#$&* course MTH174 5/16 11:10 am ********************
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********************************************* Question: problem 8.4.3 (3d edition 8.3.3) (previously 8.2.6) moment of 2 meter rod with density `rho(x) = 2 + 6x g/m YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are trying to find the moment of a 2 meter rod with density rho(X)=2 + 6x g/m We begin by taking a sample slice at point xi that is delta xi thick Moment = mass*distance So the moment of the slice would be (2+6x)(xi) We do a reimann sum approaching 0 of (2+6x)(xi) To set up integral from 0 to 2 of (2+6x)dx We integrate each separately and get 2x +6x^2. We now need to integrate again to find moment. The antiderivative is 2x^3 + x^2 from 0 to 2 We then get 2(2)^3 + 2^ 2 -)2(0)^3 + (0^2)) So we have 20 gram/meter as our final answer.
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Given Solution: The mass of an increment of length `dx, with sample point x_i, is (2 + 6 x_i ) `dx. The moment is mass * distance from axis of rotation. Assuming axis of rotation x = 0: The moment of the mass in the increment is (2 + 6 x_i) * x_i. Setting up a Riemann sum and allowing the increment to approach zero, we obtain the integral moment = int(x(2+6x), x, 0, 2). Thus the integrand is 2x + 6 x^2. An antiderivative is F(x) = x^2 + 2 x^3, so the definite integral is moment = int(x(2+6x), x, 0, 2) = F(2) - F(0) = 20. The moment of the typical increment has units of mass/unit length * length * distance from axis, or (g / m) * m * m = g * m. The units of the integral are therefore g * m, and the moment of this object is 20 g * m (i.e., 20 gram * meters). ADDITIONAL INFORMATION (finding center of mass): To get the center of mass relative to x = 0 (this was not requested here but you should know how to do this), divide the moment about x = 0 by the mass of the object: The mass of the object is easily found to be int((2+6x), x, 0, 2) (see above for the mass increment, which leads to the Riemann sum then to the integral). The center of mass is therefore center of mass = moment / mass = int(x(2+6x), x, 0, 2) / int((2+6x), x, 0, 2). The integrand for the denominator is 2 + 6 x, antiderivative G(x) = 2x + 3 x^2 and definite integral G(2) - G(0) = 16 - 0 = 16, meaning that the object has mass 16 grams (more specifcally the units of the denominator will be units of mass / unit length * length = mass, or in this case g / m * m = g). So the center of mass is at x = 20 g * m /(16 g) = 5/4 (g * m) / g = 5/4 g. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: problem 8.4.14 (4th edition 8.4.12 3d edition 8.3.12) mass between graph of f(x) and g(x), f > g, density `rho(x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We want to find the mass between the graph of f(x) and g(x) where f is greater than g and the given density if rho(x) We find a cross section at point xi that is delta xi thick. The area would be [f(x) - g(x)] *delta xi We know mass is area * density so Mass = [f(x) - g(x)] *delta xi * rho(x) So the reimann sum would be limit as x approaches 0 of [f(x) - g(x)] *delta xi * rho(x) So integral would run from a to b of [f(x) - g(x)]dx rho(x) So rho(x) times integral from a to b of [f(x) - g(x)]dx
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Given Solution: First you find the mass of a typical increment of width `dx, with sample point x within the interval. The mass is just area * density. The area of the region is height * width, or approximately (f(x) - g(x) ) * `dx. The density is `rho(x) so you get the approximation `dm = area * density = (f(x) - g(x) ) * 'dx * `rho(x) = `rho(x) (f(x) - g(x) ) * 'dx. The Riemann sum is the sum of all such mass increments for a partition of the interval, and at the interval width `dx approaches 0 this sum approaches the integral int( rho(x) * (f(x) - g(x)), x, a, b). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### do we evaluate the integral and plug in a and b or just leave it as integral from a to b of rho(x)[f(x) - g(x) dx]????
