#$&*
course Phy 231 Submitted: August 30, 20313 at approx. 1:25 pm 005. Calculus.............................................
Given Solution: 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. Self-critique (if necessary): OK. Self-critique Rating: OK. ********************************************* Question: `q005. Suppose that every week of the current millennium you go to the jeweler 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 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. Your solution: 1. If you buy the exact same amount each week then the graph would be in a straight line. 2. If you are purchasing 1 more gram of gold each week then you did the last week, it doesn’t matter how much more you purchase and bury in your backyard your graph will always be increasing. 3. You are still buying gold every week so there is no way your graph can decrease it will still increase but it will increase slower. confidence rating #$&*:: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: 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. Self-critique (if necessary): I could have put in my answer that the graph will approach a horizontal asymptote but I had not thought of that or realized that that was true. Self-critique Rating: 3 ********************************************* Question: `q006. Suppose that every week you go to the jeweler 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 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. the 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. Your solution: 1. If it's the same amount each week it would be a straight horizontal 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 straight line. 3. You are still buying gold every week so there is no way your graph can decrease it will still increase but it will increase slower. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: This set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time. Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount each week, is constant. The graph would be a horizontal straight line. Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a rising straight line because the increase in the rate is the same from one week to the next. Question 3: Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero. Self-critique (if necessary): On the third question I did not realize that the rate would never reach zero. ------------------------------------------------ Self-critique Rating: 3
``q007. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second? Your solution: 100 - 2 t + .01 t^2 t = 30, t = 40 and t = 60 100 - 2 (30) + .01 (30)^2 = 100 - 60 + 0.01(900) = 100 - 60 + 9 = 49 100 - 2 (40) + .01 (40)^2 = 100 - 80 + 0.01(1600) = 100 - 80 + 16 = 36 100 - 2 (60) + .01 (60)^2 = 100 - 120 + 0.01(3600) = 100 - 120 + 36 = 16 The depth is changing more rapidly on the second interval than the first. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: 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. Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: OK. ********************************************* Question: `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? Your solution: 10 - .1 t t = 10 and t = 20 10 - .1 (10) = 9cm/sec 10 - .1 (20) = 8cm/sec You would expect the water level to change by 8.5 cm/sec during this interval. 8.5 cm/sec because that is the average of the two rates above. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: At t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec. At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec. The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm. The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm. Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions.. The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 8.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm. Self-critique Rating: OK. ********************************************* Question: `q009. Sketch the line segment connecting the points (2, -4) and (6, 4), and the line segment connecting the points (2, 4) and (6, 1). The first of these lines if the graph of the function f(x), the second is the graph of the function g(x). Both functions are defined on the interval 2 <= x <= 6. Let h(x) be the function whose value at x is the product of the values of these two functions. For example, when x = 2 the value of the first function is -4 and the value of the second is 4, so when x = 2 the value of h(x) is -4 * 4 = -16. Answer the following based just on the characteristics of the graphs you have sketched. (e.g., you could answer the following questions by first finding the formulas for f(x) and g(x), then combining them to get a formula for h(x); that's a good skill but that is not the intent of the present set of questions). What is the value of h(x) when x = 6? Is the value of h(x) ever greater than its value at x = 6? What is your best description of the graph of h(x)? Your solution: I have no idea how to even start this problem. confidence rating #$&*:: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Question: `q010. A straight line segment connects the points (3,5) and (7,9), while the points (3, 9) and (7, 5) are connected by a curve which decreases at an increasing rate. From each of the four points a line segment is drawn directly down to the x axis, so that the first line segment is the top of a trapezoid and the second a similar to a trapezoid but with a curved 'top'. Which trapezoid has the greater area? Your solution: The trapezoid with the curved top would have the greater area because of the curve. Both trapezoids would have very similar areas but because of the curve the trapezoid with the points (3,9) and (7,5) would have a greater area. confidence rating #$&*:: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Question: `q011. Describe the graph of the position of a car vs. clock time, given each of the following conditions: The car coasts down a straight incline, gaining the same amount of speed every second The car coasts down a hill which gets steeper and steeper, gaining more speed every second The car coasts down a straight incline, but due to increasing air resistance gaining less speed with every passing second Describe the graph of the rate of change of the position of a car vs. clock time, given each of the above conditions. Your solution: I have no idea how to even start this problem. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Question: `q012. If at t = 100 seconds water is flowing out of a container at the rate of 1.4 liters / second, and at t = 150 second the rate is 1.0 liters / second, then what is your best estimate of how much water flowed out during the 50-second interval? Your solution: 150-100 = 50 1.4 - 1.0 = 0.4/2 = 0.2 liters/second 0.2 liters/second is the average so 0.2 liters flowed out during the 50 second interval. confidence rating #$&*:: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: The other problems that were similar to this one you had to find the average for so that is why I took the average. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: `gr31#$&* course Phy 231 Submitted: August 30, 20313 at approx. 1:25 pm 005. Calculus
