#$&* course Mth 163 004. `query 4
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Given Solution: ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see what you did there with the binomial, I am getting very comfortable with my algebra. Teach me more. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhere f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) = 2^x f (2) = 2^(2) = 4 F(-a) = 2^(-a) *Making a note here, a negative exponent follows the rule a^-n = 1/a^n = ½^a F(x+3) = 2^(x+3) = 2^x+3 F(x)+3 = 2^(x) + 3 = 2^x + 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(-a) = 2^(-a) = 1 / 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery functions given by meaningful names. What are some of the advantages of using meaningful names for functions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This allows us to most effectively communicate to one outside observer the proper definition of a quantity. Values like “y” or “x” are too ambiguous and one would not be able to draw much even from a series of mathematical calculations. Replace y and x with depth and time, and suddenly we see an entire spectrum of possibility. We might also have a calculation that has many variables, which may make it necessary to keep track of the nature and properties of every quantity involved. We might use depth, time, velocity, etc. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.** ONE MORE STUDENT RESPONSE: It is much easier to keep track of what you are doing if you use meaningful names, particularly in multistep procedures. When working with one three different functions, I could call them f(x) = provides the value of the “x” coordinate for any given “y” g(x) = original value of the data for the “x” coordinate for any given y h(x) = the difference between the calculated value for x and the value for x in the data for any given value of y Therefore: f(x) - g(x) = h(x) True, but anything but easy to follow However if I used the designations below, it would be much easier to keep track of what I was doing. Graph(x) = provides the value of the “x” coordinate for any given “y” Data(x) = original value of the data for the “x” coordinate for any given y Resid(x) = the difference between the calculated value for x and the value for x in the data for any given value of y Therefore: Graph(x) - Data(x) = Resid(x) True, and you know at a glance what it is trying to tell you. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)? Your solution: value(t) = $1000 (1.07)^t value(0) = 1000(1.07)^0 = 1000 Value(2) = 1000 (1.07)^(2) = 1144.9 Value(t-3) = 1000(1.07)(t+3) = 1000 * 1.07^t+3 Value(t+3/value(t)) = 1000(1.07)^(t+3) / (1000(1.07^t)) = 1.07^t+3 / 1.07^t = 1.07^3 = 1.23 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Substitute very carefully and show your steps: value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat did you get for illumination(distance)/illumination(2*distance)? Show your work on this one. Your solution: Where illumination(distance) = 50 / distance^2: illumination(1) illumination(2) illumination(3) illumination(distance)/illumination(2*distance). Lum(1) = 50 / (1)^2 = 50 Lum(2) = 50 / (2)^2 = 12.5 Lum(3) = 50 / (3)^2 = 5.6 Lum(distance) / illumination (2*distance) = 50 / (distance)^2 / (2*distance) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???I was correct in how I set up my fractions although I didn’t square the denominator, which I didn’t think had to be done. The illumination (2*distance) is outside the parenthesis of the function illumination(distance), meaning that the quantity (distance) is substituted for (distance)^2. After we write this we divide by illumination (2*distance). Where have I gone wrong here??? ------------------------------------------------ Self-critique Rating:
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Given Solution: ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is your estimate of value of x for which f(x) = 60? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 or 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was able to see that the lines do not connect linearly and would therefore cause the x value to be slightly smaller. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is your estimate of the value f(7)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 32.7 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is your estimate of the difference between f(7) and f(9)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(7) looks at about 31 or 32, f(9) looks at 25 or 26, so a difference of roughly 6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x) = 70 the x coordinate looks at about 2.9 F(x) = 30 the x coordinate looks at about 7.8 A difference of about 5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for each of the following: (Question above was incomplete - assuming you were going to use the same examples from the f(x) nottions and the generalized modeling process.) The temperature at time t = 3. Y = T(3) The temperature at time t = 5. Y = T(5) The change in temperature between t = 3 and t = 5. Y = T(5-3) The average of the temperatures at t = 3 and t = 5. Y = T((5+3) / 2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y = T((5+3) / 2) Y = T(5-3) Y = T(5) Y = T(3) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had a suspicion that the difference in time 5 and 3 would be represented with their own individual parenthesis separated by a minus sign, I was right in subtracting the latter from the former. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30? Your solution: y = f(x) 150 = f(x) (80)-(30) = f(x) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was thinking in terms of the exercises in the notation and generalized modeling process worksheet where it said to use y = f(x). I see that I also messed up the order of subtractions. ????The temperature on a graph would be the y and the time would be the x, so is the method I attempted to use at all valid???
