assignment 6 query

course mth 163

could you send me a random fully worked example on how to find the average rate change with midpoint formulas? the last two problems i worked the regular way. I could not come up with a conclusion

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_랜 assignment #006 HѴVʖ滴Ň~ Precalculus I 02-08-2006

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21:29:03 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 --> linear;y=x;y=mx+b;straight line quadratic;y=x^2;y=ax^2+bx+c;parabola expoinential;y=2^x=y=A*2^(kx)+c;increasing/decreasing horizontal asymptote power;y=x^p;y=A(x-h)^p+c;negative;horizontal and vertical asymptotesy=c and s=h; peven/odd graph symmetric or antisymmetric

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21:29:13 ** 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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21:31:20 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)?

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RESPONSE --> it is a power function the graph would be symmetric.

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21:36:03 ** 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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21:45:45 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 --> 20 cm per secod

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21:47:37 ** 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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21:52:47 What is the average rate of depth change from t = 60 to t = 80?

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RESPONSE --> change in depth is -200 change in clock time is 20 average rate of depth change is -180

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21:54:41 ** 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 --> i had everything right until the clerical error in the change in depth. why would the 78 have double negatives?

Change in depth is final depth - original depth. If the original depth is itself negative then the term '- original depth' with have a double negative.

As another example if the temperature goes from -2 degrees to -9 degrees, it should be clear that it goes down by 7 degrees so the change is -7 degrees. This is consistent with the calculation, which would be final temp - initial temp = -9 deg - (-2 deg) = -9 deg + 2 deg = -7 deg.

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21:57:41 describe your graph of y = .02t^2 - 5t + 150

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RESPONSE --> i am getting help with this i have the steps confused but my graph is a parabola with a vertex at -125,0 and crosses the x axis at 124,1 and again at -126,-1

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21:58:04 describe your graph of y = .02t^2 - 5t + 150

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22:00:47 ** 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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22:02:24 describe the pattern to the depth change rates

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RESPONSE --> they are in equal 20 cm intervals

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22:03:42 ** 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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22:08:58 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 --> i have no idea i cant find this examples in my notes

You might be too concerned about whether you're doing the right thing. You might be letting your uncertainty about whether you're right prevent you from making an attempt.

Any attempt is a good thing; even if it's not correct you learn something.

When your attempt is not correct, the self-critique process of the qa or query is designed to get you back on track.

In what follows I'm addressing this process, and pointing out what you might be missing more than the specifics of the solution. I think this will be more helpful than anything else I can tell you at this point, and I think that with your obviously good background, intelligence and work habits, if you modify your approach according to what I'm telling you here you're going to do very well in this course. Naturally I'll be doing my best to help you along the way.

Problem 2 on your worksheet is stated

2. For the time interval from t = 40 sec to t = 60 sec, the 'midpoint time' is t = 50 sec. The term 'midpoint time' is very sensible here; 50 sec is exactly halfway between 40 sec and 60 sec, and is therefore the midpoint of the time interval. Sketch the graph corresponding to each calculation:

Determine the average rate of depth change for the 1-second time interval centered at this midpoint. That is, determine the average rate of depth change between t = 49.5 sec and t = 50.5 sec.

Determine the average rate of change for the six-second time interval centered at the midpoint.

What pattern do you observe?

The first two paragraphs tell you in fairly good detail what to do.

So in the process of asking this question, it would be a good idea to include the original problem statement, and tell me specifically what you do and do not understand about each word and/or phrase in the statement.

In the process of doing this, several things might happen.

In the best case you might well figure out what the problem is asking you to do, and that would be an outstanding learning experience for you.

In the worst case you won't be able to figure out much, but you will become much more familiar with the vocabulary, and you will consciously or unconsciously (very likely both) become much better prepared to understand the solution of the present problem, and other related problems.

In most cases you'll come out somewhere in between.

But in any case, you will give me the information I need to be able to tell you exactly what you need to know.

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22:11:12 ** 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 --> i dont understand where this came from, can you explain this to me?

Problem 2 is a continuation of problem 1 and of the preceding discussion in the worksheet.

The function you should have had in mind when you got to Problem 2 is the function y = .02t^2 - 5t + 150 .

In the statement of problem 2, 'this midpoint' refers to the midpoint t = 50 seconds of the preceding instruction.

The specific statement is 'Determine the average rate of depth change for the 1-second time interval centered at this midpoint. That is, determine the average rate of depth change between t = 49.5 sec and t = 50.5 sec.'

Thus the statement in the solution given here:

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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22:14:00 what is the average rate of change for the six-second time interval centered at the midpoint.

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RESPONSE --> ?????????

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22:17:12 ** 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 --> ok, this is why i am lost not enough info and examples to know if i am doing the problem right.

The instructions

'For the time interval from t = 40 sec to t = 60 sec, the 'midpoint time' is t = 50 sec. The term 'midpoint time' is very sensible here; 50 sec is exactly halfway between 40 sec and 60 sec, and is therefore the midpoint of the time interval. Sketch the graph corresponding to each calculation:

Determine the average rate of depth change for the 1-second time interval centered at this midpoint. That is, determine the average rate of depth change between t = 49.5 sec and t = 50.5 sec.

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were intended to show you what the terminology meant.

The given solution to the 1-second interval was intended to clarify the terminology in case you didn't understand it the first time.

A detailed self-critique at that point would hopefully allow you to make a reasonable attempt at the solution to this question, even if you hadn't been able to solve it earlier.

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22:17:33 What did you observe about your two results?

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RESPONSE --> I have no clue how to get the variables and this lost me

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22:17:52 ** 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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23:18:09 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 --> ???????????????? i have plugged the values into the equation, I have no idea.

You've plugged in the values, so you should include those details here.

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23:18:45 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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23:19:40 what is the average rate of change for the six-second time interval centered at the midpoint.

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RESPONSE --> dont know

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23:20:23 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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"

The best thing I do here is suggest that you share more of the details of your thinking process so I can help you address your difficulties.

I'll be more than glad to answer further questions.