course Mth 174 y?{???F???L???assignment #012?????????????
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21:25:20 Query problem 9.2.8 (3d editin 9.1.6) (was 9.4.6) first term and ratio for y^2 + y^3 + y^4 + ...
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RESPONSE --> The first term in this case is y^2. The ratio is y. I found this by considering the series to be like the general equation: a + ar + ar^2 where a is the first term and r is the ratio.
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21:26:34 either explain why the series is not geometric or give its first term and common ratio
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RESPONSE --> This series is geometric because it has a first term (a = y^2) and a common ratio (r = y).
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21:28:53 how do you get the common ratio?
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RESPONSE --> To find the common ratio I looked at each term listed and examined how each one differed. The first term was y^2, and according to the general pattern of a geometric series, the first term is equal to a. Each consecutive term changed by a factor of y, so the ratio must be y.
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21:31:02 what do you get when you factor out y^2? How does this help you determine the first term?
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RESPONSE --> Ah, that would make a bit more logical sense. When y^2 is factored out, the sequence becomes: 1 + y + y^2 + y^3 + ... This makes it pretty clear that it is a geometric series and that the first term is equal to what was just factored (y^2) and that the ratio is y.
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22:56:34 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 --> a) A ball dropped from a height h will take t=(1/4) 'sqrt(h). To prove this, I noticed that this was in reference to time. The previous questions have been in relation to distance, so I needed something that would relate time and distance. I remembered the acceleration, velocity, and distance functions. I used the distance function along with g = -32ft/sec^2 to get: distance = - (32 / 2)t^2 + v0 + h where h is the initial position and v0 is the initial velocity. Since initial velocity is 0 at h and the distance to the ground when it hits the ground is 0, this leaves us with the equation: 0 = - (32 / 2)t^2 + h Rearranging to solve for h in terms of t: h = 16t^2 h / 16 = t^2 t = 'sqrt( h / 16 ) t = 'sqrt( h ) / 4 t = (1/4) 'sqrt( h ) So the time until the ball hits the floor can be found in terms of height alone. b) I can't figure this part out. Since the series is a finite series, I think I should somehow use the formula: a(1 - x^n) / (1 - x) But I can't get anything out of it. Dividing (1/4) 'sqrt( h ) out of the equation gives a ratio of: r = 2 * 'sqrt(3/4) * ( 1 / ( 1 - 'sqrt( 3/4) ) ) But this doesn't match anything that I can come up with.
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23:11:10 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 --> I am not quite sure. In 28 I came up with the closed form of the total vertical distance traveled: 10( 1 - (3/4)^n ) / (1 - 3/4) This looks related, but I am not sure how.
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23:16:22 What geometric series gives the time and how does this geometric series yield the above result?
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RESPONSE --> Hmm. In the series, the first term would be (1/4)'sqrt(10) And each successive term would be something like: (1/4)'sqrt( 10 * (3/4)^n ) But I am not sure where to go from there.
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23:17:04 How far does the ball travel on the nth bounce?
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RESPONSE --> On the nth bounce, the distance traveled would be: 10(3/4)^n
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23:18:03 How long does it takes a ball to complete the nth bounce?
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RESPONSE --> Not quite sure. The only thing I can think of is: (1/4) 'sqrt( 10 (3/4)^n )
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23:31:48 Query 9.2.21 (was p 481 #6) convergence of 1 + 1/5 + 1/9 + 1/13 + ...
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RESPONSE --> Hmm. I am not sure which one this is. In my book, 9.2.21 is: sum ( (1/3)^n, n=4 to 20 ) and I am supposed to find the sum. As for this problem... I don't see a recurring sequence, a number which will divide 1/5, 1/9, and 1/13. In other words, I don't think this is a geometric series.
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23:53:15 with what integral need you compare the sequence and did it converged or diverge?
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RESPONSE --> Hmm. I guess, then, the idea is something like: 1 / ( 2n + 1 ) The integral of this could be compared with the integral of: 1 / 2n The antiderivative of this function is ln(2n). Since this antiderivative grows without bound (diverges) and the antiderivative of 1 / ( 2n+1 ) is greater than it, the original function diverges as well.
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23:59:47 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 --> Hmm. Again, not quite sure. Thinking of it in terms of a graph would give a descending slope because the first rectangle would be equal to 1, the next equal to 1/5, the next 1/9, and so on. Looking at it like that would seem to indicate that the series would converge on 1, but then the integral is still increasing by however small an amount as the series continues.
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