Query 6

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course MTH 163

Dead tired, ill. Sparing details out of courtesy and considerate. Work is sloppy with this assignment.October 17th, @ 10:45 PM

006. `query 6

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Question: `qQuery 4 basic function families

What are the four basic functions?

What are the generalized forms of the four basic functions?

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Your solution:

- Linear - y= mx + b.

- Quadratic - y = ax^2 + bx + c

- Exponential and Power

confidence rating #$&*: 2

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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. **

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Self-critique (if necessary):

- As I’ve stated before, linear and quadratics are my favorite graphs, since I’ve worked with them so much. But I’ve noted that the Exponential and Power graphs carry the forms y = A*2^ (kx) + c and A (x - h )^ p + c, respectively.

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Self-critique Rating:

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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)?

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Your solution:

- A determines how much the graph is stretched. ‘h’ determines the movement of the y-directions, and k determines the direction of movements on the x axis.

confidence rating #$&*: 2

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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. **

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Self-critique (if necessary):

- So in summary, we see A affecting the stretch, h affecting the y, and k affecting the x. Not that simple, I know, but I need to make sure I have a solid understanding before I get hip-deep in something I can’t readily comprehend.

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Your statement is a good start, and gives you a usable context within which to understand the remaining details.

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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

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Your solution:

- I start with labeling the function model, depth(t) = .02t^2 - 5t + 150 and making the connection between depth(y) to f(x) in order to orient myself. So, I replace t with 20 and 40 in two separate equations, solving to find that depth(20) = 58 and depth (40) = -18. I should note here that to get these values I follow the order the numbers come in (exempli gratia, 32 - 200 + 150 = -18) Then I find the difference between the two, starting with greater depth (40) subtracted from depth(20).

- -18 - 58 = -76

- Then I subtract the difference in times from 40 and 20, = 20.

- Thus, I estimate that the average rate of depth change is decreasing 76 units for every 20 seconds.

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This would be -76 units every 20 seconds, assuming that t is in seconds.

So the rate would be

ave rate = change in depth / change in clock time = -76 unit / (20 seconds) = -3.8 units / second.

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confidence rating #$&*: 2

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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 **

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Self-critique (if necessary): For some reason a bad connection was made in this problem, but I’ve noted that the real solution is -3.8; I just had to take it a step further. But for the sake of being thorough, could you elaborate more on this specific question and others like it?

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Self-critique Rating:

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The average rate of change of quantity A with respect to quantity B is

average rate = change in A / change in B.

In this case you wanted to find the average rate of change of depth with respect to clock time, so the A quantity is depth and the B quantity is clock time.

You found the change in depth, and the change in clock time.

All that was left was to divide the two.

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Question: `qWhat is the average rate of depth change with respect to clock time from t = 60 to t = 80?

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Your solution:

- Using the same method as above, I get -4.8 as the average rate of depth change.

- CORRECTION:

-

confidence rating #$&*: 3

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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 **

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Self-critique (if necessary): Math error. Everything was done exactly as you had; but my mental math is a bit rusty, and to complete the metaphor, I’m trying to knock out some of the rust. But I understand where this is going.

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Question: `qdescribe your graph of y = .02t^2 - 5t + 150

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Your solution:

- Since it’s of the form y = ax^2 + bx + c, we can see it’s parabolic; which is huge, I might add. I had to double check my graph to be sure I was doing something wrong, but to get it to fit I had to scale each x and y unit to 1000 maximum and minimum, with 100 per each unit to get a semi-good fit. Otherwise, I see that it’s somewhat offset from the y axis and carries two zeroes.

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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.**

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Self-critique (if necessary):

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Question: `qdescribe the pattern to the depth change rates

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Your solution:

- They increase by .8

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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. **

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Self-critique (if necessary):

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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?

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Your solution:

- -3

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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. **

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Self-critique (if necessary):

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Question: `qwhat is the average rate of change with respect to clock time for the six-second time interval centered at the midpoint.

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Your solution:

- -3.

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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. **

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Self-critique (if necessary):

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Question: `qWhat did you observe about your two results?

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Your solution:

- They are the same, interestingly enough; which somewhat compares to the last pair of questions regarding the +8 chang.e

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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. **

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Self-critique (if necessary):

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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?

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Your solution:

- Please elaborate entire question and solution.

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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.**

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Self-critique (if necessary):

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Self-critique Rating:

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At what clock time would a 1-second interval centered at t = 50 seconds begin, and at what clock time would it end?

What is the depth at the beginning clock time, and at the ending clock time?

What therefore is the change in depth?

What is the change in clock time?

What therefore is the requested average rate?

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Question: `qwhat is the average rate of change for the six-second time interval centered at the midpoint.

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Your solution:

-

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Practically the same questions as before:

At what clock time would a 6-second interval centered at t = 50 seconds begin?

etc.

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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. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qwhat is the average rate of change for the six-second time interval centered at the midpoint.

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Your solution:

-

@&

Practically the same questions as before:

At what clock time would a 6-second interval centered at t = 50 seconds begin?

etc.

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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. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qwhat is the average rate of change for the six-second time interval centered at the midpoint.

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Your solution:

-

@&

Practically the same questions as before:

At what clock time would a 6-second interval centered at t = 50 seconds begin?

etc.

*@

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. **

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Self-critique (if necessary):

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Self-critique Rating:

#*&!#*&!

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You're doing well here. Since you've done most of the groundwork, my notes should clarify the points of confusion.

I hope you're feeling better, and that you've been able to get some rest.

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