#$&* course Mth 163 1/26/13 around 6 p.m. 004.
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Given Solution: f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not realize I needed to find the vertex and the quadratic function, but I understand how they were obtained. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we substitute f(a) in to the equation we get f(x) = a^2 +4 If we substitute f(x+2) into the equation we get f(x) = (x+2) ^2 +4 = (x+2) (x+2) +4. We must use FOIL and get x^2 +4x +8. If we substitute f(x+h) into the equation we get f(x) = (x+h) ^2 +4 = (x+h) (x+) +4. We expand this and get x^2 +2xh +h^2 +4. If we substitute f(x+h) - f(x) into the equation we can use the equation just obtained for f(x+h) and the equation earlier so we would get [x^2 +2xh +h^2 +4] - x^2 - 4. When we expand this we get 2xh + h^2. If we substitute [ f(x+h) - f(x) ] / h into the equation we can use the expression we just obtained and divide it by h. So we get 2xh + h^2 /h. This gives us 2x + h. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h. You should have written these expressions out, and the following should probably be represented on your paper in form similar to that given here: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x) = 5 * x1+7 F(x) = 5 * x2 + 7 F (x) = [5 *x2 + 7] + [5 *x1 -7] / (x2 * x1). Then to expand we get 5 *x2 + 5 * x1/ x2 - x1. Then we can factor out a 5 of the top and get 5 (x2 + x1) / (x2 - x1) = 5. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5. Compare what you have written down with the expressions below: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we are trying to find f (x) = -3 we can substitute -3 in for f(x) and get -3 = 5x +7. We then subtract 7 from -3 and get -10 = 5x. Then divide both sides by 5 and get x = -2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 Exercises 1-2 1. Of course you know, or at least suspect, that your function models won't always be quadratic. Obtain the indicated expressions for the following functions, none of which are quadratic: Where f(x) = x^3, find f(-2), f(2), f(-a), f(a), f(x-4) and f(x) - 4. F(x) = (-2) ^3 = -8 F(x) = (2) ^3 = 8 F(x) = (-a) ^3 F(x) = (a) ^3 F(x) = (x-4) ^3 = x^3 - 4^3 = x^3 - 64 F(x) = (x) ^3 - 4 Where f(x) = 2^x, find f(-2), f(2), f(-a), f(a), f(x+3) and f(x) + 3. F(x) = 2^-2 = 0.25 F(x) = 2^2 = 4 F(x) = 2^-a F(x) = 2^a F(x) = 2^(x+3) F(x) = 2^x + 3 2. You should also be aware that we can use letters for our functions other than f. We can denote a function by the notation g(x), h(x), y(x), Q(x), Z(x), changeRate(x), investmentValue(x), or whatever. Each of these expressions indicates a quantity that varies with x. And we don't have to use the letter x to stand for the variable. We can have expressions like A(t), V(z), accumulatedTotal(investmentStrategy), or gradeEarned(timeDevoted). In each case we say that the quantity in the parentheses is the variable. Obtain expressions for the following: Where value(t) = $1000 (1.07)^t value(0) = $1000 value(1) = $1070 value(2) = $1144.9 value(t+3) = $1000 (1.07) ^ t+3 value(t+3)/value(t). 1000 (1.07) ^ t+3 / $1000 (1.07)^t Where illumination(distance) = 50 / distance^2: illumination(1) = 50 illumination(2) = 12.5 illumination(3) = 5.5555 illumination(distance)/illumination(2*distance). = (50/distance ^2) / (50/2*distance^2) Exercises 3-4 3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Use your graph to estimate the following: The value of x for which f(x) = 60. = about 4 The value f(7). = 30 The difference between f(7) and f(9). About 2 The difference in x values between the points where f(x) = 70 and where f(x) = 30. About -4.5 4. If a temperature vs. clock time function is given by y = temperature = T(t), then what is the symbolic expression for each of the following: The temperature at time t = 3. = T3 The temperature at time t = 5. = T5 The change in temperature between t = 3 and t = 5. = T3 - T5 The average of the temperatures at t = 3 and t = 5. = (T3 + T5) / 2 What equation would we solve to find the clock time when the model predicts a temperature of 150? = Temperature = 150t How would we go about finding the length of time required for the temperature to fall from 80 to 30? Temperature = (80-30) * t Modeling Exercises 5. Questions about your depth vs. time model I am not sure what you are talking about??? I never sketched this model??? Or was asked to. For your model of depth vs. time, based on in-class measurements, answer the following questions and give detailed reasons for your answers as well as sketches. Use the f(x) notation at every opportunity as you give and reason out your answers: For how long was the depth between 34 and 47 centimeters? By how much did the depth change between t = 23 seconds and t = 34 seconds? On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds? On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds? 6. A hypothetical depth vs. time model based on three points, none of which are actual data points Sketch a graph of the following depth vs. time data: ( 0 sec, 96 cm) (10 sec, 89 cm) (20 sec, 68 cm) (30 sec, 65 cm) (40 sec, 48 cm) (50 sec, 49 cm) (60 sec, 36 cm) (70 sec, 41 cm) These data were obviously taken by someone with either bad instruments or a high degree of incompetence. However, on the average they might well give a good quadratic model. Sketch the data and sketch a smooth curve that doesn't touch any data points but comes as close as possible to the data points, on the average. Your curve will go pretty much through the middle of the data set. Pick three points on this curve, approximately equally spaced in the y direction, and use them as a basis for constructing a function model. (20, 68) (40, 48) (60, 36) Determine the average deviation for your model, and graph your function. How close is your model to the curve you sketched earlier? How well does your function seem to model the data? I was unclear of this question. I plotted those three points and did a trendline and got the equation y = -0.8x + 82.6667 and the R^2 = 0.9796. 