course Phy 231 I goofed on a couple of the questions for assignment 5...you'll notice one where I didn't scroll down all the way on the question, and one that I just flat out got wrong. This submission contains 2 assignments completed today - assignment 2 and assignment 5 from q_a_initial problems ???^?m??????Y????
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15:12:00 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> Right off, we know from the equation that the line intercepts the y-axis at (0, -4) and has a slope of 3. Plotting out the x and y coordinates in a table as in the example yields a line that does indeed pass through the y-axis at (0, -4), but does not immediately give a point for the x-axis intercept. However, if we take 3x - 4, set it equal to zero, and solve for x, we get 3x = 4, and x = 4/3. This tells us the x-axis intercept is (4/3, 0). confidence assessment: 3
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15:12:47 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> OK - Agreed. self critique assessment: 3
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15:18:17 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> The steepness does not change. The slope of 3 is constant. It remains the same throughout the graph. confidence assessment: 3
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15:18:29 The graph forms a straight line with no change in steepness.
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RESPONSE --> OK - Agreed. self critique assessment: 3
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15:22:36 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> The slope of the graph is 3. We know this because of the slope-intercept form of the equation ---> y=mx + b, where m = the slope of the line, and b = the y-interecept. We can also find the slope by taking two points from our table and using the formula (y2-y1)/(x2-x1). Two point from our table are (0, -4) and (1, -1). If we use the formula, we get (-1-(-4)) / (1-0), which equals 3/1 = 3. confidence assessment: 3
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15:23:11 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> OK - Agreed. self critique assessment: 3
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15:26:44 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph is first decreasing, and then increasing after it passes through the origin (0,0). The steepness does change - it starts out very steep, and becomes flatter as it nears the origin. Once passing through the origin, it starts flatter, and becomes steeper; a mirror image of the path from the left side of the y-axis. The graph begins by decreasing at a decreasing rate, then passes through the origin, then begins increasing at an increasing rate. confidence assessment: 3
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15:29:55 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.
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RESPONSE --> I understand what is being stated, however, what about instances where the x value is negative, such as (-1, 1), (-2, 4), (-3, 9), etc? I agree with the comments about increasing at an increasing rate, but what about the points to the left of the y-axis? Perhaps I misunderstood the question? self critique assessment: 2
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15:31:35 `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> Oh, now I see my error from the previous question. I failed to properly read the first instruction. When x is between -3 and 0, the graph is decreasing at a decreasing rate. The steepness of the graph does change - it becomes flatter as the graph approaches the origin (as y approaches 0). confidence assessment: 3
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15:31:58 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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RESPONSE --> OK - Agreed. self critique assessment: 3
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15:38:01 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> I also just realized that I didn't put the table values in the last two problems - perhaps I should have done that. Table values are as follows for this equation: (0,0), (1,1), (2, `sqrt(2)), (3,`sqrt(3)). The steepness of the graph does change - it becomes flatter as x grows larger. The graph is increasing at a decreasing rate. confidence assessment: 3
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15:38:28 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.
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RESPONSE --> OK - agreed. self critique assessment: 3
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15:42:39 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The tables values are as follows: (0,5), (1, 5/2), (2,5/4), (3,5/8). The graph is decreasing. The steepness of the graph is decreasing. The graph is therefore decreasing at a decreasing rate. confidence assessment: 3
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15:42:58 ** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **
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RESPONSE --> OK - agreed. self critique assessment: 3
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15:47:02 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph of y vs. t would be increasing because it would be upward sloping. The graph would be increasing at an increasing rate due to the rate of speed becoming faster and faster and therefore the distance between me and car becoming larger and larger. confidence assessment: 2
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15:47:21 ** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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RESPONSE --> OK - agreed. self critique assessment: 3
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?????…~????? assignment #005 005. Calculus qa initial problems 01-28-2009
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15:53:49 `q001. The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10, 29). Between which two points do you think the graph is steeper, on the average? Why do we say 'on the average'?
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RESPONSE --> The graph is steepest between (10,29) and (7,17). We know this because we can determine the slope is: (29-17) / (10-7) = 4 between the last two points; and is: (17-5) / (7-3) = 3 between the first two points. A greater slope would indicate the curve is steepest between these last two points. I am not totally sure why we say ""on the average"" but I believe it has something to do with the function not being a straight line and therefore does not have a constant slope between those sets of points. confidence assessment: 2
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15:54:12 Slope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints.
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RESPONSE --> OK - agreed. self critique assessment: 3
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16:28:50 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? 3. Will it ever exceed the number of particles in the known universe? 4. Is there any number it will never exceed? 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?
