the graph of y = 2 x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. If we were to sketch all the graphs of the form y = a x for which .5 < a < 2, what would our sketch look like? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Graphing y = .5x & y = 2x gives us the same graphs as the last question. Knowing that these 2 equations have the slops of 0.5 & 2 we know that all function will lie between those two points. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If a =.5 then our function is y = .5 x and the basic points will be (0,0) and (1,.5), as seen in the preceding problem. Similarly if a = 2 then our function is y = 2 x, with basic points (0,0) and (1,2). For .5 < a < 2, our functions will lie between the graphs of y = .5 x and y = 2 x. Since these two functions have slopes .5 and 2, the slopes of all the graphs will lie between .5 and 2. We could represent these functions by sketching dotted-line graphs of y = .5 x and y = 2 x (the dotted lines indicating that these graphs are not included in the family, because the < sign does not include equality). We could then sketch a series of several solid lines through the origin and lying between the two dotted-line graphs. STUDENT COMMENT I don’t know where to go from this point. I graphed the closest thing I could come up with but I don’t know how to explain what it is doing. INSTRUCTOR SUGGESTION You should graph the functions y = .5 x, y = .6 x, y = .8 x, y = 1.1 x, y = 1.5 x and y = 2 x, all on the same graph. Graph each function by plotting its two basic points (the x = 0 point and the x = 1 point), then sketching the straight line through these points. Using your graphs, estimate where the graph of y = .7 x, y = 1.3 x and y = 1.8 x lie. Then insert your description, according to instructions at the end of this document, along with any other work you do in response to other suggestions made below, and resubmit this document. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q003. Describe how the graphs of y = x - 2 and y = x + 3 compare with the graph of y = x. If we were to sketch all graphs of the form y = x + c for -2 < x < 3, what would our graph look like? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Both of these functions y = x-2 & y = x+3 lie parallel to the line y = x. The graph of y = x-2 is 2 times as far from the function y =x and y = x+3 lies 3 times as far from the function y =x. To sketch the graphs of y = x + c for -2 < x < 3 we can graph the 2 functions given and draw dotted lines showing that this function will lie between -2 & 3 slopes. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph of y = x - 2 lies at every point 2 units below the corresponding point on the graph of y = x, so this graph is parallel to the graph of y = x and 2 units lower. Similarly the graph of y = x + 3 lies parallel to the graph of y = x and 3 units higher. To sketch the family y = x + c for -2 < x < 3, we first can draw dotted-line graphs of y = x - 2 and y = x + 3, then a series of several solid line graphs, all parallel to the graph of y = x, lying between the two dotted-line graphs. STUDENT COMMENT: I got a little confused with y = x + c part, but I understand the first part completely. ** The instructions said to sketch all graphs of the form y = x + c for -2 < x < 3. So for example c could be -1, 0, 1 or 2, giving us the functions y = x - 1, y = x, y = x + 1 andy x+ 2. c could also be -1.9, or .432, or 2.9, giving us functions y = x - 1.9, y = x + .432, y = x + 2.9. c can be any number between -2 and 3. These graphs are as described in the given solution. ** STUDENT COMMENT i didnt really understand how to sketch y=x+c even after reading the intructors comments in the given solution INSTRUCTOR RESPONSE Suppose you were to graph y = x + c for c values -2, -1.9, -1.8, -1.7, ..., 2.8, 2.9, 3.0. This would include 50 graphs. Each of the 50 graphs would lie .1 unit higher than the one before it. The lowest of the graphs would be the c = -2 graph, y = x - 2. The highest of the graphs would be the c = 3 graph, y = x + 3. All the graphs would be parallel. If necessary, you can graph y = x - 2, then y = x - 1.9, then y = x - 1.8. You won't want to graph all 50 lines, but you could then skip to y = x + 2.8, y = x + 2.9 and y = x + 3. STUDENT COMMENT After reading the comments above I agree that I am a little confused. INSTRUCTOR RESPONSE You need to self-critique, giving me a detailed statement of what you do and do not understand about each line and each phrase in the given solution. You should in any case follow the suggestion at the end of the given solution. Graph the indicated graphs, then insert your explanation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. Describe how the graph of y = 2 x compares with the graph of y = x. Describe how the graph of y = 2 x - 2 compares with the graph of y = 2 x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As mentioned earlier the graph of y = 2x lies 2 times as far from the x axis as y = x. It has a slope of 2. The lines of function y = 2x-2 & y = 2x lie parallel from each other. The graph of y = 2x-2 has points that lie 2 units lower than the graph of y = 2x. It has a slope of y = 2. It passes through the origin as (0, -2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph of y = 2 x lies at every point twice as far the x-axis as that of y = x. This graph passes through the points (0,0) and (1, 2), i.e., passing through the origin with slope 2. The graph of y = 2x - 2 will lie 2 units below the graph of y = 2 x. This graph will therefore have a slope of 2 and will pass-through the y axis at (0, -2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q005. Suppose we graph y = 2 x + c for all values of c for which -2 < c < 3. What with our graph look like? