#$&* course Mth 163 1/21/13 around 12:00pm Question: `q001. Sketch a set of coordinate axes representing y vs. x, with y on the vertical axis and x on the horizontal axis.
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Given Solution: The x coordinates 1, 3 and 7 match the x coordinates of the three given points, the y coordinates will be the y coordinates -2, 5 and 8, respectively, of those points. At x = 5 the precise value of x, for a perfect parabola, would be 8 1/3, or about 8.333. You are unlikely to have drawn a perfect parabola and your estimate will almost certainly differ from this value. Any estimate with .5 or so of this value would be a good estimate. Drawn with complete accuracy a parabola through these points will peak between x = 3 in and x = 7. The peak of the actual parabola will occur close to x = 6. Any estimate between x = 5 and x = 7 is pretty good, and even an estimate between x = 7 and x = 8 is not unreasonable. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q002. Using your sketch from the preceding exercise, estimate the x coordinates corresponding to y coordinates 1, 3, 5 and 7. Also estimate the x values at which y is 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When estimating the x coordinates that respond to the y coordinates we get y coordinate 1 corresponds to the x coordinate 2; y coordinate 3 corresponds to the x coordinate 2.5; y coordinate 5 corresponds to the x coordinate 3; y coordinate 7 corresponds to the x coordinate 4.5. When y = 0, x = 1.55. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The easiest way to estimate your points would be to make horizontal lines on your graph at y = 1, 3, 5 and 7. You would easily locate the points were these lines intersect your graph, then estimate the x coordinates of these points. For the actual parabola passing through the given points, y will be 1 when x = 1.7 (and also, if your graph extended that far, near x = 10). y = 3 near x = 2.3 (and near x = 9.3). y = 5 at the given point (3, 5), where x = 3. y = 7 near x = 4 (and also near x = 7.7). Any set of estimates in the vicinity of these values is good. STUDENT COMMENT: Im still not clear on this. I didnt get the same as you and my graph went to 12. I know when we estimate we will not get the same answer, but did I do this right? If not what do I do to correct it? INSTRUCTOR RESPONSE: The following commentary need not be self-critiqued, though you may certainly ask questions if you wish. However I have two reasons for including it: The estimates here were given by a good student whose graph for some reason didn't match the points on this particular problem. If this student missed his/her estimates, then most students will need some clarification at this point. The discussion and the graphs shown here should be helpful to you. You should be able to sketch these graphs, label the coordinate axes based on the coordinates of the three points, and make reasonable verification of the values given here. This situation is closely related to the work we will be doing in the first few assignments, where we will actually learn to do parabolic or quadratic curve fits. Instructor response to student: Some of your estimates were reasonable, but others weren't. Your solution was if y = 0 then x = 0 if y = 1 then x = approx. 1.7 if y = 3 then x = approx. 4.6 if y = 5 then x = approx. 8.1 if y = 7 then x = approx. 11.5 If the x coordinates 1,3 7 match the y coordinates -2,5,8, then y = 1 at x = 1.7 is a very reasonable estimate y = 3 at x = 4.6 is not a reasonable estimate. By the time x = 3, y already has a value of 5, and is increasing toward value y = 8 at x = 7. For an x value between 3 and 5 we would expect y to be greater than 5; we would not expect a y value of 3 to occur on this x interval. y = 5 at x = 8.1 is a reasonable estimate. When x = 7 we have y = 8, and it's possible to draw a reasonable graph on which y decreases to 5 by the time x = 8. y = 7 at x = 11.5 isn't consistent with your previous estimates. You already have y down to value 5 by the time x = 8.1, and there's no reason to expect the graph to turn around and go back up to 7 by the time x reaches 11.5. The figure below depicts the three data points. Your graph should have similar form. The graph below shows the parabola that actually passes through these points. Your graph won't necessarily look the same, since you probably aren't practiced at drawing an accurate parabola through three given points, and besides nobody told you to use a parabola. So it's no problem if your graph differs from this one, but this 'ideal' graph is the basis for the numbers in the given solution. For this graph, y = 0 somewhere between x = 1 and x = 3, closer to x = 1; we might estimate that this occurs at x = 1.5 (a fairly precise value for the actual parabola is about x = 1.476). You will learn a lot more about parabolas within the next couple of weeks. In fact, by Assignment 4 or 5, you will have learned how to create and use a parabolic model for any three points, and you'll see that while it's not really what you'd call easy, it's not all that difficult either. The graph below isn't parabolic, but it's still a smooth curve and passes through the given points, and your graph could well look like this. The y value at x = 5 appears to be pretty close to the x = 7 value, which is y = 8. The y = 0 value again occurs between x = 1 and x = 3; perhaps a little to the left of the point in the above graph, maybe at x = 1.4. Another reasonable graph is shown below. At x = 5 the y value is probably around 7. The graph below agrees with the estimates you gave. While it fits the overall trend of the data by 'staying in the middle', it doesn't fit any of the three given points very well. This graph has been included so you can verify that it agrees reasonably with your given values if y = 0 then x = 0 if y = 1 then x = approx. 1.7 if y = 3 then x = approx. 4.6 if y = 5 then x = approx. 8.1 if y = 7 then x = approx. 11.5 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): What is the answer tox =? When y =0??? I read where it was x = 0 and where x = somewhere between 1 & 3 ------------------------------------------------ Self-critique Rating: 1
