course PHY 122 I have already completed assignments 0 and 1 during the previous course and have reviewed my saved work on the computer. I will not resubmit them unless I need to. Please let me know.
......!!!!!!!!...................................
11:00:41 `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.
......!!!!!!!!...................................
RESPONSE --> x y -3 -13 -2 -10 -1 -7 1 -1 2 2 3 5 x intercept is 1 y intercept is -4 confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:02:19 The 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.
......!!!!!!!!...................................
RESPONSE --> I went by my graph instead of using 0. self critique assessment: 2
.................................................
......!!!!!!!!...................................
11:03:50 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
......!!!!!!!!...................................
RESPONSE --> I can't tell that the steepness changes. confidence assessment: 0
.................................................
......!!!!!!!!...................................
11:04:25 The graph forms a straight line with no change in steepness.
......!!!!!!!!...................................
RESPONSE --> This is what I thought. self critique assessment: 1
.................................................
......!!!!!!!!...................................
11:05:54 `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)?
......!!!!!!!!...................................
RESPONSE --> The slope is 3. confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:08:20 Between 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:12:50 `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?
......!!!!!!!!...................................
RESPONSE --> The graph is increasing. The steepness is also increasing at an increasing rate. confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:15:43 Graph 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. *&*&.
......!!!!!!!!...................................
RESPONSE --> I understand this pretty well. self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:24:01 `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?
......!!!!!!!!...................................
RESPONSE --> The graph is decreasing at an increasing rate, confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:25:10 From 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.
......!!!!!!!!...................................
RESPONSE --> I understand it how the steepness would be decreasing at a decreasing rate. self critique assessment: 2
.................................................
......!!!!!!!!...................................
11:27:37 `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?
......!!!!!!!!...................................
RESPONSE --> The graph is increasing at a decreasing rate. confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:28:43 If 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 respresents 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 desribed take another look at your plot and make a note in your response indicating any difficulties.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:35:37 ** 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. **
......!!!!!!!!...................................
RESPONSE --> The graph is decreasing at a decreasing rate. It is a straight line. self critique assessment: 2
.................................................
......!!!!!!!!...................................
11:37:39 `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?
......!!!!!!!!...................................
RESPONSE --> Increasing at an increasing rate. confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:37:58 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
⚑}vvݡ assignment #003 003. PC1 questions qa initial problems 08-27-2008
......!!!!!!!!...................................
11:50:21 `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
......!!!!!!!!...................................
RESPONSE --> the line from (7,17) to (10,29) is steeper based on its slope of 4. The other slope is 3. confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:51:27 The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:55:39 `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
......!!!!!!!!...................................
RESPONSE --> These two numbers and only these two numbers can be used because 2 makes the left brackets zero and -2.5 makes the right brackets zero. Zero multiplied by another number is zero. confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:56:55 If x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero. If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero. The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero. Note that (x-2)(2x+5) can be expanded using the Distributive Law to get x(2x+5) - 2(2x+5). Then again using the distributive law we get 2x^2 + 5x - 4x - 10 which simplifies to 2x^2 + x - 10. However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:59:44 `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
......!!!!!!!!...................................
RESPONSE --> x could be valued at 2, 4, or -2. confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:01:21 In order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0. 3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**
......!!!!!!!!...................................
RESPONSE --> I'm not sure if I put the negative on my 4, but I came up with the same answers. self critique assessment: 2
.................................................
......!!!!!!!!...................................
12:40:25 `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
......!!!!!!!!...................................
RESPONSE --> The trapezoid containing (10,2) and (50,4) has the greater area because although the average height is less than the first set of points, the base is greater giving it a greater total area. confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:41:09 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:49:51 * `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph: As we move from left to right the graph increases as its slope increases. As we move from left to right the graph decreases as its slope increases. As we move from left to right the graph increases as its slope decreases. As we move from left to right the graph decreases as its slope decreases.
......!!!!!!!!...................................
RESPONSE --> For y=x^2, the graph increases as the slope increases. For y=1/x, the graph decreases as the slope decreases. For y=sqrt(x), the graph increases as the slope decreases. self critique assessment: 2
.................................................
......!!!!!!!!...................................
12:53:03 For x = 1, 2, 3, 4: The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate. The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero. Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases. We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate. For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
12:57:17 `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?
......!!!!!!!!...................................
RESPONSE --> The totals for the first 3 months would be 22, 24.2, and 26.42 frogs. I can't thing of a strategy to find how many frogs in 300 months. confidence assessment: 1
.................................................
......!!!!!!!!...................................
12:58:53 08-27-2008 12:58:53 At the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs. The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic). A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.
......!!!!!!!!...................................
NOTES -------> The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic).
.......................................................!!!!!!!!...................................
13:06:51 `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
......!!!!!!!!...................................
RESPONSE --> the answers would be 1,10,100,1000. The values of x are getting smaller and smaller. You could use smaller numbers such as .0001 and .00001 to continue approaching zero. The graph would be almost perpendicular to the x axis. confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:08:27 If x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph .
