course Mth271 Ü‹Þ¤§CáË°ð¹æã£cýEìÓÀ½òDæ^µ¡ ©assignment #006
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21:34:22 If the time per swing is a x ^ .5, for the value determined previously for the parameter a, then what expression represents the number of swings per minute? How does this expression compare with the function you obtained for the number of swings per minute vs. length?
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RESPONSE --> ax^.5 =60/ a 54^.5
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21:36:39 ** Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1. Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute. Simplifying this gives f = (60 / a) * x^.5. 60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements. 60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5. **
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RESPONSE --> I am not sure really what to do......I got lost and after reading this I am still not sure why should I have grouped 60/a together
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21:37:51 query problem 8. model of time per swing what are the pendulum lengths that would result in periods of .1 second and 100 seconds?
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RESPONSE --> .1= x and y= 3.6 3.6= a .1 ^.5 3.6/.1^.5=a a= 11.38
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21:38:56 ** You would use your own model here. This solution uses T = 1.1 x^.5. You can adapt the solution to your own model. According to the model T = 1.1 x^.5 , where T is period in seconds and x is length in feet, we have periods T = .1 and T = 100. So we solve for x: For T = .1 we get: .1 = 1.2 x^.5 which gives us x ^ .5 = .1 / 1.2 so that x^.5 = .083 and after squaring both sides we get x = .083^2 = .0069 approx., representing .0069 feet. We also solve for T = 100: 100 = 1.2 x^.5, obtaining x^.5 = 100 / 1.2 = 83, approx., so that x = 83^2 = 6900, approx., representing a pendulum 6900 ft (about 1.3 miles) long. **
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RESPONSE --> Once again I have no idea.....I do not know what I am doing with it gets to this and the last part......were did the 1.2 come from?
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21:40:25 query problem 9. length ratio x2 / x1.
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RESPONSE --> x2/x1 = ax2^.5 and =ax1^.5 = (ax2/ ax1) ^.5
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21:41:11 What expressions, in terms of x1 and x2, represent the frequencies (i.e., number of swings per minute) of the two pendulums?
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RESPONSE --> Sorry I think I amswered this in the last question ......so (ax2/ax1)^.5
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21:42:12 ** The solution is to be in terms of x1 and x2. If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)). Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5. If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5 **
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RESPONSE --> I dont know where the 60/ 1.1 came from.....I thought that I just had to plug in and I lost my - on the .5 also.....once a again a lot confused
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21:43:04 What expression, in terms of x1 and x2, represents the ratio of the frequencies of the two pendulums?
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RESPONSE --> (ax2/ax1)^-.5
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21:44:15 ** We need to do this in terms of the symbols x1 and x2. If f = a x^-.5 then f1 = a x1^-.5 and f2 = a x2^-.5. With these expressions we would get f2 / f1 = a x2^-.5 / (a x1^-.5) = x2^-.5 / x1^-.5 = (x2 / x1)^-.5 = 1 / (x2 / x1)^.5 = (x1 / x2)^.5. Note that it doesn't matter what a is, since a quickly divides out of our quotient. For example if a = 55 we get f2 / f1 = 55 x2^-.5 / (55 x1^-.5) = x2^-.5 / x1^-.5 = (x2 / x1)^-.5 = 1 / (x2 / x1)^.5 = (x1 / x2)^.5. This is the same result we got when a was not specified. This shouldn't be surprising, since the parameter a divided out in the third step. **
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RESPONSE --> I shoudl have divided my a's out and I didnt....I also thought that this time I should put that - in but this time I didnt need to.
