course mth 163 I did not understand this query. I did not understand where several parts of the data came from or where the examples were in the notes. They may have been on the DVD, but unfortunately I did not have the DVD with me at the time I was working on this. I will have to look over the information again and then look at these problems again. Also, could you let me know how I am doing in this class -- grade wise?
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19:07:17 Query class notes #06 If x is the height of a sandpile and y the volume, what proportionality governs geometrically similar sandpiles? Why should this be the proportionality?
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RESPONSE --> k, which is a constant by which the y adn x value are proportional. confidence assessment: 1
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19:14:19 ** the proportionality is y = k x^3. Any proportionality of volumes is a y = k x^3 proportionality because volumes can be filled with tiny cubes; surface areas are y = k x^2 because surfaces can be covered with tiny squares. **
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RESPONSE --> I now understand what the question is asking. I understand that volume proportionality is y= kx^3 and surface area proportionality is y=kx^2 self critique assessment: 2
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19:21:34 If x is the radius of a spherical balloon and y the surface area, what proportionality governs the relationship between y and x? Why should this be the proportionality?
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RESPONSE --> It would be y = kx^2 confidence assessment: 1
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19:21:49 ** Just as little cubes can be thought of as filling the volume to any desired level of accuracy, little squares can be thought of as covering any smooth surface. Cubes 'scale up' in three dimensions, squares in only two. So the proportionality is y = k x^2. Surfaces can be covered as nearly as we like with tiny squares (the more closely we want to cover a sphere the tinier the squares would have to be). The area of a square is proportional to the square of its linear dimensions. Radius is a linear dimension. Thus the proportionality for areas is y = k x^2. By contrast, for volumes or things that depend on volume, like mass or weight, we would use tiny cubes to fill the volume. Volume of a cube is proportional to the cube of linear dimensions. Thus the proportionality for a volume would be y = k x^3. **
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RESPONSE --> self critique assessment: 3
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20:48:35 ** You saw in the class notes and in the q_a_ that the rate of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b. This is the function that should be evaluated to give you the rate. Evaluating the rate of depth change function y ' = .04 t - 3 for t = 30 we get y ' = .04 * 30 - 3 = 1.2 - 3 = -1.8. COMMON ERROR: y = .02(30)^2 - 2(30) + 6 =-36 would be the rate of depth change INSTRUCTOR COMMENT: This is the depth, not the rate of depth change. **
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RESPONSE --> I did not have a chance to answer. The response came up as I bean typing. I was reading what it says and I can't find in the notes (unless in the DVD - which is not with me at this time) where the 3kx^2 come from. I actually do not understand this whole question or explaination. I have read and reread the explaination and it doesn't make any sense to me. self critique assessment: 2
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21:01:53 modeling project 3 problem a single quarter-cup of sand makes a cube 1.5 inches on a side. How many quarter-cups would be required to make a cube with twice the scale, 3 inches on a side? Explain how you know this.
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RESPONSE --> First of all I subustituted in the quarter cup for y and 1.5 for x to find k in the equation y = kx^3: .25 = k(1.5^3) .25 = k(3.375) .074 = k Then substitute in k and x: y = (.074)(3)^3 y = 1.998 cups **I am trying to think if I would have to divide by .25 to find out how many quarter cups, but not sure** confidence assessment: 2
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21:03:11 ** You can think of stacking single cubes--to double the dimensions of a single cube you would need 2 layers, 2 rows of 2 in each layer. Thus it would take 8 cubes 1.5 inches on a side to make a cube 3 inches on a side. Since each 1.5 inch cube containts a quarter-cup, a 3 inch cube would contain 8 quarter-cups. COMMON ERROR: It would take 2 quarter-cups. INSTRUCTOR COMMENT: 2 quarter-cups would make two 1.5 inch cubes, which would not be a 3-inch cube but could make a rectangular solid with a square base 1.5 inches on a side and 3 inches high. **
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RESPONSE --> If I had divided my answer of 1.998 by .25 I would have gotten about 8 cups. I over thought the problem. self critique assessment: 2
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21:03:35 What value of the parameter a would model this situation? How many quarter-cups does this model predict for a cube three inches on a side? How does this compare with your previous answer?
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RESPONSE --> confidence assessment:
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21:07:25 ** The proportionality would be y = a x^3, with y = 1 (representing one quarter-cup) when x = 1.5. So we have 1 = a * 1.5^3, so that a = 1 / 1.5^3 = .296 approx. So the model is y = .2963 x^3. Therefore if x = 3 we have y = .296 * 3^3 = 7.992, which is the same as 8 except for roundoff error. **
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RESPONSE --> I wasn't totally sure what the question was asking but I realize the answer was what I did to get my previous answer. self critique assessment: 2
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21:12:26 What would be the side measurement of a cube designed to hold 30 quarter-cups of sand? What equation did you solve to get this?
