#$&* course Mth163 6/29/12 5:40p 006. `query 6
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Given Solution: ** STUDENT RESPONSE: Linear is y=mx+b Quadratic is y=ax^2 + bx +c Exponential is y= A*2^ (kx)+c Power = A (x-h)^p+c INSTRUCTOR COMMENTS: These are the generalized forms. The basic functions are y = x, y = x^2, y = 2^x and y = x^p. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qFor a function f(x), what is the significance of the function A f(x-h) + k and how does its graph compare to that of f(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The A will vertically stretch the graph of the function the h will horizontally shift the graph the k will vertically shift the graph So depending on the values of these the graph will be either stretched making it apper taller or smaller and then horizontally shifted to the left or right and then it may me vertically shifted either up or down on the y axis confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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 with respect to clock time from t = 20 to t = 40 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When finding Depth(20) = .02t^2 - 5t + 150 I found Depth(20) = 58 When finding Depth(40) = .02t^2 - 5t + 150 I found Depth(40)= -18 SO dt= 20 and dy = -76 dy / dt = 3.8cm per second is the average rate of change confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** depth(20) = .02(20^2) - 5(20) + 150 = 58 depth(40) = .02(40^2) - 5(40) + 150 = -18 change in depth = depth(40) - depth(20) = -18 - 58 = -76 change in clock time = 40 - 20 = 20. Ave rate of depth change with respect to clock time = change in depth / change in clock time = -76 / 20 = -3.8 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the average rate of depth change with respect to clock time from t = 60 to t = 80? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: depth(60) = .02(60^2) - 5(60) + 150 = -78 depth(80) = .02(80^2) - 5(80) + 150 = -122 dt = 20 and dy= -44 dy / dt = -2.2cm per second is the average rate of change from 60 to 80 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 with respect to clock time = change in depth / change in clock time = -44 / 20 = -2.2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qdescribe your graph of y = .02t^2 - 5t + 150 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It is the left portion of an upward opening parabola, we can tell because it is a quadratic equation. It will have a vertex of (125, -162.5) the point to the right and left will be left= (124,-162.48) and right= (126,-162.48) and it will cross the x axis approximitley at x= 34.875 and 215.125 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graph is a parabola. y = .02t^2 - 5t + 150 has vertex at x = -b / (2a) = -(-5) / (2 * .02) = 125, at which point y = .02 (125^2) - 5(125) + 150 = -162.5. The graph opens upward, intercepting the x axis at about t = 35 and t = 215. Up to t = 125 the graph is decreasing at a decreasing rate. That is, it's decreasing but the slopes, which are negative, are increasing toward 0.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? DO I need to include the decreasing at a decreasing rate and slope is increasing. I fill like I have given a good explination already. I get confused on exactly what information you are looking for and are not looking for??? ------------------------------------------------ Self-critique Rating:
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Given Solution: ** Rates are -3.8, -3 and -2.2 for the three intervals (20,40), (40,60) and (60,80). For each interval of `dt = 20 the rate changes by +.8. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 2. ave rates at midpoint times what is the average rate of depth change with respect to clock time for the 1-second time interval centered at the 50 sec midpoint? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For depth(49.5) = .02(49.5^2) - 5(49.5) + 150 = -48.495 For depth(50.5) = .02(50.5^2) - 5(50.5) + 150 = -51.495 dt= 1 and dy = -3 dy / dt = -3 sec avrage rate of depth change with respect the clock time confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is the average rate of change with respect to clock time for the six-second time interval centered at the midpoint. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I found the depth for 47.5 to 52.5 and each 1 sec increment between to find that between each second the average depth was increasing by .04 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The 6-sec interval centered at t = 50 is 47 < t < 53. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? I did something different because the language was unclear??? Still found a pattern just a different one. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat did you observe about your two results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: SInce I did something different I found a different pattern. For questions similar to this in the future I will now know what you are looking for. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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 temperature change with respect to clock time for the 1-second time interval centered at the 50 sec midpoint? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Temp(49.5) = 75(2 ^ (-.05 * 49.5)) +25 = 38.489 Temp(50.5) = 75(2 ^ (-.05 * 50.5)) + 25 = 38.021 dt = 1 sec dy= -.468 dy / dt = -.468 is the average rate of temp change with respect to clock time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: f(50.5) = 38.03048331 deg f(49.5) = 38.49000231 deg The change is -0.4595190014 deg. The change in clock time is 1 second. So the average rate is ave rate = change in temperature / change in clock time = -.4595 deg / (1 sec ) = -.4595 deg/sec. STUDENT RESPONSE: .46 degrees/sec INSTRUCTOR COMMENT: More precisely -.4595 deg/sec, and this does not agree exactly with the result for the 6-second interval. Remember that you need to look at enough significant figures to see if there is a difference between apparently identical results.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is the average rate of change for the six-second time interval centered at the midpoint. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Temp(47) = 39.711 Temp(53) = 36.950 dy / dt = -.460 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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/sec. It differs from the average rate -.4595 deg/min, calculated over the 1-second interval, by more than -.0008 deg/sec. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!