course Mth 271 ƒËÞÖÙzöóçÍê´Uõò–”ûÜassignment #006
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13:52:31 Query class notes #06 If x is the height of a sandpile and y the volume, what proportionality governs geometrically similar sandpiles? Why should this be the proportionality?
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RESPONSE --> y= ax^3 or a= y/ x^3 .....because this is the forumula for proportionalitys when dealing with volume
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13:53:46 ** the proportionality is y = k x^3. Any proportionality of volumes is a y = k x^3 proportionality because volumes can be filled with tiny cubes; surface areas are y = k x^2 because surfaces can be covered with tiny squares. **
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RESPONSE --> I said this but I didnt not explain the volume part of it .....I should have said the part about volume being filled with cubes.
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13:54:34 If x is the radius of a spherical balloon and y the surface area, what proportionality governs the relationship between y and x? Why should this be the proportionality?
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RESPONSE --> y= ax^2 because we are talkng about surface area
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13:56:52 ** Just as little cubes can be thought of as filling the volume to any desired level of accuracy, little squares can be thought of as covering any smooth surface. Cubes 'scale up' in three dimensions, squares in only two. So the proportionality is y = k x^2. Surfaces can be covered as nearly as we like with tiny squares (the more closely we want to cover a sphere the tinier the squares would have to be). The area of a square is proportional to the square of its linear dimensions. Radius is a linear dimension. Thus the proportionality for areas is y = k x^2. By contrast, for volumes or things that depend on volume, like mass or weight, we would use tiny cubes to fill the volume. Volume of a cube is proportional to the cube of linear dimensions. Thus the proportionality for a volume would be y = k x^3. **
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RESPONSE --> I needed to go in a lot more detail and talk about the radius and it being linear. .....Actually to be honest I am not quit sure what you mean when you say that the square and the radius are both linear? Thanks.
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13:58:59 Explain how you would use the concept of the differential to find the volume of a sandpile of height 5.01 given the volume of a geometrically similar sandpile of height 5, and given the value of k in the y = k x^3 proportionality between height and volume.
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RESPONSE --> y= ax^3 y= 3ax^2 we know that x= 5 y= 75 a the difference between 5.01 and 5= .01 =.75
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13:59:59 ** The class notes showed you that the slope of the y = k x^3 graph is given by the rate-of-change function y' = 3 k x^2. Once you have evaluated k, using the given information, you can evaluate y' at x = 5. That gives you the slope of the line tangent to the curve, and also the rate at which y is changing with respect to x. When you multiply this rate by the change in x, you get the change in y. The differential is 3 k x^2 `dx and is approximately equal to the corresponding `dy. Since `dy / `dx = 3 k x^2, the differential looks like a simple algebraic rearrangement `dy = 3 k x^2 `dx, though what's involved isn't really simple algebra. The differential expresses the fact that near a point, provided the function has a continuous derivative, the approximate change in y can be found by multiplying the change in x by the derivative). That is, `dy = derivative * `dx (approx)., or `dy = slope at given point * `dx (approx), or `dy = 3 k x^2 `dx (approx). The idea is that the derivative is the rate of change of the function. We can use the rate of change and the change in x to find the change in y. The differential uses the fact that near x = 5 the change in y can be approximated using the rate of change at x = 5. Our proportionality is y = k x^3. Let y = f(x) = k x^3. Then y' = f'(x) = 3 k x^2. When x = 5 we have y' = f'(5) = 75 k, whatever k is. To estimate the change in y corresponding to the change .01 in x, we will multiply y ' by .01, getting a change of y ' `dx = 75 k * .01. } SPECIFIC EXAMPLE: We don't know what k is for this specific question. As a specific example suppose our information let us to the value k = .002, so that our proportionality is y = .002 x^3. Then the rate of change when x is 5 would be f'(5) = 3 k x^2 = 3 k * 5^2 = 75 k = .15 and the value of y would be y = f(5) = .002 * 5^3 = .25. This tells us that at x = 5 the function is changing at a rate of .15 units of y for each unit of x. Thus if x changes from 5 to 5.01 we expect that the change will be change in y = (dy/dx) * `dx = rate of change * change in x (approx) = .15 * .01 = .0015, so that when x = 5.01, y should be .0015 greater than it was when x was 5. Thus y = .25 + .0015 = .2515. This is the differential approximation. It doesn't take account of the fact that the rate changes slightly between x=5 and x = 5.01. But we don't expect it to change much over that short increment, so we expect that the approximation is pretty good. Now, if you evaluate f at x = 5.01 you get .251503. This is a little different than the .2515 approximation we got from the differential--the differential is off by .000003. That's not much, and we expected it wouldn't be much because the derivative doesn't change much over that short interval. But it does change a little, and that's the reason for the discrepancy. The differential works very well for decently behaved functions (ones with smooth curves for graphs) over sufficiently short intervals.**
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RESPONSE --> I am not sure about this one after looking at your work.......should I have multipilied .01 and 75 ....
