course Mth 163 kɑ~ʍԄwƝxpƺɸassignment #006
......!!!!!!!!...................................
09:14:55 Query 4 basic function families What are the four basic functions? What are the generalized forms of the four basic functions?
......!!!!!!!!...................................
RESPONSE --> The four basic functions are: y = x y = x^2 y = 2^x y = x^p The generalized fors of the basic functions are: y = mx + b y = ax^2 + bx + c y = A * 2^ (kx) + c y = A ( x - h) ^ p+c confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:16:24 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I didn't saw whether it was linear or power or any other, but it is important to remember that part of the function also. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:42:20 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)?
......!!!!!!!!...................................
RESPONSE --> I can say that I believe the function f(x-h) + k would be vertically shifted to the right if the value for k is positive. It would be vertically shifted to the left is the value for k is negative. That is compared to the graph of f(x). I'm not sure what to say the sifnificance of the function A f(x-h) + k is other than the it shifts the graph. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:47:27 07-09-2008 09:47:27 ** 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. **
......!!!!!!!!...................................
NOTES -------> I didn't know that A was part of the function but I see now. So k is the y shift because it is added to the y value, raising the graph by k units. H is the x shift, when x is replaced by x - h the y values on the table shift ""foward"" by h units. A is the multiplier and it stretches the y value and moves it that many times far from the x axis. So for the function A f(x-h) + k, we have a x axis shift, y axis shift, and a stretch
.......................................................!!!!!!!!...................................
09:47:28 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:07:40 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
......!!!!!!!!...................................
RESPONSE --> We must first find the depth for the function with the given time values: When t = 20: depth(t) = .02*20^2 - 5*20 + 150 = 8 - 250 = -242 depth(t) = .02*40^2 - 5*40 + 150 = 32 - 350 = -318 So the average rate of depth change from t = 20 to t = 40 would be the change in depth over the change in time: 318 - 242 = -76 cm 40 - 20 = 20 sec. We would have a average rate of change at -76 cm/20 sec. which would could divide out to -3.8 cm/sec. Not sure if I did the multiplication right confidence assessment: 1
.................................................
......!!!!!!!!...................................
10:09:38 ** 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 **
......!!!!!!!!...................................
RESPONSE --> I did the depth wrong but came up with the same 76 difference between the values. I see what I did wrong though, it was the part of -5(20) + 150 and -5(40) + 150. I stil got the -3.8 cm/sec though. self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:15:25 What is the average rate of depth change from t = 60 to t = 80?
......!!!!!!!!...................................
RESPONSE --> If we use the depth(t) = .02t^2 - 5t + 150 we find: t = 60, depth(t) = .02*60^2 - 5*60 + 150 = -78 t = 80, depth(t) = .02*80^2 - 5*80 + 150 = -122 The difference -122 - (-78) = -44 The time difference is 80 - 60 or 20 sec So we would get -44cm / 20 sec = -2.2 cm/sec confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:17:08 ** 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 **
......!!!!!!!!...................................
RESPONSE --> I remembered how to do the multiplication right this time and came to the correct rate of depth change. self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:23:24 describe your graph of y = .02t^2 - 5t + 150
......!!!!!!!!...................................
RESPONSE --> If we use the values we found earlier with the t values and depth(t) values we get coordinates: (20, 58), (40, -18), (60, -78), and (80, -122). The graph would be decreasing at a decreasing rate, starting in the positive quadrant and going into the negative quadrant. Since the differences in the depth(t) decrease from: 58 - (-18) = 76 -18 - (-78) = 60 -78 - (-122) = 44 The difference values decrease, so the time change stays the same but the depth value slowly decreases, causing the line of the graph to have a curve into it confidence assessment: 2
.................................................
rۧaߙ assignment #006 006. `query 6 Precalculus I 07-09-2008
......!!!!!!!!...................................