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Given Solution: ** You want to think of this as a simple product, just area * density. The area of the region is height * width, or approximately (f(x) - g(x) ) * `dx. The density is `rho(x) so you get the approximation mass = area * density = (f(x) - g(x) ) * 'dx * `rho(x) = `rho(x) (f(x) - g(x) ) * 'dx. Note that `dx stands for delta-x, a finite but small interval and that it's f - g, not f + g. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: problem 8.5.13 (4th edition 8.5. 12) (3d edition 8.4.12) 8.3.6 cylinder 20 ft high rad 6 ft full of water
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13:19:02 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: So we have a cylinder 20 ft high with radius of 6 ft that is full of water and we want to raise to 30 ft. We first take a strip delta yi thick at point yi. The area of the circular part is pi r^2 or 36 pi Density is rhog so rhog A is 62.4*36pi which is 7057.27 We know that work is force*distance So rhog(a) * distance Rhog(a) * (30-yi) So reimann sum would be from values of 0 to 20 of rhog*A * (30-yi) Integral would be from 0 to 20 of rhog * A (30-y) We pull the rhog * a in front of the integral and we know that value is 7057.27 so we have 7057.27 * integral from 0 to 20 of (30-y) Which woud be 7057.27(30 - ½ y^2) as the antiderivative So plugging in and evaluating we get 7057.27(30*20 - .5*20^2) which is 2,822,908 ft per pound confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Given Solution: ** We partition the interval 0 <= r <= 10 cm between the center of the record and its rim. An small interval of a partition will correspond to an interval of r. The part of the record for which the radius is within the partition consists of a thin ring of the disk. For example if 3.4 cm < r < 3.5 cm is an interval of the partition, then the corresponding region of the disk is the ring which is also described by 3.4 cm < r < 3.5 cm. This ring lies between the circle r = 3.4 cm and r = 3.5 cm; its 'width' is .1 cm and its 'average circumference' is somewhere between 2 pi * 3.4 cm (the circumference of the 'inner' circle) and 2 pi * 3.5 cm (the circumference of the 'outer' circle). The 'area density' of the record (mass / unit area) is area density = total mass / total area = 50 grams / total area = 50 grams / (pi * (10 cm)^2) = .16 grams / cm^2, approx. For a partition with interval width `dr, considering a typical interval with sample point r_i*: The corresponding 'ring' would have radius r_i* and width `dr. Its area would be approximately circumference * width = 2 pi r_i* `dr. The mass of the typical slice would be area * density = .16 * 2 pi r_i* `dr = 1.0 r_i* `dr, approx.. The speed of a point on the slice would be dist / time = 2 pi r_i* (33 1/3 / (60 sec)) = 3.4 r_i*, with speed in cm/s when radius is in cm. The KE of the slice is therefore .5 m v^2 = .5 ( 1.0 r_i* `dr) * (3.4 r_i*)^2, with KE in gram cm^2 / s^2. The Riemann sum of all KE contributions would, as `dr -> 0, approach the an integral which represents the total KE: total KE = integral of .5 ( 1.0 r) (3.4 r)^2 with respect to r, from r = 0 to r = 10. The simplified form of this expression is approximately 6 r^3; integrating from r = 0 to r = 10 we get approximately 15,000 gram cm^2 / s^2. The process shown here is correct, but the calculations represented here are not numerically accurate to any degree of precision; you should compare with your results with the results obtained here and if necessary rework these calculations using more accurate values. The value of the integral, using .16 * 2 pi r in place of the approximation 1.0 r, is 14 526.72443 g cm^2, to 