course Mth163 Thank-you for your help. I still don't think I'm grasping this quite as weel, thats the disadvantage to not being in a class room. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: A common response is that it takes 2 one-ft. squares to make a 2-foot square. However, below thought shows that this isn't the case. If we put 2 one foot squares side by side we get a one-foot by two-foot rectangle, not a square. If we put a second such rectangle together with the first, so that we have 2 rows with 2 squares in a row, then we have a two-foot square. Thus we see that it takes 4 squares one foot on a side to make a square 2 ft. on a side. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q002. How many cubes one foot on a side would it take to construct a cube two feet on a side? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: you have four cubes and you double it by adding four more cubes giving you a total of eight cubes. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We could begin by constructing two rows with two cubes in a row, which would sit on a square two feet by two feet. However this would not give is a cube two feet on a side, because at this point the figure we have constructed is only one foot high. So we have to add a second layer, consisting of two more rows with two cubes a row. Thus we have 2 layers, each containing 2 rows with 2 cubes in a row. Each layer has 4 cubes, so our two layers contain 8 cubes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q003. How many squares one foot on a side would it take to construct a square three feet on a side? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: for a square three feet you have three cubes per row with three rows, giving you a total of nine square feet on a side Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We would require three rows, each with 3 squares, for a total of 9 squares. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q004. How many cubes one foot on a side would take to construct a cube three feet on a side? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: with nine cubes on a layer and three layers, you would have 27 cubes total Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This would require three layers to make a cube three feet high. Each layer would have to contain 3 rows each with three cubes. Each layer would contain 9 cubes, so the three-layer construction would contain 27 cubes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q005. Suppose one of the Egyptian pyramids had been constructed of cubical stones. Suppose also that this pyramid had a weight of 100 million tons. If a larger pyramid was built as an exact replica, using cubical stones made of the same material but having twice the dimensions of those used in the original pyramid, then what would be the weight of the larger pyramid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if the second pyramid has twice the dimensions and a cubical stone has eight cubes in it, then the weight would be 8 xs 100 million tons. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Each stone of the larger pyramid has double the dimensions of each stone of the smaller pyramid. Since it takes 8 smaller cubes to construct a cube with twice the dimensions, each stone of the larger pyramid is equivalent to eight stones of the smaller. Thus the larger pyramid has 8 times the weight of the smaller. Its weight is therefore 8 * 100 million tons = 800 million tons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): should have solved it on out. Self-critique Rating: ********************************************* Question: `q006. Suppose that we wished to paint the outsides of the two pyramids described in the preceding problem. How many times as much paint would it take to paint the larger pyramid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if the second pyramid is twice the dimensions of the first one, making the second one consisting of 8 tiny cubes, giving one side of that cube 4 faces. I would say it would take 4 times as much paint than the first one. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The outside of each pyramid consists of square faces of uniform cubes. Since the cubes of the second pyramid have twice the dimension of the first, their square faces have 4 times the area of the cubes that make up the first. There is therefore 4 times the area to paint, and the second cube would require 4 times the paint &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q007. Suppose that we know that y = k x^2 and that y = 12 when x = 2. What is the value of k? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: in y = kx^2, plugging in y = 12 and x = 2 to solve for k: 12 = k * 2^2 .. k = 3 Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To find the value of k we substitute y = 12 and x = 2 into the form y = k x^2. We obtain 12 = k * 2^2, which we simplify to give us 12 = 4 * k. The dividing both sides by 410 reversing the sides we easily obtain k = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q008. Substitute the value of k you obtained in the last problem into the form y = k x^2. What equation do you get relating x and y? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if k = 3 then your equation would look like: y = 3 x^2 Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We obtained k = 3. Substituting this into the form y = k x^2 we have the equation y = 3 x^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q009. Using the equation y = 3 x^2, determine the value of y if it is known that x = 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 3 x^2, if x = 5 we solve for y: y = 3 *5^2 y = 75 Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If x = 5, then the equation y = 3 x^2 give us y = 3 (5)^2 = 3 * 25 = 75. