course Mth173 ??????D?D????assignment #008
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15:28:06 `q001. Note that there are 5 questions in thie assignment. Sketch a graph of the function y ' = .1 t - 6 for t = 0 to t = 100. Describe your graph.
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RESPONSE --> I have a line starting at point (0, -6) and ending at point (100, 4). With points {(10, -5), (20, -4), (30, -3), (40, -2), (50, -1), (60, 0), (70, 1), (80, 2), (90, 3), (100, 4)} lying on the line. confidence assessment: 3
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15:30:11 The graph of this function has an intecept on the y' axis at (0,-6) and an intercept on the x axis at (60,0). The graph is a straight line with slope .1. At t = 100 the graph point is (100,4).
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RESPONSE --> Yes the slope is .1 as indicated by the y=mx+b form when .1 is the slope and -6 is the y intercept. y= .1t-6 or where m = slope and b= yintercept. self critique assessment: 2
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15:43:06 `q002. Now sketch the graph of a function y which starts at the point (0, 100), and whose slope at every t is equal to the value of y ' at t. Thus, for example, at t = 0 we have y ' = .1 * 0 - 6 = -6, so our graph of y will start off a the t = 0 point (0,100) with a slope of -6, and the graph begins by decreasing rather rapidly. But the slope won't remain at -6. By the time we get to t = 10 the slope will be y ' = .1 * 10 - 6 = -5, and the graph is decreasing less rapidly than before. Then by the time we get to t = 20 the slope will be y ' = . 1 * 20 - 6 = -4, etc.. If you sketch a graph of y vs. t with these calculated slopes at the corresponding t values, what does the graph look like? Will it be increasing or decreasing, and will it be doing so at an increasing, decreasing or constant rate? Is the answer to this question different for different parts of the graph? If so over what intervals of the graph do the different answers apply?
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RESPONSE --> by integrating to y=ax^2 +bx +c form we get y=(.05)t^2 + (-6)t +c solve for c 100 = (.05)(0)^2 +(-6)(0) +c 100 = 0+0+c 100=c y= (.05)t^2 + (-6)t +100 using the function above we can plug in the points. We know that the graph begins at (0,100) and progresses downward with an increasing slope, and from the last question that at t=60 the slope is zero so that must be the vertex of the parabola plugging that into the function we get the point (60, -80) and then the slope becomes positive and the parabola begins to go upwards. confidence assessment: 3
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15:51:50 The graph will have slopes -6 at t = 0, -5 at t = 10, -4 at t = 9, etc.. At least for awhile, the slope will remain negative so the graph will be decreasing. The negative slopes will however become less and less steep. So the graph be decreasing at a decreasing rate. It's obvious that the slopes will eventually reach 0, and since y' is the slope of the y vs. t graph it's clear that this will happen when y' = .1 t - 6 becomes 0. Setting .1 t - 6 = 0 we get t = 60. Note, and think about about the fact, that this coincides with the x-intercept of the graph of y' vs. t. At this point the slope will be 0 and the graph will have leveled off at least for an instant. Past t = 60 the values of y' will be positive, so the slope of the y vs. t graph will be positive and graph of y vs. t will therefore be increasing. The values of y' will also be increasing, so that the slopes of the y vs. t graph will be increasing, and we can say that the graph will be increasing at an increasing rate.
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RESPONSE --> I think that when solving for C in the answer i may have thrown off my x-intercept, because the xintercept is 60 because at that point y' =0 correct? self critique assessment: 2
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15:54:43 `q003. The graph of y vs. t corresponding to the given rate function y ' = .1 t - 6 has slope -6 at the point (0,100). This slope immediately begins changing, and becomes -5 by the time t = 10. However, let us assume that the slope doesn't change until we get to the t = 10 point. This assumption isn't completely accurate, but we're going to see how it works out. If the slope remained -6 for t = 0 to t = 10, then starting at (0, 100) what point would we reach when t = 10?
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RESPONSE --> starting out at (0,100) when t=0 and going to t=10 we are progressing 10 sec and the rate of change is -6/sec so we would have a drop of 60 therefore our new point would be at (10,40) confidence assessment: 2
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15:55:13 The slope of the graph is the ratio slope = rise / run. If the slope remains at -6 from t = 0 to t = 10, then the difference between 10 is the run. Thus the run is 10 and the slope is -6, so the rise is rise = slope * run = -6 * 10 = -60. The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40).
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RESPONSE --> okay self critique assessment: 3
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15:57:24 `q004. We see that we reach the point (10, 40) by assuming a slope of -6 from t = 0 to t = 10. We have seen that at t = 10 the slope will be y ' = .1 * 10 - 6 = -5. If we maintain this slope for the interval t = 10 to t = 20, what will be the coordinates of the t = 20 point?
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RESPONSE --> starting at (10, 40) and progressing to t= 20 at a slope of -5. rise= slope * run rise = (-5) (10) rise = -50 40 -50 = -10 New point: (20, -10) confidence assessment: 3
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15:57:41 The run from t = 10 to t = 20 is 10. With a slope of -5 this implies a rise of rise = slope * run = -5 * 10 = -50. Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10).
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RESPONSE --> okay self critique assessment: 3
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16:04:52 `q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results.
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RESPONSE --> by using the prevous method and the data from the first question we have the points. {(0,100), (10, 40), (20, -10), (30, -50), (40, -80), (50, -100), (60, -110), (70, -110)} because the slope at t=60 is zero there isn't a change from t=60 to t=70. The graph goes down and then has a sharp turn at which it is flat, had we gone further with the data the graph would represent a parabola without a curve, only straight lines making it up. confidence assessment: 2
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16:05:22 The slope at t = 20 is y ' = .1 * 20 - 6 = -4. From t = 20 to t = 30 the run is 10, so the rise is rise = slope * run = -4 * 10 = -40. Starting from (20,-10) a rise of -40 and a run of 10 takes us to (30, -50). The slope at t = 30 is y ' = .1 * 30 - 6 = -3. From t = 30 to t = 40 the run is 10, so the rise is rise = slope * run = -3 * 10 = -30. Starting from (30,-50) a rise of -30 and a run of 10 takes us to (40, -80). The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100). The slope at t = 50 is y ' = .1 * 50 - 6 = -1. From t = 50 to t = 60 the run is 10, so the rise is rise = slope * run = -1 * 10 = -10. Starting from (50,-100) a rise of -10 and a run of 10 takes us to (60, -110). The slope at t = 60 is y ' = .1 * 70 - 6 = -0. From t = 60 to t = 70 the run is 10, so the rise is rise = slope * run = -0 * 10 = 0. Starting from (60,-110) a rise of and a run of 10 takes us to (70, -110).
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RESPONSE --> okay self critique assessment: 3
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