course Mth 174 ά̦IڀO؆Xү
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12:28:52 query problem 8.4.3 (8.3.4 3d edition) was 8.2.6) moment of 2 meter rod with density `rho(x) = 2 + 6x g/m
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RESPONSE --> Part a) Write a Riemann sum approximating the total mass of the rod: Since the density at a distance 'dx from the left of the rod is given by 2 + 6x. My R-sum is 2 + 6x 'dx Part b) Find the exact mass by converting the sum into an integral: I as 'dx approaches 0, from [0,2] int( 2 + 6x 'dx) = 2x + 6x^2 I then used the Fundamental Theorem to find the mass, F(b) - F(a). 2(2) + 6(2)^2 = 4 + 24 = 28. F(b) - F(a) = 28 - 0.
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12:29:18 what is the moment of the rod?
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RESPONSE --> Not real sure what the 'moment' of the rod really means.
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12:30:55 What integral did you evaluate to get a moment?
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RESPONSE --> I just realized that my original integral was wrong. I would like to use this answer to correct it. My integral should have been int(2 + 6x) = 2x + 3x^2 Which would have made the mass 16
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12:37:13 query problem 8.4.12 (was 8.2.12) mass between graph of f(x) and g(x), f > g, density `rho(x)
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RESPONSE --> With left and right bounds of x = a and x = b, upward bound of f(x), and lower bound of g(x) the mass can be found: In terms of f(x) and g(x): int(f(x)) - int(g(x)). Since f(x) is larger than g(x), the difference in area underneath the two would yield the area in between the two functions, which is the mass. In terms of Density: int[ R-sum('rho(x) 'dx)
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12:37:50 what is the total mass of the region?
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RESPONSE --> Numerically I do not know, but int(f(x)) - int(g(x)) = mass of the region
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12:38:39 What integral did you evaluate to obtain this mass?
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RESPONSE --> I didn't really find an integral to evaluate.
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12:47:33 What is the mass of an increment at x coordinate x with width `dx?
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RESPONSE --> Mass = the integral of the R-sum of 'rho(x) within the interval (x, 'dx)
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12:52:35 What is the area of the increment, and how do we obtain the expression for the mass from this area?
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RESPONSE --> I am not completely sure, but I would say the area of the increment would be smaller interval [x, 'dx] over which we can find a more solid density on which to base our R-sum. With this R-sum we can find the mass.
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12:54:51 How to we use the mass of the increment to obtain the integral for the total mass?
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RESPONSE --> mass of increment = [int from a -> b (x * 'rho(x) 'dx)] / [int from a -> b ('rho(x) 'dx)]
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13:59:51 query problem 8.5.13 (8.4.10 3d edition; was 8.3.6) cylinder 20 ft high rad 6 ft full of water
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RESPONSE --> Problem 12 asked to find how much work is required to pump all of the water over the rim, which is from 10ft to 20ft. Problem 13 asks how much work is required to pump all of the water from the same, full cylinder to a point 10 ft over the cylinder. I first cut out a slice of the cylinder with a width of 12 and a height of 'dh. I then found the Volume of this slice = 'pi * r^2 * 'dh = 'pi * w^2 * 'dh The Density of water is 1000 kg/m^2. Force of gravity on the slice = Density * g * Volume = 1000 * g * 'pi * w^2 * 'dh. Since we are trying to move every piece of water to 10 ft above the cylinder, that includes moving that water at the bottom of the cylinder also, we must move all of the water 30 ft (20ft of the cylinder + 10 ft above). Work = Force * Distance Force = 1000 * g * 'pi * w^2 * 'dh Distance = (30 - h) I then found w in terms of h: (w / h = 12 / 20) w = .6 *h Work = 1000 * g * 'pi * .36h^2 * (30 - h) * 'dh I then summed up the slices and took the limit as 'dh -> 0. R-sum [360 * g * 'pi * (30h^2 - h^3)] When g = 9.8 R-sum [360 * g * 'pi * (30h^2 - h^3)] From b = 30 and a = 0; int [360 * g * 'pi * (30h^2 - h^3)] = 'pi * 360 * p * ((30/3)h^3 - (1/4)h^4) = 'pi * 360 * 9.8 * ((30/3)h^3 - (1/4)h^4) = 'pi * 3528 * ((30/3)h^3 - (1/4)h^4) F(b) - F(a) = 748188874.5 = Work I looked it up, and this is wrong, but I am not sure where I went wrong.
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13:59:56 how much work is required to pump all the water to a height of 10 ft?
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13:59:58 What integral did you evaluate to determine this work?
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14:02:19 Approximately how much work is required to pump the water in a slice of thickness `dy near y coordinate y? Describe where y = 0 in relation to the tank.
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RESPONSE --> Work = lim('dh -> 0) R-sum [ Force * Distance] = lim('dh -> 0) R-sum [ (Density * g * Volume) * Distance]
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14:02:33 Explain how your answer to the previous question leads to your integral.
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14:26:19 query problem 8.3.18 CHANGE TO 8.5.15 work to empty glass (ht 15 cm from apex of cone 10 cm high, top width 10 cm)
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RESPONSE --> I do not see this question in the book. I am not sure what to do.
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14:26:22 how much work is required to raise all the drink to a height of 15 cm?
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14:26:25 What integral did you evaluate to determine this work?
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14:26:28 Approximately how much work is required to raise the drink in a slice of thickness `dy near y coordinate y? Describe where y = 0 in relation to the tank.
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14:26:32 How much drink is contained in the slice described above?
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14:26:37 What are the cross-sectional area and volume of the slice?
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14:26:39 Explain how your answer to the previous questions lead to your integral.
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14:27:14 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> This whole chapter so far has been pretty new and difficult. I am going to have to burn through the weekend pretty hard to make sure I am prepared for this test.
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