#$&* course Mth163 july 9,2013 15:00 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** If the y vs. x table yields a decent linear graph, you'll just find the slope and vertical-axis intercept and you're pretty much done. Your equation in this case will be y = m x + b. If your table doesn't yield a linear graph, then you'll check the graphs of log(y) vs. x, y vs. log(x) and log(y) vs. log(x) to see if any of these tables yields a linear graph. You won't have set the system up the same way it's presented here, but your setup should be consistent with the following: Start with your table of y vs. x values. Your table might have headings x y with values listed in the columns below. Add columns for log(x) and log(y), so your headings now read x y log(x) log(y) You can easily fill in the values of log(x) and log(y). Use can use the base-10 log, the natural log, or the log to any other base. Usually a special base won't be helpful to you so you will keep it simple by using base-10 or natural log. Now you can just copy the columns to create the necessary tables. The first table might be log(y) vs. log(x), in which case it would have headings log(x) log(y). The second might be log(y) vs. x, with headings x log(y). The third might be y vs. log(x), with headings log(x) y. Having filled in the columns to create each table, you will want to see which table, if any, gives you a linear graph. Hopefully you are able to do some estimates using mental arithmetic to quickly eliminate one or more possibilities. You then graph any of the possibilities you haven't eliminated. (If your mental arithmetic isn't reliable you'll probably end up having to graph all possibilities). Any graph which can be reasonably well fit with a straight line will give you a linearization. Finding the slope m and the vertical-axis intercept b you will obtain an equation for the line. If the log(y) vs. x graph is linear, then the corresponding equation is log(y) = m x + b. To express y as a function of x, this equation would then be solved for y. If the log(y) vs. log(x) graph is linear, then the corresponding equation is log(y) = m log(x) + b. To express y as a function of x, this equation would then be solved for y. If the y vs. log(x) graph is linear, then the corresponding equation is y = m log(x) + b. If none of these graphs is linear, then you would have to get creative and consider other possibilities. However, in this course it will be sufficient if you simply apply this method correctly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `q Linearizing Data and Curve Fitting Problem 1. table for y = 2 t^2 vs. t, for t = 0 to 3, linearize. Give your table and the table for sqrt(y) vs. t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& y = 2 t^2 vs. t T y 0 0 .5 .5 1 2 1.5 8 2 2.5 2.5 12.5 3 18 sqrt(y) vs. t t y 0 0 .5 .7 2 1.4 4.5 2.1 8 2.8 12.5 3.5 18 4.2 &&&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The table for y vs. t is t y 0 0 1 2 2 8 3 18 The table for sqrt(y) vs t, with sqrt(y) given to 2 significant figures, is t sqrt(y) 0 0 1 1.4 2 2.8 3 4.2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Is the first difference of the `sqrt(y) sequence constant and nonzero? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& The first difference is, however the rest are not constant The first difference of the y values on the sqrt(Y) is .707 same number as second I subtracted the first two values of the new table which were 0,.707 &&&& ********************************************* Question: `q The sqrt(y) sequence is 0, 1.4, 2.8, 4.2. The first-difference sequence is 1.4, 1.4, 1.4, which is constant and nonzero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& As you saw in my numbers earlier I did add .5’s so that may have deviated my data if I just took the values of the whole numbers I end up with the same sequence as you &&&& ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Give your values of m and b for the linear function that models your table. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& .35=m .7=b &&&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The points (t, sqrt(y) ) are (0,0), (1, 1.4), (2, 2.8), (3, 4.2). These points are fit by a straight line thru the origin with slope 1.4, so the equation of the line is sqrty) = 1.4 t + 0, or just sqrt(y) = 1.4 t. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I used 1.4(from the y value of the sqrt fn) =m(2)+b And 2.8=m8+b And then I just solved for m and b &&&
