course Mth163 Submitted June 20, 2009 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All graphs are identical parabolas. The c changes the position of each graph. If c = -3 it verti8cally shifts 1 unit below the original graph for each point. C = -2, will be two units below per point, and c = -1, will be I unit below. Where the c = 1, it shifts 1 unit up per each point from the original graph, c = 2, is 2 units up, and c = 3 is three units up. If c = 0, the graph stays the same. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph of the c= -3 function y = x^2 - 3 will lie 3 units lower than the graph of y = x^2. The graph of the c= -2 function y = x^2 - 2 will lie 2 units lower than the graph of y = x^2. The progression should be obvious. The graph of the c= 3 function y = x^2 + 3 will lie 3 units higher than the graph of y = x^2. The final graph will therefore show a series of 7 functions, with the lowest three units below the parabolic graph of y = x^2 and the highest three units above the graph of this function. Each graph will lie one unit higher than its predecessor. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): A series of seven functions, with the lowest 3 units below and the highest, 3 units above. Each graph lying 1 unit higher than its predecessor. Self-critique Rating: ********************************************* Question: `q003. The function y = (x -1)^3 is of the form y = (x -k)^3 with k = 1. What function would this form give us for k = 3? How would the graph of this function compare with that of y = x^3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of the function y = (x 3)^3, horizontally shifts 3 units to the right. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Recall how the graph of y = (x-1)^3 lies one unit to the right of the graph of y = x^3. The k = 3 function y = (x -3)^3 will lie 3 units to the right of the graph of y = x^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Make sure you say to the right of ? graph in this instance, y = x^3 graph. Self-critique Rating: ********************************************* Question: `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the function y = (x k)^3 when the values of k = 2, 3, 4, Each graph of each function will horizontally shift to the right the amount of units that k is equal to. For example k = 2, the graph will shift 2 units, k = 3, the graph will shift 3, k = 4, the graph will shift 4 units per each point. Each graph will be 1 unit more to the right than the one before it. Confidence Assessment: positive. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Both graphs of the functions y = 2^x and y = 2*2^x asymptote coming closer to the x axis but never touching it. y = 2*2^x is vertically 2 unit per each point from the graph of y = 2^x. Confidence Assessment: positive. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As we saw earlier, the graph of y = 3 * 2^x lies 3 times as far from the x-axis as a graph of y = 2^x and is at every point three times as steep. We would therefore expect the A = 2 function y = 2 * 2^x to lie 2 times is far from the x-axis as the graph of y = 2^x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q006. Suppose we wish to graph the functions y = A * 2^x for values A ranging from 2 to 5. If we graph all such functions on the same set of coordinate axes, what will the final graph look like? It is suggested that you actually sketch your graph and describe your sketch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graphs for the function y = a * 2^x, with values of a = 2, 3, 4, 5 are the same shape just 2 times, 3 times, 4 times, and 5 times as steep as the graph of the function y = 2^x. Confidence Assessment: positive. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: These graphs will range from 2 times as far to 5 times as far from the x-axis as the graph of y = 2^x, and will be from 2 to 5 times as steep. The y intercepts of these graphs will be (0,2), (0, 3), (0, 4), (0,5). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Should include the y intercepts for each graph. Self-critique Rating: ********************************************* Question: `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find a slope you must find the difference of the rise which is the y coordinates of 8 and 12. 12 8 = a rise of 4 Then the difference of the run or x coordinates 3 and 9. 9 3 = a run of 6. divide rise / run = 4 / 6 = 2/3 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666.... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Can show decimal value as well Self-critique Rating: ********************************************* Question: `q008. What are the coordinates of the t = 5 and t = 9 points on the graph of y = 2 t^2 + 3? What is the slope of the straight line between these points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 2t^2 +3, t= 5 y = 2* 5^2 + 3 = 53 and y = 9 y = 2 * 9^2 + 3 = 165 165 53 = 112 as your run 9 5 = 4 as your rise rise / run = 112/ 4 = 28 as your slope. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The t = 5 value is y = 2 * 5^2 + 3 = 2 * 25 + 3 = 50 + 3 = 53. The t = 9 value is similarly calculated. We obtain y = 165. The rise between these points is therefore 165-53 = 112. The run is from t = 5 to t = 9, a run of 9 - 5 = 4. This slope of a straight line connecting these points is therefore rise/run = 112/4 = 28. