#$&* course Mth 173 7/27 8 008. Approximate depth graph from the rate function*********************************************
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Given Solution: `aThe graph of this function has an intecept on the y' axis at (0,-6) and an intercept on the x axis at (60,0). The graph is a straight line with slope .1. At t = 100 the graph point is (100,4). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): At first I thought that I had to upload a sketched graft, but after looking at the solution I found out that I only had to describe the graph that I sketched on graph paper. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q002. This problem is a continuation of the preceding, in which y ' = .1 t - 6. Now sketch the graph of a function y which starts at the point (0, 100), and whose slope at every t is equal to the value of y ' at t. Thus, for example, at t = 0 we have y ' = .1 * 0 - 6 = -6, so our graph of y will start off a the t = 0 point (0,100) with a slope of -6, and the graph begins by decreasing rather rapidly. But the slope won't remain at -6. By the time we get to t = 10 the slope will be y ' = .1 * 10 - 6 = -5, and the graph is decreasing less rapidly than before. Then by the time we get to t = 20 the slope will be y ' = . 1 * 20 - 6 = -4, etc.. If you sketch a graph of y vs. t with these calculated slopes at the corresponding t values, what does the graph look like? Will it be increasing or decreasing, and will it be doing so at an increasing, decreasing or constant rate? Is the answer to this question different for different parts of the graph? If so over what intervals of the graph do the different answers apply? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph will be decreasing at a decreasing rate until it gets to x = 60 when y = 0. After the slope is positive. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT QUESTION The graph will decrease at a decreasing rate however the slope is decreasing at a constant rate? I believe this is where I get confused because if the slope is constant how does it cause the graph to decrease at a decreasing rate? INSTRUCTOR RESPONSE Be careful to distinguish between the two graphs. • We have a graph of y ' vs. t, and we have a graph of y vs. t. They are two separate graphs with different, though related, characteristics. • A graph of y ' vs. t is a straight line. • y ' gives us the slope of the y vs. t graph (a little more precisely, the value of y ' evaluated some value of t is the slope of the y vs. t graph at that same value of t). Whatever is true about the y ' vs. t graph, it true of the slope of the y vs. t graph. • y ' is changing (as you say it is increasing slightly as we move to the right). So the slope of the y vs. t graph is increasing. The y vs. t graph will therefore not be a straight line. • At the current point, the value of y ' is in fact negative, so the slope of the y vs. t graph is negative. That is, the y vs. t graph is decreasing. • As you have pointed out the y ' vs. t graph is 'slightly increasing', in your words. So the slope of the y vs. t graph is increasing. The y vs. t graph is therefore decreasing, but with an increasing slope. The slope of the y vs. t graph is negative, but is increasing (however slowly) toward 0. Its rate of decrease is therefore decreasing, and we can make the following equivalent statements: • The graph is decreasing at a decreasing rate. • The graph is decreasing but its slope is increasing. • The slope is negative but increasing. • The graph is concave up. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): A little confused. Are both y = at^2 +bt +c and y’ = 2at + b the slope??? ------------------------------------------------ Self-critique Rating: 3
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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: Rise = slope * run = -5 * 10 = -50 40 - 50 = -10 (20, -10) confidence rating #$&*: Confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe run from t = 10 to t = 20 is 10. With a slope of -5 this implies a rise of rise = slope * run = -5 * 10 = -50. Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t(30) -4 * 10 = -40 -10 - 40 = -50 (30, -50) t(40) -3 * 10 = -30 -50 - 30 = -80 (40, -80) t(50) -2 * 10 = -20 -80 - 20 = -100 (50, -100) t(60) -1 * 10 = -10 -100 - 10 = -110 (60, -110) t(70) 0 * 10 = 0 -110 - 0 = -110 (70, -110) confidence rating #$&*: Confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 * 60 - 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " 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: t(30) -4 * 10 = -40 -10 - 40 = -50 (30, -50) t(40) -3 * 10 = -30 -50 - 30 = -80 (40, -80) t(50) -2 * 10 = -20 -80 - 20 = -100 (50, -100) t(60) -1 * 10 = -10 -100 - 10 = -110 (60, -110) t(70) 0 * 10 = 0 -110 - 0 = -110 (70, -110) confidence rating #$&*: Confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 * 60 - 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!