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Given Solution: `a** Squaring both sides of sqrt(y) = 1.4 t we get y = 1.96 t^2. The original function was y = 2 t^2. Our values for the sqrt(y) function were accurate to only 2 significant figures. To 2 significant figures 1.96 would round off to 2, so the two functions are identical to 2 significant figures. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ------------------------------------------------ Self-critique rating:
#$&* course Mth163 july 9,2013 15:00 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** If the y vs. x table yields a decent linear graph, you'll just find the slope and vertical-axis intercept and you're pretty much done. Your equation in this case will be y = m x + b. If your table doesn't yield a linear graph, then you'll check the graphs of log(y) vs. x, y vs. log(x) and log(y) vs. log(x) to see if any of these tables yields a linear graph. You won't have set the system up the same way it's presented here, but your setup should be consistent with the following: Start with your table of y vs. x values. Your table might have headings x y with values listed in the columns below. Add columns for log(x) and log(y), so your headings now read x y log(x) log(y) You can easily fill in the values of log(x) and log(y). Use can use the base-10 log, the natural log, or the log to any other base. Usually a special base won't be helpful to you so you will keep it simple by using base-10 or natural log. Now you can just copy the columns to create the necessary tables. The first table might be log(y) vs. log(x), in which case it would have headings log(x) log(y). The second might be log(y) vs. x, with headings x log(y). The third might be y vs. log(x), with headings log(x) y. Having filled in the columns to create each table, you will want to see which table, if any, gives you a linear graph. Hopefully you are able to do some estimates using mental arithmetic to quickly eliminate one or more possibilities. You then graph any of the possibilities you haven't eliminated. (If your mental arithmetic isn't reliable you'll probably end up having to graph all possibilities). Any graph which can be reasonably well fit with a straight line will give you a linearization. Finding the slope m and the vertical-axis intercept b you will obtain an equation for the line. If the log(y) vs. x graph is linear, then the corresponding equation is log(y) = m x + b. To express y as a function of x, this equation would then be solved for y. If the log(y) vs. log(x) graph is linear, then the corresponding equation is log(y) = m log(x) + b. To express y as a function of x, this equation would then be solved for y. If the y vs. log(x) graph is linear, then the corresponding equation is y = m log(x) + b. If none of these graphs is linear, then you would have to get creative and consider other possibilities. However, in this course it will be sufficient if you simply apply this method correctly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `q Linearizing Data and Curve Fitting Problem 1. table for y = 2 t^2 vs. t, for t = 0 to 3, linearize. Give your table and the table for sqrt(y) vs. t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& y = 2 t^2 vs. t T y 0 0 .5 .5 1 2 1.5 8 2 2.5 2.5 12.5 3 18 sqrt(y) vs. t t y 0 0 .5 .7 2 1.4 4.5 2.1 8 2.8 12.5 3.5 18 4.2 &&&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The table for y vs. t is t y 0 0 1 2 2 8 3 18 The table for sqrt(y) vs t, with sqrt(y) given to 2 significant figures, is t sqrt(y) 0 0 1 1.4 2 2.8 3 4.2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Is the first difference of the `sqrt(y) sequence constant and nonzero? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& The first difference is, however the rest are not constant The first difference of the y values on the sqrt(Y) is .707 same number as second I subtracted the first two values of the new table which were 0,.707 &&&& ********************************************* Question: `q The sqrt(y) sequence is 0, 1.4, 2.8, 4.2. The first-difference sequence is 1.4, 1.4, 1.4, which is constant and nonzero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& As you saw in my numbers earlier I did add .5’s so that may have deviated my data if I just took the values of the whole numbers I end up with the same sequence as you &&&& ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Give your values of m and b for the linear function that models your table. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& .35=m .7=b &&&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The points (t, sqrt(y) ) are (0,0), (1, 1.4), (2, 2.8), (3, 4.2). These points are fit by a straight line thru the origin with slope 1.4, so the equation of the line is sqrty) = 1.4 t + 0, or just sqrt(y) = 1.4 t. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I used 1.4(from the y value of the sqrt fn) =m(2)+b And 2.8=m8+b And then I just solved for m and b
