course MTH 174 My only real question is about the average value in the #82 / #81 / #60 problem. Everything else seemed straight forward. |hrWassignment #001
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16:13:11 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 -->
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16:13:49 What was your value for the integral of f'?
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RESPONSE --> 0
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16:15:08 What was the value of f(0), and of f(7)?
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RESPONSE --> f(0) = -1 and f(7) = -1
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16:19:35 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 is increasing from 0 to 2, decreasing from 2 to 3, increasing from 3 to 4, decreasing from 4 to 6, increasing from 6 to 7. All connections between the points are line segments and are therefore straight and are not concave up or concave down.
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16:20:05 Was the graph of f(x) continuous?
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RESPONSE --> Yes, the graph of f(x) is continuous.
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16:24:10 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 which are absent in f'(x) are the points at which the line segments meet in f(x). There is no slope on a point, only on a curve whether that curve is straight or not.
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16:26:05 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) is a line segment of the constant slope at the given value of f'(x) over that interval.
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16:30:11 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 areas were quadrilaterals of corresponding area: 2, -1, 2, -4, 1.
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16:32:20 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 --> This seems consistent with my answer.
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16:33:52 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 --> I believe this is question #18 in the second edition that I am using.
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16:37:15 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 inflow was greater than the outflow between March and July with the greatest difference between the two occuring in April.
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16:45:36 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 increased the fastest between March and April. It increased most slowly between April and July. This is show by the area between the two curves.
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16:46:34 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 --> This makes sense to me.
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16:46:51 Query Section 6.2 #38
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RESPONSE -->
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16:52:38 antiderivative of f(x) = x^2, F(0) = 0
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RESPONSE --> This seems to be problem 26 in the second edition. An antiderivative is F(x) = 1/3 * x^3 + C. When F(0) = 0, the only answer for C is 0
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16:54:14 What was your antiderivative? How many possible answers are there to this question?
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RESPONSE --> The antiderivative I had was 1/3 * x^3 + C where C has an infinite amount of possible values.
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16:54:59 What in general do you get for an antiderivative of f(x) = x^2?
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RESPONSE --> 1/3 * x^3 + C is the general form.
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16:55:19 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 --> This seems consistent with my answers.
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17:03:56 Query Section 6.2 #55 (3d edition #56) indef integral of t `sqrt(t) + 1 / (t `sqrt(t))
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RESPONSE --> This was number 37 in the second edition. I changed everything to exponents: integral t * t^(1/2) + t^-1 * t^(-1/2) dt integral t^(3/2) + t(-3/2) dt 2/5 * t^(5/2) - 2t^(-1/2) + C
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17:04:06 What did you get for the indefinite integral?
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RESPONSE --> 2/5 * t^(5/2) - 2t^(-1/2) + C
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17:05:26 What is an antiderivative of t `sqrt(t)?
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RESPONSE --> 2/5 * t^(5/2) + C
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17:05:40 What is an antiderivative of 1/(t `sqrt(t))?
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RESPONSE --> -2t^(-1/2) + C
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17:06:04 What power of t is t `sqrt(t)?
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RESPONSE --> 1 + 1/2 = 3/2
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17:06:39 What power of t is 1/(t `sqrt(t))?
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RESPONSE --> -1 + -1/2 = -3/2
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17:06:54 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 --> This seems consistent with my answer.
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17:22:19 Query Section 6.2 #68 (3d edition #69) def integral of sin(t) + cos(t), 0 to `pi/4
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RESPONSE --> This seems to be #50 in the second edition. integral from 0 to `pi/4 of sin(t) + cos(t) from 0 - cos(t) + sin (t) + C is the indefinite form - cos(0) + sin(0) = -1 + 0 = -1 -cos(`pi/4) + sin(`pi/4) = - `sqrt(2)/2 + `sqrt(2)/2 = 0 The value of the definite integral is 0 - -1 = 1
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17:22:38 What did you get for your exact value of the definite integral?
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RESPONSE --> 0 - -1 = 1
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17:23:01 What was your numerical value?
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RESPONSE --> It is the same: 0 - -1 = 1
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17:23:45 What is an antiderivative of sin(t) + cos(t)?
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RESPONSE --> The indefinite form is - cos(t) + sin (t) + C
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17:25:18 Why doesn't it matter which antiderivative you use?
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RESPONSE --> The C that is used for evaluating both ends will be the same and will therefore cancel out no matter what value of C is chosen.
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17:25:44 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 --> This seems consistent with my results.
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18:08:00 Query Section 6.2 #82 (#81 3d edition) v(x) = 6/x^2 on [1,c}; find c
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RESPONSE --> This seems to be #60 in the second edition. The indefinite form of the integral 6/x^2 is: -6/x +C The definite form is: -6/c - -6/1 -6/c + 6 The average value of the function is defined by 1/(b - a) * integral from a to b of f(x). This means that: 1/(c - 1) * -6/c + 6 = 1
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18:08:06 What is your value of c?
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RESPONSE --> 6
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18:08:57 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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18:11:06 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 --> This does not match my answer. I wonder if the wording is different between the versions or if I am missing something. The exact wording of my question is ""The average value of the function v(x) = 6/x^2 on the interval [1,c] is equal to 1. Find the value of c.""
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18:16:33 Extra Problem (formerly from Section 6.2 #44): What is the indefinite integral of e^(5+x) + e^(5x)
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RESPONSE --> This seems to be #44 in my book. Utilizing the information that d/dx e^(5x) = 5 * e^(5x), the indefinite integral of e^(5+x) + e^(5x) is 1/5 * e^(5x) + e^(5+x) + C
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18:16:52 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 --> This seems consistent with my answer.
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