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q010. If it is known that y = k x^3 and that when x = 4, y = 256, then what value of y will correspond to x = 9? To determine your answer, first determine the value of k and substitute this value into y = k x^3 to obtain an equation for y in terms of x. Then substitute the new value of x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: first we solve for k from the first set of numbers: 256 = k * 4^3 .256 = k * 64 .k = 4 then we plug in k value in with the x value of 9 to solve for y: y = 4 * 9^3 y = 2916 Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To we first substitute x = 4, y = 256 into the form y = k x^3. We obtain the equation 256 = k * 4^3, or 256 = 64 k. Dividing both sides by 64 we obtain k = 256 / 64 = 4. Substituting k = 4 into the form y = k x^3, we obtain the equation y = 4 x^3. We wish to find the value of y when x = 9. We easily do so by substituting x equal space 9 into our new equation. Our result is y = 4 * 9^3 = 4 * 729 = 2916. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q011. If it is known that y = k x^-2 and that when x = 5, y = 250, then what value of y will correspond to x = 12? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: to find value of k we plug in the first set of #s: 250 = k * 5 ^-2 .. 250 = k * .04 .k = 6250, then we plug that in with the second x value to solve for y: y = 6250 * 12^-2 . y = 43.403 rounded Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting x = 5 and y = 250 into the form y = k x^-2 we obtain 250 = k * 5^-2. Since 5^-2 = 1 / 5^2 = 1/25, this becomes 250 = 1/25 * k, so that k = 250 * 25 = 6250. Thus our form y = k x^-2 becomes y = 6250 x^-2. When x = 12, we therefore have y = 6250 * 12^-2 = 6250 / 12^2 = 6250 / 144 = 42.6, approximately. *** ok, I did this the same way you did but every time I calculate the answer I get is 43.40277778 which I rounded to 43.403 011. `query 11 ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = k x^3 proportionally governs each sand pile because it can be made up off many cubes. Confidence rating: mosty ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): volumes made up of cubes, y = k x^2 is surface area because it can be covered with tiny squares. Self-critique Rating: ********************************************* Question: `qIf 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I would think y = k x^2 would be what you need to govern the relationship between y and x because it covers up the surface area. Confidence rating: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): sphere needs to be covered with the tinier squares and the radius is linear giving you y = k x^2 for the dimensions. Self-critique Rating: ********************************************* Question: `qExplain how you would use the concept of the differential to find the volume of a sandpile of height 5.01 given the volume of a geometrically similar sandpile of height 5, and given the value of k in the y = k x^3 proportionality between height and volume. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In my notes I see notes on y( volume) andx^3 being dimensions and y being proportional to x^3 assuming that all cubes grow proportionally, but I have looked on both class notes and worksheets and cannot find the concept of the differential to find volume. Where do I find this?
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Given Solution: ** The class notes showed you that the slope of the y = k x^3 graph is given by the rate-of-change function y' = 3 k x^2. Once you have evaluated k, using the given information, you can evaluate y' at x = 5. That gives you the slope of the line tangent to the curve, and also the rate at which y is changing with respect to x. When you multiply this rate by the change in x, you get the change in y. The differential is 3 k x^2 `dx and is approximately equal to the corresponding `dy. Since `dy / `dx = 3 k x^2, the differential looks like a simple algebraic rearrangement `dy = 3 k x^2 `dx, though what's involved isn't really simple algebra. The differential expresses the fact that near a point, provided the function has a continuous derivative, the approximate change in y can be found by multiplying the change in x by the derivative). That is, `dy = derivative * `dx (approx)., or `dy = slope at given point * `dx (approx), or `dy = 3 k x^2 `dx (approx). The idea is that the derivative is the rate of change of the function. We can use the rate of change and the change in x to find the change in y. The differential uses the fact that near x = 5 the change in y can be approximated using the rate of change at x = 5. Our proportionality is y = k x^3. Let y = f(x) = k x^3. Then y' = f'(x) = 3 k x^2. When x = 5 we have y' = f'(5) = 75 k, whatever k is. To estimate the change in y corresponding to the change .01 in x, we will multiply y ' by .01, getting a change of y ' `dx = 75 k * .01. } SPECIFIC EXAMPLE: We don't know what k is for this specific question. As a specific example suppose our information let us to the value k = .002, so that our proportionality is y = .002 x^3. Then the rate of change when x is 5 would be f'(5) = 3 k x^2 = 3 k * 5^2 = 75 k = .15 and the value of y would be y = f(5) = .002 * 5^3 = .25. This tells us that at x = 5 the function is changing at a rate of .15 units of y for each unit of x. Thus if x changes from 5 to 5.01 we expect that the change will be change in y = (dy/dx) * `dx = rate of change * change in x (approx) = .15 * .01 = .0015, so that when x = 5.01, y should be .0015 greater than it was when x was 5. Thus y = .25 + .0015 = .2515. This is the differential approximation. It doesn't take account of the fact that the rate changes slightly between x=5 and x = 5.01. But we don't expect it to change much over that short increment, so we expect that the approximation is pretty good. Now, if you evaluate f at x = 5.01 you get .251503. This is a little different than the .2515 approximation we got from the differential--the differential is off by .000003. That's not much, and we expected it wouldn't be much because the derivative doesn't change much over that short interval. But it does change a little, and that's the reason for the discrepancy. The differential works very well for decently behaved functions (ones with smooth curves for graphs) over sufficiently short intervals.