10 significant figures. This is a ridiculously precise value, considering that the radius of a pressed disk is consistent to only within perhaps +-.001 cm, with even more significant uncertainty in the mass. A more reasonable figure would be 14 500 g cm^2 +- 100 g cm^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###I felt ok about everything except for where the 3.4 came from in The speed of a point on the slice would be dist / time = 2 pi r_i* (33 1/3 / (60 sec)) = 3.4 r_i*, with speed in cm/s when radius is in cm. Is 3.4 just an estimated point on the slice???? If so how do you choose the value 3.4???? ------------------------------------------------ Self-critique Rating:3
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Given Solution: The given solution solves the following problem: A conical glass is 10 cm high, and at the top its width is 10 cm. How much work is required to empty the glass by raising the liquid through a straw to a height of 15 cm from the base? Solution: ** The diameter of the top of the cone is equal to the vertical distance y from the apex to the top. At height y the diameter of the cone is easily seen to be equal to y, so the cross-section at height y has radius .5 y and therefore area A = `pi ( .5 y ) ^ 2 = `pi / 4 * y^2. A slice of thickness `dy at height y has approximate volume A * `dy = `pi/4 * y^2 * `dy. This area is in cm^3 so its mass is equal to the volume and the weight in dynes is 980 * mass = 245 `pi y^2 `dy. This weight is raised from height y to height 15, a distance of 15 - y. So the work to raise the slice is force * distance = 245 `pi y^2 `dy * ( 15 - y ) = 245 ( 15 - y ) `pi y^2 `dy = 245 `pi ( 15 y^2 - y^3 ) `pi `dy Slices go from y = 0 to y = 10 cm so the integral is 245 `pi ( 15 y^2 - y^3 ) `pi dy, evaluated from y = 0 to y = 10. We get 245 `pi * (5 y^3 - y^4 / 4) evaluated between 0 and 10. The result is 245 `pi * 2500 ergs, close to 2 million ergs. Most calculations were mentally so check the precise numbers. The process is correct. ** . STUDENT COMMENT: I am stuck at a point close to the end on this problem. The integral I have is from 0 to 10 'rho g A (15-y) dy INSTRUCTOR RESPONSE: ** Good, but A is a function of y because the glass is tapering. A = `pi r^2. What is the radius r at height y? Just draw a picture--two straight lines for the outline of the glass--and use proportionalities. Draw some similar triangles if necessary. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###I feel like my answer is too large a number. Im thinking I messed up on this step So integral is 0 to 15 rhog *a (15-y) When I used area I used area of the 16pi I think I may should have used the area of the slice but I cant figure out what that would be???? Please let me know where I messed up??? ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course MTH174 5/16 11:10 am ********************
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********************************************* Question: problem 8.4.3 (3d edition 8.3.3) (previously 8.2.6) moment of 2 meter rod with density `rho(x) = 2 + 6x g/m YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are trying to find the moment of a 2 meter rod with density rho(X)=2 + 6x g/m We begin by taking a sample slice at point xi that is delta xi thick Moment = mass*distance So the moment of the slice would be (2+6x)(xi) We do a reimann sum approaching 0 of (2+6x)(xi) To set up integral from 0 to 2 of (2+6x)dx We integrate each separately and get 2x +6x^2. We now need to integrate again to find moment. The antiderivative is 2x^3 + x^2 from 0 to 2 We then get 2(2)^3 + 2^ 2 -)2(0)^3 + (0^2)) So we have 20 gram/meter as our final answer.
