#$&* course Phy 231 I never received a response to one of my questions from this homework assignment. I submitted this question midday on Friday. In order to keep my assignment from being considered late I am going ahead and submitting the assignment with my best work/solutions. Question: `q001. There are 12 questions in this document.
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Given Solution: Slope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints. 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? 3. Will it ever exceed the number of particles in the known universe? 4. Is there any number it will never exceed? 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There wasn’t a question #2 - I even went back to the web address and couldn’t find #2. confidence rating #$&*: N/A - No Question ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): N/A - No Question ------------------------------------------------ Self-critique Rating: N/A - No Question ********************************************* Question: `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'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first trapezoid is significantly smaller than the second. Reasoning for this answer can be found in the area of the trapezoids. The first trapezoid has an area of ((7-3)(5)+(7-3)(1/2)(4)) = 28. The second trapezoid has an area of ((50-10)(2)+(50-10)(1/2)(2)) = 120. These areas were calculated by using the formulas for area (e.g. a square’s area = base * height, and a triangle’s area = (1/2) * base * height.) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The answer could have been more vague than I answered it…..I followed all the calculation to the end while the given answer did not. I was thrown off by the questions wording of: “justify your answer with something more precise.” ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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 basis of your reasoning. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The segment connection x = 2 and x = 5 would be steeper than the segment connecting x = -1 and x = 7. The reason for my answer comes from the definition of steep. Generically, my definition of steep is: the top of something is drastically higher than the bottom of that something for a, usually, short lateral distance. The slope of the line at x = 2 is already significantly steep, and when connected to the point x = 5 (which also has a steep tangent line) the connecting slope is expectantly steep. However, the slope at x = -1 is negative because the function is decreasing from [-infinity, 0]. When this is paired with x = 7’s positive slope, the outcome from pairing x = -1’s longer lateral distance from x = 7, causes the slope to be less than that of the other line segment. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): The previous question made a vague mathematical answer; therefore I didn’t see it necessary to be as precise on this question. However, now that I see your given answer includes the equation for slope (rise / run), I would amend my answer to include answer that are easily proven. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. If you constructed a graph of the number of grams of gold in your backyard vs. the number of weeks since the new millennium, the points would lie on a rising straight line because your addition is linear - hence the name of the graph of a linear function. 2. However, if you bought one more gram than the previous week (every week), the graph would be an constantly increasing function going towards infinity because your addition is no longer linear - it is dependent on the amount you bought the week before. The function is exponential - hence the name of the graph of an exponential function. 3. If, instead, you bought half the amount you bought the previous week (every week), the graph would be constantly increasing but at a decreasing rate because you would always be adding gold to your collection, just at smaller and smaller amounts. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a1. 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. If you graphed the rate at which the gold is accumulating vs. the number of weeks the resulting graph would be a horizontal rate. It is easier to think about the answer to this question as being the slope of the previous graph. All-in-all, the amount of gold you add each week (the rate) is constant, so the graph would be constant or horizontal (unchanging). 2. This time, the resulting graph would be a straight line that has a positive slope. Again, it is easier to think about this question as the slope of the previous question. All-in-all, the rate at which you add gold is always consistently increasing and therefore would be a straight line with a positive slope. 3. This time, the resulting graph would be a line that has a negative slope starting on the y-axis but never going below the x-axis. The rate of the addition of the amount of gold added to your backyard is never zero, just half of what it was before, so there is a horizontal asymptote at y = 0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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. STUDENT COMMENT: I feel like I am having trouble visualizing these graphs because every time for the first one I picture an increasing straight line INSTRUCTOR RESPONSE: The first graph depicts the amount of gold you have in your back yard. The second depicts the rate at which the gold is accumulating, which is related to, but certainly not the same as, the amount of gold. For example, as long as gold is being added to the back yard, the amount will be increasing (though not necessarily on a straight line). However if less and less gold is being added every year, the rate will be decreasing (perhaps along a straight line, perhaps not). FREQUENT STUDENT RESPONSE This is the same as the problem before it. No self-critique is required. INSTRUCTOR RESPONSE This question is very different that the preceding, and in a very significant and important way. You should have self-critiqued; you should go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####. The extra effort will be more than worth your trouble. These two problems go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. The same is true of the last question in this document. STUDENT COMMENT Aha! Well you had me tricked. I apparently misread the question. Please don’t do this on a test! INSTRUCTOR RESPONSE I don't usually try to trick people, and wasn't really trying to do so here, but I was aware when writing these two problems that most students would be tricked. My real goal: The distinction between these two problems is key to understanding what calculus is all about. I want to at least draw your attention to it early in the course. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question:``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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The depths are as follows (let d = depth of water): t = 30, d = 49; t = 40, d = 36; t = 60, d = 16. The average depth of the water is changing more rapidly on the interval t = 30 to t = 40. Proof of this can be found in the slope between the points. The slope for the first line segment (t = 30 and t = 40) is: ((36-49)/(40-30)) = -1.3 cm/s. The slope of the second segment (t = 40 and t = 60) is: ((16-36)/(60-40)) = -1 cm/s. All-in-all, a slope of -1.3 cm/s is steeper than -1.0 cm/s. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: At t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the rate at which water descends is given by the equation: 10 - .1t, then the rate at which water is descending at t = 10 is: 10 - (1/10)(10) = 9 cm/s. The rate at which water is descending at t =20 is: 10 - (1/10)(20) = 8 cm/s. To calculate the expected water level change during this 10 second interval, I would average the 9 cm/s and 8 cm/s because the rate wouldn’t be any more or less than these two. The change in the water level would be approximately: ((9+8)/2) * 10 = 85 cm. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT RESPONSES The following, or some variation on them, are very common in student comments. They are both very good questions. Because of the importance of the required to answer this question correctly, the instructor will typically request for a revision in response to either student response: • I don't understand how the answer isn't 1 cm/s. That's the difference between 8 cm/s and 9 cm/s. • I don't understand how the answer isn't 8.5 cm/s. That's the average of the 8 cm/s and the 9 cm/s. INSTRUCTOR RESPONSE A self-critique should include a full statement of what you do and do not understand about the given solution. A phrase-by-phrase analysis of the solution is not unreasonable (and would be a good idea on this very important question), though it wouldn't be necessary in most situations. An important part of any self-critique is a good question, and you have asked one. However a self-critique should if possible go further. I'm asking that you go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####, before submitting it. The extra effort will be more than worth your trouble. This problem, along with questions 5 and 6 of this document, go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. You should review the instructions for self-critique, provided at the link given at the beginning of this document. STUDENT COMMENT The question is worded very confusingly. I took a stab and answered correctly. When answering, """"How much would you therefore expect the water level to change during this 10-second interval?"""" It is hard to tell whether you are asking for what is the expected change in rate during this interval and what is the changing """"water level."""" But now, after looking at it, with your comments, it is clearer that I should be looking for the later. Thanks! INSTRUCTOR RESPONSE 'Water level' is clearly not a rate. I don't think there's any ambiguity in what's being asked in the stated question. The intent is to draw the very important distinction between the rate at which a quantity changes, and the change in the quantity. It seems clear that as a result of this question you understand this and will be more likely to make such distinctions in your subsequent work. This distinction is at the heart of the calculus and its applications. It is in fact the distinction between a derivative and an integral. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####Though my answer is correct, I should have calculated the most and least amount of change in depth. This gives parameters and is more precise. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When x = 6 h(x) = 4 * 1 = 4. The value of h(x) is never greater than its value at x = 6. This can be seen when viewing the positive portions of f(x) and g(x). The negative portion is disregarded because only one function has a negative portion (i.e. a negative multiplied by a positive will always yield a negative). The positive portions of both graphs cross paths and go off in opposite directions and therefore, the largest numbers that can be used to create a product are 6 and 1 for the defined x. The graph of h(x) would be negative from x = 2 to x = 4 and positive from x = 4 to x = 6.The maximum would occur at x = 6 and the minimum would occur at x = 2 because these are the points where f(x) and g(x) have the largest (absolute value) to contribute to the multiplication operation. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q010. A straight line segment connects the points (3,5) and (7,9), while the points (3, 9) and (7, 5) are connected by a curve which decreases at an increasing rate. From each of the four points a line segment is drawn directly down to the x axis, so that the first line segment is the top of a trapezoid and the second a similar to a trapezoid but with a curved 'top'. Which trapezoid has the greater area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The trapezoid with points (3,9) and (7,5) has a greater area because both utilize the same dimensions and therefore an equal area, but the second trapezoid has that area plus what some. More specifically, both trapezoids areas have a square bottom and a triangle top - both squares and triangles have the same dimensions (and therefore the same area). However, the trapezoid with the curved top has addition area beyond what the triangle represents - an area bounded by the curve and the triangle’s hypotenuse. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Description of the graph of the position of a car vs. clock time (d-t graph) AND velocity of the car vs. time (v-t graph): *please note that all v-t graphs would include the curve or line starting somewhere above 0 m/s since the car is already moving when the data is collected to be graphed* 1. The description clearly states that the v-t graph must be a linear graph since ‘gaining the same amount of speed’ indicates that the velocity is changing evenly and, therefore, the v-t graph must show a ‘linear’ change in velocity with respect to time. The corresponding d-t graph’s curve would need to have a slope that constantly increased since the slope of the d-t graph corresponds to the v-t graph’s y-values. The curve of the d-t graph would resemble an exponential curve. 2. The description clearly states that the line of the v-t graph must be an exponential curve in order to show that the velocity (y-values) is changing more quickly than the seconds passing. If the car’s velocity is increasing exponentially then the car’s position must change correspondingly (i.e. if the car is traveling faster and faster then the car is covering more and more distance in shorter amount of time). Therefore the d-t graph must also be an exponential curve. 3.If the car is gaining less and less speed (approaching, but never crossing over, a gain of 0 m/s) with each passing second then the velocity is increasing at an decreasing rate. The v-t graph would then need to be curve with positive slope that decreases exponentially as time approaches infinity. The corresponding d-t graph would then need to show a curve of a positive function that increases at a decreasing rate since the distance is always become greater than before just at a smaller rate each time. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find how much water flowed out of a container between 100 seconds and 150 seconds I would first determine the amount that would have flowed out during each rate if the rate stayed the same for the 50 seconds. I would then state that the amount of water that left the container would have to be between the two. The first rate at 100 seconds was 1.4 liters / second…..if this rate had occurred for 50 seconds, then 70 liters of water would have left the container (1.4 * 50, the unit of seconds cancel each other when the multiplication is applied and leaves the answer in units of liters). However, if the second rate had occurred for the entire 50 seconds, then 50 liters of water would have left the container (1.0 * 50, the unit of seconds cancel each other when the multiplication is applied and leaves the units of liters). Therefore the correct answer must be between the two calculated values since the rate never rose over 1.4 or below 1.0, that is: the amount of water that left the container, w, is 50