course MTH 271 10/04, 1500 assignment #008008. Approximate depth graph from the rate function
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20:11:58 `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 function y'=.1t - 6 produces a linear graph, that increases at a consistent rate. The y-intercept of the graph is (0, -6). The x-intercept of the graph is (60,0). confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:22:08 `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 of y vs. t with the calculated slopes will be a parabola, pointing downward. The rates of change will be different for different intervals of the graph. From t=0 to t=60, the graph will be decreasing at a decreasing rate. At t=60, the rate of change is 0, forming the point of the parabola. For t=60 and greater values of t, the graph will be increasing at an increasing rate. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:27:02 `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 --> Using the point-slope form of (y - y1)=m(x-x1), we are able to find the equation of the line. y-100=-6(x-0) y-100=-6x y=-6x + 100 Given an x value (or, in this case, a t value) of 10: y= -6(10) + 100 y=-60 + 100 y=40 When t=10, y=40 for a point of (10,40). confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:31:08 `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 --> m=-5 point= (10,40) Using point-slope form: y-40=-5(x -10) y - 40= -5x + 50 y= -5x + 90 ...we find the new equation for the line with the new slope. Given an x value of 20: y=-5(20) + 90 y=-100 + 90 y=-10 At t=20, y=-10. (20,-10) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:38:31 `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 (10,40), m=-5. (20,-10) To find the rate of change at 20: y'=.1(20) - 6 y'= 2-6 y'=-4 y--10= -4(x - 20) y+10=-4x + 80 y=-4x + 70 y(30)=-4(30) + 70 y(30)=-50 (30,-50) y'(30)=.1(30) - 6= -3 y- -50=-3(x-30) y+50=-3x + 90 y=-3x + 40 y(40)=-3(40) + 40 y(40)=-80 (40,-80) y'(40)=.1(40) -6 = -2 y- -80=-2(x-40) y+80=-2x + 80 y=-2x y(50)=-2(50) y(50)=-100 (50,-100) y'(50)=.1(50) -6 =-1 y- -100= -1(x - 50) y+100=-x + 50 y=-x - 50 y(60)=-(60) -50 y(60)=-110 (60,-110) y'(60)=.1(60) - 6 y'(60)=0 y- -110=0(x-60) y+110=0 y=-110 (70,-110) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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