course Mth 163 Ÿæùë’¢ã’ÝÝV…ª|ÜL˜Éøæâassignment #011 011. `query 11 Precalculus I 02-26-2008
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23:29:14 ** 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 --> 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 self critique assessment: 3
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23:30:09 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 = kx^2 confidence assessment: 1
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23:30:46 ** 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 --> 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. self critique assessment: 3
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23:33:18 ** 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 --> 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. self critique assessment: 3
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23:34:36 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 --> y = .02(30^2) - 3(30) + 6 y = 18 - 90 + 6 y = -66 confidence assessment: 3
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23:35:29 ** 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 --> 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. I did rate of depth change. self critique assessment: 3
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23:37:00 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 single quarter-cups because it would be cubed. confidence assessment: 1
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23:37:44 ** 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 --> 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. self critique assessment: 3
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23:38:24 ** 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 --> 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 self critique assessment: 3
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23:39:22 ** 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 --> 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. self critique assessment: 3
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23:46:19 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 = .001x^3 for 1/4 cup used confidence assessment: 3
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23:46:55 ** 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 --> 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 self critique assessment: 3
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23:48:36 query problem 4. number of swings vs. length data. Which function fits best?
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RESPONSE --> y = ax ^-.3 confidence assessment: 1
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23:48:59 ** 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 --> 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. self critique assessment: 3
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23:51:00 problem 7. time per swing model. For your data what expression represents the number of swings per minute?
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RESPONSE --> n confidence assessment: 0
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23:51:19 ** 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 --> 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. self critique assessment: 3
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23:51:33 If the time per swing in seconds is y, then what expression represents the number of swings per minute?
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RESPONSE --> x confidence assessment: 3
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23:51:51 ** 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 --> 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. self critique assessment: 3
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23:53:37 If the time per swing is a x ^ .5, for the value determined previously for the parameter a, then what expression represents the number of swings per minute? How does this expression compare with the function you obtained for the number of swings per minute vs. length?
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RESPONSE --> f = s/y confidence assessment: 0
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23:53:58 ** Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1. Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute. Simplifying this gives f = (60 / a) * x^.5. 60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements. 60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5. **
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RESPONSE --> Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute. Simplifying this gives f = (60 / a) * x^.5. 60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements. 60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5. self critique assessment: 3
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23:55:16 ** You would use your own model here. This solution uses T = 1.1 x^.5. You can adapt the solution to your own model. According to the model T = 1.1 x^.5 , where T is period in seconds and x is length in feet, we have periods T = .1 and T = 100. So we solve for x: For T = .1 we get: .1 = 1.2 x^.5 which gives us x ^ .5 = .1 / 1.2 so that x^.5 = .083 and after squaring both sides we get x = .083^2 = .0069 approx., representing .0069 feet. We also solve for T = 100: 100 = 1.2 x^.5, obtaining x^.5 = 100 / 1.2 = 83, approx., so that x = 83^2 = 6900, approx., representing a pendulum 6900 ft (about 1.3 miles) long. **
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RESPONSE --> This solution uses T = 1.1 x^.5. You can adapt the solution to your own model. According to the model T = 1.1 x^.5 , where T is period in seconds and x is length in feet, we have periods T = .1 and T = 100. So we solve for x: For T = .1 we get: .1 = 1.2 x^.5 which gives us x ^ .5 = .1 / 1.2 so that x^.5 = .083 and after squaring both sides we get x = .083^2 = .0069 approx., representing .0069 feet. We also solve for T = 100: 100 = 1.2 x^.5, obtaining x^.5 = 100 / 1.2 = 83, approx., so that x = 83^2 = 6900, approx., representing a pendulum 6900 ft (about 1.3 miles) long. self critique assessment: 3
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23:56:37 ** The solution is to be in terms of x1 and x2. If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)). Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5. If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5 **
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RESPONSE --> If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)). Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5. If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5 self critique assessment: 3
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23:57:01 ** We need to do this in terms of the symbols x1 and x2. If f = a x^-.5 then f1 = a x1^-.5 and f2 = a x2^-.5. With these expressions we would get f2 / f1 = a x2^-.5 / (a x1^-.5) = x2^-.5 / x1^-.5 = (x2 / x1)^-.5 = 1 / (x2 / x1)^.5 = (x1 / x2)^.5. Note that it doesn't matter what a is, since a quickly divides out of our quotient. For example if a = 55 we get f2 / f1 = 55 x2^-.5 / (55 x1^-.5) = x2^-.5 / x1^-.5 = (x2 / x1)^-.5 = 1 / (x2 / x1)^.5 = (x1 / x2)^.5. This is the same result we got when a was not specified. This shouldn't be surprising, since the parameter a divided out in the third step. **
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RESPONSE --> We need to do this in terms of the symbols x1 and x2. If f = a x^-.5 then f1 = a x1^-.5 and f2 = a x2^-.5. With these expressions we would get f2 / f1 = a x2^-.5 / (a x1^-.5) = x2^-.5 / x1^-.5 = (x2 / x1)^-.5 = 1 / (x2 / x1)^.5 = (x1 / x2)^.5. Note that it doesn't matter what a is, since a quickly divides out of our quotient. For example if a = 55 we get f2 / f1 = 55 x2^-.5 / (55 x1^-.5) = x2^-.5 / x1^-.5 = (x2 / x1)^-.5 = 1 / (x2 / x1)^.5 = (x1 / x2)^.5. This is the same result we got when a was not specified. This shouldn't be surprising, since the parameter a divided out in the third step. self critique assessment: 3
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