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14:01:39 What would be the rate of depth change for the depth function y = .02 t^2 - 3 t + 6 at t = 30? (instant response not required)
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RESPONSE --> deriv... y= 2at+ b y= 2(.02)t + -3 t= 30 y= -1.8
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14:02:26 ** You saw in the class notes and in the q_a_ that the rate of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b. This is the function that should be evaluated to give you the rate. Evaluating the rate of depth change function y ' = .04 t - 3 for t = 30 we get y ' = .04 * 30 - 3 = 1.2 - 3 = -1.8. COMMON ERROR: y = .02(30)^2 - 2(30) + 6 =-36 would be the rate of depth change INSTRUCTOR COMMENT: This is the depth, not the rate of depth change. **
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RESPONSE --> I know that we need to find the deriv.....and that allowed me to get the right answer...
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«û¯áŠH¶qŠàŸüɺQ¢¦¯¾£y‚ÎÅ assignment #006 í¾Ôµ¸³•‹„½Ï³}ÛÚ£Ž×xÑ¥Lä™›ÂÐ Applied Calculus I 06-23-2006
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18:45:59 modeling project 3 problem a single quarter-cup of sand makes a cube 1.5 inches on a side. How many quarter-cups would be required to make a cube with twice the scale, 3 inches on a side? Explain how you know this.
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RESPONSE --> 3*4 ........3 in. wide 3 in. cube 3 in. deep 3 in. high ......=12 and 12/1.5= 8 ....we know this because of the hint on the model
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18:46:58 ** You can think of stacking single cubes--to double the dimensions of a single cube you would need 2 layers, 2 rows of 2 in each layer. Thus it would take 8 cubes 1.5 inches on a side to make a cube 3 inches on a side. Since each 1.5 inch cube containts a quarter-cup, a 3 inch cube would contain 8 quarter-cups. COMMON ERROR: It would take 2 quarter-cups. INSTRUCTOR COMMENT: 2 quarter-cups would make two 1.5 inch cubes, which would not be a 3-inch cube but could make a rectangular solid with a square base 1.5 inches on a side and 3 inches high. **
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RESPONSE --> OK...... I came up with the right answer but I am not sure if I did it the same you did.....Did I do this right? I think I used different thinking
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18:49:43 What value of the parameter a would model this situation? How many quarter-cups does this model predict for a cube three inches on a side? How does this compare with your previous answer?
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RESPONSE --> y= ax^3 1= a 1.5 ^3 a= 1/ 1.5^3 a= .296 y= .296 * 3^3= 8 This is right on with my other answer......Should I have thought about doing this the other time?
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18:49:53 ** The proportionality would be y = a x^3, with y = 1 (representing one quarter-cup) when x = 1.5. So we have 1 = a * 1.5^3, so that a = 1 / 1.5^3 = .296 approx. So the model is y = .2963 x^3. Therefore if x = 3 we have y = .296 * 3^3 = 7.992, which is the same as 8 except for roundoff error. **
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RESPONSE --> I got it
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18:51:27 What would be the side measurement of a cube designed to hold 30 quarter-cups of sand? What equation did you solve to get this?