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Given Solution: The mass of an increment of length `dx, with sample point x_i, is (2 + 6 x_i ) `dx. The moment is mass * distance from axis of rotation. Assuming axis of rotation x = 0: The moment of the mass in the increment is (2 + 6 x_i) * x_i. Setting up a Riemann sum and allowing the increment to approach zero, we obtain the integral moment = int(x(2+6x), x, 0, 2). Thus the integrand is 2x + 6 x^2. An antiderivative is F(x) = x^2 + 2 x^3, so the definite integral is moment = int(x(2+6x), x, 0, 2) = F(2) - F(0) = 20. The moment of the typical increment has units of mass/unit length * length * distance from axis, or (g / m) * m * m = g * m. The units of the integral are therefore g * m, and the moment of this object is 20 g * m (i.e., 20 gram * meters). ADDITIONAL INFORMATION (finding center of mass): To get the center of mass relative to x = 0 (this was not requested here but you should know how to do this), divide the moment about x = 0 by the mass of the object: The mass of the object is easily found to be int((2+6x), x, 0, 2) (see above for the mass increment, which leads to the Riemann sum then to the integral). The center of mass is therefore center of mass = moment / mass = int(x(2+6x), x, 0, 2) / int((2+6x), x, 0, 2). The integrand for the denominator is 2 + 6 x, antiderivative G(x) = 2x + 3 x^2 and definite integral G(2) - G(0) = 16 - 0 = 16, meaning that the object has mass 16 grams (more specifcally the units of the denominator will be units of mass / unit length * length = mass, or in this case g / m * m = g). So the center of mass is at x = 20 g * m /(16 g) = 5/4 (g * m) / g = 5/4 g. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: problem 8.4.14 (4th edition 8.4.12 3d edition 8.3.12) mass between graph of f(x) and g(x), f > g, density `rho(x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We want to find the mass between the graph of f(x) and g(x) where f is greater than g and the given density if rho(x) We find a cross section at point xi that is delta xi thick. The area would be [f(x) - g(x)] *delta xi We know mass is area * density so Mass = [f(x) - g(x)] *delta xi * rho(x) So the reimann sum would be limit as x approaches 0 of [f(x) - g(x)] *delta xi * rho(x) So integral would run from a to b of [f(x) - g(x)]dx rho(x) So rho(x) times integral from a to b of [f(x) - g(x)]dx
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Given Solution: First you find the mass of a typical increment of width `dx, with sample point x within the interval. The mass is just area * density. The area of the region is height * width, or approximately (f(x) - g(x) ) * `dx. The density is `rho(x) so you get the approximation `dm = area * density = (f(x) - g(x) ) * 'dx * `rho(x) = `rho(x) (f(x) - g(x) ) * 'dx. The Riemann sum is the sum of all such mass increments for a partition of the interval, and at the interval width `dx approaches 0 this sum approaches the integral int( rho(x) * (f(x) - g(x)), x, a, b). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### do we evaluate the integral and plug in a and b or just leave it as integral from a to b of rho(x)[f(x) - g(x) dx]????
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Given Solution: ** You want to think of this as a simple product, just area * density. The area of the region is height * width, or approximately (f(x) - g(x) ) * `dx. The density is `rho(x) so you get the approximation mass = area * density = (f(x) - g(x) ) * 'dx * `rho(x) = `rho(x) (f(x) - g(x) ) * 'dx. Note that `dx stands for delta-x, a finite but small interval and that it's f - g, not f + g. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: problem 8.5.13 (4th edition 8.5. 12) (3d edition 8.4.12) 8.3.6 cylinder 20 ft high rad 6 ft full of water
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13:19:02 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: So we have a cylinder 20 ft high with radius of 6 ft that is full of water and we want to raise to 30 ft. We first take a strip delta yi thick at point yi. The area of the circular part is pi r^2 or 36 pi Density is rhog so rhog A is 62.4*36pi which is 7057.27 We know that work is force*distance So rhog(a) * distance Rhog(a) * (30-yi) So reimann sum would be from values of 0 to 20 of rhog*A * (30-yi) Integral would be from 0 to 20 of rhog * A (30-y) We pull the rhog * a in front of the integral and we know that value is 7057.27 so we have 7057.27 * integral from 0 to 20 of (30-y) Which woud be 7057.27(30 - ½ y^2) as the antiderivative So plugging in and evaluating we get 7057.27(30*20 - .5*20^2) which is 2,822,908 ft per pound confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Given Solution: ** We partition the interval 0 <= r <= 10 cm between the center of the record and its rim. An small interval of a partition will