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Given Solution: To find the profit for a selling price of x = 4 dollars, we would look at the x = 4 point on the graph. This point is easily located by sketching a vertical line through x = 4. Projecting over to the y-axis from this point, you should have obtained an x value somewhere around 6 of 7, depending on how you drew the graph. This represents a profit of about 6 or 7 cents. The profit is the y value, so to obtain the selling price x corresponding to a profit of y = 7 we sketch the horizontal line at y = 7, which as in a preceding exercise will give us x values of about 4 or 5 (y = 7 would also occur at x = 7.7, approx., if the entire parabola was drawn). STUDENT COMMENT: I think I'm doing my graphs wrong. Can you help? INSTRUCTOR RESPONSE: See also my notes on the preceding problems, especially the ones with the graphs. The graph depicted below doesn't fit the original points very well, but we'll use it to illustrate the situation. As before you should review the coordinates of the three given points and make sense of them on this graph. The 'red' line is a vertical line through x = 4. This line is drawn first. The 'green' horizontal line is then drawn through the point where the x = 4 line intersects the graph. You should verify for yourself that the 'green' line intersects the y axis around y = 6 (the actual coordinate is about y = 6.2, but you aren't expected to be able to verify the value that accurately; you should however be able to tell that the coordinate is closer to 6 than to 5 or 7, based on the coordinates of the three given points). This would tell us that if the selling price is around 4 units, the profit is around 6 units. To find the selling price that will bring a profit of 7 units, we first draw the horizontal 'blue' line through y = 7. Then we draw the vertical 'purple' line, through the point where our horizontal line intersects the graph. This line intersects the x axis somewhere between x = 4 and x = 5. It's hard to tell whether the x value is closer to 4 than to 5, so we might estimate that this occurs around x = 4.5 (the actual value as depicted is about x = 4.43). This would tell us that to achieve a profit of 7 units, the selling price should be about 4.5 units. Your estimates will of course differ from these, depending on how you drew your graph. For example, if the parabola is taken as the 'perfect' model for the data, the results might vary from these by as much as a unit. Worth noting, even though you probably don't understand it yet: For a number of reasons you will soon understand, it turns out that the parabola is a very reasonable model for estimating how price and profit are related. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did my numbers backwards. I put selling price on x and item on y and then looked to where y equaled 4 for some reason??? I am not sure why I did that, but I understand now. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q004. On another set of coordinate axes, plot the points (-3, 4) and (5, -2). Sketch a straight line through these points. We will proceed step-by-step obtain an approximate equation for this line: First substitute the x and y coordinates of the first point into the form y = m x + b. What equation do you obtain when you make this substitution? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When making this substitution I get the equation: 4 = m*-3 + b. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting x = -3 and y = 4 into the form y = m x + b, we obtain the equation 4 = -3 m + b. We can reverse the right- and left-hand sides to get -3 m + b = 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q005. Substitute the coordinates of the point (5, -2) into the form y = m x + b. What equation do you get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I get the equation: -2 = 5*m + b. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting x = 5 and y = -2 into the form y = m x + b, we obtain the equation -2 = 5 m + b. Reversing the sides we have 5 m + b = -2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. You have obtained the equations -3 m + b = 4 and 5 m + b = -2. Use the method of elimination to solve these simultaneous equations for m and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When we add these equations together we must make one negative in order to eliminate b. I chose to change the second equation giving me 2 = -5m -b. When we eliminate b we get 6 = -8m. Divide 6 by -8 and get m = -0.75. Then plug this into either equation, I chose the first, and you get 4 = -0.75 * -3 + b. When you multiply those together you get 4 = 2.25 + b. Then subtract 2.25 from 4 and get b = 1.75. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Starting with the system -3 m + b = 4 5 m + b = -2 we can easily eliminate b by subtracting the equations. If we subtract the first equation from the second we obtain -8 m = 6, with solution m = -3/4. Substituting this value into the first equation we obtain (-3/4) * -3 + b = 4, which we easily solve to obtain b = 7/4. To check our solution we substitute m = -3/4 and b = 7/4 into the second equation, obtaining 5 ( -3/4) + 7/4 = -2, which gives us -8/4 = -2 or -2 = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q007. Substitute your solutions b = 7/4 and m = -3/4 into the original form y = m x + b. What equation do you obtain? What is the significance of this equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When I substitute my answers into the y=mx+b formula I get: y = -3/4 x + 7/4 This equation is significant because it tells you what the slop and y intercept are and shows a straight line through the points (-3, 4) and (5, -2). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting b = 7/4 and m = -3/4 into the form y = m x + b, we obtain the equation y = -3/4 x + 7/4. This is the equation of the straight line through the given points (-3, 4) and (5, -2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 NOTES: A Mathematical Model: a mathematical rule, or function, that allows us to compute the depth for any given time, or the time for any given depth. The quadratic formula: If y = a x^2 + bx + c, then y = 0 if, and only if, x = [ -b +- `sqrt(b^2 - 4ac) ] / (2a). Depth function: depth = a t^2 + bt + c Exercises: Here are some data for the temperature of a hot potato vs. time: Time (minutes) Temperature (Celsius) 0 95 10 75 20 60 30 49 40 41 50 35 60 30 70 26 Graph these data below, using an appropriate scale: Pick three representative points and circle them. Write the equations that result from the assumption that the appropriate mathematical model is a quadratic function y = a t^2 + b t + c. 1600a + 40b + c = 41 2500a + 50b + c = 35 3600a + 60b + c = 30 Eliminate c from your equations to obtain two equations in a and b. When we eliminate c we get: 900a + 40b = -6 & 2000a + 60b = -17 Solve for a and b. A = 0.00516 B = - 0.03375 Write the resulting model for temperature vs. time. The resulting model would then be y = 0.00516t^2 - 0.03375t + 34.094 Make a table for this function: Time (minutes) Model Function's Prediction of Temperature 0 0 10 34.3735 20 35.483 30 37.733 40 42.20 50 36.3065 60 52.445 70 59.1155 Sketch a smooth curve representing this function on your graph. Expand your table to include the original temperatures and the deviations of the model function for each time: Time (minutes) Temperature (Celsius) Prediction of Model Deviation of Observed Temperature from Model 0 95 10 75 20 60 30 49 40 41 50 35 60 30 70 26 I am not sure of how to do this??? I think I got my numbers wrong somewhere along the way??? Find the average of the deviations. Comment on how well the function model fits the data. (Note: the model might or might not do a good job of fitting the data. Some types of data can be fit very well by quadratic functions, while some cannot). 001. `query1 ********************************************* Question: `qQuery Introduction to General Themes; Examples (no summary needed) What were some of the things in this introduction that you found interesting or surprising? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The interesting and surprising things in this section were that we can use every day, real world models for a given function, finding real numbers to feed into the equations, and thinking of questions in a different way. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Continue to the next question ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qQuery Introductory Flow Experiment (no summary needed) Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate? Support your conclusion. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The dept is changing at a slower and slower rate or a decreasing rate because if you time the depth at equal increments you will find that it takes longer. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If you time the water at equal time intervals you should find that the depth changes by less and less with each new interval. If you timed the depth at equal intervals of depth you should find that each interval takes longer than the one before it. Either way you would conclude that water flows from the hole at a decreasing rate. The reason is that as the water depth decreases the pressure forcing the water out of the hole decreases. ** STUDENT COMMENT: I dont even know how to critique myself on this answer. I did not get it right and did not know how to express how to support my answer (which was wrong). After reading the given solution, I can see what you are looking for in respect to supporting my solutions. I feel that I need to do more to learn how to chart this correctly. This is a topic I will discuss with my tutor. INSTRUCTOR RESPONSE You have stated that you understand the given solution, and given the pattern of the work you have submitted I believe it is likely that you do. In cases where you're not sure you completely understand, it's preferable that you make more specific statements that demonstrate your understanding. I can evaluate and give you feedback on such statements. The key statements in this document are the following: If you timed the depth at equal intervals of depth you should find that each interval takes longer than the one before it. Either way you would conclude that water flows from the hole at a decreasing rate. I don't know whether it's necessary on this problem, but always remember that you are welcome to insert additional statements into a copy of the posted document, mark them with &&&& and submit for further feedback. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat does the graph of depth vs. clock time look like? Is it increasing or decreasing? Does the rate of increase or decrease speed up or slow down? Does your graph intercept the y axis? Does it intercept the x axis? How would you describe its overall shape? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of the depth vs. clock time is decreasing and its rate decreases or slows down. The graph doesnt intercept the y axis because it starts at the y axis. The overall shape is a downward parabola on the left side. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graph will start on the positive y axis and will decrease at a decreasing rate. The shape of the graph is the left-hand side of a parabola that opens upward. The right-hand half of the parabola does not correspond to the flow. The left-hand half of the parabola, which corresponds to the flow, gets less and less steep with increasing clock time, matching the fact that that the rate of decrease is slowing. At the instant the containers empties, the water will be at the level of the hole. If depth is measured relative to the hole, then at the instant depth will reach zero. The corresponding graph point will lie on the t axis and will correspond to the vertex of the parabola. ** STUDENT ANSWER and self-critique: It decreases at a decreasing rate. The graph starts on the y axis, but does not intercept the x axis. It slopes to the right. (self-critique:) I really need to work more on this. This given solution is difficult to follow. INSTRUCTOR COMMENT: Your answer was completely consistent with the first sentence in the given solution, which read 'The graph will start on the positive y axis and will decrease at a decreasing rate'. So you clearly understand that statement, and you gave a good answer to the question. The given solution continues with more details, which you should try to understand and should address in your self-critique. Taking the given solution one statement at a time: The shape of the graph is the left-hand side of a parabola that opens upward. Do you know what a parabola is? This is something you should know from your prerequisite courses, but experience shows that most students have at best a somewhat vague and unspecific understanding of what a parabola is. It's not something that sticks with students from high school courses (and it's not clear that understanding of this term is necessary for passing the SOL, in which case it would not even be taught in most courses). If you don't know, you would ideally state this in your self-critique so I know to clarify it for you. This term will also be clarified in detail in upcoming assignments. For this reason, it's best you should begin to think about it here, and the way to do that would be by asking about it. For reference, a partial picture of a parabola is displayed below: The right-hand half of the parabola does not correspond to the flow. It would be a good idea to have sketched a parabola, opening upward and passing through the positive y axis, according to the given description. With the sketch it's easier to think about why the right-hand side of the parabola doesn't correspond to the flow (for the fairly obvious reason that the right-hand side is increasing and the depth of the water is decreasing; however this would be less obvious without a sketch). The left-hand half of the parabola, which corresponds to the flow, gets less and less steep with increasing clock time ... It would again be useful to have a sketch to refer to. It should be clear that the left-hand side of a parabola which opens upward gets less and less steep as you move to the right. You might or might understand that the vertical coordinate of the graph corresponds to the depth of the water, with depth increasing as you move upward in vertical direction, while the horizontal coordinate corresponds to the clock time, which increases as you move horizontally to the right. ... matching the fact that that the rate of decrease is slowing The rate of decrease is represented by the steepness of the graph. The graph (at least its left-hand half) gets less and less steep as you move to the right. At the instant the containers empties, the water will be at the level of the hole. If depth is measured relative to the hole, then at the instant depth will reach zero. Students often do not understand what it means for the depth to be 'measured relative to the hole'. In this case it would be appropriate to ask about the meaning of this phrase. (The meaning is that the position of the hole would be taken as the 0-point for the measurement, with the upward direction positive so that, for example, a depth of 10 cm means that the water level is 10 cm above the hole). If depth is measured relative to the hole, then since water stops leaking out when the water level is at the hole, the depth would at that point be 0. This would first occur at the instant in time when the depth reaches 0. The corresponding graph point will lie on the t axis and will correspond to the vertex of the parabola. Anyone who does not know what 'the vertex of a parabola' means would certainly be expected to ask. (The vertex of a parabola which opens upward is the lowest point on the parabola. As discussed above this point indicates the depth of the water and the clock time at which the water first reaches the hole. To the right of the vertex, the graph starts rising again and no longer corresponds to the flow situation.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 163 1/21/13 around 12:00pmI forgot to put my access code in the last one. Sorry. Question: `q001. Sketch a set of coordinate axes representing y vs. x, with y on the vertical axis and x on the horizontal axis.