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Given Solution: ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I set up my functions correctly and I’ve got the right idea, but there was no need to guess the actual points on the graph, which confirmed my suspicion that we weren’t quite given enough information to actually solve for the clock time. In its most simplified form we have: F(t1) = 34 F(t2) = 47 F(t2-t1) ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qBy how much did the depth change between t = 23 seconds and t = 34 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(34) - f(23) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This would be f(34) - f(23). If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, you would find that f(34) = 50.6 and f(23) = 60.8 so f(34) - f(23) = 50.6 - 60.8 = -10.2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This stuff gets my brain jumbled when I work on it for a while but I’m getting better at sorting everything out. I will keep working and I believe everything will click in due time. ------------------------------------------------ Self-critique Rating:
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Given Solution: ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in t / change in depth = 11 s / [ f(34) - f(23) ]. If f(t) gives depth in cm this result will be in seconds / centimeter, the ave number of seconds required for depth to change by 1 cm. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that 11 s / [ f(34) - f(23) ] = 11 s / (-10.2 cm) = -1.08 sec / cm, indicating that depth decreases by 1 cm in 1.08 seconds. ** If your solution matches this one, you solved the problem as it was intended and did very well. However, after this solution had been in use for years, a sharp-eyed student noticed that the problem is actually not well-posed. Consider the following: VALID STUDENT OBJECTION (problem is actually not well-posed) Solution shows decrease (indicating direction) rather than measure of change. Question would actually ask for an absolute value - increase and or decrease would not be indicated? INSTRUCTOR RESPONSE: Very good. Technically the question isn't well-posed. The change is the change and not the magnitude of the change, so we're asking for a positive change in depth. The phrasing about the time interval is 'how long ... to change', which implies a positive time interval. The positive change doesn't occur in a positive time interval; a negative time interval doesn't answer the question as phrased. The question would have been well-posed had it asked 'How long does it take for the depth to decrease by 1 cm. ... etc.'. I'm going to leave the phrasing of the question as is, and add this note to the solution. Most students who answer the question correctly will then have an opportunity to consider the idea of a 'well-posed problem'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Seeing these problems when we are given a function and a set of data points, and then asked to relate them in various ways actually came about easier to me then what we are doing here. Breaking everything down into its individual parts has oddly made things more confusing, feels as though I’m working backwards. It is also possible that I am not working as sharply at 6am this morning then previously, in any case, I push onward. ------------------------------------------------ Self-critique Rating:
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Given Solution: ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From a set of data of a depth vs. time experiments I would select three points and input them into the function y = ax^2 + bx + c and solve to find 3 linear equations, which can be solved for parameters a, b, and c. Once these parameters are found we can validate out graph by substituting the known values of x and y into the graph and gauge its accuracy by comparing it to our data. Sometimes it will fit well, other times not. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I sketched a graph using the depth vs. time data:(0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Oh okay for simplicity sake I will use the points in the solution: (0, 96) (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), (70, 41) ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat 3 data point did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (10, 89), (40, 48), (60, 36), confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat was your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (10, 89), (40, 48), (60, 36), 89 = a(10)^2 + b (10) + c 89 = 100a + 10b + c 48 = a(40)^2 + b(40) + c 48 = 1600a + 40b + c 36 = a(60)^2 + b(60) + c 36 = 3600a + 60b + c We solve and obtain: a = 0.02 b = -2.13 c = 108.8 depth(t) = 0.02t^2 - 2.13t + 108 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089t^2 - 1.4992t + 98.8544. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the average deviation for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: for the function depth(t) = 0.02t^2 - 2.13t + 108 T Y F(t) deviation 0 96 108 12 10 89 89.5 0.5 20 68 74.2 6.2 30 65 62.1 2.9 40 48 54.8 6.8 50 49 52.3 3.3 60 36 53 17 70 41 56.9 15.9 Average deviation: 8.1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I added a column 2 columns to the chart given in our assignment one labeled 'Model data' and the other 'Deviation'. I then subtracted the model data readings from the corresponding given data to get the individual deviations. I then averaged out the numbers in the deviation column. The average deviation turned out to be 3.880975.** The given points are (0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). If the model is f(t) = .0089t^2 - 1.4992t + 98.8544 then we can create the following table: t y f(t) deviation 0 96 98.544 2.544 10 89 84.442 4.558 20 68 72.12 4.12 30 65 61.578 3.422 40 48 52.816 4.816 50 49 45.834 3.166 60 36 40.632 4.632 70 41 37.21 3.79 For example when t = 30 the data point is (30, 65). The function value is f(30) = 61.578. The deviation between the function value and the data point is | 65 - 61.578 | = 3.422. Note that the function values are calculated to a ridiculous number of significant figures. Since the original data are given only to whole-number values, it would be more appropriate to round the values of f(t) to the nearest whole number, giving us the table t y f(t) deviation 0 96 99 3 10 89 84 5 20 68 72 4 30 65 62 3 40 48 53 5 50 49 46 3 60 36 41 5 70 41 37 4 The average deviation would be the average of the deviations. Adding the deviations up we get 32. Dividing this by 9, the number of data points, we find that the average deviation is about ave dev = 32 / 9 = 3.6, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow close is your model to the curve you sketched earlier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I sketched a number of curvesand I’m not quite sure which one this applies to, perhaps our first one where f(2) = 80, f(5) = 40 and f(10) = 25. It doesn’t fit this one very well tho… confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation ** "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qHow close is your model to the curve you sketched earlier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I sketched a number of curvesand I’m not quite sure which one this applies to, perhaps our first one where f(2) = 80, f(5) = 40 and f(10) = 25. It doesn’t fit this one very well tho… confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation ** "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!