004. `query 4 ********************************************* Question: `qWhere f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x) = (-2) ^3 = -8 F(x) = (-a) ^3 F(x) = (x-4) ^3 = x^3 - 4^3 = x^3 - 64 F(x) = (x) ^3 - 4 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 understand my errors and understand the correct answers. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhere f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x) = 2^(2) = 4 F(x) = 2^(-a) F(x) = 2^(x+3) F(x) = 2^x + 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(-a) = 2^(-a) = 1 / 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery functions given by meaningful names. What are some of the advantages of using meaningful names for functions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The advantages of using meaningful names for functions is because it makes more sense to using something like time = instead of y = and if you haven’t looked at your work for a few days you can go back and see what you were actually talking about if you use meaningful names. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 3 ********************************************* Question: `qWhat were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Where value(t) = $1000 (1.07)^t value(0) = $1000 value(1) = $1070 value(2) = $1144.9 value(t+3) = $1000 (1.07) ^ t+3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat did you get for illumination(distance)/illumination(2*distance)? Show your work on this one. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Where illumination(distance) = 50 / distance^2: illumination(1) = 50 illumination(2) = 12.5 illumination(3) = 5.5555 illumination(distance)/illumination(2*distance). = (50/distance ^2) / (50/2*distance^2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I constructed this graph by using the coordinates (2,80), (5, 40), & (10, 25). When I did this I got a curved line that I think would be power or exponential. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is your estimate of value of x for which f(x) = 60? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Approximately f(4). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is your estimate of the value f(7)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(7) = approximately 30 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is your estimate of the difference between f(7) and f(9)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for each of the following: (Question above was incomplete - assuming you were going to use the same examples from the f(x) nottions and the generalized modeling process.) The temperature at time t = 3. The temperature at time t = 5. The change in temperature between t = 3 and t = 5. The average of the temperatures at t = 3 and t = 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The temperature at time t = 3. = T3 The temperature at time t = 5. = T5 The change in temperature between t = 3 and t = 5. = T3 - T5 The average of the temperatures at t = 3 and t = 5. = (T3 + T5) / 2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: What equation would we solve to find the clock time when the model predicts a temperature of 150? = Temperature = 150t How would we go about finding the length of time required for the temperature to fall from 80 to 30? Temperature = (80-30) * t confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand my errors and the correct answers. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was unclear about what graph this was referring to??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): The question made it seem like we had to have a graph completed which I didn’t so I did not try to solve this because I assumed I had missed something. ------------------------------------------------ Self-critique Rating: 3
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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): I again was unsure about the graph to use??? ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qOn the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same answers as above confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Same as above ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qOn the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same as above confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Same as above ------------------------------------------------ Self-critique Rating: 3
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat 3 data point did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used (20, 68), (40, 48), & (60, 36) points. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat was your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My function model was y = -0.8x + 82.6667 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I am not sure if I did this right???
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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): I am still unsure of what to do??? ------------------------------------------------ Self-critique Rating: 0 ********************************************* Question: `qHow close is your model to the curve you sketched earlier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not know what model I was supposed to sketch earlier??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was very confused with the depth vs. time model that was referred to and then confused at the end of what to do??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 do not know what model I was supposed to sketch earlier??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was very confused with the depth vs. time model that was referred to and then confused at the end of what to do??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: #*&!
