course MTH 174 ???y?????€????assignment #001
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21:44:40 Query Section 6.1 #15 f'=1 on (0,2), -1 on (2,3), 2 on (3,4), -2 on (4,6), 1 on (6,7); f(3) = 0
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RESPONSE --> (a) (b) f(0) = 1 and f(7) = -1 (c) f(b) - f(a) = (-1) - 1 = -2 f(b) = The integral of f'(x) from 0 to 7 + f(a) = -1
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21:45:05 What was your value for the integral of f'?
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RESPONSE --> -2
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21:46:24 What was the value of f(0), and of f(7)?
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RESPONSE --> f(0) = 1, f(7) = -1
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21:50:12 Describe your graph of f(x), indicating where it is increasing and decreasing and where it is concave up, where it is straight and where it is concave down.
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RESPONSE --> It is increasing from x = 2 to 3 and from 3 to 4, and from 5 to 6. It is decreasing from x = 0 to 2 and from 4 to 5. It is concave down from 2 to 5. It is concave up from 0 to 3 and 4 to 6.
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21:50:42 Was the graph of f(x) continuous?
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RESPONSE --> Yes, it was continuous.
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21:51:54 How can the graph of f(x) be continuous when the graph of f'(x) is not continuous?
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RESPONSE --> The points of discontinuity indicate where the slope is undefined on the graph, hence the peaks.
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21:53:55 What does the graph of f(x) look like over an interval where f'(x) is constant?
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RESPONSE --> A cosine function.
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21:56:23 What were the areas corresponding to each of the four intervals over which f'(x) was constant? What did each interval contribute to the integral of f'(x)?
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RESPONSE --> Straight lines with a constant slope correspond to each of the four intervals over which f'(x) was constant. Each interval contributed to the constant slope of a line over that interval.
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21:57:26 the change in f from x=0 to x=2 is 2 (area beneath line segment from x=0 to x=1), then from x=2 to x=3 is -1, then from x=3 to x=4 is +2, then from x=4 to x=6 is -4, then from x=6 to x=7 is +1. If f(3) = 0 then f(4) = 0 + 2 = 2, f(6) = 2 - 4 = -2 and f(7) = f(6) + 1 = -1. Working back from x=3, f(2) = 0 - (-1) = 1 and f(0) = 1 - 2 = -1. The integral is the sum of the changes in f ' which is 2 - 1 + 2 - 4 + 1 = 0. Alternatively since f(0) = -1 and f(7) = -1 the integral is the difference f(7) - f(0) = -1 - (-1) = 0.
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RESPONSE --> OK
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22:06:54 Query Section 6.1 #25 outflow concave up Jan 93 -Sept, peaks Oct, down somewhat thru Jan 94; inflow starts lower, peaks May, down until Jan; equal abt March and late July
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RESPONSE --> OK
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22:10:40 When was the quantity of water greatest and when least? Describe in terms of the behavior of the two curves.
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RESPONSE --> The quantity was greatest in July, because between probably Feb. and Mar. up until July the rate of inflow was greater than outflow. The quantity was least in Jan. '94, because between July and Jan. the outflow was greater than the inflow.
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22:14:00 When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves.
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RESPONSE --> The quantity of water was increasing fastest around April when there was the greatest difference between inflow and outflow and most slowly increasing around July when the inflow was closest to outflow.
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22:14:15 Between any two dates the corresponding outflow is represented by the area under the outflow curve, and the inflow by the area under the inflow curve. When inflow is greater than outflow the quantity of water in the reservoir will be increasing and when the outflow is greater than the inflow quantity of water will be decreasing. We see that the quantity is therefore decreasing from January 93 through sometime in late February, increasing from late February through the beginning of July, then again decreasing through the end of the year. The reservoir will reach a relative maximum at the beginning of July, when the outflow rate overtakes the inflow rate. The amount of water lost between January and late February is represented by the difference between the area under the outflow curve and the area under the inflow curve. This area corresponds to the area between the two graphs. The amount of water gained between late February and early July is similarly represented by the area between the two curves. The latter area is clearly greater than the former, so the quantity of water in the reservoir will be greater in early July than on Jan 1. The loss between July 93 and Jan 94, represented by the area between the two graphs over this period, is greater than the gain between late February and early July, so the minimum quantity will occur in Jan 94. The rate at which the water quantity is changing is the difference between outflow and inflow rates. Specifically the net rate at which water quantity is changing is net rate = inflow rate - outflow rate. This quantity is represented by the difference between the vertical coordinate so the graphs, and is maximized around late April or early May, when the inflow rate most greatly exceeds the outflow rate. The net rate is minimized around early October, when the outflow rate most greatly exceeds the inflow rate. At this point the rate of decrease will be maximized.
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RESPONSE --> OK
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22:20:26 Query Section 6.2 #38
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RESPONSE --> f(x) = [(2*`sqrt(x^3)) \ 3] + C
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22:21:18 antiderivative of f(x) = x^2, F(0) = 0
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RESPONSE --> OK
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22:23:45 What was your antiderivative? How many possible answers are there to this question?
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RESPONSE --> I listed my antiderivative as the answer to Section 6.2 question # 38. There were infinite many answers to the question. What did that last statement / declaration have to do with anything?
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22:24:50 What in general do you get for an antiderivative of f(x) = x^2?
