course Mth 173 e|๓QwʎStudent Name:
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22:50:35 `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 --> On the average the graph is steeper between the point (7,17) and the point (10,29). I determined this by first finding the difference between the two points. I subtracted 7 from 10 and 17 from 29 to find the difference between the first two points which resulted in (4,12). I subtracted 7 from 10 and 17 from 29 to find the difference between the second two points which resulted in (3,12). I then used the slope formula which is Slope=Rise/Run. The average slope between the first two points was 3 and the average slope between the second two points was 4. The reason that I dictate it as ""average"" slope is because if the function is a curve the slope can change constantly between the two points.
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w}աwЎS݈yۅŠ Student Name: assignment #005 005. Calculus
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22:54:23 `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 -->
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22:56:16 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 --> I believe that I answere this problem correctly, however I did make an error when documenting my work. I displayed the wrong numbers in my math when finding the difference between the two points.
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23:17:01 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 --> I begin by plugging the different values in for x and solving. 1/(2.1-2) = 1/.1 = 10 1/(2.01-2) = 1/.01 = 100 1/(2.001-2) = 1/.001 = 1000 1/(2.0001-2) = 1/.0001 = 10000 1. As this process continues and x continues to get closer to 2 the value of 1/(x-2) continues to increase. 2. The value will exceed a billion when the value of x becomes anything smaller than 2.000001. The value will also exceed one trillion billions. 3. The value of 1/(x-2) is infinite and can always get larger. 4. There is not a number that the value cannot exceed. 5. On the graph of y=1/(x-2) the graph looks like a vertical line in the vicinity of x=2. This is due to the fact the line will continue to get closer to 2 but will never touch it.
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23:20:53 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 --> In reviewing the answer to the question I find the opposite side of the viewpoint I just took towards 1/(x-2) This is that if x is less than 2 then y will be negative and reach towards -infinity.
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00:42:21 `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 --> When I observe the trapezoid created by the points (3,5) and (7,9) I looks smaller than the trapezoid created by the points (10,2) and (50,4). I can make the first two points create a slightly larger rectangle with the points (3,9) and (7,9) and I can make the second two points create a slightly smaller rectangle with the points (10,2) and (50,2). If I do this the area of the first rectangle is 36 and the area of the second rectangle is 80 which tells me that the second trapezoid ((10,2)(50,4)) has more than twice the area of the first trapezoid ((3,5)(7,9)).Using this method I know that the second trapezoid ((10,2)(50,4)) has the greater area.
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00:45:25 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 --> The answer uses average widths and average altitude which is similar to what I did. I actually overworked myself and could have made the whole process easier than I did.
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01:46:17 `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 --> First I take each x and plug it into f(x) = x^2 x=2, f(x)=2^2 = 4 x=5, f(x)=5^2 = 25 x=-1,f(x)=-1^2=1 x=7,f(x)=7^2=49 This gives me four points being (2,4) (5,25) (-1,1) (7,49) I can then take the slope formula and determine the slope for each line Slope of (2,4) and (5,25) = rise (25-4=21)/run(5-2=3) Therefore the slope between(2,4) and (5,25) is 7 Slope of (-1,1) and (7,49)= rise (49-1=48)/run(7-(-1)) Therefore the slope between(-1,1) and (7,49) is 6 Therefore the line segment connecting the points (2,4) and (5,25) on the graph of f(x)=x^2 is steeper that the line segment connecting the points(-1,1) and (7,49)
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01:47:09 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 --> Completed
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01:51:43 `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 --> If you placed the same number of grams of gold in your back yard each week and then graph it since Jan 2000 the line on the graph will be a rising straight line.
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02:00:50 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 --> I misunderstood the question. I didnt realize I was suppossed to look at the amount of gold purchased each week from every angle (the same each week, more each week, and less each week). I do understand the thinking behind this though.
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02:06:05 `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 --> If you buy 1 more gram of gold each week that you go to purchase it than you did the previous week then then the line on the graph will be a line which rises faster and faster. If, each week you bury half the amount you did the previous week then the line on the graph will fall but it will fall more and more slowly.
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02:06:56 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 --> Completed
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02:20:50 7. 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 --> First I will plug the different values for time or t into the depth of water in a container =100-2t+.01 t^2 t=30, 100-2(30)+.01 (30)^2 = depth=49 t=40, 100-2(40)+.01 (40)^2 = depth=36 t=60, 100-2(60)+.01 (60)^2 = depth=16 I can then find the differences between the three depths 49-36=13 and 36-16=20, next I must take into consideration that twice as much time passes between the second and third interval than between the first and second interval 20*1/2=10 Therefore I find that on average the depth is changing more rapidly between the times of 30 and 40 seconds than it is between the times of 40 and 60 seconds.
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02:21:31 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 --> Completed
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02:29:24 8. 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 --> depth in container = 10-.1t t=time in seconds at 10 seconds= 10-.1(10)=9 cm/s at 20 seconds= 10-.1(20)=8 cm/s Therefore the average flow between 10 and 20 seconds is 8.5 cm/s so during this 10 second interval I would expect the water level to decreas 8.5 cm
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02:31:04 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 messed up in this problem because I forgot the last step of multiplying the average decrease in depth of 8.5 cm/s by the elapsed time of 10 seconds 8.5*10=85 which would have given me a change of 85 cm
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