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Given Solution: `a** Squaring both sides of sqrt(y) = 1.4 t we get y = 1.96 t^2. The original function was y = 2 t^2. Our values for the sqrt(y) function were accurate to only 2 significant figures. To 2 significant figures 1.96 would round off to 2, so the two functions are identical to 2 significant figures. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& To get the new log(y) values I put old x values on top / old y value Example : Log(1)/log(12)=0 When I did this my new t values were 7 21 63 189
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Given Solution: `a** A table for the function is t y = 7 ( 3^t) 0 7 1 21 2 63 3 189 The table for log(y) vs. t is t log(7 ( 3^t)) 0 0..85 1 1.32 2 1.80 3 2.28/ Sequence analysis on the log(7 * 3^t) values: sequence 0.85 1.32 1.80 2.28 1st diff .47 .48 .48 The first difference appears constant with value about .473. log(y) is a linear function of t with slope .473 and vertical intercept .85. We therefore have log(y) = .473 t + .85. Thus 10^(log y) = 10^(.473 t + .85) so that y = 10^(.473 t) * 10^(.85) or y = (10^.473)^t * (10^.85), which evaluating the power of 10 with calculator gives us y = 2.97^t * 7.08. To 2 significant figures this is the same as the original function y = 3 * 7^t. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I did not log(7 ( 3^t)) group it all together
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Given Solution: `a** For the simulated data the y values are .14, 5.66, 23.2, 52, 82.2 and 135.5. The square roots of these values are 0.374; 2.38, 4.82; 7.21; 9.34, 11.64. Plotting these square roots vs. t = 0, 1, 2, 3, 4, 5 we obtain a nearly straight-line graph. The best-fit linear function to sqrt(y) vs. x gives us sqrt(y) = 2.27•t + 0.27. Your function should be reasonably close to this but will probably not be identical. Squaring both sides we get y = 5.1529•t^2 + 1.2258•t + 0.0729. If the small term .0729 is neglected we get y = 5.15 t^2 + 1.23 t. Because of the 1.23 t term this isn't a particularly good approximation to y = 5 t^2.** BRIEF SUMMARY: If `sqrt(y) = 2.27 x + .05, then squaring both sides gives us y = (2.27 x + .05)^2, which when expanded gives you something fairly close to y = 5 t^2, but not all that close. STUDENT COMMENT ok i dont really understand where the 1.23 came from in the first place INSTRUCTOR RESPONSE: The 1.23 arises when you square the binomial. (a + b)^2 = a^2 + 2 a b + b^2. So (2.27 x + .05)^2 = 5.15 t^2 + 1.23 t + 0.073 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I used the first set of given y= 5t^2 sqrt(y) data sets &&& ------------------------------------------------ Self-critique rating: ********************************************* Question: `q problem 9. Assuming exponential follow the entire 7-step procedure for given data set Give your x and y data. Show you solution. Be sure to give the average deviation of your function from the given data? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& X 1 2 3 4 5 Sqrt(y) .65 .54 .46 .39 .32 .26 Average of the deviations were .1133
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Given Solution: `aFor (t, y) data set (0,.42), (1,.29),(2,.21),(3,.15),(4,.10),(5,.07) we get log(y) vs. t: t log(y) 0 -.375 1 -.538 2 -.678 3 -.824 4 -1 5 -1.15 A best fit to this data gives log(y) = - 0.155•x - 0.374. Solving we get 10^log(y) = 10^(- 0.155•t - 0.374) or y = 10^-.374 * (10^-.155)^t or y = .42 * .70^t. The columns below give t, y as in the original table, y calculated as y = .42 * .70^t and the difference between the predicted and original values of y: 0 0.42 0.42 0 1 0.29 0.294 -0.004 2 0.21 0.2058 0.0042 3 0.15 0.14406 0.00594 4 0.1 0.100842 -0.000842 5 0.07 0.0705894 -0.0005894 The deviations in the last column have an average value of -.00078. This indicates that the model is very good. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I should have taken the log instead of the sqrt &&& ------------------------------------------------ Self-critique rating: ********************************************* Question: `q problem 11. determine whether the log(y) vs. t or the log(y) vs. log(t) transformation works. Complete the problem and give the average discrepancy between the first function and your data. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& .x y log(x) log(y) 0.5 0.7 -0.301 -0.155 1 0 .97 -1 -0.0132 1.5 1.21 0.176 0.0828 2 1.43 0.301 0.155 2.5 1.56 0.398 0.193 Log(y) vs t x y log(x) log(y) 2 2.3 0.301 0.362 4 5 0.602 0.699 6 11.5 0.778 1.0607 8 25 0.903 1.398 Log y vs t