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Given Solution: 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. Self-critique (if necessary): OK. Self-critique Rating: OK. ********************************************* Question: `q005. Suppose that every week of the current millennium you go to the jeweler 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 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. Your solution: 1. If you buy the exact same amount each week then the graph would be in a straight line. 2. If you are purchasing 1 more gram of gold each week then you did the last week, it doesn’t matter how much more you purchase and bury in your backyard your graph will always be increasing. 3. You are still buying gold every week so there is no way your graph can decrease it will still increase but it will increase slower. confidence rating #$&*:: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: 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. Self-critique (if necessary): I could have put in my answer that the graph will approach a horizontal asymptote but I had not thought of that or realized that that was true. Self-critique Rating: 3 ********************************************* Question: `q006. Suppose that every week you go to the jeweler 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 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. the 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. Your solution: 1. If it's the same amount each week it would be a straight horizontal 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 straight line. 3. You are still buying gold every week so there is no way your graph can decrease it will still increase but it will increase slower. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: 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 each 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 rising straight 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. Self-critique (if necessary): On the third question I did not realize that the rate would never reach zero. ------------------------------------------------ Self-critique Rating: 3 ``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? Your solution: 100 - 2 t + .01 t^2 t = 30, t = 40 and t = 60 100 - 2 (30) + .01 (30)^2 = 100 - 60 + 0.01(900) = 100 - 60 + 9 = 49 100 - 2 (40) + .01 (40)^2 = 100 - 80 + 0.01(1600) = 100 - 80 + 16 = 36 100 - 2 (60) + .01 (60)^2 = 100 - 120 + 0.01(3600) = 100 - 120 + 36 = 16 The depth is changing more rapidly on the second interval than the first. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: At t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average. Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: OK. ********************************************* Question: `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval? Your solution: 10 - .1 t t = 10 and t = 20 10 - .1 (10) = 9cm/sec 10 - .1 (20) = 8cm/sec You would expect the water level to change by 8.5 cm/sec during this interval. 8.5 cm/sec because that is the average of the two rates above. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: 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 8.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm. Self-critique Rating: OK. ********************************************* Question: `q009. Sketch the line segment connecting the points (2, -4) and (6, 4), and the line segment connecting the points (2, 4) and (6, 1). The first of these lines if the graph of the function f(x), the second is the graph of the function g(x). Both functions are defined on the interval 2 <= x <= 6. Let h(x) be the function whose value at x is the product of the values of these two functions. For example, when x = 2 the value of the first function is -4 and the value of the second is 4, so when x = 2 the value of h(x) is -4 * 4 = -16. Answer the following based just on the characteristics of the graphs you have sketched. (e.g., you could answer the following questions by first finding the formulas for f(x) and g(x), then combining them to get a formula for h(x); that's a good skill but that is not the intent of the present set of questions). What is the value of h(x) when x = 6? Is the value of h(x) ever greater than its value at x = 6? What is your best description of the graph of h(x)? Your solution: I have no idea how to even start this problem. confidence rating #$&*:: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Question: `q010. A straight line segment connects the points (3,5) and (7,9), while the points (3, 9) and (7, 5) are connected by a curve which decreases at an increasing rate. From each of the four points a line segment is drawn directly down to the x axis, so that the first line segment is the top of a trapezoid and the second a similar to a trapezoid but with a curved 'top'. Which trapezoid has the greater area? Your solution: The trapezoid with the curved top would have the greater area because of the curve. Both trapezoids would have very similar areas but because of the curve the trapezoid with the points (3,9) and (7,5) would have a greater area. confidence rating #$&*:: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Question: `q011. Describe the graph of the position of a car vs. clock time, given each of the following conditions: The car coasts down a straight incline, gaining the same amount of speed every second The car coasts down a hill which gets steeper and steeper, gaining more speed every second The car coasts down a straight incline, but due to increasing air resistance gaining less speed with every passing second Describe the graph of the rate of change of the position of a car vs. clock time, given each of the above conditions. Your solution: I have no idea how to even start this problem. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Question: `q012. If at t = 100 seconds water is flowing out of a container at the rate of 1.4 liters / second, and at t = 150 second the rate is 1.0 liters / second, then what is your best estimate of how much water flowed out during the 50-second interval? Your solution: 150-100 = 50 1.4 - 1.0 = 0.4/2 = 0.2 liters/second 0.2 liters/second is the average so 0.2 liters flowed out during the 50 second interval. confidence rating #$&*:: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: The other problems that were similar to this one you had to find the average for so that is why I took the average. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: