#$&* course Mth 173 10/1 00:30
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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. ** 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? ** 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Explain 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: We know the volume of the sand pile for the height of the sand pile being 5, and thus know the proportionality constant k. The given function is y = k x^3 which is a volume function, thus the differential of this function would give us the rate at which volume is changing with respect to height of the sand pile. The differential function will be y’ = 3k * (x^2). Now since we know the value of k and x (length of the sand pile as 5 units) we can calculate the rate at which the volume is changing when height is 5 units. Considering the rate to not vary much we can consider the average rate between the interval of 5 units length and 5.01 units length. Now we know that the average rate of volume change in that height interval = (change in volume in that time interval) / (change in height in that interval) We know change in the height in that interval which will be = 5.01 units - 5 units = 0.01 units and the average rate of volume in that interval is also know which is y’(5). Thus the average change in volume in that interval = (average rate) * (change in height during that interval). Thus the volume at 5.01 units height will be = volume at 5 units height + change in height. But this volume will be the approximate volume and not the actual volume as the rate of volume change changes in that interval. The approximation improves as the curve becomes smoother and the interval decreases. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): 3 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q What 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: Given depth function with respect to time y = 0.02 t^2 - 3t + 6 thus the differential of the depth function will give the rate of change of depth function with respect time. Thus the rate of depth change with respect to time will be y’ = dy / dt = 0.02 * d( t^2) / dt - 3 * d(t)/dt + 6 y’ = dy / dt = 0.04t - 3 thus the rate of depth change function y’ = 0.04t - 3 thus the rate of depth change at t = 30 units will be y’(30) = 0.04 * 30 - 3 = 1.2 - 3 = - 1.8 depth units / time units confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You saw in the class notes and in the q_a_ that the rate of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b. This is the function that should be evaluated to give you the rate. Evaluating the rate of depth change function y ' = .04 t - 3 for t = 30 we get y ' = .04 * 30 - 3 = 1.2 - 3 = -1.8. COMMON ERROR: y = .02(30)^2 - 2(30) + 6 =-36 would be the rate of depth change INSTRUCTOR COMMENT: This is the depth, not the rate of depth change. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qmodeling project 3 problem a single quarter-cup of sand makes a cube 1.5 inches on a side. How many quarter-cups would be required to make a cube with twice the scale, 3 inches on a side? Explain how you know this. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given a quarter-cup makes a sand cube of 1.5 inches a side. We need to make a cube of 3 inches side. In order to make a cube of 3 inches, we need to increase the length, breadth as well as the height to 3 inches to make a cube of 3 inches. We stack another cube of 1.5 inches to increase the length to 3 inches, but breadth and height remaining to 1.5 inches. On stacking another 2 cubes to the original 2 cubes the length and breadth both become 3 inches but in order to increase the height to 3 inches the entire stacking of the cubes is to be doubled. Thus a total of 7 cubes is added to make a cube of 3 inches, thus a total of 8 cubes of 1.5 inches constitute a 3 inches cube. Since 1 quarter-cup of sand make a cube, thus in order to make 8 cubes 8 quarter-cup of sand is required to make a cube of 3 inches on a side. confidence rating #$&*: 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: OK ********************************************* 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 given function y = a * x^3 where y is the number of quarter cups and x is the side of the cube. We have 1 quarter cup making a cube of 1.5 inches. Thus 1 = a * (1.5^3), 1 = a * 3.375 , thus a = 1 / 3.375 = 0.296296 quarter cups / cubic inches We also know that 8 quarter cups make a cube of side 3 inches Thus 8 = a * (3 ^ 3), 8 = a * 27, thus a = 0.296296 quarter cups / cubic inches Since the value of a for both the conditions is the same thus the function is consistent and the value of a = 0.296296 quarter cups / cubic inches confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The proportionality would be y = a x^3, with y = 1 (representing one quarter-cup) when x = 1.5. So we have 1 = a * 1.5^3, so that a = 1 / 1.5^3 = .296 approx. So the model is y = .2963 x^3. Therefore if x = 3 we have y = .296 * 3^3 = 7.992, which is the same as 8 except for roundoff error. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat would be the side measurement of a cube designed to hold 30 quarter-cups of sand? What equation did you solve to get this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we know the function which is y = a * ( x ^ 3 ) where a = 0.296296 as calculated before and y is the quarter cups and x inches of the cube. Thus in order to obtain the side length in inches to hold 30 quarter cups of sand we use the function to obtain the value of x when value of y = 30 quarter cups Thus 30 = 0.296296 * ( x^3 ) 101.250 = x^3 Thus x = 4.6608 inches which is the side of the cube with length in inches to hold 30 quarter cups of sand. