course MTH 174 12:30 p.m. 7/11/10. I didn't realize until today that I hadn't submitted this. I had done the assignment but just forgot to send it in. ѺԒaЭȮͣassignment #012
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11:58:49 Query problem 9.2.8 (3d edition 9.1.6) (was 9.4.6) first term and ratio for y^2 + y^3 + y^4 + ...
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RESPONSE --> Here a= y^2 and the ratio is y.
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11:59:23 either explain why the series is not geometric or give its first term and common ratio
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RESPONSE --> The series is geometric and again: a = y^2 and the common ratio is y.
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12:00:18 how do you get the common ratio?
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RESPONSE --> The common ratio was found by dividing out a, which is y^2 in this problem.
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12:01:41 what do you get when you factor out y^2? How does this help you determine the first term?
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RESPONSE --> After factoring out y^2 you have: 1+ y+ y^2+y^3 + y^4+... This tells you that the first term, a, is y^2 and leaves the common ratio.
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12:02:34 Query problem 9.2.29 (3d edition 9.1.24) (was 9.4.24) bouncing ball 3/4 ht ratio
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RESPONSE --> The height of the n-th bounce is found by: h= 10 (3/4)^n.
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12:11:04 how do you verify that the ball stops bouncing after 1/4 `sqrt(10) + 1/2 `sqrt(10) `sqrt(3/2) (1 / (1-`sqrt(3/4)) sec?
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RESPONSE --> This is done first by showing that the time it takes for the ball drop from a given height Using s= (-g*t^2)/2 + h. Since the position we want is zero we set this equation equal to zero and solve fort, getting: t= 1/4*(sqrt(h)). Then we have to realize that this is the amount of time it takes for the ball to fall through a height but for the total time we also have to include the bounce up from the floor to the height. This can be closely estimated to be the same as the amount of time it takes to fall, so we double the time for the bounce time series. The other thing we need to see is that the first drop has no bounce up associated with its height so it is added on its own. Forming the geometric series using this information gives the formula we that we are given for the total time.
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12:19:45 What geometric series gives the time and how does this geometric series yield the above result?
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RESPONSE --> The series is the first drop time plus twice the series for the next drop times, we use a = 1/4(sqrt(10*3/4)),and the common ratio of sqrt(3/4). In closed form this gives us: 1/4sqrt(10) + 2[(1/4)(sqrt10*3/4)(1-sqrt(3/4)^n)]/(1-sqrt(3/4)). This equation yields the above results when n grows very large sqrt(3/4)^n becomes very small so the equation takes as a limit (with some simplification as well): 1/4sqrt(10) + (1/2)(sqrt(10)sqrt(3/4)[1/(1-sqrt(3/4))].
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12:20:33 How far does the ball travel on the nth bounce?
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RESPONSE --> The drop distance is 10(3/4)^n and to account for the upward bounce we multiply this by 2.
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12:21:39 How long does it takes a ball to complete the nth bounce?
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RESPONSE --> Here the drop bounce time is 1/4(sqrt(10)(sqrt(3/4)^n). The time for the total bounce is also found by multiplying this by 2.
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12:24:06 Query 9.2.21 (was p 481 #6) convergence of 1 + 1/5 + 1/9 + 1/13 + ...
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RESPONSE --> This can be expressed as SUM( 1/(4x-3), 1, inf). This can be compared with the integral: INT( 1/(4x-3), 1, inf).
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12:26:54 07-11-2010 12:26:54 with what integral need you compare the sequence and did it converged or diverge?
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NOTES -------> The integral is: INT(1/(4x-3), 1, inf). This integral does not converge. This is validated both by performing the integration that gives: 1/4( ln| 4x-3|) which doesn't converge as well as comparing it to 1/x^p with p=1.
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12:26:59 with what integral need you compare the sequence and did it converged or diverge?
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RESPONSE --> The integral is: INT(1/(4x-3), 1, inf). This integral does not converge. This is validated both by performing the integration that gives: 1/4( ln| 4x-3|) which doesn't converge as well as comparing it to 1/x^p with p=1.
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12:29:49 Explain in terms of a graph how you set up rectangles to represent the series, and how you oriented these rectangles with respect to the graph of your function in order to prove your result.
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RESPONSE --> In order for our comparison to work the series must be greater than the comparison integral which means that for this function we have a LEFT type of setup with the left edge of the rectangles lying on the line of the function so that a LEFT approximation would be overestimated meaning our sum was greater than the function and since the function diverges so to must our series.
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