course MTH 174 RVܩKÃassignment #009
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19:34:29 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 --> a) Riemann Sum: `sigma 2 + 6x g/m * dx b) integral from 0 to 2 (2 + 6x) dx 2x + 6x^2 / 2 from 0 to 2 2(2) + 6(2)^2 / 2 - 2(0) + 6(0)^2 / 2 4 + 12 = 16 g
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19:46:29 what is the moment of the rod?
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RESPONSE --> The moment of the rod would be: integral from 0 to 2 x(2 + 6x) dx integral from 0 to 2 (2x + 6x^2) dx 2x^2 / 2 + 6x^3 / 3 from 0 to 2 x^2 + 2x^3 from 0 to 2 2(2)^2 + 2(2)^3 - 2(0)^2 + 2(0)^3 8 + 16 = 24 g * m is the moment of the rod
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19:46:45 What integral did you evaluate to get a moment?
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RESPONSE --> integral from 0 to 2 x(2 + 6x) dx
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20:05:00 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 --> Area is integral from a to b of [f(x) - g(x)]dx Mass is integral from a to b of `rho(x) * [f(x) - g(x)] dx
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20:05:08 what is the total mass of the region?
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RESPONSE --> Mass is integral from a to b of `rho(x) * [f(x) - g(x)] dx
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20:05:17 What integral did you evaluate to obtain this mass?
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RESPONSE --> Mass is integral from a to b of `rho(x) * [f(x) - g(x)] dx
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20:06:31 What is the mass of an increment at x coordinate x with width `dx?
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RESPONSE --> Mass of an increment of x with width `dx is `rho(x) * [f(x) - g(x)] `dx
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20:07:35 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 --> Area of the increment is [f(x) - g(x)] * `dx The area of this increment times the density function will determine the mass of the area.
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20:08:47 How to we use the mass of the increment to obtain the integral for the total mass?
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RESPONSE --> The mass of the increment is used as the basis for the Riemann sum of all masses and a limit is taken giving the integral to be evaluated.
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20:19:59 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 --> Slicing the water in strips parallel to x axis: Volume of a strip is `pi * (6ft)^2 * `dy Force on a strip 62.4 lb/ft^3 * `pi * 36ft^2 * `dy Work done in pumping the water strip: 62.4 lb/ft^3 * `pi * 36ft^2 * (30 - y) * `dy The integral will be from 0 to 20 62.4 lb/ft^3 * `pi * 36ft^2 * (30 - y) dy 62.4 lb/ft^3 * `pi * 36ft^2 * [30(20) - (20)^2 / 2] 62.4 lb/ft^3 * `pi * 36ft^2 * [600 - 200] = 898560 * `pi ft-lb.
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20:20:32 how much work is required to pump all the water to a height of 10 ft?
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RESPONSE --> Work required is 898560 * `pi ft-lb.
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20:20:48 What integral did you evaluate to determine this work?
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RESPONSE --> The integral will be from 0 to 20 62.4 lb/ft^3 * `pi * 36ft^2 * (30 - y) dy
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20:21:50 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 done in pumping the water strip: 62.4 lb/ft^3 * `pi * 36ft^2 * (30 - y) * `dy y=0 is the bottom of the tank.
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20:23:17 Explain how your answer to the previous question leads to your integral.
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RESPONSE --> The work is done over the distance 30 - y and is the distance in the force * displacement equation for work.
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20:38:17 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 --> g = 980cm/s^2 Slicing the water in slices parallel to the x-axis: For a slice: Volume = `pi [1/2 (8 - y)]^2 `dy Force = 1.2 * `pi [1/2 (8 - y)]^2 `dy Work is: 1.2 * `pi [1/2 (8 - y)]^2 * 980 * (15 - y) * `dy Integral from 0 to 8 1.2 * `pi [1/2 (8 - y)]^2 * 980 * (15 - y) dy Multiplying out the integrand and pulling out factors to be in front of the integral: 294 * `pi * integral 960 - 304y + 79y^2 - y^3 294 * `pi * [960y - 304y^2 / 2 + 79y^3 / 3 - y^4 / 4] 294 * `pi * [960 * 8 - 304 * 8^2 / 2 + 79 * 8^3 / 3 - 8^4 / 4] 294 * `pi * [7680 - 9728 + 13482 * 2/3 - 1024] = 9615585 dynes * 10^-5 N/1dyne = approx 96.15N
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20:38:39 how much work is required to raise all the drink to a height of 15 cm?
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RESPONSE --> 9615585 dynes * 10^-5 N/1dyne = approx 96.15N
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20:39:12 What integral did you evaluate to determine this work?
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RESPONSE --> 294 * `pi * integral 960 - 304y + 79y^2 - y^3 dy
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20:40:13 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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RESPONSE --> Work for one slice is: 1.2 * `pi [1/2 (8 - y)]^2 * 980 * (15 - y) * `dy y = 0 is the bottom of the glass.
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20:40:44 How much drink is contained in the slice described above?
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RESPONSE --> Volume of slice = `pi [1/2 (8 - y)]^2 `dy
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20:41:54 What are the cross-sectional area and volume of the slice?
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RESPONSE --> Volume of slice is `pi [1/2 (8 - y)]^2 `dy Cross-sectional area of the slice `pi [1/2 (8 - y)]^2
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20:46:41 Explain how your answer to the previous questions lead to your integral.
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RESPONSE --> Volume = `pi [1/2 (8 - y)]^2 `dy Force = 1.2 * `pi [1/2 (8 - y)]^2 `dy Work is: 1.2 * `pi [1/2 (8 - y)]^2 * 980 * (15 - y) * `dy The work is the force applied through the interval. The Force is the Volume * density. Volume is Cross-sectional area * `dy Cross-sectional area is found by using similar triangles to find the width of the strip. I then took half that for the radius.
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20:46:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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