#$&* course mth 174 4/16 9:50 pm Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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18:28:51
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Note: Problem numbering is according to the problems as presented in Problem Assignments . The numbering will differ from that in your text. ********************************************* Question: Section 6.1, Problem 5 5th edition Problem 14 4th edition Problem 5 [[6.1.5 (previously 6.1 #12)]] f '(x) =1 for x on the interval (0,2), -1 on (2,3), 2 on (3,4), -2 on (4,6), 1 on (6,7) f(3) = 0 What was your value for the integral of f '? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When looking at the graph of f’ in the text I see the following Between 0 and 2 the area is 2*1=2 so the slope is 2/2=1???
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Given Solution: 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. Let me know if you disagree with or don't understand any of this and I will explain further. Let me know specifically what you do and don't understand. ** ** Alternative solution: Two principles will solve this problem for you: 1. The definite integral of f' between two points gives you the change in f between those points. 2. The definite integral of f' between two points is represented by the area beneath the graph of f' between the two points, provided area is understood as positive when the graph is above the x axis and negative when the graph is below. We apply these two principles to determine the change in f over each of the given intervals. Answer the following questions: What is the area beneath the graph of f' between x = 0 and x = 2? What is the area beneath the graph of f' between x = 3 and x = 4? What is the area beneath the graph of f' between x = 4 and x = 6? What is the area beneath the graph of f' between x = 6 and x = 7? What is the change in the value of f between x = 3 and x = 4? Since f(3) = 0, what therefore is the value of f at x = 4? Now that you know the value of f at x = 4, what is the change in f between x = 4 and x = 6, and what therefore is the value of f at x = 6? Using similar reasoning, what is the value of f at x = 7? Then using similar reasoning, see if you can determine the value of f at x = 2 and at x = 0.** STUDENT QUESTION: I did not understand how to obtain the value of f(0), but I found that f(7) was 10 by adding all the integrals together INSTRUCTOR RESPONSE: The total area is indeed 10, so you're very nearly correct; however the integral is like a 'signed' area--areas beneath the x axis make negative contributions to the integral--and you added the 'absolute' areas &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### It took me a long time to wrap my head around this problem. I think I have got the solution and steps down now. On a test, will we have a graph illustration as in the book or will we have any points to plot the f’ graph??? If the instructions only say, “f '(x) =1 for x on the interval (0,2), -1 on (2,3), 2 on (3,4), -2 on (4,6), 1 on (6,7)” with no illustration or listed points, I have a difficult time envisioning the f’graph in order to determine the area underneath the points to I can determine the slope of the f graph???? ------------------------------------------------ Self-critique Rating:3
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18:37:09
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confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graph of f(x) is increasing, with slope 1, on the interval (0,2), since f'(x) = 1 on that interval, decreasing, with slope -1, on the interval (2,3), where f'(x) = -1, increasing, with slope +2, on the interval (3,4), where f'(x) = +2, decreasing, with slope -2, on the interval (4,6), where f'(x) = -2, and increasing, with slope +1, on the interval (6,7), where f'(x) = +1. The concavity on every interval is zero, since the slope is constant on every interval. Since f(3) = 0, f(4) = 2 (slope 2 from x=3 to x=4), f(6) = -2 (slope -2 from x = 4 to x = 6), f(7) = -1 (slope +1 from x=6 to x=7). Also, since slope is -1 from x=2 to x=3, f(2) = +1; and similar reasoning shows that f(0) = -1. ** ** The definite integral of f'(x) from x=0 to x=7 is therefore f(7) - f(0) = -1 - (-1) = 0. ** ** Basic principles: 1. The slope of the graph of f(x) is f'(x). So the slope of your f graph will be the value taken by your f' graph. 2. Note that if the slope of the f graph is constant for an interval that means that the graph is a straight line on the interval. Using these principles answer the following questions: What is the slope of the f graph between x = 0 and x = 2? What is the slope of the f graph between x = 3 and x = 4? What is the slope of the f graph between x = 4 and x = 6? What is the slope of the f graph between x = 6 and x = 7? Given that f(3) = 0 and using the value of the slope of the f graph between x = 3 and x = 4, describe the f graph between these two points. Using similar information describe the graph for each of the other given intervals. Also answer the following: What would have to be true of the f' graph for the f graph to be concave up? Same question for concave down. **
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18:37:09
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### What would have to be true of the f' graph for the f graph to be concave up? Same question for concave down. ** In order for f to be concave up, F’ would have to be rise at an inconsistent rate or slope??? In order for f to be concave down, F’ would have to fall at an inconsistent rate or slope??? I am not sure if the above answer is correct????
