#$&* course Mth 163
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Given Solution: ** STUDENT RESPONSE: Linear is y=mx+b Quadratic is y=ax^2 + bx +c Exponential is y= A*2^ (kx)+c Power = A (x-h)^p+c INSTRUCTOR COMMENTS: These are the generalized forms. The basic functions are y = x, y = x^2, y = 2^x and y = x^p. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qFor a function f(x), what is the significance of the function A f(x-h) + k and how does its graph compare to that of f(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The k represents the y shift, the h is the x shift, A is our stretch. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Okay so A is a vertical stretch, I should have specified. ------------------------------------------------ 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: Depth(20) = 0.2(20)^2 - 5(20) + 150 = 130 Depth(40)= 0.2(40)^2 - 5(40) + 150 = 270 Change in depth = 270 - 130 = 140 Change in clock time = 40 - 20 = 20 Average rate of change in depth with respect to clock time = 140/20 = 7(unit) / sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** depth(20) = .02(20^2) - 5(20) + 150 = 58 depth(40) = .02(40^2) - 5(40) + 150 = -18 change in depth = depth(40) - depth(20) = -18 - 58 = -76 change in clock time = 40 - 20 = 20. Ave rate of depth change with respect to clock time = change in depth / change in clock time = -76 / 20 = -3.8 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I don’t know how we got such different answers….. oh okay I looked back over this I left out a decimal place in 0.02. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the average rate of depth change with respect to clock time from t = 60 to t = 80? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’ll assume we’re using the generalized model of the quadratic function from before: Depth(60) = 0.02(60^2) - 5(60) + 150 = -78 Depth(80) = 0.02(80^2) - 5(80) + 150 = - 122 Change in depth / change in clock time = (-122 - (-78)) / (80-60) = -44/ 20 = - 2.2 (unit of depth) / sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: The vertex of the graph can be found by x = -b / (2a) = 5 / (2*0.02) = 125 Substituting this value into the function gives us our y value: Y = 0.02(125^2) - 5 (125) + 150 = -162.5 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): I didn’t do as well with this one, if I solved the function for the value of t using the quadratic formula, I could find the points at which y = 0 (if such points existed) and if they did, subtract the value of the vertex (-b/2a) to find the two crucial points to the left and right of the vertex, the A value is then how many units to go up, is that correct? Ah, your values of 35 and 215 must be the two values at which y = 0? The vertex and these two points should be enough to describe the graph how we need to, but is there anything else? ------------------------------------------------ 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: The midpoint is centered between 49.5 and 50.5 which is 50. Solving the depth values for 50.5 and 49.5 we get: Depth(49.5) = -48.5 Depth(50.5) = -51.5 Change in depth = -3 / 1 = -3 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: Depth(53) = 58.8 Depth(47) = -40.8 58.8 - 40.8 = 18 / 6 = 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The 6-sec interval centered at t = 50 is 47 < t < 53. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat did you observe about your two results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rates between the two points are the same, meaning that every point will be like this. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): This property is unique to quadratic functions but what about linear functions? Aren’t they defined in a somewhat similar way, with the slope being the same from one point to the next? I may be confusing that from physics where the slope of a linear graph represents the average rate of change of velocity with respect to clock time, which will be the same between any two points if the object accelerates uniformly? ------------------------------------------------ Self-critique 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): I kept the entire decimal in my calculator but I’ve never seen a situation where it was necessary for us to report so many decimals.. ------------------------------------------------ Self-critique 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): Ah so we need the extra decimal places to see the difference because it is so small