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Given Solution: `a** The first table gives us x y log(x) log(y) 0.5 0.7 -0.30103 -0.1549 1 0.97 0 -0.01323 1.5 1.21 0.176091 0.082785 2 1.43 0.30103 0.155336 2.5 1.56 0.39794 0.193125 log(y) vs. x is not linear. log(y) vs. log(x) is linear with equation log(y) = 0.5074 log(x) - 0.0056. Applying the inverse transformation we get 10^log(y) =10^( 0.5074 log(x) - 0.0056) which we simplify to obtain y = 0.987•x^0.507. The second table gives us x y log(x) log(y) 2 2.3 0.30103 0.361728 4 5 0.60206 0.69897 6 11.5 0.778151 1.060698 8 25 0.90309 1.39794 log(y) vs. x is linear, log(y) vs. log(x) is not. From the linear graph we get log(y) = 0.1735x + 0.0122, which we solve for y: 10^log(y) = 10^(0.1735x + 0.0122) or y = 10^.0122 * 10^(0.1735•x) = 1.0285 * 1.491^x. ** STUDENT QUESTION: Im not sure how to come up with this 10^log(y) =10^( 0.5074 log(x) - 0.0056) which we simplify to obtain y = 0.987•x^0.507. INSTRUCTOR RESPONSE: I assume you understand why log(y) vs. x is linear, while log(y) vs. log(x) is not, and why log(y) = 0.1735x + 0.0122. So solve log(y) = q we would use the fact that log(y) and 10^y are inverse functions, so that 10^(log(y)) = y. The equation log(y) = q implies the equation 10^(log(y)) = 10^q, which becomes y = 10^q. We apply the same strategy to the present equation. log(y) = 0.1735x + 0.0122 becomes 10^(log(y)) = 10^(0.1735 x + 0.0122), which becomes y = 10^(0.1735 x + 0.0122). Applying the laws of exponents this becomes y = 10^(0.1735 x) * 10^0.0122. Since 10^(0.1735 x) = (10^0.1735) ^x = 1.491^x, and 10^0.0122 = 1.0285, we have y = 1.0285 * 1.491^x . 10^log(y) =10^( 0.5074 log(x) - 0.0056) becomes y = 0.987•x^0.507 by a very similar series of steps: 10^log(y) = y 10^(0.5074 log(x)) = (10^(log(x)) )^ 0.5074 = x ^ 0.5074 10^-0.0122 = 0.987 so the equation 10^log(y) =10^( 0.5074 log(x) - 0.0056) becomes 10^(log y) = 10^(.5074 log(x)) * 10^-0.0056, which in turn becomes y = 0.987•x^0.507. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& log(y) = q implies the equation 10^(log(y)) = 10^q, which becomes y = 10^q. does this mean our q is our t or x value??? &&&&
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Given Solution: `a** The table is x y = f(x) 0 0 0.5 0.25 1 1 1.5 2.25 2 4 Reversing columns we get the following partial table for the inverse function: x f ^ -1 (x) 0 0 0.25 0.5 1 1 2.25 1.5 4 2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q Describe your graph consisting of the smooth curves corresponding to both functions. How are the pairs of points positioned with respect to the y = x function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& on the first one my graph have slight dip or curve near the beginning and on the opposite it has a slight upward curve of hill near the beginning. They mirror each other. &&&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The curve of the original function is increasing at an increasing rate, the curve for the inverse function is increasing at a decreasing rate. The curves meet at (0, 0) and at (1, 1). The line connecting the pairs of points passes through the y = x line at a right angle, and the y = x line bisects each connecting line. So the two graphs are mirror images of one another with respect to the line y = x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& The curve of the original function is increasing at an increasing rate, the curve for the inverse function is increasing at decreasing rate. The curves meet at (0, 0) and at (1, 1). The line connecting the pairs of points passes through the y = x line at a right angle, and the y = x line bisects each connecting line. Yes, my graph does all of the above &&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q 8. If we reversed the columns of the 'complete' table of the squaring function from 0 to 12, precisely what table would we get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& Our reversed table would give us the table for the square root function y = sqrt(x) &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Our reversed table would give us the table for the square root function y = sqrt(x). The y = x^2 and y = sqrt(x) functions are inverse functions for x >= 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q 9. If we could construct the 'complete' table of the squaring function from 0 to infinity, listing all possible positive numbers in the x column, then why would we be certain that every possible positive number would appear exactly one time in the second column? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& Because you would eventually get the sqrt of the number you took the sqrt of from the first coulomb out of the sqrts &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The second column consists of all the squares. In order for a number to appear in the second column the square root of that number would have to appear in the first. Since every possible number appears in the first column, then no matter what number we select it will appear in the second column. So every possible positive number appears in the second column. If a number appears twice in the second column then its square root would appear twice in the first column. But no number can appear more than once in the first column. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q What number would appear in the second column next to the number 4.31 in the first column? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 18.58 &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The table is the squaring function so next to 4.31 in the first column, 4.31^2 = 18.5761 would appear in the second. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q What number would appear in the second column next to the number `sqrt(18) in the first column? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 4.24 &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The square of sqrt(18) is 18, so 18 would appear in the second column. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q What number would appear in the second column next to the number `pi in the first column? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& The square of pi or 9.8569…. &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** pi^2 would appear in the second column. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q What would we obtain if we reversed the columns of this table? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& sqrt(pi^2) &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aOur table would have the square of the second-column value in the first column, so the second column would be the square root of the first column. Our function would now be the square-root function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q What number would appear in the second column next to the number 4.31 in the first column of this table? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& Sqrt(4.31)=2.07 &&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** you would have sqrt(4.31) = 2.076 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q What number would appear in the second column next to the number `pi^2 in the first column of this table? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& sqrt(pi^2) So Pi &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The number in the second column would be pi, since the first-column value is the square of the second-column value. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q What number would appear in the second column next to the number -3 in the first column of this table? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& -3^2 9 &&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** -3 would not appear in the first column of the reversed table of the squaring function, since it wouldn't appear in the second column of that table. ** STUDENT COMMENT: (student gave the answer 1.73 i) oh wow that was really tricky INSTRUCTOR RESPONSE: sqrt(-3) = 1.73 i is a very good answer; if the domain and range of the function include the complex numbers, this would in fact be a 3-significant-figure approximation of the number corresponding to -3. Since we're dealing here with the real numbers, though, -3 never appears in the second column of the x^2 function, so it won't appear in the first column of the inverse function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I got confused with which column was which I should write it down So since it is imagery it would not pass zero so therefore it would not be on our table, since x>=0 &&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& X log(2)=log(18) Divide each side by log (2) . x=log(18)/log(2)= approx. 4.17 &&&
