Student Name: assignment #008 008. Approximate depth graph from the rate function
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20:35:22 `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 --> The graph of this function would have a y-intercept of (0, -6), and a slope of .1. This is a linear function. At t = 0, y' = -6 At t = 100, y' = 4
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20:35:30 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 --> OK.
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20:36:55 `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 --> The graph will be decreasing at a decreasing rate.
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20:39:47 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 --> OK, but I was forgetting about the values greater than t = 60 when y' will become positive and increase at an increasing rate.
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20:40:20 `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 --> The point would be (10, 40) if the slope stayed at -6 between the points t = 0 and t = 10
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20:40:25 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 --> OK.
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20:41:24 `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 --> The starting point is (10, 40). If we maintain the slope of -5 between the points t = 10 and t = 20, the resulting point would be 50 units lower (10 * -5 = -50) and 10 units to the right. (20, -10)
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20:41:28 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 --> OK.
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20:46:14 `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 --> At the point t = 20, the slope is y' = .1 * 20 - 6 = -4. Meaning that the resulting point will be 40 units lower than the starting point (20, -10). The resulting point is (30, -50) At the point t = 30, the slope is y' = .1 * 30 - 6 = -3. Meaning that the resulting point will be 30 units lower than the starting point (30, -50). The resulting point is (40, -80) At the point t = 40, the slope is y' = .1 * 40 - 6 = -2. Meaning that the resulting point will be 20 units lower than the starting point (40, -80). The resulting point is (50, -100) At the point t = 50, the slope is y' = .1 * 20 - 6 = -1. Meaning that the resulting point will be 10 units lower than the starting point (50, -100). The resulting point is (60, -110) At the point t = 60, the slope is y' = .1 * 60 - 6 = 0. Meaning that the resulting point will be 0 units lower than the starting point (60, -110). The resulting point is (70, -110)
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20:46:47 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 --> OK.
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Pãyƙmɲ}ӧg assignment #008 008. `query 8 Applied Calculus I 02-19-2006
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21:09:35 What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?
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RESPONSE --> f(z) = 2^z and g(t) = 3t - 5
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21:09:43 CORRECT STUDENT RESPONSE: f(z)=2^z and g(t)=3t-5, so that f(g(t)) = 2^g(t) = 2^(3t-5).
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RESPONSE --> ok.
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21:10:45 19:11:56
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RESPONSE --> ?
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21:12:34 prob 1.3.66 temperature conversion
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RESPONSE --> problem 1.3.66 in my book asks to sketch parallel/perpendicular lines to given lines
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21:15:48 What linear equation relates Celsius to Fahrenheit?
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RESPONSE --> deg. F = 1.8 (deg. C) + 32
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21:16:04 CORRECT STUDENT RESPONSE: degrees Fahrenheit=1.8(degrees Celsius)+32, or F = 1.8 C + 32. INSTRUCTOR COMMENT: Since each Fahrenheit degree is 1.8 Celsius degrees a graph of F vs. C will have slope 1.8. Since F = 32 when C = 0 the graph will have vertical intercept at (0, 32) so the y = m x + b form will be F = 1.8 C + 32.
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RESPONSE --> OK.
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21:21:10 How did you use the boiling and freezing point temperatures to get your relationship?
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RESPONSE --> If you use the points at which water boils as coordinates on a coordinate plane, you get the points (0, 32) and (100, 212). If you find the slope of the line connecting these two points, it is 1.8. Put this into the y = mx + b form and you get 32 = 1.8 (0) + b solve for b and we get: b = 32 This leaves me with the equation: y = 1.8x + 32 or F = 1.8C + 32
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21:21:16 ** A graph of Fahrenheit vs. Celsius temperatures gives us the two (x,y) points (o,32) and (100,212). We use these two points to find the slope m=(y2-y1)/(x2-x1) = (212 - 32) / 100 = 1.8. Now we insert the coordinates of the point (0,32) and into the point-slope form y = 1.8 x + b of a line to get 32 = 1.8 * 0 + b. We easily solve to get b = 32. {So the equation is y = 1.8 x + 32, or F = 1.8 C + 32. **
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RESPONSE --> OK.
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