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RESPONSE --> I got 4.66 in. by using the y=kx^3: First of all I multiplied 30 quarter cups by .25 to know the total... 7.5 = .074x^3 101.351 = x^3 take the cube root of 101.351 to get x x= 4.66 confidence assessment: 2
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21:12:43 ** You are given the number of quarter-cups, which corresponds to y. Thus we have 30 = .296 x^3 so that x^3 = 30 / .296 = 101, approx, and x = 101^(1/3) = 4.7, approx..**
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RESPONSE --> self critique assessment: 3
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21:33:47 query problem 2. Someone used 1/2 cup instead of 1/4 cup. The best-fit function was y = .002 x^3. What function would have been obtained using 1/4 cup?
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RESPONSE --> This info may again be on the DVD, which I do not have with me at the present moment. I don't fully understand. confidence assessment: 0
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21:37:05 ** In this case, since it takes two quarter-cups to make a half-cup, the person would need twice as many quarter-cups to get the same volume y. He would have obtained half as many half-cups as the actual number of quarter-cups. To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3. **
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RESPONSE --> After reading the explaination, this makes sense. I guess I was trying to make things more complicated that it is. self critique assessment: 2
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22:45:23 query problem 4. number of swings vs. length data. Which function fits best?
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RESPONSE --> of the ones listed: y= a x^-.3 confidence assessment: 1
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22:46:30 ** If you try the different functions, then for each one you can find a value of a corresponding to every data point. For example if you use y = a x^-2 you can plug in every (x, y) pair and solve to see if your values of a are reasonably consistent. Try this for the data and you will find that y = a x^-2 does not give you consistent a values?very (x, y) pair you plug in will give you a very different value of a. The shape of the graph gives you a pretty good indication of which one to try, provided you know the shapes of the basic graphs. For this specific situation the graph of the # of swings vs. length decreases at a decreasing rate. The graphs of y = a x^.p for p = -.3, -.4, -.5, -.6 and -.7 all decrease at a decreasing rate. In this case you would find that the a x^-.5 function works nicely, giving a nearly constant value of a.
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RESPONSE --> None of this makes total sense. I will watch the DVD when I get a chance to see if I understand any better. self critique assessment: 2
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22:47:33 problem 7. time per swing model. For your data what expression represents the number of swings per minute?
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RESPONSE --> 60 T confidence assessment: 1
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22:50:03 ** The model that best fits the data is a x^-.5, and with accurate data we find that a is close to 55. The model is pretty close to # per minute frequency = 55 x^-.5. As a specific example let's say we obtained counts of 53, 40, 33 and 26 cycles in a minute at lengths of 1, 2, 3 and 4 feet, then using y = a x^-.5 gives you a = y * x^.5. Evaluating a for y = 53 and x = 1 gives us a = 53 * 1^.5 = 53; for y = 40 and x = 2 we would get a = 40 * 2^.5 = 56; for y = 34 and x = 3 we get a = 33 * 3^.5 = 55; for y = 26 and x = 4 we get a = 26 * 4^.5 = 52. Since our value of a are reasonably constant the y = a x^.5 model works pretty well, with a value of a around 54. The value of a for accurate data turns out to be about 55.**
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RESPONSE --> Honestly, I am going to have to look bak over this material. I don't understand where much of this information came from. After I take a second look I will let you know if I understand any better. self critique assessment: 2
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22:51:06 If the time per swing in seconds is y, then what expression represents the number of swings per minute?
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RESPONSE --> y = 55x^.5 confidence assessment: 1
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22:52:39 ** To get the number of swings per minute you would divide 60 seconds by the number of seconds in a swing (e.g., if a swing takes 2 seconds you have 30 swings in a minute). So you would have f = 60 / y, where f is frequency in swings per minute. COMMON ERROR: y * 60 INSTRUCTOR COMMENT: That would give more swings per minute for a greater y. But greater y implies a longer time for a swing, which would imply fewer swings per minute. This is not consistent with your answer. **
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RESPONSE --> This particular explaination does make sense. I understand how they arrived at the answer. self critique assessment: 2
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22:54:09 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 --> y = (55x^.5)/60 They are inversely related. confidence assessment: 1
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22:55:06 ** 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 --> Again, I will have to read the notes again. self critique assessment: 2
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23:00:25 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 --> Still confused.....Will reread the notes and try these again just to make sure I understand. confidence assessment: 1
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23:00:43 ** 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 --> self critique assessment: 3
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23:02:34 query problem 9. length ratio x2 / x1.
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RESPONSE --> confidence assessment: 0
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23:03:00 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 --> confidence assessment: 0
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23:03:25 ** 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 --> same as most other problems---I will reread over the notes. self critique assessment: 1
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