course Mth 174 I believe I understand how Query works now. If I understand correctly, the 1st question
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17:39:09 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) I found each point on f by using the slopes (f ') and distances (delta x's) and going both directions (to 0 and to 7) from 3. This resulted in a graph with points (0,-1), (2,1), (3,0), (4,2), (6,-2), and (7,-1). (b) Thus, f(0) and f(7) both equal -1. (c) I then found the antiderivatives of the various triangles and one square geometrically (0 to 1 = -1/2; 1 to 3 = 1; 3 to 5 = 2, 5 to 6 = -1, and 6 to 7 = -1 1/2). These summed to a total for 0 to 7 = 0. I also used the Fundamental Theorem of Calculus to find the antiderivative by subtracting f(0) from f(7), thus -1 - (-1) =0, which gave the same result as the geometric method.
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17:39:38 What was your value for the integral of f'?
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RESPONSE --> Answered in previous response: 0
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17:40:07 What was the value of f(0), and of f(7)?
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RESPONSE --> Answered in first response: -1 for both
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17:44:22 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 --> The graph of f(x) is increasing from x=0 to x=2, from x=3 to x=4, and from x=6 to x=7. It is decreasing from x=2 to x=3 and x=4 to x=6. The graph is made of all lines, so it is straight everywhere.
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17:53:23 Was the graph of f(x) continuous?
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RESPONSE --> The graph of f(x) is continuous--it is defined everywhere from x=0 to x=7, and the limit of f(x), as x approaches any point on the graph is equal to f or that point.
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18:49:53 How can the graph of f(x) be continuous when the graph of f'(x) is not continuous?
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RESPONSE --> The fact that the graph of f'(x) is not continuous only means that the graph of f(x) will not form a smooth curve at the points where f'(x) is not continuous. As long as f(x) is differentiable at every point, it is continuous.
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18:53:21 What does the graph of f(x) look like over an interval where f'(x) is constant?
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RESPONSE --> If f(x) is a constant over an interval, the slope of f(x) over that interval is 0, thus the graph of f'(x) over that interval will be the x-axis.
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19:00:47 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 --> The area for the interval x=0 to x=2 is 0. The area for the interval x=2 to x=3 is 1/2. The area for the interval x=3 to x=4 is 1. The area for the interval x=4 to x=6 is 0. The area for the interval x=6 to x=7 is -1 1/2. The integral of f'(x) can be found by summing the areas for each interval, since the integral for the total interval (x=0 to x=7) is the total area between the x-axis and the curve. The sum of the areas is 0.
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19:06:36 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 --> I believe this is the same as my method, as I used the value of f' times the interval of x (the area) to find the change in f(x) on that interval, and then I worked both ways from 3.
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19:08:16 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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19:30:34 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 --> I combined the outflow and inflow curves to make one net flow curve. This curve is concave up from Jan '93 to about March, then concave down from March until July, and concave up from July to Jan '94. This curve has zeroes at Mar (local minimum) and Jul (local and global maximum), and continues negative from Jul through Jan '94 (global minimum). Looking at the two curves separately, the outflow is greater than the inflow in Jan '93 (the quantity of water is decreasing), then gradually approach each other until they cross (are equal) at Mar (creating a zero net flow). Since they inflow then is greater than the outflow (the quantity of water is increasing), the zero at Mar is a local minimum. The inflow over outflow increases until about May, then decreases to another zero net flow in July, after which the outflow is greater than the inflow the rest of the time (thus the July zero is a local maximum. It is also a global maximum since the area between the curves from Mar to Jul (increasing quantity) is greater than the area between the curve from Jan '93 to Mar--which means that the water level in Jan '93 must have been less in Jan '93 than in Jul. Finally, since there is a net outflow (outflow curve greater than inflow) from Jul through Jan '94, and the area between the curves in that interval is greater than the area between the curves from Jan '93 to Jul, the quantity of water decreased more than the net increase from Jan '93 to Jul. Thus Jan '94 must be a global minimum.
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19:39:26 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 about May since the difference between inflow and outflow (with inflow greater) was greatest at that point. The quantity of water was increasing most slowly just to the inside (net inflow) of the points where the curves crossed (zero net flow) at about Mar and July.
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19:42:29 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 --> Okay. My response concurs in all except the very last point. I interpreted the second part of the last question to mean when was the quantity increasing the slowest, not decreasing the most. I concur that th maximum net outflow is Oct, where the difference between the curves (outflow greater) is greatest.
