course mth 163 ~Swă}֩G썱|J|Fassignment #006
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04:12:12 Query 4 basic function families What are the four basic functions? What are the generalized forms of the four basic functions?
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RESPONSE --> the four basic functions are: linear y = mx+b quadratic y =x^2 exponential y = 2^x power y = x^p
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04:12:31 ** 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. **
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RESPONSE --> ok
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04:14:02 For 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)?
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RESPONSE --> A is a vertical stretch, x is a horizontal shift, and k is a vertical shift. The graph of this would be the graph of f(x) with those changes taken into consideration.
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04:15:56 ** 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. **
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RESPONSE --> ok
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04:17:57 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
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RESPONSE --> the average rate would be the change in depth/change in time, so it's (-18-58)/(20), which gives us -3.8 cm/sec
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04:18:13 ** 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 **
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RESPONSE --> ok
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04:19:18 What is the average rate of depth change from t = 60 to t = 80?
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RESPONSE --> change in depth (-122 - -78) = -44 divided by the change in time, which is 20 gives us -2.2 cm/sec
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04:19:22 ** 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 **
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RESPONSE --> ok
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04:20:17 describe your graph of y = .02t^2 - 5t + 150
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RESPONSE --> it starts at 150 and decreases at a slower and slower rate over time. presumably, it's a parabola, since the equation is quadratic.
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04:23:24 ** 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.**
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RESPONSE --> ok. I left out the vertex, but I know how to get it
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04:23:59 describe the pattern to the depth change rates
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RESPONSE --> the rate changes by .8 cm/sec each time.
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04:24:24 ** 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. **
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RESPONSE --> ok
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04:25:17 query 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?
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RESPONSE --> -3/1 = -3 cm/sec
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04:26:27 ** 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. **
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RESPONSE --> ok
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04:28:49 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> depth(47.5) = -42.4 depth(52.5) = -57.4 (-57.4 - -42.4) / 6 = -2.5 cm/s
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04:30:12 ** 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. **
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RESPONSE --> Yeah, I guess it would have to be 47 to 53 for 50 to be at the center, since 6 is an even number. I guess that threw me off.
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04:30:29 What did you observe about your two results?
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RESPONSE --> Well, they should have been the same
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04:30:42 ** 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. **
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RESPONSE --> OK.
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04:33:50 query 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?
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RESPONSE --> depth(49.5) = 38.49 depth (50.5) = 38.0305 (38.0305 - 38.49) / 1 = .4595 degress celsius/min
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04:34:06 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.**
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RESPONSE --> ok
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04:37:05 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> depth(47) = 39.711 depth(53) = 36.949 (36.949 - 39.711) / 6 = -.4603 degrees celsius/min
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04:37:20 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. **
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RESPONSE --> okay
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