correspond to an interval of r. The part of the record for which the radius is within the partition consists of a thin ring of the disk. For example if 3.4 cm < r < 3.5 cm is an interval of the partition, then the corresponding region of the disk is the ring which is also described by 3.4 cm < r < 3.5 cm. This ring lies between the circle r = 3.4 cm and r = 3.5 cm; its 'width' is .1 cm and its 'average circumference' is somewhere between 2 pi * 3.4 cm (the circumference of the 'inner' circle) and 2 pi * 3.5 cm (the circumference of the 'outer' circle). The 'area density' of the record (mass / unit area) is area density = total mass / total area = 50 grams / total area = 50 grams / (pi * (10 cm)^2) = .16 grams / cm^2, approx. For a partition with interval width `dr, considering a typical interval with sample point r_i*: The corresponding 'ring' would have radius r_i* and width `dr. Its area would be approximately circumference * width = 2 pi r_i* `dr. The mass of the typical slice would be area * density = .16 * 2 pi r_i* `dr = 1.0 r_i* `dr, approx.. The speed of a point on the slice would be dist / time = 2 pi r_i* (33 1/3 / (60 sec)) = 3.4 r_i*, with speed in cm/s when radius is in cm. The KE of the slice is therefore .5 m v^2 = .5 ( 1.0 r_i* `dr) * (3.4 r_i*)^2, with KE in gram cm^2 / s^2. The Riemann sum of all KE contributions would, as `dr -> 0, approach the an integral which represents the total KE: total KE = integral of .5 ( 1.0 r) (3.4 r)^2 with respect to r, from r = 0 to r = 10. The simplified form of this expression is approximately 6 r^3; integrating from r = 0 to r = 10 we get approximately 15,000 gram cm^2 / s^2. The process shown here is correct, but the calculations represented here are not numerically accurate to any degree of precision; you should compare with your results with the results obtained here and if necessary rework these calculations using more accurate values. The value of the integral, using .16 * 2 pi r in place of the approximation 1.0 r, is 14 526.72443 g cm^2, to 10 significant figures. This is a ridiculously precise value, considering that the radius of a pressed disk is consistent to only within perhaps +-.001 cm, with even more significant uncertainty in the mass. A more reasonable figure would be 14 500 g cm^2 +- 100 g cm^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###I felt ok about everything except for where the 3.4 came from in The speed of a point on the slice would be dist / time = 2 pi r_i* (33 1/3 / (60 sec)) = 3.4 r_i*, with speed in cm/s when radius is in cm. Is 3.4 just an estimated point on the slice???? If so how do you choose the value 3.4???? ------------------------------------------------ Self-critique Rating:3
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Given Solution: The given solution solves the following problem: A conical glass is 10 cm high, and at the top its width is 10 cm. How much work is required to empty the glass by raising the liquid through a straw to a height of 15 cm from the base? Solution: ** The diameter of the top of the cone is equal to the vertical distance y from the apex to the top. At height y the diameter of the cone is easily seen to be equal to y, so the cross-section at height y has radius .5 y and therefore area A = `pi ( .5 y ) ^ 2 = `pi / 4 * y^2. A slice of thickness `dy at height y has approximate volume A * `dy = `pi/4 * y^2 * `dy. This area is in cm^3 so its mass is equal to the volume and the weight in dynes is 980 * mass = 245 `pi y^2 `dy. This weight is raised from height y to height 15, a distance of 15 - y. So the work to raise the slice is force * distance = 245 `pi y^2 `dy * ( 15 - y ) = 245 ( 15 - y ) `pi y^2 `dy = 245 `pi ( 15 y^2 - y^3 ) `pi `dy Slices go from y = 0 to y = 10 cm so the integral is 245 `pi ( 15 y^2 - y^3 ) `pi dy, evaluated from y = 0 to y = 10. We get 245 `pi * (5 y^3 - y^4 / 4) evaluated between 0 and 10. The result is 245 `pi * 2500 ergs, close to 2 million ergs. Most calculations were mentally so check the precise numbers. The process is correct. ** . STUDENT COMMENT: I am stuck at a point close to the end on this problem. The integral I have is from 0 to 10 'rho g A (15-y) dy INSTRUCTOR RESPONSE: ** Good, but A is a function of y because the glass is tapering. A = `pi r^2. What is the radius r at height y? Just draw a picture--two straight lines for the outline of the glass--and use proportionalities. Draw some similar triangles if necessary. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###I feel like my answer is too large a number. Im thinking I messed up on this step So integral is 0 to 15 rhog *a (15-y) When I used area I used area of the 16pi I think I may should have used the area of the slice but I cant figure out what that would be???? Please let me know where I messed up??? ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!