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Given Solution: `a** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base 2}(18) = log(18) / log(2). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q 2 ^ (4x) = 12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 4x log(2)=log(12) 4x=log(12)/log(2) x=[log(12)/log(2)]/4 . x=.896 &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as 4 x = log{base 2}(12) = log(12) / log(2). ** STUDENT QUESTION: Mr.Smith, I don’t understand how you have this laid out. Could give more of a step by step detail so that I might understand it INSTRUCTOR RESPONSE: Sure. Step-by-step: • b^x = a is expressed in logarithmic form as x = log{base b}(a) 2 ^ (4x) = 12 is of the form b^x = a, but with b = 2, x replace by 4x and a = 12. Thus the form x = log{base b}(a) becomes 4x = log{base 2}(12). log{base 2}(12) = log(12) / log(2). Thus 4x = log(12) / log(2). You weren't asked to solve this for x, but had you been asked the solution would be found by dividing both sides by 4, which gives us x = log(12) / (4 log(2) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q 5 * 2^x = 52 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& Log(5)*x log(2)=log(52)
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Given Solution: `a** You get 2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I should have combed my numerators I can combine them in fashion because they are both logs that were not a part of the base?? My new x would therefore be x=3.37851 right?? &&& ------------------------------------------------ Self-critique rating: ********************************************* Question: `q 2^(3x - 4) = 9. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 3x log(2)- 4 log(2)= log(9) 3x log(2)=log(9)+4 log(2) 3x=[ log(9)+4 log(2)]/log(2) x={[ log(9)+4 log(2)]/log(2)}/3 or x=[ log(9/3)+4 log(2/3)]/log(2) .x=2.39 x=-.755 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You get 3x - 4 = log 9 / log 2 so that 3x = log 9 / log 2 + 4 and x = ( log 9 / log 2 + 4 ) / 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& Put the plus sign in the wrong place &&& Question: `q 14. Solve each of the following equations: 2^(3x-5) + 4 = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 3x log(2) -5 log(2) + log(4)=0
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Given Solution: `a** You get log(-4)/log(2)=3x - 5. However log(-4) is not a real number so there is no solution. Note that 2^(3x-5) cannot be negative so the equation is impossible. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& How did the 4 become negative???
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Given Solution: `a** You get 2^(1/x) = 3 so that 1/x = log(3) / log(2) and x = log(2) / log(3) = .63 approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& How did you flip the very last part??? x = log(2) / log(3) &&&
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Given Solution: `a** 2^x * 2^(1/x) is the same as 2^(x + 1/x) so you get x + 1/x = log{base 2}(15). Multiplying both sides by x we get x^2 + 1 = log{base 2}(15). This is quadratic. We rearrange to get x^2 - log{base 2}(15) x + 1 = 0 then use quadratic formula with a=1, b=-log{base 2}(15) and c=1. Our solutions are x = 0.2753664762 OR x = 3.631524119. ** STUDENT COMMENT I don't think I would've made the correlation with a quadratic after working with exponential functions for so long.. I wasn't sure how to combine 1/x + x, either, which I'm guessing should be simple but it's eluding me at the moment. INSTRUCTOR RESPONSE You want to solve the equation, and the most efficient method is to multiply both sides by x. However to add 1/x + x you put both terms over the common denominator x. You do this by multiplying the second term, which has no denominator, by x / x. We get (1/x) + x * (x / x) = (1/x) + (x^2 / x) = (1 + x^2) / x. Again we wouldn't do that here, but that's how it would be done if we wanted to add the two terms. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):
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Given Solution: `a** You take the 1/4 power of both sides to get 2^x = 5^(1/4) so that x = log(5^(1/4) ) / log(2) = 1/4 log(5) / log(2). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: Show the tables you would get when attempting to linearize the data set x y 10 3 20 10 30 29 and indicate which is most likely linear. You may use the following approximations: ln(1) = 0, ln(20) = 3.0, ln(10) = 2.3, ln(29) = 3.4 (you haven't been given the value of ln(30) but it should be easy to make a sufficiently accurate commonsense estimate of its value, to within +-0.1 of its actual value, given the information provided). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 2.3=m10+b 3.0=m20+b Solve for m and b . m=.07 b=1.6 .y=.07t+1.6 X 10 20 30 Y 2.3 3.0 3.4
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