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19:51:25 Query Section 6.2 #38
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RESPONSE -->
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19:56:49 antiderivative of f(x) = x^2, F(0) = 0
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RESPONSE --> I think this is the answer to #34, not #38. #34 asks for the antiderivative of f(x) = 2x, F(0) = 0. This produces F(x) = x^2.
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20:05:30 What was your antiderivative? How many possible answers are there to this question?
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RESPONSE --> Assuming, we're talking about #34, the general antiderivative of f(x) = 2x is x^2 + C, since F(0) = 0, 0^2 + C = 0, thus C = 0, so the antiderivative is F(x) = 2x. There is only possible solution since C must = 0. For #38: The general antiderivative is F(x) = (2/3)x^(3/2) + C. If F(0) = 0, 0 = (2/3)*0^(3/2) + C, thus C = 0, so the antiderivative is F(x) = (2/3)x^(3/2). Again, there is only one possible solution since C must = 0.
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20:09:34 What in general do you get for an antiderivative of f(x) = x^2?
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RESPONSE --> If f(x) = x^2, the general antiderivative is (x^3)/3 + C.
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20:10:26 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 -->
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20:11:18 Query Section 6.2 #55 (3d edition #56) indef integral of t `sqrt(t) + 1 / (t `sqrt(t))
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RESPONSE -->
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20:14:56 What did you get for the indefinite integral?
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RESPONSE --> The indefinite integral is (2/5)t^(5/2) - 2t^(-1/2) + C.
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20:16:47 What is an antiderivative of t `sqrt(t)?
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RESPONSE --> An antiderivative of t*sqrt(t) is (2/5)t^(5/2).
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20:21:42 What is an antiderivative of 1/(t `sqrt(t))?
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RESPONSE --> An antiderivative of 1/(t*sqrt(t)) is -2t^(-1/2).
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20:23:11 What power of t is t `sqrt(t)?
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RESPONSE --> t*sqrt(t) is t to the (3/2) power.
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20:24:04 What power of t is 1/(t `sqrt(t))?
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RESPONSE --> 1/(t*sqrt(t)) is t to the (-3/2) power.
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20:24:42 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 --> OK
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20:32:07 Query Section 6.2 #68 (3d edition #69) def integral of sin(t) + cos(t), 0 to `pi/4
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RESPONSE -->
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20:36:19 What did you get for your exact value of the definite integral?
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RESPONSE --> The definite integral is (-cos t + sin t) evaluated from 0 to pi/4. The value of this integral is 1.
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20:36:51 What was your numerical value?
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RESPONSE --> The numerical value is 1.
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20:38:01 What is an antiderivative of sin(t) + cos(t)?
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RESPONSE --> An antiderivative of sin(t) + cos(t) is -cos(t) + sin(t).
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20:41:57 Why doesn't it matter which antiderivative you use?
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RESPONSE --> It doesn't matter which antiderivative is used because no matter what value C takes on, when the antiderivative is evaluated (F(b) - F(a)), C is subtracted from itself, thus always = 0 (canceling out).
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20:42:21 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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20:43:05 Query Section 6.2 #82 (#81 3d edition) v(x) = 6/x^2 on [1,c}; find c
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RESPONSE -->
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20:57:43 What is your value of c?
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RESPONSE --> My value of c is 6. I used the average value of f rule: Average value of f = (1/(b-a)) times [the integral of f(x)dx, evaluated from a to b] Multiplying both sides by (b-a) gives: Avg. Value * (b-a) = Integral of f(x)dx (from a to b) Substituting c for b, 1 for a, and 1 for the average value (givens) and finding the antiderivative results in: c - 1 = -6/x, evaluated from 1 to c, thus: c - 1 = -6/c - (-6/1), Which is simplified to: c^2-7c+6 = 0 This equation has two solutions, c= 1 and c=6. c=1 is not possible since it would result in dividing by 0 in the average value rule. Thus, c=6.
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21:01:14 In symbols, what did you get for the integral of 6 / x^2 over the interval [1, c]?
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RESPONSE --> Answer given in previous response: ((-6)/c) - ((-6)/1) = ((-6)/c) +6
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21:18:50 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 don't understand the part of the solution that begins with ..the result must be 1... If we set the value of F(c) - F(1) = 1, we're saying that the area under v(x) from 1 to c = 1, which to me means the total value of v(x) on the interval is 1. But the question said that the average value of v(x) on that interval is 1. Therefore, it makes sense to me to use the average value rule.
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21:33:55 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 is: e^(5+x) + (1/5)e^(5x).
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21:34:38 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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