course mth271 8 open QAIf your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aThe 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With the slopes as a negative number the slope would be decreasing, at a decreasing rate if I am look at this right the 1st point is (0,100) with a slope of –6 and at T=10 the slope is –5 and the average of this 10sec of the graph is –5.5 cm/sec and the change over the 10sec will be –55 thus point (10,45) now we look at point T=20 from the given above we see that the rate of change here is –4 cm/sec the average of T=10 and T=20 is –4.5 cm/sec and over the 10sec we see a –45 cm of change taken us to (20,0) . The graph of this starts on the y axis at (0,100) going into quadrant 1with a negative slope passing through the x axis at point (20,0) at T=60 the instantaneous change of rate =0 and then the rate or change becomes positive and increasing at an increasing rate. If I did this right the equation for this from the derivative Y`=.1t-6 we get y=.05t^2-6t+100 . confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With the given that we start at T=0 at point (0,100) and our rate of change stays at –6 to T=10 we would see –6 rate /sec for 10sec taking us to point (10,40) confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As before we are dealing with 10sec and a rate of –5 for a total of –50 this will take us to point (20,-10) confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We had got to point (20,-10) as we follow this pattern at T=20 our slope is –4 over the 10 sec we come up with –40 then we take this from –10 we end up at point (30,-50) flowing this we get points (40,-80);(50,-100);(60,-110);(70,-110) The graph of this will be strait lines which have a negative slope or decreasing slope at a decreasing rate and having no slope between T=60 and T=70 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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).