#$&* course mth 174 time 655 PMdate 9-20-10 Calculus II
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18:28:51
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Note: Problem numbering is according to the problems as presented in Problem Assignments . The numbering will differ from that in your text. ********************************************* Question: Section 6.1, Problem 5 5th edition Problem 14 4th edition Problem 5 [[6.1.5 (previously 6.1 #12)]] f '(x) =1 for x on the interval (0,2), -1 on (2,3), 2 on (3,4), -2 on (4,6), 1 on (6,7) f(3) = 0 What was your value for the integral of f '? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The solution I arrived at is arrived at by plotting the points and then finding the sums I think this is correct my final solution was a result around 0. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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? 0 What is the area beneath the graph of f' between x = 3 and x = 4? 12 What is the area beneath the graph of f' between x = 4 and x = 6? 26 What is the area beneath the graph of f' between x = 6 and x = 7? 42 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? 2 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? 10 Using similar reasoning, what is the value of f at x = 7? 14 Then using similar reasoning, see if you can determine the value of f at x = 2 and at x = 0.** -1,-2 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): Ok I see that this is going to get very complicated very quick but I think I can handle it. I understand where the integrals come from and what they mean I think but I want to know how you arrive at the solution for a f(3)=18 or whatever and what is the easiest way to calculate this?
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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? 1 What is the slope of the f graph between x = 3 and x = 4? +2 What is the slope of the f graph between x = 4 and x = 6? -2 What is the slope of the f graph between x = 6 and x = 7? +1 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. The slope is -2 according to the equation. 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. ** The graph of the concave up woulb have a slope of something like this -2+2+1 and the concave down bould be like +1+2-1-2 or something like that
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: Was the graph of f(x) continuous? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes it was continuous on the specified intervals between pi and pi/2. Or whatever the limit is. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: How can the graph of f(x) be continuous when the graph of f ' (x) is not continuous?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because the graph of f’(x) is the simplified cut down almost approximation for the short time interval of its derivative meaning. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: Section 6.1, Problem 10: The graph of outflow vs. time is concave up Jan 1993 -Sept, peaks ub October, then decreases somewhat thru Jan 1994; the inflow starts lower than the outflow, peaks in May, then decreases until January; inflow is equal to outflow around the middle of March and again in late July. **** When was the quantity of water greatest and when least? Describe in terms of the behavior of the two curves. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The quantity of water will be greatest in July and least in Feb. ** 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? The time interval when the slope of the graph is positive Over time interval(s) is the amount of water decreasing? ** When the slope of the graph is 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? The antiderivative or change in quantity. What aspect of which graph gives you the rate at which water is flowing out of the reservoir? The outflow slope 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? It has to be increasing at a linear almost expontential 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? The graphs have to be a positive slope with it slowly decreasing in slope. At this time . 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? The slope is decreasing and it has to be increasingly decreasing at this time. 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? ** The water level has to have an outflow that is decreasing at a linear or mybe an expontnetial rate.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: Section 6.2, Problem 5 [[6.2.5 (previously 6.2 #26)]] antiderivative of f(x) = x^2, F(0) = 0
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the antiderivative of this is F(x)=x^3/3 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 I didn’t solve for the f(x)=0 but I do however understand that there is a C in the antiderivative formula and this is very important to finding the values of the x for these change in quantity formulas. ------------------------------------------------ Self-critique rating #$&*2.5 ********************************************* Question: Section 6.2, Problem 8 [[(previously 6.2 #56)]] indef integral of t `sqrt(t) + 1 / (t `sqrt(t)) **** What did you get for the indefinite integral? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The indefinite integral for this is 2/5 *x^5/2 +2*sqrt x and the formula I used was fx*g’x+gxf’x dx=fxgx
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Given Solution: ** The function can be written t^(3/2) + t^(-3/2). Both are power functions of the form t^n. Antiderivative is 2/5 * t^(5/2) - 2 t^(-1/2) + c or 2/5 t^(5/2) - 2 / `sqrt(t) + c. **
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11:39:51 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I think I got most of it right. ------------------------------------------------ Self-critique rating #$&*2.5 ********************************************* Question: 6.2.9 (previously 6.2 #50) definite integral of sin(t) + cos(t), 0 to `pi/4 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The solution to this integral is 1.
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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 ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: Why doesn't it matter which antiderivative you use?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because it the change in quantity that you are trying to obtain and either way it will result in a difference. confidence rating #$&* 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): Ok this makes more sense than the garbage that I thought. It makes no difference because the limit restricts answers on one side to the other so they come out the same . neat,
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The antiderivative is -6/x so the we sub in c-1*1for the avg length. Then I set up and solve using the calculator’s quadratic and get 6 and 1.
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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): Ok wow if that was a start intro I cant imagine how lengthy these are going to be, But the explanation helped.
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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): Ok I think that I used the wrong formula to calculate this I used the indefinite integral formula, but I think that I messed up somewhere can you help?
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