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RESPONSE --> (x^3 / 3) + C
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22:26:44 An antiderivative of x^2 is x^3/3. The general antiderivative of x^2 is F(x) = x^3/3 + c, where c can be anything. There are infinitely many possible specific antiderivative. However only one of them satisfied F(0) = 0. We have F(0) = 0 so 0^3/3 + c = 0, or just c = 0. The antiderivative that satisfies the conditions of this problem is therefore F(x) = x^3/3 + 0, or just F(x) = x^3/3.
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RESPONSE --> I believe I read a statement similiar to: antiderivative of f(x) = x^2 ... I would have understood: Find the antiderivative of f(x) where f(x) = x^2...
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23:24:13 Query Section 6.2 #55 (3d edition #56) indef integral of t `sqrt(t) + 1 / (t `sqrt(t))
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RESPONSE --> indef integral of t `sqrt(t) + 1 / (t `sqrt(t)) = indef integral of t `sqrt(t) plus indef integral 1 / (t `sqrt(t)). indef integral of t `sqrt(t) = indef integral of t^(3/2) = (t^(5/2) / (5/2) = ((t^(5/2) / 1) * (2/5) = 2 `sqrt(t^5)
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23:24:49 What did you get for the indefinite integral?
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RESPONSE --> The answer was typed in the last question.
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23:25:24 What is an antiderivative of t `sqrt(t)?
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RESPONSE --> The answer was entered previously.
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23:25:40 What is an antiderivative of 1/(t `sqrt(t))?
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RESPONSE --> The answer was entered previously.
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23:26:28 What power of t is t `sqrt(t)?
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RESPONSE --> t `sqrt(t) = t * t^(1/2) = t^(3/2)
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23:27:07 What power of t is 1/(t `sqrt(t))?
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RESPONSE --> t^(-3/2)
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23:27:38 The function can be written t^(3/2) + t^(-3/2). Both are power functions of the form t^n. Antiderivative is 2/5 * t^(5/2) - 2 t^(-1/2) + c or 2/5 t^(5/2) - 2 / `sqrt(t) + c.
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RESPONSE --> I neglected to enter the constant of integration, C
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23:35:27 Query Section 6.2 #68 (3d edition #69) def integral of sin(t) + cos(t), 0 to `pi/4
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RESPONSE --> def integral of sin(t) + cos(t), 0 to `pi/4 = (sin(`pi/4) - cos(`pi/4)) - (sin(0) - cos(0)) = 1
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23:35:33 What did you get for your exact value of the definite integral?
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RESPONSE --> 1
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23:35:45 What was your numerical value?
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RESPONSE --> 1
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23:36:06 What is an antiderivative of sin(t) + cos(t)?
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RESPONSE --> sin(t) - cos(t)
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23:36:43 Why doesn't it matter which antiderivative you use?
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RESPONSE --> The derivative of a constant is always 0.
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23:36:51 An antiderivative is -cos(t) + sin(t), as you can see by taking the derivative. Evaluating this expression at `pi/4 gives -`sqrt(2)/2 + `sqrt(2)/2 = 0. Evaluating at 0 gives -1 + 0 or -1. The antiderivative is therefore 0 - (-1) = 1. The general antiderivative is -cos(t) + sin(t) + c, where c can be any number. You would probably use c = 0, but you could use any fixed value of c. Since c is the same at both limits of the integral, it subtracts out and has no effect on the value of the definite integral.
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RESPONSE --> OK
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01:14:35 Query Section 6.2 #82 (#81 3d edition) v(x) = 6/x^2 on [1,c}; find c
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RESPONSE --> 1 = 1 / (c - 1) times the integral of v(x) from 1 to c. The antiderivative of v(x) = 6 * x^-1 / -1 = -6 / x = V(x) the integral of v(x) from 1 to c = V(c) - V(1) = (-6 / c) + 6 1 = 1 / (c - 1) * ((-6 / c) + 6) (c - 1) = (-6 / c) + 6 c - 7 = -6 / c c^2 - 7c + 6 = 0 (c - 1)(c - 6) = 0 c cannot be 1 because it would not allow for an interval. c = 6
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01:14:48 What is your value of c?
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RESPONSE --> The value of c is 6.
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01:15:19 In symbols, what did you get for the integral of 6 / x^2 over the interval [1, c]?
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RESPONSE --> (-6 / c) + 6
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01:21:05 An antiderivative of 6 / x^2 is F(x) = -6 / x. Evaluating between 1 and c and noting that the result must be 1 we get F(c) - F(1) = -6/c- (-6/1) = 1 so that -6/c+6=1. We solve for c: -6/c=1-6 6/c=-5 -6=-5c c=6/5.
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RESPONSE --> I used the average value theorem on page 260 of the text: Average value of f from a to b = (1 / (b - a) times the integral of f(x) from a to b. I do not understand your solution.
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01:24:48 Extra Problem (formerly from Section 6.2 #44): What is the indefinite integral of e^(5+x) + e^(5x)
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RESPONSE --> The indefinite integral of e^(5+x) + e^(5x) is e^(5+x) + e^(5x) + C.
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01:25:28 The derivative of e^(5+x) is, by the Chain Rule, (5+x)' * e^(5+x) = 1 * e^(5 + x) = e^(5 + x) so this function is its own antiderivative. The derivative of e^(5x) is (5x) ' * e^(5x) = 5 * e^(5x). So to get an antiderivative of e^(5x) you would have to use 1/5 e^(5x), whose derivative is e^(5x). **
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RESPONSE --> OK
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