course MTH 174 ۍʥE犉assignment #003 |hrW Physics II 09-21-2008
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20:27:51 query 6.5 #8 Galileo: time for unif accel object to traverse dist is same as if vel was ave of init and final; put into symbols and show why true
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RESPONSE --> a) `d = delta `dt = time `ds = space traversed v0 = velocity at rest vf = higest velocity (vf + v0)/2 = mean velocity He is saying that `ds / `dt = (vf + v0)/2 b) dv/dt = -g = constant ds/dt = v = -gt + v0 s = -1/2 g t^2 + v0 t +s0 We can find the time that it takes to hit the ground where s = 0ft and s0 = 100ft. 0 = 1/2 (-32 ft/s^2) t^2 + (0 ft/s) t + 100ft -100ft = -16ft/s^2 t^2 t = sqrt(-100ft/-16ft/s^2) t = +- 2.5s (Using positive time, t = 2.5s) The final velocity is -32 ft/s^2 * 2.5s + 0ft/s = -80 ft/s The mean velocity is ( -80 ft/s + 0 ft/s) / 2 = -40 ft/s Falling 100 ft in 2.5s: -100 ft / 2.5s = -40 ft/s c) Originally found using antiderivatives: dv/dt = -g = constant ds/dt = v = -gt + v0 s = -1/2 g t^2 + v0 t + s0 Note `d = delta Remember that t is really a small `dt: Rearranging these: vf = -g`dt + v0 `dv = -g`dt and s = -1/2 g `dt^2 + v0 `dt + s0 `ds = -1/2 g `dt^2 + v0 `dt Galileo said `ds / `dt = (vf + v0)/2 We can verify that this is true when acceleration is uniform (or constant) by substituting in for vf: `ds / `dt = [(-g`dt + v0) + v0]/2 `ds / `dt = (-g`dt + 2v0)/2 `ds / `dt = -1/2 g`dt + v0 `ds = -1/2 g `dt^2 + v0`dt s = -1/2 g`dt^2 + v0`dt + s0 This verifies that his assertion is correct.
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20:29:04 how can you symbolically represent the give statement?
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RESPONSE --> `d = delta `dt = time `ds = space traversed v0 = velocity at rest vf = higest velocity (vf + v0)/2 = mean velocity He is saying that `ds / `dt = (vf + v0)/2
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20:29:31 How can we show that the statement is true?
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RESPONSE --> Originally found using antiderivatives: dv/dt = -g = constant ds/dt = v = -gt + v0 s = -1/2 g t^2 + v0 t + s0 Note `d = delta Remember that t is really a small `dt: Rearranging these: vf = -g`dt + v0 `dv = -g`dt and s = -1/2 g `dt^2 + v0 `dt + s0 `ds = -1/2 g `dt^2 + v0 `dt Galileo said `ds / `dt = (vf + v0)/2 We can verify that this is true when acceleration is uniform (or constant) by substituting in for vf: `ds / `dt = [(-g`dt + v0) + v0]/2 `ds / `dt = (-g`dt + 2v0)/2 `ds / `dt = -1/2 g`dt + v0 `ds = -1/2 g `dt^2 + v0`dt s = -1/2 g`dt^2 + v0`dt + s0 This verifies that his assertion is correct.
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20:32:09 How can we use a graph to show that the statement is true?
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RESPONSE --> On a graph of velocity vs time, slope of the graph between two velocities will be linear when acceleration is contsant. The area of that trapezoid will be the same as a rectangle with the mean height of the velocities and the same time.
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20:39:46 query problem 7.1.22 (3d edition #18) integral of `sqrt(cos(3t) ) * sin(3t)
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RESPONSE --> integral sqrt(cos (3t)) * sin (3t) dt integral (cos (3t))^(1/2) * sin (3t) dt Substitution: u = cos (3t) du/dt = -sin (3t) * 3 du = -3sin (3t) dt -1/3 du = sin (3t) dt integral u^(1/2) * -1/3 du -1/3 * 2/3 u^(3/2) + C -2/9 cos(3t))^(3/2) + C
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20:41:06 what did you get for the integral and how did you reason out your result?