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* 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: Since someone used ½ cup instead of ¼ cup thus the function would fit the model with y values in multiple of ½. Now since ½ is the double value of ¼ and we want a function which should fit for input values of ¼ instead of ½. The value we obtain with input values of ½ is the double we obtain with input values of ¼. Thus doubling the y function will give us the function with input value of 1/4. Thus the function will be y = 0.004 x^3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** In this case, since it takes two quarter-cups to make a half-cup, the person would need twice as many quarter-cups to get the same volume y. He would have obtained half as many half-cups as the actual number of quarter-cups. To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery problem 4. number of swings vs. length data. Which function fits best? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We take the values of x and y and since we observe that the y value is decreasing thus it must be a decreasing function. We try substituting the various values of x and y and substitute them in various expressions to see whether the expression holds true or not. We start with y = a x^( -.2 ) proceeding to y = ax^( -.3 ) , y = ax ^ ( -.4), y = ax ^(-.5) and see whether the data is consistent with any of the function or not. Thus a for the first function is a = y * x ^(.2) but the a value of a for the data is not consistent with this function, next the same happens with the function y = a*x^( -.3) and ax^ ( -.4 ) but when checked for the function a * x^(-.5) the variations for a are minimum and the function is consistent. Thus for the number of swings vs. length data the function y = a * x^(-.5) fits best. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For each function you can substitute the x and y data for each point, then solve for the constant a. If the values of a are not relatively constant, then the function is not constant. For example, you can try the function y = a x^-2. You can plug in every (x, y) pair and solve of the resulting equations for a. Try this for the data and you will find that y = a x^-2 does not give you consistent a values—every (x, y) pair you plug in will give you a very different value of a. If you know the shapes of the basic graphs, you can compare them with a graph of the data and get a pretty good indication of which functions to try. 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. So you could try these functions, in any reasonable order: You might start with the y = a x^-.3 function. The values of the parameter a will probably vary quite a bit from data point to data point, and if so this function won't be a good candidate. You could then try the y = a x^-.4 function. The values of the parameter a will likely still vary significantly, but probably not as much as for the y = a x^-.3 function. You could then try the y = a x^-.5 function. The values of the parameter a will likely still vary a bit, but probably not by a whole lot. If you have good data this function will probably be your best candidate. You could then try the y = a x^-.6 function. The values of the parameter a will likely vary more than they did for the y = a x^-.5 function, confirming that the y = a x^-.5 function is the best candidate so far. The y = a x^-.7 function will probably result in quite a bit of variation in a, and is not likely to be the best function. STUDENT COMMENT: this concept is a little fuzzy to me. Im not quite sure what you mean when you say that ax^-.5 results in a nearly constant value. INSTRUCTOR RESPONSE: If the proportionality is y = a x^-2, then if we solve for a we get a = y x^2. If you evaluate a = y * x^2 for each of your data points, then if y = a x^-2 is a good model you should get about the same value of a for each point. For example if x values 2, 4, 7, 10 give you respective y values .5, .3, .2, .1, then your a values would be a = .5 * 2^2 = 2 a = .3 * 4^2 = 4.8 a = .2 * 7^2 = 9.8 a = .1 * 10^2 = 10 These values appear to be increasing. So the data don't appear to be consistent with the form y = a x^-2. Another proportionality might yield relatively constant values for a. If so, then your data would be consistent with that proportionality. For example y = a x^-1, or y = a x^-.5 might give you good results. It is also possible that the data aren't consistent with any power-function proportionality. Check this out with the data for this problem, and see what you find. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Just informing I could not perform the experiment but by seeing the data in the solutions I performed the steps and further questions ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qproblem 7. time per swing model. For your data what expression represents the number of swings per minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Performing similar steps as performed in the previous steps we obtain that the function for times per swing vs. length as y = a * x ^ ( - 0.5 ) where a is approximated to 55. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The model that best fits the data is a x^-.5, and with accurate data we find that a is close to 55. The model is pretty close to # per minute frequency = 55 x^-.5. As a specific example let's say we obtained counts of 53, 40, 33 and 26 cycles in a minute at lengths of 1, 2, 3 and 4 feet, then using y = a x^-.5 gives you a = y * x^.5. Evaluating a for y = 53 and x = 1 gives us a = 53 * 1^.5 = 53; for y = 40 and x = 2 we would get a = 40 * 2^.5 = 56; for y = 34 and x = 3 we get a = 33 * 3^.5 = 55; for y = 26 and x = 4 we get a = 26 * 4^.5 = 52. Since our value of a are reasonably constant the y = a x^.5 model works pretty well, with a value of a around 54. The value of a for accurate data turns out to be about 55.