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:12:14 * `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?
......!!!!!!!!...................................
RESPONSE --> E= 800(24^2) self critique assessment: 1
.................................................
......!!!!!!!!...................................
13:13:34 For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
......!!!!!!!!...................................
RESPONSE --> I just didn't multiply everything out. Otherwise, I was correct. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:16:24 * `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
......!!!!!!!!...................................
RESPONSE --> E=800(3t+9)^2 self critique assessment: 1
.................................................
......!!!!!!!!...................................
13:20:12 Since v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).
......!!!!!!!!...................................
RESPONSE --> Might want to check your math on the 800 * 54t. self critique assessment: 2
.................................................
jҪ݊vϐ[~ assignment #001 001. Rates qa rates 08-27-2008
......!!!!!!!!...................................
13:22:34 `q001. You should copy and paste these instructions to a word processor for reference. However you can always view them, as well as everything else that has appeared in this box, by clicking the 'Display Everything' button. 1. For the next question or answer, you click on 'Next Question / Answer' button above the box at top left until a question has been posed. Once a question has been posed you are to answer before you click again on this button. 2. Before clicking for an answer, type your best answer to the current question into the box to the right, then clip on the 'Enter Answer' button. 3. After entering your answer you will click on 'Next Question / Answer' to view the answer to the question. Do not tamper with the information displayed in the left-hand box. 4. If your answer was incorrect, incomplete or would otherwise require revision, you will enter a self-critique. If you learned something from the answer, you need to restate it in your own words in order to reinforce your learning. If there is something you feel you should note for future reference, you should make a note in your own words. Go to the response box (the right-hand box) and type in a self-critique and/or notes, as appropriate. Do not copy and paste anything from the left-hand box, since that information will be saved in any case. 5. If you wish to save your response to your Notes file you may choose to click on the 'Save As Notes' button rather than the 'Enter Answer' button. Doing so will save your work for your future reference. Your work will be saved in a Notes file in the c:\vhmthphy folder. The title of the Notes file will also include the name you gave when you started the program. 6. After clicking either the 'Enter Response' or the 'Save as Notes' button, click on 'Next Question / Answer' and proceed in a similar manner. In the right-hand box briefly describe your understanding of these instructions, then click 'Enter Answer'.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:22:50 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 3
.................................................
i݉vڢҞ assignment #002 002. Volumes qa areas volumes misc 08-27-2008
......!!!!!!!!...................................
13:28:35 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
......!!!!!!!!...................................
RESPONSE --> v=3*5*7 v=105cm^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:29:24 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:29:51 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
......!!!!!!!!...................................
RESPONSE --> 96m^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:30:36 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:31:50 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
......!!!!!!!!...................................
RESPONSE --> 800m^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:32:11 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:34:50 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
......!!!!!!!!...................................
RESPONSE --> 2355cm^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:36:27 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.
......!!!!!!!!...................................
RESPONSE --> I don't understand why you don't multiply out pi. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:43:35 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
......!!!!!!!!...................................
RESPONSE --> A metal can where h=10cm and a radius of 3cm has a volume of v=9pi*10 or 90pi cm^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:45:07 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:48:50 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
......!!!!!!!!...................................
RESPONSE --> v=1/3 Bh =1/3*50 * 60=1000cm^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:49:09 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:50:11 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
......!!!!!!!!...................................
RESPONSE --> 60m^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:50:25 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:54:11 `q008. What is a volume of a sphere whose radius is 4 meters?
......!!!!!!!!...................................
RESPONSE --> v=4/3pi(r^2) v=4/3pi(16) v=64/3pi m^3 v=67m^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:54:54 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.
......!!!!!!!!...................................
RESPONSE --> I squared the radius instead of cubing it. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:58:19 `q009. What is the volume of a planet whose diameter is 14,000 km?
......!!!!!!!!...................................
RESPONSE --> v=4/3 pi(7000^3) confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:58:59 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:59:53 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
......!!!!!!!!...................................
RESPONSE --> Basically v=A & h confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:00:11 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:01:23 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
......!!!!!!!!...................................
RESPONSE --> We take 1/3 of the volume of a rectangular solid or regular cylinder. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:01:36 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:02:20 `q012. Summary Question 3: What is the formula for the volume of a sphere?
......!!!!!!!!...................................
RESPONSE --> V= 4/3pi(r^3) confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:02:33 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
14:04:51 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> Some of these formulas I have retained throughout the years. I picture them in my head when needed. I have tutored some college and high school students throughout the years, so I get refresher courses in some areas. confidence assessment: 2
.................................................
ZǵX֤뢂} assignment #002 002. Volumes qa areas volumes misc 08-27-2008 ⹔Yj^Qp assignment #002 002. Volumes qa areas volumes misc 08-27-2008
......!!!!!!!!...................................
14:07:37 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
......!!!!!!!!...................................
RESPONSE --> 105 cm^3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:08:02 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
|JO}hѕq_ assignment #001 001. Areas qa areas volumes misc 08-27-2008 "