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RESPONSE --> cos 3t was clearly inside another function to the 1/2 power and additionally, the derivative of cos is -sin which allows the substitution to take place.
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20:46:13 query problem 7.1.20 (3d edition #21) antiderivative of x^2 e^(x^3+1)
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RESPONSE --> integral of x^2 e^(x^3+1) Subtituting u = x^3 + 1: du/dx = 3x^2 du = 3x^2 dx 1/3 du = x^2 dx integral of e^u 1/3 du 1/3 e^u + C 1/3 e^(x^3 + 1) + C
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20:46:31 what is the antiderivative?
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RESPONSE --> 1/3 e^(x^3 + 1) + C
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20:46:44 What substitution would you use to find this antiderivative?
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RESPONSE --> x^3 + 1
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21:03:20 query problem 7.1.37 (3d edition #35) antiderivative of (t+1)^2 / t^2
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RESPONSE --> integral of (t+1)^2 /t^2 The first thing I did was to try u = t+1 but that gave du=dt, which is not helpful. Next I expanded (t+1)^2 to see what that would do. integral of (t^2 + 2t + 1)/t^2 integral of t^2/t^2 + 2t/t^2 + 1/t^2 These can be separated and dealt with. integral of t^2/t^2 dt is the integral of 1 dt = t + C integral of 1/t^2 = -1/t + C The middle term can use a substitution for t^2 to solve it and takes care of the 2t on top: u = t^2 du/dt = 2t du = 2t dt integral of 1/u du is ln |u| + C so substituting back in we find ln |t^2| + C Adding them all together we have: t + ln |t^2| - 1/t + C
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21:03:32 what is the antiderivative?
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RESPONSE --> t + ln |t^2| - 1/t + C
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21:03:58 What substitution would you use to find this antiderivative?
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RESPONSE --> The substitution that was needed in the middle term for u = t^2
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21:09:29 query 7.1.64 (3d edition #60). int(1/(t+7)^2, t, 1, 3)
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RESPONSE --> I didn't see this one in my text, but here goes: integral (1/(t + 7)^2, t, 1,3) u = t+7 du/dt = 1 du = dt integral u^-2 du -u^-1 from 1 to 3 is: -1/3 - -1/1 = 2/3
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21:09:37 What did you get for the definite integral?
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RESPONSE --> 2/3
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21:10:10 What antiderivative did you use?
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RESPONSE --> -1/u + C
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21:10:47 What is the value of your antiderivative at t = 1 and at t = 3?
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RESPONSE --> at 1, the value is -1/1 and at 3 the value is -1/3
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21:12:16 query 7.1.86. World population P(t) = 5.3 e^(0.014 t).
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RESPONSE --> I don't believe this one was assigned and I do not see it in my text. I am not sure what I need to do. I am proceeding to the next question.
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21:18:41 What were the populations in 1990 and 2000?
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RESPONSE --> P(t) = 5.3 e^(0.014 t) I am assuming t is in years, but using 1990 as t would put the population at 6.66 * 10^12 people which is 6.6 trillion people, which I don't think is quite correct. P(1990) = 5.3 e^(0.014 * 1990) = 6.66 * 10^12 P(2000) = 5.3 e^(0.014 * 2000) = 7.67 * 10^12
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21:30:33 What is the average population between during the 1990's and how did you find it?
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RESPONSE --> Using the average value of the function to be 1/(2000-1990) * integral of 5.3 e^(0.014 t) from 1990 to 2000 u = 0.014t du/dt = 0.014 du/0.014 = dt 1/10 * 5.3/0.014 * integral of e^u from 1990 to 2000 1/10 * 5.3/0.014 * [e^(0.014 * 2000) - e^(0.014 *1990)] = 7.15 * 10^12 I am probably using the wrong times for my the data to be accurate, but I believe the rest of the work is correct.
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21:33:11 What is the value of your antiderivative at t = 1 and at t = 3?
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RESPONSE --> The value at t=1 is 384, at t=3 the value is 395
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21:33:18 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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