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Given Solution: The x coordinates 1, 3 and 7 match the x coordinates of the three given points, the y coordinates will be the y coordinates -2, 5 and 8, respectively, of those points. At x = 5 the precise value of x, for a perfect parabola, would be 8 1/3, or about 8.333. You are unlikely to have drawn a perfect parabola and your estimate will almost certainly differ from this value. Any estimate with .5 or so of this value would be a good estimate. Drawn with complete accuracy a parabola through these points will peak between x = 3 in and x = 7. The peak of the actual parabola will occur close to x = 6. Any estimate between x = 5 and x = 7 is pretty good, and even an estimate between x = 7 and x = 8 is not unreasonable. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q002. Using your sketch from the preceding exercise, estimate the x coordinates corresponding to y coordinates 1, 3, 5 and 7. Also estimate the x values at which y is 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When estimating the x coordinates that respond to the y coordinates we get y coordinate 1 corresponds to the x coordinate 2; y coordinate 3 corresponds to the x coordinate 2.5; y coordinate 5 corresponds to the x coordinate 3; y coordinate 7 corresponds to the x coordinate 4.5. When y = 0, x = 1.55. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The easiest way to estimate your points would be to make horizontal lines on your graph at y = 1, 3, 5 and 7. You would easily locate the points were these lines intersect your graph, then estimate the x coordinates of these points. For the actual parabola passing through the given points, y will be 1 when x = 1.7 (and also, if your graph extended that far, near x = 10). y = 3 near x = 2.3 (and near x = 9.3). y = 5 at the given point (3, 5), where x = 3. y = 7 near x = 4 (and also near x = 7.7). Any set of estimates in the vicinity of these values is good. STUDENT COMMENT: Im still not clear on this. I didnt get the same as you and my graph went to 12. I know when we estimate we will not get the same answer, but did I do this right? If not what do I do to correct it? INSTRUCTOR RESPONSE: The following commentary need not be self-critiqued, though you may certainly ask questions if you wish. However I have two reasons for including it: The estimates here were given by a good student whose graph for some reason didn't match the points on this particular problem. If this student missed his/her estimates, then most students will need some clarification at this point. The discussion and the graphs shown here should be helpful to you. You should be able to sketch these graphs, label the coordinate axes based on the coordinates of the three points, and make reasonable verification of the values given here. This situation is closely related to the work we will be doing in the first few assignments, where we will actually learn to do parabolic or quadratic curve fits. Instructor response to student: Some of your estimates were reasonable, but others weren't. Your solution was if y = 0 then x = 0 if y = 1 then x = approx. 1.7 if y = 3 then x = approx. 4.6 if y = 5 then x = approx. 8.1 if y = 7 then x = approx. 11.5 If the x coordinates 1,3 7 match the y coordinates -2,5,8, then y = 1 at x = 1.7 is a very reasonable estimate y = 3 at x = 4.6 is not a reasonable estimate. By the time x = 3, y already has a value of 5, and is increasing toward value y = 8 at x = 7. For an x value between 3 and 5 we would expect y to be greater than 5; we would not expect a y value of 3 to occur on this x interval. y = 5 at x = 8.1 is a reasonable estimate. When x = 7 we have y = 8, and it's possible to draw a reasonable graph on which y decreases to 5 by the time x = 8. y = 7 at x = 11.5 isn't consistent with your previous estimates. You already have y down to value 5 by the time x = 8.1, and there's no reason to expect the graph to turn around and go back up to 7 by the time x reaches 11.5. The figure below depicts the three data points. Your graph should have similar form. The graph below shows the parabola that actually passes through these points. Your graph won't necessarily look the same, since you probably aren't practiced at drawing an accurate parabola through three given points, and besides nobody told you to use a parabola. So it's no problem if your graph differs from this one, but this 'ideal' graph is the basis for the numbers in the given solution. For this graph, y = 0 somewhere between x = 1 and x = 3, closer to x = 1; we might estimate that this occurs at x = 1.5 (a fairly precise value for the actual parabola is about x = 1.476). You will learn a lot more about parabolas within the next couple of weeks. In fact, by Assignment 4 or 5, you will have learned how to create and use a parabolic model for any three points, and you'll see that while it's not really what you'd call easy, it's not all that difficult either. The graph below isn't parabolic, but it's still a smooth curve and passes through the given points, and your graph could well look like this. The y value at x = 5 appears to be pretty close to the x = 7 value, which is y = 8. The y = 0 value again occurs between x = 1 and x = 3; perhaps a little to the left of the point in the above graph, maybe at x = 1.4. Another reasonable graph is shown below. At x = 5 the y value is probably around 7. The graph below agrees with the estimates you gave. While it fits the overall trend of the data by 'staying in the middle', it doesn't fit any of the three given points very well. This graph has been included so you can verify that it agrees reasonably with your given values if y = 0 then x = 0 if y = 1 then x = approx. 1.7 if y = 3 then x = approx. 4.6 if y = 5 then x = approx. 8.1 if y = 7 then x = approx. 11.5 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): What is the answer tox =? When y =0??? I read where it was x = 0 and where x = somewhere between 1 & 3 ------------------------------------------------ Self-critique Rating: 1