#$&* course Mth 163 1/26/13 around 6 p.m. 004.
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Given Solution: f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not realize I needed to find the vertex and the quadratic function, but I understand how they were obtained. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we substitute f(a) in to the equation we get f(x) = a^2 +4 If we substitute f(x+2) into the equation we get f(x) = (x+2) ^2 +4 = (x+2) (x+2) +4. We must use FOIL and get x^2 +4x +8. If we substitute f(x+h) into the equation we get f(x) = (x+h) ^2 +4 = (x+h) (x+) +4. We expand this and get x^2 +2xh +h^2 +4. If we substitute f(x+h) - f(x) into the equation we can use the equation just obtained for f(x+h) and the equation earlier so we would get [x^2 +2xh +h^2 +4] - x^2 - 4. When we expand this we get 2xh + h^2. If we substitute [ f(x+h) - f(x) ] / h into the equation we can use the expression we just obtained and divide it by h. So we get 2xh + h^2 /h. This gives us 2x + h. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h. You should have written these expressions out, and the following should probably be represented on your paper in form similar to that given here: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x) = 5 * x1+7 F(x) = 5 * x2 + 7 F (x) = [5 *x2 + 7] + [5 *x1 -7] / (x2 * x1). Then to expand we get 5 *x2 + 5 * x1/ x2 - x1. Then we can factor out a 5 of the top and get 5 (x2 + x1) / (x2 - x1) = 5. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5. Compare what you have written down with the expressions below: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we are trying to find f (x) = -3 we can substitute -3 in for f(x) and get -3 = 5x +7. We then subtract 7 from -3 and get -10 = 5x. Then divide both sides by 5 and get x = -2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 Exercises 1-2 1. Of course you know, or at least suspect, that your function models won't always be quadratic. Obtain the indicated expressions for the following functions, none of which are quadratic: Where f(x) = x^3, find f(-2), f(2), f(-a), f(a), f(x-4) and f(x) - 4. F(x) = (-2) ^3 = -8 F(x) = (2) ^3 = 8 F(x) = (-a) ^3 F(x) = (a) ^3 F(x) = (x-4) ^3 = x^3 - 4^3 = x^3 - 64 F(x) = (x) ^3 - 4 Where f(x) = 2^x, find f(-2), f(2), f(-a), f(a), f(x+3) and f(x) + 3. F(x) = 2^-2 = 0.25 F(x) = 2^2 = 4 F(x) = 2^-a F(x) = 2^a F(x) = 2^(x+3) F(x) = 2^x + 3 2. You should also be aware that we can use letters for our functions other than f. We can denote a function by the notation g(x), h(x), y(x), Q(x), Z(x), changeRate(x), investmentValue(x), or whatever. Each of these expressions indicates a quantity that varies with x. And we don't have to use the letter x to stand for the variable. We can have expressions like A(t), V(z), accumulatedTotal(investmentStrategy), or gradeEarned(timeDevoted). In each case we say that the quantity in the parentheses is the variable. Obtain expressions for the following: Where value(t) = $1000 (1.07)^t value(0) = $1000 value(1) = $1070 value(2) = $1144.9 value(t+3) = $1000 (1.07) ^ t+3 value(t+3)/value(t). 1000 (1.07) ^ t+3 / $1000 (1.07)^t Where illumination(distance) = 50 / distance^2: illumination(1) = 50 illumination(2) = 12.5 illumination(3) = 5.5555 illumination(distance)/illumination(2*distance). = (50/distance ^2) / (50/2*distance^2) Exercises 3-4 3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Use your graph to estimate the following: The value of x for which f(x) = 60. = about 4 The value f(7). = 30 The difference between f(7) and f(9). About 2 The difference in x values between the points where f(x) = 70 and where f(x) = 30. About -4.5 4. If a temperature vs. clock time function is given by y = temperature = T(t), then what is the symbolic expression for each of the following: The temperature at time t = 3. = T3 The temperature at time t = 5. = T5 The change in temperature between t = 3 and t = 5. = T3 - T5 The average of the temperatures at t = 3 and t = 5. = (T3 + T5) / 2 What equation would we solve to find the clock time when the model predicts a temperature of 150? = Temperature = 150t How would we go about finding the length of time required for the temperature to fall from 80 to 30? Temperature = (80-30) * t Modeling Exercises 5. Questions about your depth vs. time model I am not sure what you are talking about??? I never sketched this model??? Or was asked to. For your model of depth vs. time, based on in-class measurements, answer the following questions and give detailed reasons for your answers as well as sketches. Use the f(x) notation at every opportunity as you give and reason out your answers: For how long was the depth between 34 and 47 centimeters? By how much did the depth change between t = 23 seconds and t = 34 seconds? On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds? On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds? 6. A hypothetical depth vs. time model based on three points, none of which are actual data points Sketch a graph of the following depth vs. time data: ( 0 sec, 96 cm) (10 sec, 89 cm) (20 sec, 68 cm) (30 sec, 65 cm) (40 sec, 48 cm) (50 sec, 49 cm) (60 sec, 36 cm) (70 sec, 41 cm) These data were obviously taken by someone with either bad instruments or a high degree of incompetence. However, on the average they might well give a good quadratic model. Sketch the data and sketch a smooth curve that doesn't touch any data points but comes as close as possible to the data points, on the average. Your curve will go pretty much through the middle of the data set. Pick three points on this curve, approximately equally spaced in the y direction, and use them as a basis for constructing a function model. (20, 68) (40, 48) (60, 36) Determine the average deviation for your model, and graph your function. How close is your model to the curve you sketched earlier? How well does your function seem to model the data? I was unclear of this question. I plotted those three points and did a trendline and got the equation y = -0.8x + 82.6667 and the R^2 = 0.9796. 