course Mth 174 r????y????????j??assignment #001
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22:37:53 What was the value of f(0), and of f(7)?
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RESPONSE --> my integral of f was -1
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22:40:51 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 was increasing being -1 and 1 on the y-axis, then it was drecreasing from 1 to 0 on the y-axis. It went from 0 up to 2 on the y-axis so it was increasing. Then it was decreasing from 2 to -2 on the y-axis. Finally it was increasing from -2 to -1. it was concave up from 0 to 2 on the x-axis. then it was concave down from 2 to 3. concave up from 3 to 4. then concave down from 4 til 6 all on the x-axis.
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22:40:59 Was the graph of f(x) continuous?
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RESPONSE --> no it was not continuous
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22:41:26 How can the graph of f(x) be continuous when the graph of f'(x) is not continuous?
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RESPONSE --> because it is the antiderviative so it can be continuous without the antiderviative being continuous
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22:42:21 What does the graph of f(x) look like over an interval where f'(x) is constant?
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RESPONSE --> i dont know
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??????????·??·?? assignment #001 ????·???????g????w??? Physics II 09-12-2007
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22:45:30 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 -->
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v_?????|??X???????U·??·assignment #001 ????·???????g????w??? Physics II 09-12-2007
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22:51:31 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 --> The integral of f is 0. the values of f(0) and f(7) are both -1 the graph of f(x) was increasing at the integrals of 0 to 2 on the x-axis, 3 to 4 on the x-axis, 6 to 7 on the x-axis. It was decreasing at the integrals of 2 to 3 on the x-axis, 4 to 6 on the x-axis. yes it was continuous It can be continuous because the graph of f'(x) is the derviative of the f(x) graph so it will change most of the time. the area from x=0 to x=2 is 2, from x=2 to x=3 it is -1, then from x=3 to x=4 it is +2, then from x=4 to x=6 it is -4 and then from x=6 to x=7 it is +1.
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22:54:19 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 great it is at July 1993 and minimum is in January 1994. The two graphs are opposite of each other crossing one another every once in a while. it was increasing the fastest in May 1993 and the slowest in Oct 1993. The graphs of these two stay pretty close together but never cross.
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22:55:25 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 --> alright i understand now that the changing of water quantity is represented by the equation net rate = inflow rate - outflow rate.
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22:58:24 Query Section 6.2 #38
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RESPONSE --> the antiderviative in this case will be 2x^(3/2) / 3 which is the only possibility
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22:58:55 antiderivative of f(x) = x^2, F(0) = 0
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RESPONSE --> x^(3) / 3
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22:59:27 What was your antiderivative? How many possible answers are there to this question?
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RESPONSE --> my antiderivative was x^3 / 3 and it is the only possibility
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22:59:43 What in general do you get for an antiderivative of f(x) = x^2?
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RESPONSE --> you get x^3 / 3
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23:00:32 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 --> ok i understand that i left off my +c when i was giving the general antiderivative answer and got that there are infinite many possiblities.
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23:02:14 Query Section 6.2 #55 (3d edition #56) indef integral of t `sqrt(t) + 1 / (t `sqrt(t))
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RESPONSE --> the indefinite integral here is 2t^(5/2) / 5 - 2t^(-1/2) + C
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23:02:53 What did you get for the indefinite integral?
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RESPONSE --> i got 2t^(5/2) / 5 - 2t^(-1/2) + C
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23:03:23 What is an antiderivative of t `sqrt(t)?
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RESPONSE --> the antiderivative is 2t^(5/2) / 5
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23:03:42 What is an antiderivative of 1/(t `sqrt(t))?
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RESPONSE --> the antiderivative is 2t^(-1/2)
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23:03:55 What power of t is t `sqrt(t)?
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RESPONSE --> to the 4th
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23:04:07 What power of t is 1/(t `sqrt(t))?
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RESPONSE --> it would be -1/2
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23:04:23 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 i understand that is what i got
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23:06:53 Query Section 6.2 #68 (3d edition #69) def integral of sin(t) + cos(t), 0 to `pi/4
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RESPONSE --> the definite integral here is -cost + sint and the answer here is -cos(pi/4) + cos(0) + sin(pi/4) - sin(0)
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23:08:33 What did you get for your exact value of the definite integral?
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RESPONSE --> i got the answer 1.0137
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23:08:43 What was your numerical value?
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RESPONSE --> my numerical value was 1.0137
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23:08:57 What is an antiderivative of sin(t) + cos(t)?
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RESPONSE --> it is -cost + sint
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23:09:21 Why doesn't it matter which antiderivative you use?
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RESPONSE --> because it could change the whole problem depending on which sin or cos you use here
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23:10:01 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 --> aight i still again forgot my + c when giving the general antiderivative but other than that i believe i got the answer pretty right.
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23:12:48 Query Section 6.2 #82 (#81 3d edition) v(x) = 6/x^2 on [1,c}; find c
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RESPONSE --> c will be -4 because the when 1 is used it is 5 so the other side should have ot be added up to be -4 so 5 = -4 would equal 1.
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23:12:57 What is your value of c?
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RESPONSE --> my value of c is -4
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23:13:19 In symbols, what did you get for the integral of 6 / x^2 over the interval [1, c]?
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RESPONSE --> my integral is 6x - x^1.5 / 1.5
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23:14:23 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 --> ok i totally got that wrong the antiderivative of this was -6 / x and c ended up being 6/5. i started off wrong by getting the antiderivative wrong but now i understand how they got the answer that they did.
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23:16:56 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 is not the problem that is in my book of 6.2 #44 because my book reads (x^3 -2)dx, so i dont know where they indefinite integral problem came from.
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23:17:42 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 --> alright i understand it now but i still dont know where that problem came from.
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