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we were to graph this we would have graphs that lie c units from the graph of y =2x. All of the graphs will be straight lines passing through the y axis. The graphs will have slopes between -2 & 3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Each graph will lie c units vertically from the graph of y = 2 x, therefore having slope 2 the passing through the y-axis at the point (0, c). The family of functions defined by y = 2 x + c will therefore consist of a series of straight lines each with slope 2, passing through the y-axis between (0, -2) and (0, 3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q006. Sketch two points, not particularly close to one another, with one point in the second quadrant and the other in the first, with clearly different y values. Label the first point (x1, y1) and the second (x2, y2). Draw a straight line passing through both of these points and extending significantly beyond both. In terms of the symbols x1, x2, y1, and y2, what is slope of this straight line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No matter what points are picked the rise will be y2-y1 and the run will b x2-x1. The slope is therefore (y2-y1)/ (x2-x1). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise of a line is from y = y1 to y = y2, a rise of y2-y1. The run is similarly found to be x2-x1. The slope is therefore slope = (y2-y1) / (x2-x1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q007. On the sketch you made for the preceding problem, and add a point (x, y) on your straight line but not between the two points already labeled, and not too close to either. What is the slope from (x1, y1) to (x, y)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope would be y-y1 / x-x1. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope from (x1, y1) to (x, y) is slope = rise/run = (y - y1) / (x - x1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q008. Should the slope from (x1, y1) to (x, y) be greater than, equal to or less than the slope from (x1, y1) to (x2, y2)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slopes must be equal because all slopes on a line are equal. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope between any two points of a straight line must be the same. The two slopes must therefore be equal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q009. The slope from (x1, y1) to (x, y) is equal to the slope from (x1, y1) to (x2, y2). If you set the expressions you obtained earlier for the slopes equal to one another, what equation do you get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slopes are (y2 - y1) / (x2 - x1) and (y - y1) / (x - x1). If we set these equations equal to each other we get (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slopes are (y2 - y1) / (x2 - x1) and (y - y1) / (x - x1). Setting the two slopes equal we obtain the equation (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1). STUDENT COMMENT mine is the opposite but i think i would be the same INSTRUCTOR RESPONSE: Your solution was (y2 - y1) / (x2 - x1), if appropriate signs of grouping are inserted to reflect your obvious intent. The signs of both your numerator and denominator would be opposite the signs of the given solution (i.e., y2 - y1 = - (y1 - y2), and x2 - x1 = - (x1 - x2)). When divided the result would therefore be identical (negative / negative is positive). So your solution is completely equivalent to the given solution. However note that you need to group numerator and denominator. y2-y1/x2-x1 means divide y1 by x2, subract that result from y2 then subtract x1 from that. Not what you intended, though I know what you meant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q010. Setting the two slopes equal you should have obtained (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1). Solve this equation for y. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this equation for y we first have to multiply both sides by x-x1. This gives us (y - y1) = (y2 - y1) / (x2 - x1) * (x - x1). Add y1 to both sides and get y = (y2-y1)/ (x2-x1) * (x-x1) + y1 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Starting with (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1), we wish to isolate y on the left-hand side. We therefore begin by multiplying both sides by (x - x1) to obtain (y - y1) = (y2 - y1) / (x2 - x1) * (x - x1). We could then add y1 to both sides to obtain y = (y2 - y1) / (x2 - x1) * (x - x1) + y1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q011. What is the slope of the straight line defined by the points (2, 5) and (6, 8)? In terms of x and y, what is the slope of the straight line from (2, 5) to the unspecified point (x, y)? It (x, y) lies on the straight line from (2, 5) to (6, 8), then the slope from (2, 5) to (x, y) must be equal to the slope from (2, 5) to (6, 8). Your answers to the preceding two questions therefore constitute two expressions for the slope of the line, one expression being a simple fraction and the other a symbolic expression in terms of x and y. Set these two expressions equal to get an equation involving x and y. What is your equation? Solve your equation for y. What is your result? Do the coordinates (6, 8) satisfy your equation? Should they? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of the line with points (6,8) & (2, 5) would be 8-5/ 6-2. Simplifying this gives us ¾ = 0.75. The slope of a straight life from (2,5) to (x,y) would be y - 5 / x-2. Setting these two equations equal to each other gives us (y-5)/ (x-2) = 0.75 Solving for y we get y = (x-2) *.75 +5 The coordinates (6,8) would make the equation 8 = (6-2) *.75 +5. These points do satisfy the equation as they should.