19:44:05 Query 4 basic function families What are the four basic functions? What are the generalized forms of the four basic functions?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
19:44:07 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
19:44:10 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)?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
19:44:12 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
19:44:15 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
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
19:44:17 ** 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 **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
19:44:20 What is the average rate of depth change from t = 60 to t = 80?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
19:44:23 ** 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 **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
19:45:20 describe your graph of y = .02t^2 - 5t + 150
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
20:00:26 ** 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.**
......!!!!!!!!...................................
RESPONSE --> The vertex is 125, I need to remember to use the formula to find the vertex of a graph. So if we find that x = -b / (2a) = -(-5) / 2*.02) = 125 we can find the y value for the vertex as well which helps us with the shape of the graph. So y = .02(125^2) - 5(125) + 150 = -162.5. The graph does open upward but I'm not sure what is meant by the fact that the graph is decreasing at a decreasing rate. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:10:09 describe the pattern to the depth change rates
......!!!!!!!!...................................
RESPONSE --> We know that the time changes from 20 to 40 to 60 to 80, all with a difference of 20 seconds. The depths change in different amounts, we go from 58 to -18 to -78 to - 122. We've got differences of: 58 - (-18) = 76 -18 - (-78) = 60 -76 - (-122) = 44 We can see that all three numbers have a difference of 16. confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:11:54 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I was looking at the time and depth changes. I see that I need to compare the rates, -3.8, -3, and -2.2. I do see that there is a change in the rate by +.8 which indicates a steady rate change. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:50:33 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?
......!!!!!!!!...................................
RESPONSE --> So if we use 1 second time intervals for our time centered around 50 second we would use 49 and 51. depth(t) = .02*(49^2) - 5*49 + 150 = -46.98 depth(t) = .02*(51^2) - 5*51 + 150 = -52.98 So we have a 2 sec difference between 51 and 49. We have a 6 cm difference(-52.98 - (-46.98)). We find an average rate of depth change of 6cm/2 sec or 3 cm per second. confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:53:47 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I thought that the 1 second time interval went both ways around the 50 seconds. I still got the -3 cm/sec. as the average rate of change. self critique assessment: 3
.................................................
what is the average rate of change for the six-second time interval centered at the midpoint.
......!!!!!!!!...................................
RESPONSE --> With a six-second time interval centered at midpoint 50 we would have values of 53 and 47. So depth (t) = .02*(53^2) - 5*53 + 150 = -58.82 depth (t) = .02*(47^2) - 5*47 + 150 = -40.82 So depth(53) - depth(47) = -58.82 - (-40.82) = -18 And since we have a 6 second interval I believe we would divide by 6 to find our average rate of change. -18 cm / 6 sec = -3 cm/sec confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:20:03 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:22:40 What did you observe about your two results?
......!!!!!!!!...................................
RESPONSE --> The two results were the same average rate of change. This would indicate that the rate of change is steady from 47 to 53 seconds. confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:23:30 07-10-2008 17:23:30 ** 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. **
......!!!!!!!!...................................
NOTES -------> That's interesting that for a quadrant function the rate is the same for all intervals having the same midpoint.
.......................................................!!!!!!!!...................................
17:23:31 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:42:26 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?
......!!!!!!!!...................................
RESPONSE --> So with a 1 second time interval centered around 50 we have to times of 50.5 and 49.5. depth(50.5) = 75(2^(-.05*50.5)) + 25 = 38.03056 depth(49.5) = 75(2^(-.05*49.5)) + 25 = 38.49001 depth(50.5) - depth(49.5) = -0.45945/1 = -.045945cm/sec confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:45:51 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.**
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:52:58 what is the average rate of change for the six-second time interval centered at the midpoint.
......!!!!!!!!...................................
RESPONSE --> With a six second interval we have: depth(53) = 75(2^(-.05*53)) + 25 = 36.94901 depth(47) = 75(2^(-.05*47)) + 25 = 39.710959 depth(53) - depth(47) = -2.761949 To get the average we have -2.761949/6 = -0.460325 confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:56:04 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................