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qIf the time per swing in seconds is y, then what expression represents the number of swings per minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given that time per swing in second is y. Thus time per swing in min would be y / 60. Thus the number of swings per minute will be = 60 / y which is nothing but the frequency of the pendulum confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* 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: Given time per swing function is y = a * x ^ 0.5 The value of a taken from the solution is 1.1 As seen in the previous function The number of swings per min = 60 / ( time per swing ) = 60 / ( a * x ^ 0.5 ) = (60 / 1.1) * x ^ ( -0.5) The number of swings per min function = 54.5454 * x ^ ( -0.5) This expression is in accordance with the expression mentioned previously where the function was approximated to y = 55 * x ^ ( -0.5) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1. Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute. Simplifying this gives f = (60 / a) * x^(.-5). 60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements. 60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery problem 8. model of time per swing what are the pendulum lengths that would result in periods of .1 second and 100 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As we know the times per swing vs. length model function is y = 1.1 * ( x ) ^ 0.5 , where y is the time period in sec and x is the length in feet thus for y = 0.1 the corresponding x is 0.1 = 1.1 * ( x ) ^ 0.5 0.090909 = x ^ 0.5 x = 0.008264 feet thus y = 100 sec the corresponding x is 100 = 1.1 * ( x ) ^ 0.5 90.90909 = x ^ 0.5 x = 8264.462645 feet confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery problem 9. length ratio x2 / x1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: The given expression for the frequency of the pendulum is F = ( 60 / 1.1 ) * x ^ -0.5 Thus for length x1 the frequency would be F1 = ( 60 / 1.1 ) * x1 ^ -0.5 And for length x2 the frequency would be F2 = ( 60 / 1.1 ) * x2 ^ -0.5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The solution is to be in terms of x1 and x2. If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)). Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5. If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat expression, in terms of x1 and x2, represents the ratio of the frequencies of the two pendulums? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In the previous question we saw that F1 = ( 60 / 1.1 ) * x1 ^ -0.5 and F2 = ( 60 / 1.1 ) * x2 ^ -0.5 Thus the ratio of the 2 frequencies will be F1 / F2 = ( x1 / x2 ) ^ -0.5 = ( x2 / x1 ) ^ 0.5 This is the expression for the ratio of the frequencies in terms of the lengths of the pendulum. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* 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: The frequency equation for the pendulum problem is F = 54.54545 * ( x^ -.5) The frequency at length 2.4 feet = 35.2089 1/min The frequency at length 2.6 feet = 33.8277 1/min The change in the frequency = - 1.3812 1/min The change in the length = 0.2 feet Thus the rate at which the frequency changes with respect to time = -6.906 1 / min*feet For frequency change between 2 and 3 feet The change in the length = 3 - 2 = 1 feet Thus the change of frequency in that interval = - 6.906 * 1 1/min = -6.906 1/min Whereas now using the frequency equation The frequency at length 2 feet = 38.569 1/min The frequency at length 3 feet = 31.492 1/min Thus the change in the frequency = -7.077 1/min The difference between the frequency at length 2 and 3 feet by using the formula differs from that calculated by using the average rate of frequency change between 2.4 and 2.6 feet is because the frequency equation is not a linear equation. The rate of change of frequency change continuously changes throughout the function, thus the frequency change will change. Similarly, For lengths 4.4 and 4.6 The frequency at length 4.4 feet = 26.0035 1/min The frequency at length 4.6 feet = 25.4319 1/min The change in the frequency = - .5716 1/min The change in the length = 0.2 feet Thus the rate at which the frequency changes with respect to time = -2.858 1 / min*feet For frequency change between 4 and 5 feet The change in the length = 5 - 4 = 1 feet Thus the change of frequency in that interval = - 2.858 * 1 1/min = -2.858 1/min Whereas now using the frequency equation The frequency at length 4 feet = 27.273 1/min The frequency at length 5 feet = 24.393 1/min Thus the change in the frequency = -2.88 1/min The difference between the frequency at length 4 and 5 feet by using the formula differs from that calculated by using the average rate of frequency change between 4.4 and 4.6 feet is because the frequency equation is not a linear equation. The rate of change of frequency change continuously changes throughout the function, thus the frequency change will change. The difference of frequency between 2 and 3 feet length and between 4 and 5 feet length differ because the frequency is different at different lengths and the rate at which it changes also is different at different lengths, thus for the same length gap the frequency change is different. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* Question: `q query problem 1.2.24 5th; 1.2.19 4th formula for exponential function through left (1,6) and (2,18) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Text problem 1.2.24 and 1.2.19 submitted through text_06 assignment. Exponential function through (1,6) and (2,18) Let an exponential function y = a * b ^ x Since it passes through (1,6 ) and (2,18) thus it satisfies the equation as well thus 6 = a * b and 18 = a * b^2 Dividing the 2nd equation by the 1st we get 18 / 6 = a * b^2 / a * b 3 = b By substituting the value of b = 3 in the equation we obtain the value of a 6 = a * 3 a = 2 thus the exponential function will be y = 2 * ( 3^x ) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** An exponential function has one of several forms, including y = A * b^x and y = A * e^(kx). Using y = A * b^t and substituting the t and y coordinates of the two points gives us 6 = A * b^1 18 = A * b^2. Dividing the second equation by the first we get 3 = b^(2-1) or b = 3. Substituting this into the first equation we get 6 = A * 3^1 so A = 2. Thus the model is y = 2 * 3^t . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!