course phy201
1/18/2010 4:49
If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
002. Describing Graphs
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Question: `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions.
Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points.
Now make a table for and graph the function y = 3x - 4.
Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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Your solution:
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this graph is linear
confidence rating: 2
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Given Solution:
`aThe graph goes through the x axis when y = 0 and through the y axis when x = 0.
The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3.
The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4).
Your graph should confirm this.
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Self-critique (if necessary):
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Self-critique Rating:3
Did not do the x and y intercepts, but I understand how it is done.
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Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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Your solution:
Yes, It is increasing
confidence rating: 2
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Given Solution:
`aThe graph forms a straight line with no change in steepness.
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Self-critique (if necessary): Ok, I may not understand what exactly it meant by steepness, I was thinking since it was increasing it would also be getting steeper?????
When you walk up a hill, typically as you approach the top the slope starts to level off--it gets less steep.
When you go up a ramp the steepness stays the same until you get to the end of the ramp.
When you start climbing a hill, typically it gets steeper for awhile, the stays at about a constant slope, then gets less steep toward the top.
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Self-critique Rating:1
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Question: `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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Your solution:
confidence rating:
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Given Solution:
`aBetween any two points of the graph rise / run = 3.
For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3.
Note that 3 is the coefficient of x in y = 3x - 4.
Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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Self-critique (if necessary): I am not understanding this part …….“For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3.” I understand that the co-efficient is 3 and if you replace x=2 you get 2 for the anwer, I guess I am a little confused about what the 3*2-4+2 is and the run being 8-2=6 I am confused here…is ther a formual I am forgetting?????????????
3 * 2 - 4 is what you get when you substitute 2 for x. 3 * 2 - 4 = 2, by order of operations (muliply 3 by 2, then subtract 4).
Slope = rise / run.
The rise between two graph points is the change in the y coordinate. The run is the change in the x coordinate.
Our function is y = 3 x - 4.
When x = 2, we substitute 2 for x to get y = 3 * 2 - 4, which is equal to 2.
When x = 8, we substitute 8 for x to get y = 3 * 8 - 4, which is equal to 20.
The graph therefore contains the points (2, 2) and (8, 20).
You should have made a graph showing these points. If not you should do so now.
As you go from point to point your y coordinate goes from 2 to 20. So the 'rise' between the points is 20 - 2 = 18.
Your x coordinate goes from 2 to 8. So the 'run' between the points is 8 - 2 = 6.
The slope is rise / run = 18 / 6 = 3.
The numbers 2 and 8, which were used for the x values, were chosen arbitrarily. Any other two x values would have given you different coordinates, likely with different rise and run. However whatever two x values you use, you will get the same slope. The slope of this graph is constant, and is equal to 3.
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Self-critique Rating:1
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Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing?
Does the steepness of the graph change and if so, how?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution: The graph is increasing and the steepness would be increasing on the negative axis
confidence rating:2
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Given Solution:
`aGraph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right.
The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate
STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate?
INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1?
In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.
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Self-critique (if necessary):????????Does that mean that the steepness will also increase, or is that not answered because we were to stop at zero????? I am a little hazy on what the steepness is
The hill analogy I used above might be helpful.
Formally, steepness could be defined as the magnitude of the slope, i.e., the absolute value of the slope.
Two graphs with respective slopes 4 and -4 would be equally steep; both would have slope of magnitude 4. Both of these graphs would be steeper than, say a graph with slope 3 or -3.
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Self-critique Rating:2
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Question: `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0.
Would you say that the graph is increasing or decreasing?
Does the steepness of the graph change and if so, how?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution: I think it is increasing at a decreasing rate
confidence rating: 2
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Given Solution:
`aFrom left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing.
Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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Self-critique (if necessary):
Ok, I understand the decreasing at a decreasing reate but I am still unclear of the steepness??????????
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Self-critique Rating:2
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Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing?
Does the steepness of the graph change and if so, how?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution:
The graph is increasing at an increasing rate , the steepness is not changed
confidence rating:2
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Given Solution:
`aIf you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing.
The graph would be increasing at a decreasing rate.
If the graph represents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing.
If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as described take another look at your plot and make a note in your response indicating any difficulties.
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Self-critique (if necessary):
Ok, I think I see why it is increasing at a decreasing rate, because the numbers of the y values are decreasing, but when reading from left to right the numbers are increasing, I am still unsure why the steepness is decreasing, I see why going from right to left, but the graph looks linear???????????????????????????
The y value increases, but it changes by less and less for every succeeding x value. So the graph is increasing, but by less and less with each step. It's increasing but at a decreasing rate.
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Self-critique Rating:1
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Question: `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing?
Does the steepness of the graph change and if so, how?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution:
This graph is decreasing with an increasing rate
confidence rating: 1
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Given Solution:
`a** From basic algebra recall that a^(-b) = 1 / (a^b).
So, for example:
2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4.
5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc.
The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time.
The graph is therefore decreasing at a decreasing rate. **
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Self-critique (if necessary):
I am not sure I think I should have looked at the graph from left to right??????
That's correct.
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Self-critique Rating:1
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Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster.
If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution:
Y the car is moving further and further away so the distance is increasing, t = time in seconds and the car is moving faster and faster so the time is decreasing in seconds…..I would say that the graph of the car would be increasing at an decreasing rate
confidence rating: 1
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Given Solution:
`a** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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Self-critique (if necessary): ok The distance is increasing as time increases, the car going faster and faster is really not relevant I am assuming
If the distance increases by more and more with each second, the speed is increasing, and vice versa. The two are equivalent.
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Self-critique Rating:2
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&#Be sure to see my note(s), inserted at various places in this document, and let me know if you have questions. &#