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21:45:52 query problem Challenge Problem for Calculus-Bound Students: how much would the frequency change between lengths of 2.4 and 2.6 feet
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RESPONSE --> y= a (2.4)^-.5 and y= a(2.6)^-.5 I can not solve this to find out because I dont know a or y
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21:47:27 ** STUDENT SOLUTION: Note that we are using frequency in cycles / minute. I worked to get the frequency at 2.4 and 2.6 y = 55.6583(2.4^-.5) = 35.9273 and y = 55.6583(2.6^-.5)= 34.5178. subtracted to get -1.40949 difference between 2.4 and 2.6. This, along with the change in length of .2, gives average rate -1.409 cycles/min / (.2 ft) = -7.045 (cycles/min)/ft , based on the behavior between 2.4 ft and 2.6 ft. This average rate would predict a change of -7.045 (cycles/min)/ft * 1 ft = -7/045 cycles/min for the 1-foot increase between 2 ft and 3 ft. The change obtained by evaluating the model at 2 ft and 3 ft was -7.2221 cycles/min. The answers are different because the equation is not linear and the difference between 2.4 and 2.6 does not take into account the change in the rate of frequency change between 2 and 2.4 and 2.6 and 3 for 4.4 and 4.6 y = 55.6583(4.4^-.5) y = 55.6583(4.6^-.5) y = 26.5341 y = 25.6508 Dividing difference in y by change in x we get -2.9165 cycles/min / ft, compared to the actual change -2.938 obtained from the model. The answers between 4-5 and 2-3 are different because the equation is not linear and the frequency is changing at all points. **
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RESPONSE --> I really dont know after looking at this were the 55 came from
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21:49:32 1.1.20 (was 1.1.18 are (0,4), (7,-6) and (-5,11) collinear?
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RESPONSE --> 7, -6 and -5, 11 the distance = 20.8 then (0,4) ( -5,11) = 8.6 (0,4) (7,-6) = 12.2 12.2+ 8.6= 20.8
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21:49:49 ** The distance between (-5,11) and (7,-6) is approximately 20.81: d3 = sqr rt [(7+5)^2 + -(6 +11)^2] d3 = sqr rt 433 d3 = 20.81 Using the distance formula the distances between (-5,11) and (0,4) is 8.6 and the distance between (0,4) and (7, -6) is 12.2. 'Collinear' means 'lying along the same straight line'. If three points are collinear then the sum of the distances between the two closer pairs of points will equal the distance between the furthest two. Since 8.6 + 12.2 = 20.8 the points are on the same straight line. **
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RESPONSE --> I understand this stuff pretty good
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21:51:24 1.1.24 find x | dist (2,-1) to (x,2) is 5What value of x makes the distance 5?
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RESPONSE --> 5= sqrt(x-2^2 + 3^2) 25= (x-2)^2 + 9 16= (x-2)^2 4= (x-2) x= 6
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21:52:13 ** The expression for the distance from (2, -1) to (x, 2) is = sqrt((x-2)^2 + (2+1)^2). This distance is to be 5, which gives us the equation5 = sqrt((x-2)^2 + (2+1)^2) Starting with the equation 5 = sqrt((x-2)^2 + (2+1)^2) we first square both sides to get 25 = (x-2)^2 + 9 or (x-2)^2 = 16. Solutions are found by taking the square root of both sides, keeping in mind that (x-2)^2 doesn't distinguish between positive and negative values of x - 2. We find that (x - 2) = +_ sqrt(16) = +- 4. (x-2) = 4 gives us the solution x = 6 and (x-2) = -4 gives us the solution x = -2. **
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RESPONSE --> I forgot about the +- but i got one out of two....I understand that 2+-4= -2
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21:53:06 1.1.36 (was 1.1.34 percent increase in Dow What are the requested percent increases?
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RESPONSE --> change= 200 so 200/ 10600= .02
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21:54:05 ** Aug 99 to Nov 99 Change in value = 10600 - 10600 = 0 At a percent of the initial value we have 0/10600 = 0, or 0% increase May 2000 to September 2000: change in value = 10600-10800 = -200 As percent of initial value: -200/10600 = .019 or 1.9% approx.. **
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RESPONSE --> My 200 should have been - ........why in your does -200/10600= .019 ......whould it not be a - ? THanks.
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