#$&* course Mth 173 10/7 1:30
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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 given differential function is y’ = 0.1t - 6 which is the slope function The function y can be found by integrating the y’ function which gives us the result: y = 0.05 t^2 - 6t + c which passes through the point (0, 100). By substituting this point in the function we obtain the arbitrary value c. Thus 100 = 0.05 * (0^2) - 6 * 0 + c, c = 100. Thus the function thus obtained is y = 0.05 t^2 - 6t + 100. Thus according to the integral rules the y function is a parabola whose vertex is (60, -80). Since the function is defined within the interval t = 0 and t = 100. The value of the function y(0) = 0.05*0*0 - 6*0 + 100 = 100 and the value of the function at 100 is y(100) = 0.05 * 100 * 100 - 6 * 100 + 100 = 500 - 600 + 100 = 0. Thus we see that the function y decreases from t = 0 to t = 60 and then increases from t = 60 to t = 100. Since the slope of the function is given by the linear equation y’ = 0.1t - 6 and since the slope is a linear function thus the slope is ever increasing from t = 0 to t = 100. Thus we see that for the interval t = 0 to t = 60 the function y decreases at a decreasing rate (since the slope is increasing), whereas for the t = 60 to t = 100 interval the graph of the function y increases at an increasing rate (again due to the increasing slope) Now if you differentiate the slope function y’ = 0.1t - 6, we get the rate at which the slope changes, y” = 0.1 is the rate at which the slope changes. Thus the slope increases at a constant rate of 0.1. thus in brief the graph of the function y is a parabola whose vertex is ( 60 , -80 ) , which is decreasing from 0 to 60 , increasing from 60 to 100 and the slope of which is negative from [ 0, 60) , 0 at t = 60 , and positive from ( 60 , 100 ], the slope is ever increasing with a constant rate of 0.1. Whereas the graph decrease at an decreasing rate for the t = 0, t= 60 interval and increases at an increasing rate for the t = 60, t = 100 interval. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Given the point (0, 100 ). The slope at this point is -6, and according to the given data it is known that the slope remains -6 till we reach t = 10. Let y be the value corresponding to t = 10. Thus the 2 points are ( 0 , 100 ) and ( 10 , y ). Using the 2 point form the slope of the given line joining the 2 points is given by Slope = ( y - 100 ) / ( 10 - 0 ) = ( y - 100 ) / 10. We also know that the slope of the line is -6. Thus ( y - 100 ) / 10 = -6 , y - 100 = -60 and thus y = 40. Thus the point that we would reach when t = 10 with a slope of -6 is ( 10 , 40 ). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe slope of the graph is the ratio slope = rise / run. If the slope remains at -6 from t = 0 to t = 10, then the difference between 10 is the run. Thus the run is 10 and the slope is -6, so the rise is rise = slope * run = -6 * 10 = -60. The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. We see that we reach the point (10, 40) by assuming a slope of -6 from t = 0 to t = 10. We have seen that at t = 10 the slope will be y ' = .1 * 10 - 6 = -5. If we maintain this slope for the interval t = 10 to t = 20, what will be the coordinates of the t = 20 point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given point ( 10 , 40 ) which has a slope of -5. Now it is also told that the slope remains -5 for the t = 10 and t = 20 interval. Let y be the value of the point corresponding to t = 20 that we reach from ( 10 , 40 ) after keeping the slope -5. Thus the 2 points obtained are ( 10 , 40 ) and ( 20 , y ). Using the 2 point form the slope of the line is given by, slope = ( y - 40 ) / ( 20 - 10 ) = ( y - 40 ) / 10. Also the slope of the line is -5, thus ( y - 40 ) / 10 = -5 Thus ( y - 40 ) = -50 and y = -10. Thus the point we reach by maintaining the slope of -5 from the point ( 10 , 40 ) we obtain the point ( 20 , -10 ). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: The slope function is y ' = .1 t - 6. Thus the slope at t = 20 will be y’(20) = 0.1*20 - 6 = 2 - 6 = -4 From the point (20, -10) we draw a line with slope -4 to reach a point at t = 30. The run for the line = 30 - 20 = 10 And the slope = -4. Slope = rise / run Thus the rise = slope * run = -4 * 10 = -40. Since the y coordinate corresponding to t = 20 is -10 and the rise is -40, the y coordinate corresponding to t = 30 will be = -10 - 40 = -50 Thus with the slope of -4 we reach the point (30, -50). The slope at t = 30 will be y’(30) = 0.1*30 - 6 = 3 - 6 = -3 From the point (30, -50) we draw a line with slope -3 to reach a point at t = 40. The run for the line = 40 - 30 = 10 And the slope = -3. Slope = rise / run Thus the rise = slope * run = -3 * 10 = -30. Since the y coordinate corresponding to t = 30 is -50 and the rise is -30, the y coordinate corresponding to t = 40 will be = -50 - 30 = -80 Thus with the slope of -3 we reach the point (40, -80). Similarly the slope at t = 40 is -2 The run for the line is 10 and thus the rise will be -20 The co-ordinate (40, -80) thus will reach the co-ordinate (50, -100) Next the slope at t=50 is -1 The run for the line is 10 and thus the rise will be -10 The co-ordinate (50, -100) thus will reach the co-ordinate (60, -110) Lastly the slope at t=60 is 0 The run for the line is 10 and thus the rise will be 0 The co-ordinate (60, -110) will thus reach the co-ordinate (70, -110) Thus the graph so created will be a graph with multiple straight lines of increasing slope value. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!