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qWhat would be the rate of depth change for the depth function y = .02 t^2 - 3 t + 6 at t = 30? (instant response not required) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = .02(30)^2 3(30) + 6 .. y = -66 Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have looked through my worksheet and my class notes and I cannot find that function: y` = .04t - 3
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, Self-critique Rating: ********************************************* Question: `qWhat 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the value of a parameter is y = ax^3 with y = quarter cups = 1 and x = cube width = 1.5 we can solve for a the value of the parameter: 1 = a * 1.5^3 . a = 1/3.375 a = .296 rounded the model will be y = .296x^3 if x = 3 then we have: y = .296* 3^3 .y = 7.992 or 8 rounded Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 = .296 x^3. Therefore if x = 3 we have y = .296 * 3^3 = 7.992, which is the same as 8 except for roundoff error. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qWhat would be the side measurement of a cube designed to hold 30 quarter-cups of sand? What equation did you solve to get this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: using the previous model: y = .296 x^3 we are given the y value to plug in and solve for x; 30 = .296 x^3 ..x^3 = 101 x = 101^1/3 ..x = 4.657 Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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..** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: there are two quarter cups in ½ cup so to get the function fro the amount of quarter cup to match this amount you would have to double the amount: y = .004x^3. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery problem 4. number of swings vs. length data. Which function fits best? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: using y = ax^ .5 best fits the graph of # of swings vs length giving you a more constant rate. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 valuesevery (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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qproblem 7. time per swing model. For your data what expression represents the number of swings per minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: using model y = a x^-.5 and using some of my data points: using the rise over run theory and doing change in swings / change in length( in my case was in inches) two points: (33, 23) and (25,18) y = # of swings, x = length of string (18 23) / (25 33) = .625 # of swings per min.giving a model of: # of swings per min = .625 x ^ -5 Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, if I did my calculation right. Self-critique Rating: ********************************************* Question: `qIf the time per swing in seconds is y, then what expression represents the number of swings per minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if you have 60 seconds in a minute and you take that amount swings per second and divided that into 60 to get # of swings per minute. expression: # of swings per min. = 60 / y Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qIf 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The time per swing is y = a x^5 and # of swings per minute = 60 / y then your expression would be represented by function = 60 / (a x*-5) Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok but I still dont under stand or Im just confusing my self on this whole concept and with different data points I have nothing to compare to, and where did you get the 1.1 data from? ok even though my data was not right , using the data that you gave in the confirmed f = ax^-.5 would work because of the consistency.
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, but I did this with x ^-.5 Self-critique Rating: ********************************************* Question: `qquery problem 9. length ratio x2 / x1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qWhat expressions, in terms of x1 and x2, represent the frequencies (i.e., number of swings per minute) of the two pendulums? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if x = frequencies then you would substitute as follows: f = 1.1*x1^-.5 and f = 1.1 * x2^ -.5 Confidence rating: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok instead of 1.1 you should have used 55 or a Self-critique Rating: ********************************************* Question: `qWhat expression, in terms of x1 and x2, represents the ratio of the frequencies of the two pendulums? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if you took these terms as f1 = a x1^-.5 and f2 = a x2^-.5 the ratio would be f2 / f1 (like the rise/ run) (a x2^-.5) / ( a x1^-.5) (x2^ -.5) / ( x2^ -.5) (x2 / x1)^.5 Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery problem Challenge Problem for Calculus-Bound Students: how much would the frequency change between lengths of 2.4 and 2.6 feet YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if you took f = 55x^-.5 and x being 2.4 and 2.6 you would have: f = 55*2.4^-.5 = 35.50 f = 55 *2.6^-.5 = 34.11 the change will be : (34.50 35.50)/(2.6 2.4) -1 / .2 .-5 is the frequency change between lengths Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: "