#$&* course MTH 173 1243 18FEB2013 006. query 6
.............................................
Given Solution: ** 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. ** 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? ** 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 (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we start with: y = kx^3 and take the derivative of it using the chain rule we will get: y’ = 3kx^2 We will need to find the slope of the line at x = 5. So for this we will substitute in 5 for x and whatever value we are given for k. This will get us the slope at x=5 for this function. Since the slope of the pile at 5 will be very close to the slope at 5.01, we can take the slope information from the above problem and use that to find the volume of the second sand pile. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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 (if necessary): The answer given is quite a bit more lengthy that I had imagined it to be. It brings up several good points though. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 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) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the equation: y = .02 t^2 - 3 t + 6 we first find the derivative: Using the chain rule we get: Y = .04t - 3 Then, we plug t=30 into the equation and solve. We get: Y = -1.8 Thus, -1.8 is the rate at which the depth is changing when t = 30. This is also known as the slope. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** You saw in the class notes and in the q_a_ that the rate of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b. This is the function that should be evaluated to give you the rate. Evaluating the rate of depth change function y ' = .04 t - 3 for t = 30 we get y ' = .04 * 30 - 3 = 1.2 - 3 = -1.8. COMMON ERROR: y = .02(30)^2 - 2(30) + 6 =-36 would be the rate of depth change INSTRUCTOR COMMENT: This is the depth, not the rate of depth change. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qmodeling project 3 problem a single quarter-cup of sand makes a cube 1.5 inches on a side. How many quarter-cups would be required to make a cube with twice the scale, 3 inches on a side? Explain how you know this. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a ¼ cup of sand makes a cube 1.5 in ^3, then we will need eight cubes total to make a cube that measures 3 inches ^3. Using 8 cubes, this would be 8 quarter cups. We know this by thinking in three dimensions with four cubes on the bottom layer and four cubes on the top layer. This would make a block that measures 3 inches on each side. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If y is ¼ cup and x is 1.5, we would have y = ax^3 as: 1 = a(1.5)^3 1 = a(3.375) a = 1 / 3.375 a = .296 Now if x = 3, Y = .296(3)^3 Y = 8 This matches my previous answer. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** The proportionality would be y = a x^3, with y = 1 (representing one quarter-cup) when x = 1.5. So we have 1 = a * 1.5^3, so that a = 1 / 1.5^3 = .296 approx. So the model is y = .2963 x^3. Therefore if x = 3 we have y = .296 * 3^3 = 7.992, which is the same as 8 except for roundoff error. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat would be the side measurement of a cube designed to hold 30 quarter-cups of sand? What equation did you solve to get this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For this we can use the equation: 30 = .2963 x^3 30/.2963 = x^3 101.2487 = x^3 3root of 101.2487 = x 4.66 = x So each side of the cube would be 4.66. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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..** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the best fit function was y = .002x^3 and they used ½ cup instead of the ¼ cup, we would make y = 2 and change the function to: Y = .004x^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** In this case, since it takes two quarter-cups to make a half-cup, the person would need twice as many quarter-cups to get the same volume y. He would have obtained half as many half-cups as the actual number of quarter-cups. To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 4. number of swings vs. length data. Which function fits best? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used the 7ft string method with 550 paracord and a full water bottle for the pendulum. For lengths 2, 3, 4, 5, 6, and 7 I had results of 39, 30, 26, 23, 22, 20 respectively. From averaging the swings per minute I came up with : A = 4.489 And the equation: Y = 4.489x^-.5 fits the model the best. Trying -.3, -.4, -.6, -.7 into my equation did not work out as well as the -.5 for p. This is expected since I used -.5 to come up with the 4.489. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** For each function you can substitute the x and y data for each point, then solve for the constant a. If the values of a are not relatively constant, then the function is not constant. For example, you can try the function y = a x^-2. You can plug in every (x, y) pair and solve of the resulting equations for a. 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. If you know the shapes of the basic graphs, you can compare them with a graph of the data and get a pretty good indication of which functions to try. 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. So you could try these functions, in any reasonable order: You might start with the y = a x^-.3 function. The values of the parameter a will probably vary quite a bit from data point to data point, and if so this function won't be a good candidate. You could then try the y = a x^-.4 function. The values of the parameter a will likely still vary significantly, but probably not as much as for the y = a x^-.3 function. You could then try the y = a x^-.5 function. The values of the parameter a will likely still vary a bit, but probably not by a whole lot. If you have good data this function will probably be your best candidate. You could then try the y = a x^-.6 function. The values of the parameter a will likely vary more than they did for the y = a x^-.5 function, confirming that the y = a x^-.5 function is the best candidate so far. The y = a x^-.7 function will probably result in quite a bit of variation in a, and is not likely to be the best function. STUDENT COMMENT: this concept is a little fuzzy to me. Im not quite sure what you mean when you say that ax^-.5 results in a nearly constant value. INSTRUCTOR RESPONSE: If the proportionality is y = a x^-2, then if we solve for a we get a = y x^2. If you evaluate a = y * x^2 for each of your data points, then if y = a x^-2 is a good model you should get about the same value of a for each point. For example if x values 2, 4, 7, 10 give you respective y values .5, .3, .2, .1, then your a values would be a = .5 * 2^2 = 2 a = .3 * 4^2 = 4.8 a = .2 * 7^2 = 9.8 a = .1 * 10^2 = 10 These values appear to be increasing. So the data don't appear to be consistent with the form y = a x^-2. Another proportionality might yield relatively constant values for a. If so, then your data would be consistent with that proportionality. For example y = a x^-1, or y = a x^-.5 might give you good results. It is also possible that the data aren't consistent with any power-function proportionality. Check this out with the data for this problem, and see what you find. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 7. time per swing model. For your data what expression represents the number of swings per minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y = 52.85x^-.5 represents the number of swings per minute best. I did this by plugging in and averaging the times across my plot. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** The model that best fits the data is a x^-.5, and with accurate data we find that a is close to 55. The model is pretty close to # per minute frequency = 55 x^-.5. As a specific example let's say we obtained counts of 53, 40, 33 and 26 cycles in a minute at lengths of 1, 2, 3 and 4 feet, then using y = a x^-.5 gives you a = y * x^.5. Evaluating a for y = 53 and x = 1 gives us a = 53 * 1^.5 = 53; for y = 40 and x = 2 we would get a = 40 * 2^.5 = 56; for y = 34 and x = 3 we get a = 33 * 3^.5 = 55; for y = 26 and x = 4 we get a = 26 * 4^.5 = 52. Since our value of a are reasonably constant the y = a x^.5 model works pretty well, with a value of a around 54. The value of a for accurate data turns out to be about 55.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf the time per swing in seconds is y, then what expression represents the number of swings per minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If y = swings per second (SPS) Swings per minute (SPM) SPM = 60/SPS confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### I used some made up acronyms but could have used frequency and y in my answer. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qIf 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Here I would divide ax^.5 into 60 to get swings per minute. Y = 60/ (ax^-.5) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1. Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute. Simplifying this gives f = (60 / a) * x^(.-5). 60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements. 60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### It turns out that I could have simplified this a little further. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 8. model of time per swing what are the pendulum lengths that would result in periods of .1 second and 100 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation: Y = 1.118x^.5 For a period of .1 seconds, .1 = 1.118x^.5 .008 = x For a period of .1 seconds the length of our pendulum would be .008 feet long, An almost impossible feat by the way. For a period of 100 seconds, 100 = 1.118x^.5 8000.5 = x Thus, for a period of 100 seconds, the length of out pendulum would be roughly 8000 feet long. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 9. length ratio x2 / x1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): For x1 and x2 as x2/x1: f(x2/x1) = 1.118(x2/x1)^-.5 ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat expressions, in terms of x1 and x2, represent the frequencies (i.e., number of swings per minute) of the two pendulums? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is simply: f = 52.85(x1)^-.5 for the first length f = 52.85(x2)^-.5 for the second length confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** The solution is to be in terms of x1 and x2. If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)). Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5. If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat expression, in terms of x1 and x2, represents the ratio of the frequencies of the two pendulums? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The ratio would be x2/x1 as: f(x1/x2) = [a(x1)/a(x2)]^.5 f(x1/x2) = (x1/x2)^.5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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 (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem Challenge Problem for Calculus-Bound Students: how much would the frequency change between lengths of 2.4 and 2.6 feet YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For 2.4 feet: Y = 52.85(2.4)^-.5 Y = 34.11 For 2.6 feet: Y = 52.85(2.6)^-.5 Y = 32.78 32.78 - 34.11 = -1.34 cycles/min between the 2.4 and 2.6 foot intervals. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION: Note that we are using frequency in cycles / minute. I worked to get the frequency at 2.4 and 2.6 y = 55.6583(2.4^-.5) = 35.9273 and y = 55.6583(2.6^-.5)= 34.5178. subtracted to get -1.40949 difference between 2.4 and 2.6. This, along with the change in length of .2, gives average rate -1.409 cycles/min / (.2 ft) = -7.045 (cycles/min)/ft , based on the behavior between 2.4 ft and 2.6 ft. This average rate would predict a change of -7.045 (cycles/min)/ft * 1 ft = -7/045 cycles/min for the 1-foot increase between 2 ft and 3 ft. The change obtained by evaluating the model at 2 ft and 3 ft was -7.2221 cycles/min. The answers are different because the equation is not linear and the difference between 2.4 and 2.6 does not take into account the change in the rate of frequency change between 2 and 2.4 and 2.6 and 3 for 4.4 and 4.6 y = 55.6583(4.4^-.5) y = 55.6583(4.6^-.5) y = 26.5341 y = 25.6508 Dividing difference in y by change in x we get -2.9165 cycles/min / ft, compared to the actual change -2.938 obtained from the model. The answers between 4-5 and 2-3 are different because the equation is not linear and the frequency is changing at all points. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### My answer is not the full solution as worked out in the project because the majority of the problem was not asked in the question. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q query problem 1.2.24 5th; 1.2.19 4th formula for exponential function through left (1,6) and (2,18) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P = p0(a)^t 18 = p0(a)^2 6 = p0(a)^1 We can divide these two to get: 3 = (a)^1 3 = a Then plug 3 in for a in the second equation: 6 = p0(3)^1 2 = p0 Thus: P = 2(3)^t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** An exponential function has one of several forms, including y = A * b^x and y = A * e^(kx). Using y = A * b^t and substituting the t and y coordinates of the two points gives us 6 = A * b^1 18 = A * b^2. Dividing the second equation by the first we get 3 = b^(2-1) or b = 3. Substituting this into the first equation we get 6 = A * 3^1 so A = 2. Thus the model is y = 2 * 3^t . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!