course Mth 174 001. `query 1 Physics II 01-25-2009
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12:39:24 What was your value for the integral of f'?
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RESPONSE --> The integral of
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m?{p??????y?6V???assignment #001 ?????S??B+???Physics II 01-25-2009 ????????ax?????? assignment #001 ?????S??B+???Physics II 01-25-2009
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13:22:02 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 --> I don't quite understand what I'm finding.
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??????w????? assignment #001 ?????S??B+???Physics II 01-25-2009
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18:12:00 What was your value for the integral of f'?
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RESPONSE --> I wasn't exactly sure what to find since I have the third edition of the book and the questions are different for chapter 6. I drew a graph representing the given coordinates and f values. I found the value of the integral of f to be 1.
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18:14:15 What was the value of f(0), and of f(7)?
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RESPONSE --> f(0) = ( - infinity, 0) f(7)= (7, +infinity)
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18:19:35 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 --> My graph is increasing from - infinity to (0,2). It then descreases and is concave upto the point (2,3). It then increases and is concave down to (3,4), and then decreases to (3,0). It is then concave up to (4,6) and increases to + infinity.
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18:20:45 Was the graph of f(x) continuous?
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RESPONSE --> This graph is continuous
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18:29:33 How can the graph of f(x) be continuous when the graph of f'(x) is not continuous?
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RESPONSE --> The graph f(x) can be continuous when the graph f'(x) is not because to be continuous the graph because a continuous graph must have a limit that exists and as the limit approaches a f(x)= f(a). Also f(a) must be defined. Solving f'(x) the function changes and might not meet these requirements.
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18:32:01 What does the graph of f(x) look like over an interval where f'(x) is constant?
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RESPONSE --> When f(x) is constant, f(x) is equal to the same y value for different x values.
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20:03:40 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 --> I'm not sure but I believe that the constant areas where in between each interval. I thought the areas were what f was set equal to in the problem. Each interval contributed to the integral of f'(x) because you can look at the graph and see that the first interval (0,2)-(2,3) was increasing at a decreasing rate. The next interval, (2,3)-(3,4), first had was increasing at an increasing rate, and then increasing at a decreasing rate. The interval (3,4)-(4,6) was increasing at a decreasing rate and the interval (4,6)-(6,7) was increasing at an increasing rate. I understand how to figure out when it is increasing or decreasing, but I'm not sure if I drew the graph correctly.,
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20:12:10 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 understand these calculations, and if given a graph I I could figure them out. I believe I drew my graph wrong on this problem. I still don't understand what was ment by f=n. I thought it was the area but now I'm not sure.
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20:22:45 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 don't know very much about this graph my the description and this question is not in my book. If it is May 93 I assume the quanity of water is greatest where inflow peaks in May and least either in Jan 93 or Jan94. The curve of the outflow reaches its peak before the curve of the inflow. The water would be greatest where there is the least difference in inflow and outflow, which is where inflow peaks in may. It would be least where the curves are farther apart which would be at the beginning or end of the curves on the graph.
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20:24:48 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 quanity of water was increasing fastest in May because the peak of the incomming water has a steep slope, so the water would be increasing at an increasing rate. It would be increasing slowly just after the peak when the curve is increasing at a decreasing rate.
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20:27:58 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 --> I understand everything described here. Again, it would have been easier to figure out if I could have drawn a more accurate graph. I now see how the areas under the curves shows the changes in the curves in relation to each other.
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20:31:54 antiderivative of f(x) = x^2, F(0) = 0
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RESPONSE --> THe antiderivative of f(x) = x^2 is f(x) = x^3 / 3
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20:32:41 What was your antiderivative? How many possible answers are there to this question?
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RESPONSE --> My antiderivative was f(x) = x^3/3.
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20:33:11 What in general do you get for an antiderivative of f(x) = x^2?
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RESPONSE --> f(x) = x^3/3
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20:36:28 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 --> The general antideravitive is the function = c, where c can be anything. Only one satisfies F(0) = 0, where c is 0.
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20:39:05 Query Section 6.2 #55 (3d edition #56) indef integral of t `sqrt(t) + 1 / (t `sqrt(t))
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RESPONSE --> The integral is t^3/3 + (t^3/3 *1)
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20:39:32 What did you get for the indefinite integral?
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RESPONSE --> t^3/3 = (t^3/t *1)
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20:41:19 What is an antiderivative of t `sqrt(t)?
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RESPONSE --> If the (t) is part of the exponent than it would be t^3t/3
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20:41:57 What is an antiderivative of 1/(t `sqrt(t))?
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RESPONSE --> - t^3/3
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20:42:23 What power of t is t `sqrt(t)?
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RESPONSE --> I do not understand this question.
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20:43:11 What power of t is 1/(t `sqrt(t))?
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RESPONSE --> I do not understand what the power would be.
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20:44:28 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 solved that t is squared, not the square root. I understand the questions and answers now.
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20:46:17 What did you get for your exact value of the definite integral?
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RESPONSE --> The integral is -cos(t) = sin(t)
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20:46:34 What was your numerical value?
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RESPONSE --> numerical value?
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20:46:56 What is an antiderivative of sin(t) + cos(t)?
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RESPONSE --> -cos(t) +sin(t)
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20:47:22 Why doesn't it matter which antiderivative you use?
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RESPONSE --> Each antiderivative expresses the function differently.
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20:48:02 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 -->
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20:50:40 Query Section 6.2 #82 (#81 3d edition) v(x) = 6/x^2 on [1,c}; find c
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RESPONSE --> v(x) = 6/x^2 +c x=1 since its on (1,c) 1=6/1^2 +c 1=6 +c -5 = c
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20:50:45 What is your value of c?
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RESPONSE --> -5
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20:52:09 In symbols, what did you get for the integral of 6 / x^2 over the interval [1, c]?
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RESPONSE --> 6/ x^2 6 (x-2) x^-3/3
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20:53:41 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 -->
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20:55:49 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 integral is e^5+x * 1 + e^5x * 5
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20:57:22 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 --> I thought with the chain rule you solve for the derivative of u and then multiply it by eu.
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