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Given Solution: ** A function f(x) is continuous at x = a if the limit of the f(x), as x approaches a, exists and is equal to f(a). Is this condition fulfilled at every point of the f(x) graph? **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### ** A function f(x) is continuous at x = a if the limit of the f(x), as x approaches a, exists and is equal to f(a). I don’t follow??? I know that the function is continuous between each integral as it has a consistent slope for each integral. Does this mean that the graph is consistent??? Is this condition fulfilled at every point of the f(x) graph? ** At every point of f(x) graph, there is a constant and continuous slope, although overall there is not consistency for the entire graph??? ??? is the above correct or am I misunderstanding???
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of F(x) is continuous between each integral of x because each integral has a consistent and continuous slope??? I’m not sure why F(x) would be continuous and not F’(x) except for the fact that the above question said, “A function f(x) is continuous at x = a if the limit of the f(x), as x approaches a, exists and is equal to f(a).” Which is a statement I don’t fully understand??? confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 18:38:11
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Given Solution: ** If f ' is constant then the slope of the f(x) graph is constant, so the graph of f(x) must be linear ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### I didn’t use the word linear, but I think my description is correct???? I will remember the term linear for the future.
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**** When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves. The water increased the fastest in the month of May because the inflow curve is much more than the outflow curve. The water increased the slowest in October (when it is actually the most water lost) because the outflow curve is much more than the inflow curve. What aspect of which graph gives you the rate at which water is flowing out of the reservoir? The water is flowing out when the outflow curve is higher than the inflow curve. So when the outflow graph is at its largest y value over the y value of the inflow graph, water is flowing out at a higher rate. What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at an increasing rate? For water to be increasing at an increasing rate, the y value of the inflow curve has to be higher and higher than the corresponding y value of the outflow curve as we move left to right. What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at a decreasing rate? For water to be increasing at a decreasing rate, the inflow rate has to be greater than the outflow rate but the gap between the inflow and outflow has to be closing in. So as we move left to right the y value of the inflow curve is less and less higher than the y value of the outflow curve. What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at an increasing rate? If the water is to be decreasing at an increasing rate, the water has to be going out faster and faster than the water is coming in. So the outflow graph has to have a higher y value than the inflow graph does and this space between the outflow and the inflow has to be getting greater and greater. So as we move left to right the y value of the outflow is increasing and the y value of the inflow is decreasing. What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at a decreasing rate? ** For the water in the lake to be decreasing at a decreasing rate, the outflow graph has to be more than the inflow graph, so the outflow must have a higher y value than the inflow, but the gap between the 2 has to be getting less and less (closing in). As we move left to right the y value of the outflow is decreasing and the y value of the inflow is increasing, but the outflow curve is still above the inflow curve. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** ** When inflow is > outflow the amount of water in the reservoir will be increasing. If outflow is < inflow the amount of water will be decreasing. Over what time interval(s) is the amount of water increasing? Over time interval(s) is the amount of water decreasing? **
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**** 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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18:47:03 The curve increases most between Jan and Apr and it decreases most between July and October ** What aspect of which graph gives you the rate at which water is flowing into the reservoir? What aspect of which graph gives you the rate at which water is flowing out of the reservoir? What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at an increasing rate? What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at a decreasing rate? What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at an increasing rate? What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at a decreasing rate? **
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18:47:04
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### I hope I answered all of the components of this question. I feel pretty confident but let me know if I am wrong or missed anything???
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The antiderivative of f(x) =x^2 would be 1/3x^3 + C which can be checked to confirm that the derivative of 1/3x^3 +C is x^2. Evaluated at F(0) is 1/3(0)^3 +C= 0 so C=0 so the final answer is F(x)= 1/3x^3+0 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** An antiderivative of x^2 is x^3/3. The general antiderivative of x^2 is F(x) = x^3/3 + c, where c can be anything. There are infinitely many possible specific antiderivative. However only one of them satisfied F(0) = 0. We have F(0) = 0 so 0^3/3 + c = 0, or just c = 0. The antiderivative that satisfies the conditions of this problem is therefore F(x) = x^3/3 + 0, or just F(x) = x^3/3. **
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18:47:58 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: Section 6.2, Problem 8 [[(previously 6.2 #56)]] indef integral of t `sqrt(t) + 1 / (t `sqrt(t)) **** What did you get for the indefinite integral? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I began by re-writing the square roots as powers Indef intergral of t*t^1/2 + t^-1*t^-1/2 Indef intergral of t^3/2 + t^-3/2 I then thought about what the integral would be and found the answer below 2/5t^5/2 -2t^-1/2 +c Which can be confirmed by taking the derivative which would give you t^3/2 + t^-3/2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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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11:39:51 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok except I didn’t put parenthesis around by fraction powers. ------------------------------------------------ Self-critique Rating: ********************************************* Question: 6.2.9 (previously 6.2 #50) definite integral of sin(t) + cos(t), 0 to `pi/4 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When taking a definite integral you have to first find the antiderivative and then evaluate at the given points. The antiderivative of sin(t) would be -cos(t) which we can verify by taking the derivative of -cos(t) which is -(-sin(t)) which is sin(t). The antiderivative of cos(t) is sin(t) which we can verify by taking the derivative of sin(t) which is cos(t) So the antiderivative of the function is -cos(t)+sin(t)+c We now evaluate the antiderivative at pi/4 -cos(pi/4) + sin(pi/4)=0 We now evaluate the antiderivative at 0 -cos(0) + sin(0)= -1 So to find the definite integral we say 0-(-1) which is 0+1=1 So the answer is 1 +c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** An antiderivative is -cos(t) + sin(t), as you can see by taking the derivative. Evaluating this expression at `pi/4 gives -`sqrt(2)/2 + `sqrt(2)/2 = 0. Evaluating at 0 gives -1 + 0 or -1. The antiderivative is therefore 0 - (-1) = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ## ok except I put c at the end of the antiderivative which I think is correct???
