#$&* course Mth 163 2-14-13 1:13
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Given Solution: ** For the basic linear function f(x) = x the A = -.3 graph is obtained by vertically stretching the y = x function by factor -.3, resulting in a straight line thru the origin with slope -.3, basic points (0,0) and (1, -.3), and the A = 1.3 graph is obtained by vertically stretching the y = x function by factor 1.3, resulting in is a straight line thru the origin with slope 1.3, basic points (0,0) and (1, 1.3). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qdescribe the graphs of y = f(x) + c for c = .3 and c = -2.7 and compare; explain the comparison. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Both graphs have a slope of 1 and they are parallel. The graph of .3 is moved vertically up by .3 and -2.7 is moved down vertically by 2.7. confidence rating #$&*: 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:ok ********************************************* 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)=-1.77x1-3.87 Y2=f(x2)=-1.77x2-3.87 Slope=(-1.77x2-3.87 ,-1.77x1-3.87) / (x2-x1) = (-177(x2-x1)) / (x2-x1)= -177 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** y1 = f(x1) = -1.77 x1 - 3.87 y2 = f(x2) = -1.77 x2 - 3.87. `dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77 x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1). Thus slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77. This is the slope of the straight line, showing that these symbolic calculations are consistent. ** STUDENT QUESTION My question is how did you take -1.77 x2 + 1.77 x1 and get -1.77(x2 - x1)? I understand the x2-x1 but what happened to the 1.77? INSTRUCTOR RESPONSE This may be clearer if we work backwards: -1.77 * (x2 - x1) = -1.77 * x2 - (-1.77 * x1) = -1.77 x2 + 1.77 x1, which is the same thing as 1.77 x1 - 1.77 x2. -1.77 * (x2 - x1) was chosen as the form for the numerator, so we could easily divide it by x2 - x1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery problem 5. graphs of families for y = mx + b. Describe your graph of the family: m = 2, b varies from -3 to 3 by step 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Each graph has a slope of 2 and they are all parallel. The y intercepts move up one unit with each graph. confidence rating #$&*: 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: ********************************************* Question: `qquery problem 6. three basic points graph of y = .5 x + 1 what are your three basic points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (-1, .5) (0,1) (1, 1.5) (-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): I found the point lying to the left as well as the right. ------------------------------------------------ Self-critique 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): This is the same question as above. ------------------------------------------------ Self-critique Rating: ********************************************* 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): I set up my points with the force being the x and distance being the y because the distance the ball travels if dependent upon how much force you apply. If I set up the problem as distance as the x, my result would be the same because of the direct variation------ (that is a physics idea) The two points I chose are (.4, 1.5) and (.9, 3.5) I get the equations 1.5=.4x+b 3.5=.9x+b ------------------------------------------------ Self-critique Rating: ok ********************************************* 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: ok ********************************************* Question: `qwhat equation do you get from the slope and y-intercept? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Slope= 3.5-1.5 over .9-.4 = 2/ .5 = 4 Y intercept= 1.5=.4(4)+b = 1.5=1.6+b = -.1=b confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* Question: `qwhat is your linear regression model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using my graphing calculator and the original points I got the equation y=4.03x+.01 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Your linear regression model would be obtained using a graphing calculator or DERIVE. As a distance student you are not required to use these tools but you should be aware that they exist and you may need to use them in other courses. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat force would be required to hold the pendulum 47 centimeters from its equilibrium position? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation I got from using my calculator (y=4.03x+.01), I found that a force of 11.649 would be required to hold the pendulum at 47 cm. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* 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: Using y=4.03x+.01 I found that a force of 19.83 is needed to hold the pendulum at 80 m, but the problem was that the length of the string was only 8 meters. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: I used the equation f= .10*47+.21 and got the answer 15.41 which would be to much force to push or pull INSTRUCTOR COMMENT: ** The problem is that you can't hold a pendulum further at a distance greater than its length from its equilibrium point--the string isn't long enough. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qHow far could you hold the pendulum from its equilibrium position using a string with a breaking strength of 25 pounds? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation y=4.03x+.01and plugging 25 in for the force (in my case x) I got 100.76 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the average rate of change associated with this model? Explain this average rate in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the y=4.03x+.01 equation, I found that the slope is 4.03 confidence rating #$&*:: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium we can use any two (x, y) points to get the rate of change. In all cases we will get rate of change = change in y / change in x = .367. The change in y is the change in the force, while the change in x is the change in position. The rate of change therefore tells us how much the force changes per unit of change in position (e.g., the force increases by 15 pounds for every inch of displacement). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t exactly understand why this would be: “The change in y is the change in the force, while the change in x is the change in position” I am under the impression that the change in position happen as a RESULT of the amount of force one applies, making the y the change in position. ------------------------------------------------ Self-critique Rating: ok
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Given Solution: ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qAs you gradually pull the pendulum from a point 30 centimeters from its equilibrium position to a point 80 centimeters from its equilibrium position, what average force must you exert? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation y=4.03x+.01 and assuming force= x and distance= y 30=4.03x+.01 = x= 2999/ 403 or 7.44168 80=4.03x+.01 = x= 7999/403 or 19.8486 Average= (19.8486-7.44168) / (80-30) = 12.40692/50 or .248 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Again—I am using force= x and distance= y because the distance is a result of the force one applies. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery problem 8. flow range What is the linear function range(time)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since there were only two points, I connected a line of best fit and found two points on the line. I then plugged them into the slope formula and re-plugged them to find B. The resulting equation was Range(time)=-.952(time)+112.38 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I obtained model one by drawing a line through the data points and picking two points on the line and finding the slope between them. I then substituted this value for m and used one of my data points on my line for the x and y value and solved for b. the line I got was range(t) = -.95t + 112.38. y = -16/15x + 98 INSTRUCTOR COMMENT: This looks like a good model. According to the instructions it should however be expressed as range(time) = -16/15 * time + 98. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat is the significance of the average rate of change? Explain this average rate in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate is the change in range/ the change in time. This means that while the clock time changes, the change in range changes dependently. The rate of change is saying how fast the rate is changing with response to time. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* Question: `qWhat is the average slope associated with this model? Explain this average slope in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Slope is the average rate at which the range is changing. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* Question: `qquery problem 9. If your total wealth at clock time t = 0 hours is $3956, and you earn $8/hour for the next 10 hours, then what is your total wealth function totalWealth( t ), where t is time in hours? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Total(wealth) (t)= y Time=x It crosses the y axis at 3956 Slope=8 TotalWealth(t)=8t+3956 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* Question: `qAt what clock time will your total wealth reach $4000? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation y=8x+3956 4000=8x+3956 5.5=x confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* Question: `qWhat is the meaning of the slope of your graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Slope represents the rate at which you earn money (1 hour interval). The slope is increasing at a steady rate. 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: ok ********************************************* 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): Y= number sold X=selling price Number sold= -50(selling price) +1700 ------------------------------------------------ Self-critique Rating: ********************************************* 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