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RESPONSE --> The coorindates are as follows: (2.1, 10), (2.01, 100), (2.001, 1000), (2.0001, 10000). 1. As the value of x gets closer and closer to 2, the value of y gets larger and larger. 2. Yes, the value will equal a billion when x = 2.000000001. Yes, eventually it will exceed 1 trillion billions. 3. Yes, it will exceed the number of particles in the universe (which I'm assuming to be less than infiniti). 4. As the value of x approaches 2, the value of y approaches infiniti. The only value it will never exceed is infiniti. 5. The graph is decreasing at a decreasing rate. In the vicinity of x=2, it is a straight line, however, the graph never actually reaches x=2 (at x=2, the equal is null). self critique assessment: 3
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16:29:55 For x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000. It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts. As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it. Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0). As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds. Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity.
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RESPONSE --> OK- agreed. self critique assessment: 3
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19:22:24 `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> Area of a trapezoid is defined as: a*( (b1+b2) / 2 ), where b1 and b2 are the lengths of the 2 bases, and a is the height. So, the first trapezoid with points (3,5) and (7,9), has b1 and b2 lengths of 5 and 9 respectively. The height [a] is 4. The area is therefore equal to 4 * ( (5+9) / 2) = 4 * 7 = 28. The second trapezoid with points (10,2) and (50,4), has b1 and b2 lengths of 2 and 4 respectively. The height [a] is 40. The area is therefore equal to 40 * ( (2+4) / 2 ) = 40 * 3 = 120. The 2nd trapezoid therefore has the larger area. confidence assessment: 3
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19:23:41 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
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RESPONSE --> OK - agreed. I went to the length of actually calculating the area, rather then just providing explanation as to which had the greater area. self critique assessment: 2
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19:28:10 `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basisof your reasoning.
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RESPONSE --> The line segment connecting x=2 and x=5 points is steeper because the slope of this line is equal to 7, while the slope of the line connecting x = -1 and x=7 points is 6. confidence assessment: 3
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19:28:24 The line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7. The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6. The slope of the first segment is greater.
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RESPONSE --> OK- agreed self critique assessment: 3
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19:31:05 `q005. Suppose that every week of the current millenium you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before.. 1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. {}3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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RESPONSE --> Assuming the amount of gold obtained each week is the same, then the points should lie on a straight line, which rises at a constant rate. confidence assessment: 3
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19:34:08 1. If it's the same amount each week it would be a straight line. 2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate. 3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function.
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RESPONSE --> Interesting. I completely missed the 3rd point you made. It sounds like what you're saying is: as the total amount of gold becomes very large, the same amount of gold each week will increase the total amount by a smaller and smaller percentage. Is this correct?
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19:39:49 `q006. Suppose that every week you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your backyard. 1. If you graph the rate at which gold is accumulating from week to week vs. tne number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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RESPONSE --> Ohhh. I didn't scroll down on the last problem! I was wondering why I completely missed the reasoning. 1. It will be a straight line that increases. 2. If every week 1 extra gram is added to the amount that's buried, the line will increase at an increasing rate. 3. If every week, half the amount of gold is buried, the line will increase at a decreasing rate. confidence assessment: 3
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19:59:27 This set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time. Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount eact week, is constant. The graph would be a horizontal straight line. Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a risingstraight line because the increase in the rate is the same from one week to the next. Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero.
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RESPONSE --> Yikes, I missed that. I do understand the concept though - because we're dealing with rate, as opposed to total quantity, the dynamics of the graph are different. On question 1, since the rate is not changing, it's a level straight line. On question 2, since the rate is increasing at a constant pace, it's a straight increasing line. On question 3, since the rate is decreasing, we have a line that is decreasing, yet never reaches the asymptote of 0. self critique assessment: 2
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20:06:28 ``q007. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second?
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RESPONSE --> On average, the depth is changing more rapidly during the first time interval. From t=0 to t=30, there is a change of 51 cm (at t = 0, the depth is 100 cm, whereas at t =30, the depth is 49 cm). From t = 30 to t =60, there is a change of 33 cm (at t = 60, depth is equal to 16 cm). This indicates that the depth is increasing, but at a decreasing rate. Therefore, we can say the change is more rapid during the first time interval. self critique assessment: 3
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20:06:50 At t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average.
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RESPONSE --> OK- agreed. self critique assessment: 3
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20:13:50 `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval?
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RESPONSE --> At t=10, the water is descending at 9cm/s; at t=20, the water is descending at 8 cm/s. Therefore, during this 10 second interval, I would expect the water level to change by 10 cm because the difference between the 2 rates is 1 cm/s. Therefore, at 1 cm/s * 10s, we have 10 cms/s or 10 cm. confidence assessment: 2
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20:17:50 At t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec. At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec. The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm. The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm. Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions.. The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 9.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm.
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RESPONSE --> I was fine on the first part, but the second portion of the question is somewhat baffling to me. I'm unsure how the rate of change is the average of the 2 rates. Can you please explain this one further? self critique assessment: 1
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