004. `query 4 ********************************************* Question: `qWhere f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x) = (-2) ^3 = -8 F(x) = (-a) ^3 F(x) = (x-4) ^3 = x^3 - 4^3 = x^3 - 64 F(x) = (x) ^3 - 4 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 understand my errors and understand the correct answers. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhere f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x) = 2^(2) = 4 F(x) = 2^(-a) F(x) = 2^(x+3) F(x) = 2^x + 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(-a) = 2^(-a) = 1 / 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery functions given by meaningful names. What are some of the advantages of using meaningful names for functions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The advantages of using meaningful names for functions is because it makes more sense to using something like time = instead of y = and if you haven’t looked at your work for a few days you can go back and see what you were actually talking about if you use meaningful names. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 3 ********************************************* Question: `qWhat were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Where value(t) = $1000 (1.07)^t value(0) = $1000 value(1) = $1070 value(2) = $1144.9 value(t+3) = $1000 (1.07) ^ t+3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat did you get for illumination(distance)/illumination(2*distance)? Show your work on this one. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Where illumination(distance) = 50 / distance^2: illumination(1) = 50 illumination(2) = 12.5 illumination(3) = 5.5555 illumination(distance)/illumination(2*distance). = (50/distance ^2) / (50/2*distance^2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I constructed this graph by using the coordinates (2,80), (5, 40), & (10, 25). When I did this I got a curved line that I think would be power or exponential. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is your estimate of value of x for which f(x) = 60? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Approximately f(4). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is your estimate of the value f(7)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(7) = approximately 30 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is your estimate of the difference between f(7) and f(9)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for each of the following: (Question above was incomplete - assuming you were going to use the same examples from the f(x) nottions and the generalized modeling process.) The temperature at time t = 3. The temperature at time t = 5. The change in temperature between t = 3 and t = 5. The average of the temperatures at t = 3 and t = 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The temperature at time t = 3. = T3 The temperature at time t = 5. = T5 The change in temperature between t = 3 and t = 5. = T3 - T5 The average of the temperatures at t = 3 and t = 5. = (T3 + T5) / 2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: What equation would we solve to find the clock time when the model predicts a temperature of 150? = Temperature = 150t How would we go about finding the length of time required for the temperature to fall from 80 to 30? Temperature = (80-30) * t confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand my errors and the correct answers. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was unclear about what graph this was referring to??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): The question made it seem like we had to have a graph completed which I didn’t so I did not try to solve this because I assumed I had missed something. ------------------------------------------------ Self-critique Rating: 3
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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): I again was unsure about the graph to use??? ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qOn the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same answers as above confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Same as above ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qOn the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same as above confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Same as above ------------------------------------------------ Self-critique Rating: 3
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat 3 data point did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used (20, 68), (40, 48), & (60, 36) points. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat was your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My function model was y = -0.8x + 82.6667 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I am not sure if I did this right???
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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): I am still unsure of what to do??? ------------------------------------------------ Self-critique Rating: 0 ********************************************* Question: `qHow close is your model to the curve you sketched earlier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not know what model I was supposed to sketch earlier??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was very confused with the depth vs. time model that was referred to and then confused at the end of what to do??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 do not know what model I was supposed to sketch earlier??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was very confused with the depth vs. time model that was referred to and then confused at the end of what to do??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: #*&!