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q009. Suppose y = 2 t^2 + 3 represents the depth, in cm, of water in a container at clock time t, in seconds. At what average rate does the depth of water change between t = 5 seconds and t = 9 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the same calculations as in the above problem them rise/ run would be cm / sec. 112cm / 4sec. = 28 cm/sec. Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The t = 5 second and t = 9 second depths are easily calculated to be y = 53 cm and y = 165 cm. The depth therefore changes from 53 cm to 165 cm, a change of 165 cm - 53 cm = 112 cm, in the 4 seconds between the to clock times. The average rate of depth changes therefore 112 cm/(4 seconds) = 28 cm/second. We note that this calculation is identical to the calculation of the slope of the straight line connecting the t = 5 and t = 9 points of the graph y = 2 t^2 + 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you have the change in y(depth) and divide that by the change in t(time) you will get the average rate between these two points. Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise of the graph represents the change in the depth y and the run represents the change clock time t. The slope, which is rise/run, therefore represents change in depth/change in clock time, which is the average rate at which the depth changes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 006. `query 6 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qQuery 4 basic function families What are the four basic functions? What are the generalized forms of the four basic functions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The 4 basic functions are: quadratic y = x^2 exponential y = 2^x power y = x^p linear y = x The 4 generlized forms are: quadratic y = ax^2 + bx + c exponential y = a[2^(kx)] +c power y = a(x-h)^p + c linear y = mx + b Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: Linear is y=mx+b Quadratic is y=ax^2 + bx +c Exponential is y= A*2^ (kx)+c Power = A (x-h)^p+c INSTRUCTOR COMMENTS: These are the generalized forms. The basic functions are y = x, y = x^2, y = 2^x and y = x^p. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qFor a function f(x), what is the significance of the function A f(x-h) + k and how does its graph compare to that of f(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: h is the distance from the x that we horizontally shift, k is how far we vertically shift, and a is how many units we vertically stretch. Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: A designates the x strectch factor while h affects a y shift & k affects an x shift INSTRUCTOR COMMENTS: k is the y shift for a simple enough reason -- if you add k to the y value it raises the graph by k units. h is the x shift. The reason isn't quite as simple, but not that hard to understand. It's because when x is replaced by x - h the y values on the table shift 'forward' by h units. A is a multiplier. When all y values are multiplied by A that moves them all A times as far from the x axis, which is what causes the stretch. Thus A f(x-h) + k is obtained from f(x) by vertical stretch A, horizontal shift h and vertical shift k. The two aren't the same, but of course they're closely related. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q query introduction to rates and slopes, problem 1 ave vel for function depth(t) = .02t^2 - 5t + 150 give the average rate of depth change from t = 20 to t = 40 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t = .02t^2 5t + 150 t = 20, 40 depth(20) = .02(20)^2 5(20) + 150 depth(20) = 8 100 + 150 depth(20) = 58 depth(40) = .02(40)^2 5(40) + 150 depth(40) = 32 200 + 150 depth(40) = -18 change in depth = -18 58 = -76 change in time = 40 20 = 20 the average change in depth would be -76 / 20 =-3.8 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** depth(20) = .02(20^2) - 5(20) + 150 = 58 depth(40) = .02(40^2) - 5(40) + 150 = -18 change in depth = depth(40) - depth(20) = -18 - 58 = -76 change in clock time = 40 - 20 = 20. Ave rate of depth change = change in depth / change in clock time = -76 / 20 = -3.8 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qWhat is the average rate of depth change from t = 60 to t = 80? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: depth(t) = .02t^2 5t + 150 t = 60, 80 depth(60) = .02(60)^2 5(60) + 150 depth(60) = 72 300 +150 depth(60) = -78 depth(80) = .02(80)^2 -5(80) +150 depth(80) = 128 400 + 150 depth(80) = -122 change in depth = -122 (-78) = -44 change in time = 80-60 = 20 the average change in depth = -44 / 20 = -2.2 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** depth(60) = .02(60^2) - 5(60) + 150 = -78 depth(80) = .02(80^2) - 5(80) + 150 = -122 change in depth = depth(80) - depth(60) = -122 - (-78) = -44 change in clock time = 40 - 20 = 20. Ave rate of depth change = change in depth / change in clock time = -44 / 20 = -2.2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qdescribe your graph of y = .02t^2 - 5t + 150 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: using the vertex formula x = -b / 2a to find the vertex of y = .02t^2 5t +150 x = -(-5) / 2(.02) = -(-5) / .04 = 125 then add x in equation to find y y = .02(125)^2 5(125) + 150 y = 312.5 625 + 150 y = -162.5 The graph opens upward, because of its negative slope it is increasing towards 0 Confidence Assessment: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graph is a parabola. y = .02t^2 - 5t + 150 has vertex at x = -b / (2a) = -(-5) / (2 * .02) = 125, at which point y = .02 (125^2) - 5(125) + 150 = -162.5. The graph opens upward, intercepting the x axis at about t = 35 and t = 215. Up to t = 125 the graph is decreasing at a decreasing rate. That is, it's decreasing but the slopes, which are negative, are increasing toward 0.