course Mth 174 I guess I just continue to use the same access code, right? Thanks!
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16:17:48 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) Here is a point by point breakdown of the graph: f(0) = -1 f(1) = 0 f(2) = 1 f(3) = 0 f(4) = 2 f(5) = 0 f(6) = -2 f(7) = -1 b) f(0) = -1 and f(7) = -1 c) The simplest way I can see to do find the integral would be to find the area between the line and the x-axis, then add it all together. Here is my result: 2 - 1 + 2 - 4 + 1 = 0 I can't really think of another way unless it would be to find the upper and lower sums and then see what they are close to.
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16:18:05 What was your value for the integral of f'?
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RESPONSE --> My value was 0, as explained in c) previously.
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16:18:27 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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16:22:37 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 is concave up on the interval (0, 2.5) and (3.5, 4.5). The graph is concave down over the periods (2.5, 3.5) and (5.5, 7). There are no points on this graph where the line is straight up or down.
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16:36:03 Was the graph of f(x) continuous?
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RESPONSE --> Yes, the graph is continuous.
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16:42:39 How can the graph of f(x) be continuous when the graph of f'(x) is not continuous?
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RESPONSE --> Silly me. It can't be continuous because the derivative makes sharp changes from one point to another. This creates sharp corners on the graph of the original function, making the function non-continuous.
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16:47:40 What does the graph of f(x) look like over an interval where f'(x) is constant?
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RESPONSE --> The graph of f(x) looks like a straight line on any interval where f'(x) is constant. Hmm, I guess that invalidates what I previously said about the parts of the graph being concave up and down.
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16:56:32 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 --> Interval (0, 2): area 2. Interval (2, 3): area -1. Interval (3, 4): area 2. Interval (4, 6): area -4 Interval (6, 7): area 1. So each interval contributed the area to the integral.
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16:57:24 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 --> Ah, I see. Not only from the sum of the areas, but also using the fundamental theorem to find the value of the integral. I must remember that.
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17:14:50 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 --> b) I found that the quantity of water was largest around July. I found that it was smallest in January 1994. c) It looks like it was increasing most rapidly during May, and decreasing most rapidly during October.
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17:19:57 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 of water was greatest around July. At this point, the amount of inflow matched the amount of outflow, meaning that the overall amount of water was greatest at this point (a peak). The least quantity of water was in Jan 1994. At this point the inflow had been constantly dropping since roughly April. Even though the outflow had peaked in October, the outflow was still much greater than inflow, so the overall amount of water had been dropping since July.
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17:21:51 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 the fastest around May. At this point the inflow had peaked and was much greater than the outflow. The water was decreasing most rapidly around October. At this point the outflow was greater than the inflow and reached a peak.
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17:24:05 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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17:25:04 Query Section 6.2 #38
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RESPONSE --> I found one solution to this problem: F(x) = ( 2x^(3/2) ) / 3
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17:26:02 antiderivative of f(x) = x^2, F(0) = 0
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RESPONSE --> Oh. Well, 4.6.38 in the book is trying to find the antiderivative of f(x) = 'sqrt(x).
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17:26:57 What was your antiderivative? How many possible answers are there to this question?
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RESPONSE --> My antiderivative was: F(x) = ( 2x^(3/2) ) / 3 I believe there is only one solution for the function to have the solution F(0) = 0.
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17:28:06 What in general do you get for an antiderivative of f(x) = x^2?
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RESPONSE --> When looking at the antiderivative for x^2, I get: x^3 / 3 The general form is this: ( x^(n+1) ) / (n+1)
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17:28:43 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 --> Agh! I forgot the +C!!!
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17:29:41 Query Section 6.2 #55 (3d edition #56) indef integral of t `sqrt(t) + 1 / (t `sqrt(t))
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RESPONSE --> I found the indefinite integral to be: ( 2t^(5/3) ) / 3 - 2 / 'sqrt(t) + C.
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17:29:47 What did you get for the indefinite integral?
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RESPONSE --> See prior answer.
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17:31:14 What is an antiderivative of t `sqrt(t)?
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RESPONSE --> What I did was first contract t `sqrt(t) into t^(3/2) using exponential rules. Next, I took the antiderivative using the general form: 2t^(5/2) / 5
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17:46:07 What is an antiderivative of 1/(t `sqrt(t))?
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RESPONSE --> Again, I first contracted the function using exponential rules: t^(-3/2) Then the antiderivative is a little easier to find: -2t^(-1/2) Which can be written: -2 / 'sqrt(t)
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17:46:24 What power of t is t `sqrt(t)?
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RESPONSE --> t^(3/2)
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17:47:52 What power of t is 1/(t `sqrt(t))?
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RESPONSE --> The power is t^(-3/2).
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17:48:10 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 --> Hey, I got it. Cool. That shortcut works quite well.
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17:57:59 Query Section 6.2 #68 (3d edition #69) def integral of sin(t) + cos(t), 0 to `pi/4
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RESPONSE --> Not assigned. First I need to find the antiderivatives of each term seperately and then combine them. sin(t) = -cos(t) cos(t) = sin(t) Antiderivative: sin(t) - cos(t). Now the simplest way to find the definite integral would be to take F(b) - F(a): ( sin('pi/4) - cos('pi/4) ) - ( sin(0) - cos(0) ) = 0 - -1 = 1
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17:58:29 What did you get for your exact value of the definite integral?
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RESPONSE --> My exact value ended up being 1.
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17:59:22 What was your numerical value?
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RESPONSE --> Hmm. I guess the exact value was the equation, and the numerical value is 1.
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17:59:40 What is an antiderivative of sin(t) + cos(t)?
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RESPONSE --> The antiderivative I found was: sin(t) - cos(t).
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18:08:57 Why doesn't it matter which antiderivative you use?
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RESPONSE --> Because in this case C doesn't matter. All we are looking for is the change between 0 and 'pi/4.
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18:09:54 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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18:19:18 Query Section 6.2 #82 (#81 3d edition) v(x) = 6/x^2 on [1,c}; find c
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RESPONSE --> We can use the Fundamental Theorem to find c: F(c) - F(1) = 1 F(c) - 6 = 1 F(c) = 7 6 / x^2 = 7 x^2 = 6/7 x = 'sqrt(6/7) = c = 0.92582
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18:19:35 What is your value of c?
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RESPONSE --> 'sqrt(6/7) or 0.92582.
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20:39:20 In symbols, what did you get for the integral of 6 / x^2 over the interval [1, c]?
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RESPONSE --> Whoops, I did it wrong. I didn't realize that the function given was the derivative. However, I still don't have a good idea on how to do it.
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20:40:16 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 --> Phoey. I thought about doing this, but what why would the average derivative be equal to the value of the function?
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21:01:26 Extra Problem (formerly from Section 6.2 #44): What is the indefinite integral of e^(5+x) + e^(5x)
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RESPONSE --> I believe the indefinite integral would be: e^(5+x) + ( e^(5x) ) / 5
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21:01:41 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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