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Given Solution: ** For the basic linear function f(x) = x the A = -.3 graph is obtained by vertically stretching the y = x function by factor -.3, resulting in a straight line thru the origin with slope -.3, basic points (0,0) and (1, -.3), and the A = 1.3 graph is obtained by vertically stretching the y = x function by factor 1.3, resulting in is a straight line thru the origin with slope 1.3, basic points (0,0) and (1, 1.3). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qdescribe the graphs of y = f(x) + c for c = .3 and c = -2.7 and compare; explain the comparison. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All of the slopes were identical and had y intercepts from -2.7 to .3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graphs will have slopes identical to that of the original function, but their y intercepts will vary from -2.7 to .3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 4. linear function y = f(x) = -1.77 x - 3.87 What are your symbolic expressions, using x1 and x2, for the corresponding y coordinates y1 and y2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y1 = f(x1) = -1.77 x1 - 3.87 y2 = f(x2) = -1.77 x2 - 3.87. `dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77 x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1). The slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** y1 = f(x1) = -1.77 x1 - 3.87 y2 = f(x2) = -1.77 x2 - 3.87. `dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77 x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1). Thus slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77. This is the slope of the straight line, showing that these symbolic calculations are consistent. ** STUDENT QUESTION My question is how did you take -1.77 x2 + 1.77 x1 and get -1.77(x2 - x1)? I understand the x2-x1 but what happened to the 1.77? INSTRUCTOR RESPONSE This may be clearer if we work backwards: -1.77 * (x2 - x1) = -1.77 * x2 - (-1.77 * x1) = -1.77 x2 + 1.77 x1, which is the same thing as 1.77 x1 - 1.77 x2. -1.77 * (x2 - x1) was chosen as the form for the numerator, so we could easily divide it by x2 - x1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 5. graphs of families for y = mx + b. Describe your graph of the family: m = 2, b varies from -3 to 3 by step 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All of the slopes are 2 obviously because m is the slope and equals 2 and they will all cross the y axis at points between -3 and 3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graphs will all have slope 2 and will pass thru the y axis between y = -3 and y = 3. The family will consist of all such graphs. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 6. three basic points graph of y = .5 x + 1 what are your three basic points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = mx + b with b= 1. So the first point is y intercept is (0, 1). The others are (1, 1.5) and (-2, 0). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This is of the form y = mx + b with b= 1. So the y intercept is (0, 1). The point 1 unit to the right is (1, 1.5). The x-intercept occurs when y = 0, which implies .5 x + 1 = 0 with solution x = -2, so the x-intercept is (-2, 0). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 6. three basic points graph of y = .5 x + 1 What are your three basic points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The same points as before (0, 1), (1, 1.5), & (-2, 0). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The y intercept occurs when x = 0, at which point we have y = .5 * 0 + 1 = 1. So one basic point is (0, 1). The point 1 unit to the right of the y axis occurs at x = 1, where we get y = .5 * 1 + 1 = 1.5 to give us the second basic point (1, 1.5) }The third point, which is not really necessary, is the x intercept, which occurs when y = 0. This gives us the equation 0 = .5 x + 1, with solution x = -2. So the third basic point is (-2, 0). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 7. simple pendulum force vs. displacement What are your two points and what line do you get from the two points? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok My two points were (1,1, & .21) & (2.0, .54) ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qSTUDENT RESPONSE: The two points are (1.1, .21) and (2.0, .54). These points give us the two simultaneous equations .21- m(1.1) + b .54= m(2.0) +b. If we solve for m and b we will get our y = mx + b form. INSTRUCTOR COMMENT: I believe those are data points. I doubt if the best-fit line goes exactly through two data points. In the future you should use points on your sketched line, not data points. However, we'll see how the rest of your solution goes based on these points. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qwhat equation do you get from the slope and y-intercept? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope is m = .367 and the y-intercept is b = -.193. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: b= .21 m=.19 INSTRUCTOR COMMENT: ** b would be the y intercept, which is not .21 since y = .21 when x = 1.1 and the slope is nonzero. If you solve the two equations above for m and b you obtain m = .367 and b = -.193. This gives you equation y = mx + b or y = .367 x - .193. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is your linear regression model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Positive going up semi-steeply. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Your linear regression model would be obtained using a graphing calculator or DERIVE. As a distance student you are not required to use these tools but you should be aware that they exist and you may need to use them in other courses. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat force would be required to hold the pendulum 47 centimeters from its equilibrium position? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We must use the equation y = .367 x - .193, with y = force and x= number of cm from equilibrium, then we have x = 47 and we get force = y = .367 * 47 - .193 = 17 Therefore the force would be 17 force units confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If your model is y = .367 x - .193, with y = force and x= number of cm from equilibrium, then we have x = 47 and we get force = y = .367 * 47 - .193 = 17 approx. The force would be 17 force units. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhy would it not make sense to ask what force would be necessary to hold the pendulum 80 meters from its equilibrium position? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No. It would not make since because the string is not long enough confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: I used the equation f= .10*47+.21 and got the answer 15.41 which would be to much force to push or pull INSTRUCTOR COMMENT: ** The problem is that you can't hold a pendulum further at a distance greater than its length from its equilibrium point--the string isn't long enough. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow far could you hold the pendulum from its equilibrium position using a string with a breaking strength of 25 pounds? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used the equation y = .367 * x - .193 with y = force = 25 lbs we plug in the answers geting the equation 25 = .367 x - .193 solving for x we get that x = 69 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the model y = .367 * x - .193 with y = force = 25 lbs we get the equation 25 = .367 x - .193, which we solve to obtain x = 69 (approx.). Note that this displacement is also unrealistic for this pendulum. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the average rate of change associated with this model? Explain this average rate in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Again using the equation y = .367 * x - .193 with y = force and x = displacement. To get the rate of change we can use slope = change in y / change in x = .367. The change in y is the change in the force, while the change in x is the change in position. The rate of change then tells how much the force changes by change in position. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium we can use any two (x, y) points to get the rate of change. In all cases we will get rate of change = change in y / change in x = .367. The change in y is the change in the force, while the change in x is the change in position. The rate of change therefore tells us how much the force changes per unit of change in position (e.g., the force increases by 15 pounds for every inch of displacement). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the average slope associated with this model? Explain this average slope in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation again y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qAs you gradually pull the pendulum from a point 30 centimeters from its equilibrium position to a point 80 centimeters from its equilibrium position, what average force must you exert? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Force at 30 cm: y = .367 * 30 - .193 = 10.8 Force at 80 cm: y = .367 * 80 - .193 = 29 ave force between 30 cm and 80 cm is (10.8 + 29) / 2 = 20 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** if it was possible to pull the pendulum back this far and if the model applies you will get Force at 30 cm: y = .367 * 30 - .193 = 10.8 approx. and Force at 80 cm: y = .367 * 80 - .193 = 29 approx. so that ave force between 30 cm and 80 cm is therefore (10.8 + 29) / 2 = 20 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 8. flow range What is the linear function range(time)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Range(time) = -16/15 * time + 98. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I obtained model one by drawing a line through the data points and picking two points on the line and finding the slope between them. I then substituted this value for m and used one of my data points on my line for the x and y value and solved for b. the line I got was range(t) = -.95t + 112.38. y = -16/15x + 98 INSTRUCTOR COMMENT: This looks like a good model. According to the instructions it should however be expressed as range(time) = -16/15 * time + 98. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the significance of the average rate of change? Explain this average rate in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate of change = change in range / change in clock time. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** the average rate of change is change in range / change in clock time. The average rate of change indicates the average rate at which range in cm is changing with respect to clock time in sec, i.e., the average number of cm / sec at which the range changes. Thus the average rate tells us how fast, on the average, the range changes. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the average slope associated with this model? Explain this average slope in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average slope of this model is the rate at which the range of the flow changes--the average rate at which the position of the end of the stream changes. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** it's the average rate at which the range of the flow changes--the average rate at which the position of the end of the stream changes. It's the speed with which the point where the stream reaches the ground moves across the ground. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 9. If your total wealth at clock time t = 0 hours is $3956, and you earn $8/hour for the next 10 hours, then what is your total wealth function totalWealth( t ), where t is time in hours? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would give us y intercept 3956, since that is the t = 0 value, and slope 8, because slope represents change in total wealth / change in t, i.e., the number of dollars per hour. So we get that TotalWealth(t) = 8 * t + 3956 . confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would have y intercept 3956, since that is the t = 0 value, and slope 8, since slope represents change in total wealth / change in t, i.e., the number of dollars per hour. A graph with y-intercept b and slope m has equation y = m t + b. Thus we have totalWealth(t) = 8 * t + 3956 . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qAt what clock time will your total wealth reach $4000? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used the equation totalWealth(t) = 4000 = 8 t + 3956 T = 5.5 hours confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: To find the clock time when my total wealth will reach 4000 I solved the equation totalWealth(t) = 4000. The value I got when I solved for t was t = 5.5 hours. 4.4 hours needed to reach 4000 4000 = 10x + 3956 INSTRUCTOR COMMENT: Almost right. You should solve 4000 = 8 x + 3956, obtaining 5.5 hours. This is equivalent to solving totalWealth(t) = 4000 = 8 t + 3956, which is the more meaningful form of the relationship. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the meaning of the slope of your graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Slope = steady rate at which money is earned on an hourly basis. Just a steady increase in weath confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: GOOD STUDENT RESPONSE: The slope of the graph shows the steady rate at which money is earned on an hourly basis. It shows a steady increase in wealth. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300. What linear function numberSold(price) describes this situation? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300. What linear function numberSold(price) describes this situation? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qIf you make a graph of y = numberSold vs. x = price you have graph points (30, 200) and (28, 300). You need the equation of the straight line through these points. You plug these coordinates into the form y = m x + b and solve for m and b. Or you can use another method. Whatever method you use you get y = -50 x + 1700. Then to put this into the notation of the problem you write numberSold(price) instead of y and price instead of x. You end up with the equation numberSold(price) = -50 * price + 1700. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qIf the store must meet a quota by selling 220 units per week, what price should they set? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We use the equation of 220 = -50 * price + 1700 Solving for it we get price = 30 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If the variables are y and x, you know y so you can solve for x. For the function numberSold(price) = -50 * price + 1700 you substitute 220 for numbersold(price) and solve for price. You get the equation 220 = -50 * price + 1700 which you can solve to get price = 30, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qIf each widget costs the store $25, then how much total profit will be expected from selling prices of $28, $29 and $30? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t feel there is enough info to solve this confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: If each widget costs the store $25, then they should expect to earn a profit of 300 dollars from a selling price of $28, 250 dollars from a price of $29 and 200 dollars from a price of $30. To find this I solved the equations numberSold(28); numberSold(29), and numberSold(30). Solving for y after putting the price values in for p. They will sell 300, 250 and 200 widgets, respectively (found by solving the given equation). To get the total profit you have to multiply the number of widgets by the profit per widget. At $28 the profit per widgit is $3 and the total profit is $3 * 300 = $900; at $30 the profit per widgit is $5 and 200 are sold for profit $1000; at $29 what happens? ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I still don’t understand
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Given Solution: ** The t = 20 point is (20,64) and the t = 60 point is (60, 16), so the slope is (-48 / 20) = -1.2. This can be plugged into the form y = m t + b to get y = -1.2 t + b. Then plugging in the x and y coordinates of either point you get b = 88. y = -1.2 t + 88 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 13. quadratic depth function y = depth(t) = .01 t^2 - 2t + 100. What is `dy / `dt based on the two time values t = 30 sec and t = 40 sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For t = 30 we get y = 49 and for t = 40 we get y = 36. Therefore the slope between (30, 49) and (40,36) is (36 - 49) / (40 - 30) = -1.3. So the depth is changing at an average rate of -1.3 cm / sec confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For t = 30 we have y = 49 and for t = 40 we have y = 36. The slope between (30, 49) and (40,36) is (36 - 49) / (40 - 30) = -1.3. This tells you that the depth is changing at an average rate of -1.3 cm / sec. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is `dy / `dt based on t = 30 sec and t = 31 sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We get the same for 30 and for t = 31 we get y = 47.61. The slope is then -1.39 confidence rating #$&*: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Based on t = 30 and t = 31 the value for `dy / `dt is -1.39, following the same steps as before ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 3 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is `dy / `dt based on t = 30 sec and t = 30.1 sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We know when t = 30 y = 49 and when t = 30.1 y = 48.86. So the slope is -1.399 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: The value for 'dy / `dt based on t = 30 sec and t = 30.1 sec is -1.4 INSTRUCTOR COMMENT: ** Right if you round off the answer. However the answer shouldn't be rounded off. Since you are looking at a progression of numbers (-1.3, -1.39, and this one) and the differences in these numbers get smaller and smaller, you have to use a precision that will always show you the difference. Exact values are feasible here and shoud be used. I believe that this one comes out to -1.399. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat do you think you would get for `dy / `dt if you continued this process? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The limiting value would be -1.4 so it would be answers around that. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: An even more and more accurate slope value. I don't think it would have to continue to decrease. INSTRUCTOR COMMENT **If you look at the sequence -1.3, -1.39, -1.399, ..., what do you think happens? It should be apparent that the limiting value is -1.4 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat does the linear function tell you? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The linear function tells you that at any clock time t the rate of depth change is given by the function .02 t - 2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The function tells you that at any clock time t the rate of depth change is given by the function .02 t - 2. For t = 30, for example, this gives us .02 * 30 - 2 = -1.4, which is the rate we obtained from the sequence of calculations above. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 14. linear function y = f(x) = .37 x + 8.09 Y = .37*1 + 8.09 y = .37*2 +8.09 y= .37* 3 +8.09 Y = 8.46 y = 8.83 y = 9.2 As x increases y increase &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat are the first five terms of the basic sequence {f(n), n = 1, 2, 3, ...} for this function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first five terms are 8.46, 8.83, 9.2, 9.57, and 9.94 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The first five terms are 8.46, 8.83, 9.2, 9.57, and 9.94 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat is the pattern of these numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The numbers increase by .37 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** These numbers increase by .37 at each interval. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qIf you didn't know the equation for the function, how would you go about finding the 100th member of the sequence? How can you tell your method is valid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You could do this because you know that there are 99 times between the first number and the 100th number. So multiply 99 by .37 then add it to the first value confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You could find the 100th member by noting that you have 99 ‘jumps’ between the first number and the 100 th, each ‘jump’ being .37. Multiplying 99 times .37 and then adding the result to the 'starting value' (8.46). STUDENT RESPONSE: simply put 100 as the x in the formula .37x +8.09 INSTRUCTOR COMMENT: That's what you do if you have the equation. Given just the numbers you could find the 100th member by multiplying 99 times .37 and then adding the result to the first value 8.46. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qfor quadratic function y = g(x) = .01 x^2 - 2x + 100 what are the first five terms of the basic sequence {g(n), n = 1, 2, 3, ...}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y = .01*1^2 - 2*1 +100 = 98.01 y = 01*2^2 - 2*2 +100 = 96.04 y = 01*3^2 - 2*3+100 = 94.09 As x increases y decreases The first 5 terms are 98.01, 96.04, 94.09, 92.16, 90.25 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We have g(1) = .01 * 1^2 - 2 * 1 + 100 = 98.01 g(2) = .01 * 2^2 - 2 * 2 + 100 = 96.04, etc. The first 5 terms are therefore {98.01, 96.04, 94.09, 92.16, 90.25} &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the pattern of these numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With each interval of x, y changes .02 more than the previous number confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The changes in these numbers are -1.97, -1.95, -1.93, -1.91. With each interval of x, the change in y is .02 greater than for the previous interval. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qIf you didn't know the equation for the function, how would you go about finding the next three members of the sequence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can look at the pattern before knowing that the next three changes are -1.89, -1.87, -1.85. So we get g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** According to the pattern established above, the next three changes are -1.89, -1.87, -1.85. This gives us g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow can you verify that your method is valid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you evaluate 5, 6, & 7 we can confirm these results. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You can verify the result using the original formula; if you evaluate it at 5, 6 and 7 it should confirm your results. That's the best answer that can be given at this point. You should understand, though that even if you verified it for the first million terms, that wouldn't really prove it (who knows what might happen at the ten millionth term, or whatever). It turns out that to prove it would require calculus or the equivalent. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qquery problem 15. The difference equation a(n+1) = a(n) + .4, a(1) = 5 If you substitute n = 1 into a(n+1) = a(n) + .4, how do you determine a(2)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a(1+1) = a(1) + .4, or a(2) = a(1) + .4. Knowing a(1) = 5 you get a(2) = 5.4. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You get a(1+1) = a(1) + .4, or a(2) = a(1) + .4. Knowing a(1) = 5 you get a(2) = 5.4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q If you substitute n = 2 into a(n+1) = a(n) + .4 how do you determine a(3)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Just the same as before: You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) + .4 Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You have to do the substitution. You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) + .4 Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qIf you substitute n = 3 into a(n+1) = a(n) + .4, how do you determine a(4)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same as previously get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is a(100)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (100) would be equal to a(1) plus 99 jumps as we could got earlierof .4, or 5 + 99*.4 = 44.6. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** a(100) would be equal to a(1) plus 99 jumps of .4, or 5 + 99*.4 = 44.6. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 17. difference equation a(n+1) = a(n) + 2 n, a(1) = 4. What is the pattern of the sequence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For n = 1 we have n+1 = 2 so that the equation a(n+1) = a(n) + 2 n becomes a(2) = a(1) + 2 * 1 Since a(1) = 4 (this was given) we have a(2) = a(1) + 2 * 1 = 4 + 2 = 6. Reasoning similarly, n = 2 gives us a(3) = a(2) + 2 * 2 = 6 + 4 = 10. n = 3 gives us a(4) = a(3) + 2 * 3 = 10 + 6 = 16; etc. The sequence is 4, 6, 10, 16, 24, 34, etc. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For n = 1 we have n+1 = 2 so that the equation • a(n+1) = a(n) + 2 n becomes • a(2) = a(1) + 2 * 1 Since a(1) = 4 (this was given) we have • a(2) = a(1) + 2 * 1 = 4 + 2 = 6. Reasoning similarly, n = 2 gives us • a(3) = a(2) + 2 * 2 = 6 + 4 = 10. n = 3 gives us • a(4) = a(3) + 2 * 3 = 10 + 6 = 16; etc. The sequence is 4, 6, 10, 16, 24, 34, ... . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat kind of function do you think a(n) is (e.g., linear, quadratic, exponential, etc.)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Quadratic confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The differences of the sequence are 4, 6, 8, 10, 12, . . .. The difference change by the same amount each time, which is a property of quadratic functions. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qquery the slope = slope equation Explain the logic of the slope = slope equation (your may take a little time on this one) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope = slope equation sets the slope between two given points equal to the slope between one of those points and the variable point (x, y). Since all three points lie on the same straight line, the slope between any two of the three points must be equal to the slope between any other pair. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The slope = slope equation sets the slope between two given points equal to the slope between one of those points and the variable point (x, y). Since all three points lie on the same straight line, the slope between any two of the three points must be equal to the slope between any other pair. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 7. streamRange(t), 50 centimeters at t = 20 seconds, range changes by -10 centimeters over 5 seconds. what is your function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At t = 20 sec this would give you y = -2 * 20 + 50 = 10. But y = 50 cm when t = 20 sec. Slope is -10 cm / (5 sec) = -2 cm/s, so you have y = -2 t + b. Plug in y = 50 cm and t = 20 sec and b = 90 cm. The equation or Stream range y = -2 t + 90 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The rate at which streamRange changes is change in streamRange / change in t = -10 cm / (5 sec) = -2 cm/s. This will be the slope m of the graph. Since streamRange is 50 cm when t = 20 sec the point (20, 50) lies on the graph. So the graph passes through (20, 50) and has slope -2. The function is therefore of the form y = m t + b with m = -2, and b such that 50 = -2 * 20 + b. Thus b = 90. The function is therefore y = -2 t + 90, or using the meaningful name of the function steamRange(t) = -2t + 90 You need to use function notation. y = f(x) = -2x + 90 would be OK, or just f(x) = -2x + 90. The point is that you need to give the funcion a name. Another idea here is that we can use the 'word' streamRange to stand for the function. If you had 50 different functions and, for example, called them f1, f2, f3, ..., f50 you wouldn't remember which one was which so none of the function names would mean anything. If you call the function streamRange it has a meaning. Of course shorter words are sometimes preferable; just understand that function don't have to be confined to single letters and sometimes it's not a bad idea to make the names easily recognizable. STUDENT RESPONSE: y = -2x + 50 INSTRUCTOR COMMENT: ** At t = 20 sec this would give you y = -2 * 20 + 50 = 10. But y = 50 cm when t = 20 sec. Slope is -10 cm / (5 sec) = -2 cm/s, so you have y = -2 t + b. Plug in y = 50 cm and t = 20 sec and solve for b. You get b = 90 cm. The equation is y = -2 t + 90, or streamRange(t) = -2t + 90. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qwhat is the clock time at which the stream range first falls to 12 centimeters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Stream Range(t) = -2t + 90 So would set Stream Range(t) = 12 and solve 12 = -2t + 90 and t = 39 sec confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the correct equation streamRange(t) = -2t + 90, you would set streamRange(t) = 12 and solve 12 = -2t + 90, obtaining t = 39 sec. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 9. equation of the straight line through t = 5 sec and the t = 7 sec points of the quadratic function depth(t) = .01 t^2 - 2t + 100 What is the slope and what does it tell you about the depth function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Start with y = m t + b, y = -1.88 t + b. Solve for b. Plug in the coordinates of the t = 7 and get 90.9 = -1.88 * 7 + b so b = 104 y = depth(t) = -1.88 t + 104. confidence rating #$&*: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You have to get the whole equation. y = m t + b is now y = -1.88 t + b. You have to solve for b. Plug in the coordinates of the t = 7 point and find b. You get 90.9 = -1.88 * 7 + b so b = 104, approximately. Find the correct value. The equation will end up something like y = depth(t) = -1.88 t + 104. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 3 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qThe slope of the linear function is -1.88. This tells me that the depth is decreasing as the time is increasing at a rate of 1.88 cm per sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow closely does the linear function approximate the quadratic function at each of the given times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The deviations are for t = 3, 4, 5, 6, 7, & 8 are .08, .03. 0. -.01, 0, .03. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The deviations are for t = 3, 4, 5, 6, 7, & 8 as follows: .08, .03. 0. -.01, 0, .03. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qat what t value do we obtain the closest values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t = 6, the deviation for this is -.01 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Not counting t= 5 and t = 7, which are 0, the next closest t value is t = 6, the deviation for this is -.01. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qOn which side of the t = 5 and t = 7 points is the linear approximation closer to the quadratic function? On which side does the quadratic function 'curve away' from the linear most rapidly? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When t = 4 the approximation is closer. The quadratic function curves away toward positive x side confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** On the t = 4 side the approximation is closer. The quadratic function curves away on the positive x side. ** Query Add comments on any surprises or insights you experienced as a result of this assignment. The slope = slope helped me out a lot. Learning that I can solve a linear in different ways was helpful. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 STUDENT COMMENT I found the difference equation to be a bit challenge to comprehend (it seems it can get pretty complicated) but very exciting as well. I'm still not entirely sure what uses it will have in the future, but it seems like an important concept to have for future reference. INSTRUCTOR RESPONSE The difference equation is a way of specifying how a quantity changes, step by step. There are numerous situations in which all was know is the initial value of a quantity and the rules for how it changes. • For example when water flows from a hole in the bottom of a uniform cylinder, it is the depth of water that determines how fast it comes out. • All we know, then, is the initial depth of the water and the rule for how quickly the depth changes. • It turns out that we can approximate the behavior of the depth function using the difference equation y(n+1) = y(n) - k * sqrt(y(n)), where k is a constant number determined by the diameter of the cylinder. If you continue your study of mathematics you will eventually get to the fourth semester of the standard calculus sequence, a course entitled 'Introduction to Ordinary Differential Equations'. Most second-semester calculus courses also include a briefer introduction to the subject. Your exposure to difference equations in this course will be usefu helpful to you when you reach that point. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: STUDENT COMMENT I found the difference equation to be a bit challenge to comprehend (it seems it can get pretty complicated) but very exciting as well. I'm still not entirely sure what uses it will have in the future, but it seems like an important concept to have for future reference. INSTRUCTOR RESPONSE The difference equation is a way of specifying how a quantity changes, step by step. There are numerous situations in which all was know is the initial value of a quantity and the rules for how it changes. • For example when water flows from a hole in the bottom of a uniform cylinder, it is the depth of water that determines how fast it comes out. • All we know, then, is the initial depth of the water and the rule for how quickly the depth changes. • It turns out that we can approximate the behavior of the depth function using the difference equation y(n+1) = y(n) - k * sqrt(y(n)), where k is a constant number determined by the diameter of the cylinder. If you continue your study of mathematics you will eventually get to the fourth semester of the standard calculus sequence, a course entitled 'Introduction to Ordinary Differential Equations'. Most second-semester calculus courses also include a briefer introduction to the subject. Your exposure to difference equations in this course will be usefu helpful to you when you reach that point. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!