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): need to point out where it crosses the x axis Self-critique Rating: ********************************************* Question: `qdescribe the pattern to the depth change rates YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: in the function depth(t) = .02t^2 5t + 150 if at the following intervals t = 20, 40 then t = 40,60, then t = 60, 80 rates would be -3.8, -3, and a -2.2 with a pattern of adding .8 to each rate as the get bigger. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Rates are -3.8, -3 and -2.2 for the three intervals (20,40), (40,60) and (60,80). For each interval of `dt = 20 the rate changes by +.8. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery problem 2. ave rates at midpoint times what is the average rate of depth change for the 1-second time interval centered at the 50 sec midpoint? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if 50 is your average midpoint then your 1 sec interval would be between 49.5 sec. and 50.5 sec. to find the average depth change for this interval subtract the depth(49.5) depth(50.5) / 1 sec.= -3 /1 sec =-3 Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The 1-sec interval centered at t = 50 is 49.5 < t < 50.5. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(50.5) - depth(49.5)]/(50.5 - 49.5) = -3 / 1 = -3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qwhat is the average rate of change for the six-second time interval centered at the midpoint. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: with a mid point of 50, your 6 sec. interval will be between 47sec. and 53 sec. using formula depth(t) = .02t^2 5t + 150 to calculate depth then do depth(53) depth(47) / 6sec. depth(-58.82) depth(-40.82) / 6 sec. = -18 / 6sec. = -3 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The 6-sec interval centered at t = 50 is 47 < t < 53. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qWhat did you observe about your two results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate with different intervals was the same. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The two rates match each other, and they also match the average rate for the interval 40 < t < 60, which is also centered at t = 50. For a quadratic function, and only for a quadratic function, the rate is the same for all intervals having the same midpoint. This is a property unique to quadratic functions. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): two rates match each other.* Quadratic functions only* the rates are the same for all intervals. Self-critique Rating: ********************************************* Question: `qquery problem 3. ave rates at midpt times for temperature function Temperature(t) = 75(2^(-.05t)) + 25. What is the average rate of depth change for the 1-second time interval centered at the 50 sec midpoint? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: using the above function to solve for depth(50.5) depth(49.5) / 1 sec you have depth(38.030) depth(38.49) / 1 sec. = -0.46 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: .46 degrees/minute INSTRUCTOR COMMENT: More precisely .4595 deg/min, and this does not agree exactly with the result for the 6-second interval. Remember that you need to look at enough significant figures to see if there is a difference between apparently identical results.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok. but the answer I have is a negative and your answer is a positive. I calculated the way it was done in previous problems and I dont understand unless you switched the problem around and can you do that. Self-critique Rating: ********************************************* Question: `qwhat is the average rate of change for the six-second time interval centered at the midpoint. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Taking the same solution of the given depth and dividing it by 6 we get, - 0.077(if rounded off) Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: .46 degrees/minute INSTRUCTOR COMMENT: The 1- and 6-second results might possibly be the same to two significant figures, but they aren't the same. Be sure to recalculate these according to my notes above and verify this for yourself. The average rate for the 6-second interval is .4603 deg/min. It differs from the average rate .4595 deg/min, calculated over the 1-second interval, by almost .001 deg/min. This differs from the behavior of the quadratic. For a quadratic that the results for all intervals centered at the same point will all agree. This is not the case for the present function, which is exponential. Exact agreement is a characteristic of quadratic functions, and of no other type. ** I dont understand were I miscalculated to this problem. Could you please explain. " snd *************************