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RESPONSE --> 30= .296 x ^3 x^3= 101.4 x= 101.4 ^(1/3) = 4.66
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18:51:39 ** You are given the number of quarter-cups, which corresponds to y. Thus we have 30 = .296 x^3 so that x^3 = 30 / .296 = 101, approx, and x = 101^(1/3) = 4.7, approx..**
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RESPONSE --> No problem
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18:52:50 query problem 2. Someone used 1/2 cup instead of 1/4 cup. The best-fit function was y = .002 x^3. What function would have been obtained using 1/4 cup?
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RESPONSE --> y= .002 (2) x^3 y= .004 x^3
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18:53:26 ** In this case, since it takes two quarter-cups to make a half-cup, the person would need twice as many quarter-cups to get the same volume y. He would have obtained half as many half-cups as the actual number of quarter-cups. To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3. **
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RESPONSE --> Yes...we need twice as many cups so that would double it to .004
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18:57:29 query problem 4. number of swings vs. length data. Which function fits best?
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RESPONSE --> (7, 20) (6, 22) (5, 24) (4, 26) (3, 29) (2, 33) y= ax^-5 or y= ax^-4 because these are the average points in the middle that turns out the best when you plug in
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18:58:21 ** If you try the different functions, then for each one you can find a value of a corresponding to every data point. For example if you use y = a x^-2 you can plug in every (x, y) pair and solve to see if your values of a are reasonably consistent. Try this for the data and you will find that y = a x^-2 does not give you consistent a values-every (x, y) pair you plug in will give you a very different value of a. The shape of the graph gives you a pretty good indication of which one to try, provided you know the shapes of the basic graphs. For this specific situation the graph of the # of swings vs. length decreases at a decreasing rate. The graphs of y = a x^.p for p = -.3, -.4, -.5, -.6 and -.7 all decrease at a decreasing rate. In this case you would find that the a x^-.5 function works nicely, giving a nearly constant value of a.
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RESPONSE --> I understand this.......
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18:59:54 problem 7. time per swing model. For your data what expression represents the number of swings per minute?
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RESPONSE --> a= 54 because 24= a5^-.5
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19:00:18 ** The model that best fits the data is a x^-.5, and with accurate data we find that a is close to 55. The model is pretty close to # per minute frequency = 55 x^-.5. As a specific example let's say we obtained counts of 53, 40, 33 and 26 cycles in a minute at lengths of 1, 2, 3 and 4 feet, then using y = a x^-.5 gives you a = y * x^.5. Evaluating a for y = 53 and x = 1 gives us a = 53 * 1^.5 = 53; for y = 40 and x = 2 we would get a = 40 * 2^.5 = 56; for y = 34 and x = 3 we get a = 33 * 3^.5 = 55; for y = 26 and x = 4 we get a = 26 * 4^.5 = 52. Since our value of a are reasonably constant the y = a x^.5 model works pretty well, with a value of a around 54. The value of a for accurate data turns out to be about 55.**
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RESPONSE --> I think that I understand this one ok
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19:01:13 If the time per swing in seconds is y, then what expression represents the number of swings per minute?
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RESPONSE --> y/ 60 would give you minutes because there are 60 sec in a min.,
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19:01:38 ** To get the number of swings per minute you would divide 60 seconds by the number of seconds in a swing (e.g., if a swing takes 2 seconds you have 30 swings in a minute). So you would have f = 60 / y, where f is frequency in swings per minute. COMMON ERROR: y * 60 INSTRUCTOR COMMENT: That would give more swings per minute for a greater y. But greater y implies a longer time for a swing, which would imply fewer swings per minute. This is not consistent with your answer. **
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RESPONSE --> Well.....I think that I was on the right track then fliped it around and I dont know why
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