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In this problem it doesn’t matter which value you use for C because you are finding the -cos T and the Sin t for a specific number in this case pi/4 and 0. So whichever value you use for C would be the same for pi/4 and 0. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### Is my answer correct, I am a little confused and want to make sure I am understanding correctly???
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We first have to find the antiderivative of 6/X^2. The first thing I do is rewrite that as 6x^-2. To find the antiderivative I figured -6x^-1 which I then verified by taking the derivative which would give me 6x^-2 so I confirm the antiderivative is correct. -6X^-1 can be written as -6/x So now we want to know the average on interval (1,C) for -6/x. The interval at 1 is -6/1=-6 The interval at c is -6/C The average interval at 1- interval at C -6 - (-6/C) = 1-C -6+6/C=1-C now we multiply by c to get the c off the denominator -6C + 6= c - c^2 now move everything to the right C^2-7c+6=0 now factor (c-1)(c-6) = 0 Solve each one C=1 c= 6 We don’t know which of these C is so we start substituting in to see if one doesn’t work Int. 1 - int C -6-6=-12 Or -6-1=-7 At this point I’m not sure what to do since it appears either value of c will work. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: An antiderivative of 6 / x^2 is F(x) = -6 / x. The definite integral is equal to the product of the average value and the length of the interval. In this case average value is 1 and the interval from x = 1 to x = c has length c - 1. So the definite integral must be 1 * ( c - 1). Evaluating between 1 and c and using the above fact that the result must be 1 we get F(c) - F(1) = -6/c- (-6/1) = c - 1 so that -6/c+6=c - 1. We solve for c, first getting all terms on one side: c - 7 + 6/c = 0. Multiplying both sides by c to get c^2 - 7 c + 6 = 0. Either be factoring or the quadratic formula we get c = 6 or c = 1. If c = 1 the interval has length 0 and the definite integral is not defined. This leaves the solution c= 6. STUDENT QUESTION I had some trouble with this problem I got -6/x for antideri. So I thought that at F(1) = -6 and F(1.5) = -4 Then I got really confused for some reason used the logic F(b)- F(a) = -4 - (-6) = 2 when divided by 2 = 1. I see what you did but not so sure about the logic. INSTRUCTOR RESPONSE Note that the length of the interval between x = 1 and x = 1.5 is .5. The integral is 2, but the average value between x=1 and x=1.5 is (integral) / (length of interval) = 2 / .5 = 4, not 2. The average value of the integral must be 1. The integral of a function over an interval is equal to its average value over that interval, multiplied by the length of the interval: • ave value = definite integral / length of interval It follows immediately that • definite integral = ave value * length of interval In this case the interval has length (c - 1) and the average value must be 1. The integral must therefore be 1 * (c - 1). The integral is from x = 1 to x = c. So The integral of 6/x^2 from x = 1 to x = c must equal 1 * (c - 1). The integral of 6/x^2 from x = 1 to x = c is -6 / c - (-6 / 1) = 6 - 6/c. Thus 6 - 6/c = 1 * (c - 1). We solve to get c, and we obtain c = 6.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ## I got all of the work to where c could either be 1 or 6 But I don’t understand the part below “The definite integral is equal to the product of the average value and the length of the interval. In this case average value is 1 and the interval from x = 1 to x = c has length c - 1. So the definite integral must be 1 * ( c - 1)” I thought you subtracted the top interval from the bottom interval which would be int c - int 1 which would be either 6-(-6) or 1-(-6) I don’t understand what I am doing wrong???? ------------------------------------------------ Self-critique Rating:
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Given Solution: ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### I believe I got the correct answer, please let me know if I did not I didn’t see a combined answer with a + C##### ------------------------------------------------ Self-critique Rating:3
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