#$&* course Mth 163 3/21/13 around 12 p.m. 016. `query 16
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Given Solution: `aSTUDENT RESPONSE: my example was (4^2)(5^3) did not equal 20^6 INSTRUCTOR COMMENT ** more generally a^n * b^n = a^(n+m), which usually does not equal (ab) ^ (n * m) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhy is the follow erroneous: a^(-n) = - a^n YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a^(-n) = 1 / a^n is positive when a is positive whereas -a^n is negative when a is positive Basically -12^3 does not equal 12^-3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: 2^-3 is not equal to -2^3 INSTRUCTOR COMMENT: ** A more general counterexample: a^(-n) = 1 / a^n is positive when a is positive whereas -a^n is negative when a is positive ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhy is the following erroneous: a^n + a^m = a^(n+m) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have the example (6^3) * (6^4) = (6^7) since (6*6*6) * (6*6*6*6) = 5*5*5*5*5*5*5 = 5^7. But (6^3) + (6^7) = 6*6*6 + 6*6*6*6*6= 6*6*6 (1 + 6*6) = 6^3( 1 + 6^2), not 6^7.** confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: (5^3)+(5^4) is not equal to 5^7 INSTRUCTOR COMMENT: (5^3) * (5^4) = (5^7) since (5*5*5) * (5*5*5*5) = 5*5*5*5*5*5*5 = 5^7. However (5^3) + (5^7) = 5*5*5 + 5*5*5*5*5 = 5*5*5( 1 + 5*5) = 5^3( 1 + 5^2), not 5^7.** STUDENT QUESTION: Why do you write it out as 5*5*5(1+5*5) = 5^3(1+5^2) and how did you get that from the original problem? INSTRUCTOR RESPONSE: If you factor 5*5*5 out of 5*5*5 + 5*5*5*5*5 you get 5*5*5 ( 1 + 5*5). If you aren't sure of why, multiply that product out: 5*5*5 ( 1 + 5*5) = 5*5*5*1 + 5*5*5 * 5*5 = 5*5*5 + 5*5*5*5*5. The factorization used here can be generalized to prove that a^3 + a^4 is not equal to a^(3 + 4), and more generally that a^b + a^c is cannot be identified with a^(b + c). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: Why is the following erroneous: a^0 = 0 4^0 is not equal to 0 INSTRUCTOR COMMENT: ** a^(-n) * a^n = 0 but neither a^(-n) nor a^n need equal zero ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Why is the following erroneous: a^n * a^m = a^(n*m). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a^n * a^m = a^(n+m), not a^(n*m). Basically we can take the example (12^7)(12^2) and see that it is not equal to 12^14. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: (4^7)(4^2) is not equal to 4^14 INSTRUCTOR COMMENT: Right. Generally a^n * a^m = a^(n+m), not a^(n*m). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qproblem 2. Graph and describe Give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 1200 (2^(.12 t) ) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (0,1200),(1,1304) negative x-axis ratio=2^.12 or 163/150 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE (0,1200),(1,1304) negative x-axis ratio=163/150 INSTRUCTOR COMMENT: the precise ratio is 2^.12, which is probably pretty close to 163/150 STUDENT QUESTION: Does it matter which form you write this ratio in? INSTRUCTOR RESPONSE: If you're looking for an exact result then the fraction would be the most useful form. If, as in most applications, you're dealing with approximate numbers in the first place, then the approximation is the more desirable form. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 400 ( 1.07 ) ^ t YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (0,400),(1,428) Neg. x-axis ratio = 1.07 or 107/100 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE (0,400),(1,428) Neg. x-axis 1.07 or 107/100 is ratio INSTRUCTOR COMMENT: that ratio is correct and is of course equal to 1.07 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 250 ( 1 - .12 ) ^ t YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Basic points = (0,250),(1,220) Positive x-axis = the horizontal asymptote The ratio of y values at the basic point is 220 / 250 = .88 = 1-.12 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE The basic points are (0,250),(1,220) The positive x-axis is the horizontal asymptote The ratio of y values at the basic points is 220 / 250 = .88. INSTRUCTOR COMMENT: Note also that the ratio .88 is equal to 1-.12; .88 is the growth factor and .12 is the growth rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = .04 ( .8 ) ^ t YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Basic points = (0,.04),(1,.032) and the asymptote is the positive x-axis. The ratio is .32 / .4 = .8. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE (0,.04),(1,.032) are the basic points and the asymptote is the positive x-axis. The ratio is .32 / .4 = .8. The pattern is that the ratio is equal to b, where b is the base for the form y = A b ^ x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q problem 3. y = f(x) = 5 (1.27^x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat is the ratio between the y values at x = 0 and at x = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhat is the ratio between the y values at to x = 3.4 and x = 4.4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 1.27 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 1.27. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qVerify that the ratio of y values is again the same for your own points where x differs by 1 unit. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Points were 4.5 and 5.5 y ratio = 1.27 again confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT RESPONSE: My points were 4.5 and 5.5, and the y ratio was again 1.27 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhat is the ratio of y values when x values are separated by two units? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X values are separated by 2 units so the ratio = 1.27^(x+2) / 1.27^x = 1.27^(x+2 - x) = 1.27^2 = 1.61 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If x values are separated by 2 units then the ratio is 1.27^(x+2) / 1.27^x = 1.27^(x+2 - x) = 1.27^2 = 1.61 approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q problem 4. Ratio of y values at x = x1 and x = x1+1 What does your result tell you about how the ratio depends on the x value x1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If y = A b^x then the value at x1 is A b^x1 and value at x1 + 1 is A b ^(x1 + 1). So the ratio of these values is A b^(x1+1) / A b^x1 = b^(x1 + 1 - x1) = b^1 = b. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If y = A b^x then the value at x1 is A b^x1 and the value at x1 + 1 is A b ^(x1 + 1). The ratio of these values is A b^(x1+1) / A b^x1 = b^(x1 + 1 - x1) = b^1 = b. The ratio should have been b, where b is the base in the form y = A b^x. This is the same as in all previous examples, which shows that there is no dependence on x1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qproblem 5. y = 3 (2 ^ (.3 x) ). What is the ratio of the two basic-point y values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Basic points are x = 0 and x = 1 y values a= 3(2^.(3*0) ) = 3 and 3(2^(.3 *1) ) = 3 * 2^.3 = 3.69 approx. The ratio of these values is 3.69 / 3 = 1.23. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The basic points are the x = 0 and x = 1 points. The corresponding y values are 3(2^.(3*0) ) = 3 and 3(2^(.3 *1) ) = 3 * 2^.3 = 3.69 approx. The ratio of these values is 3.69 / 3 = 1.23. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhat is the y = A b^x form of this function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have the equation 3 (2 ^ (.3 x) ) = 3 (2^.3)x = 3 * 1.23^x This changed into the y form = A b^x where A = 3 and b = 1.23. So we have y = 3 * 1.23^x confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 3 (2 ^ (.3 x) ) = 3 (2^.3)x = 3 * 1.23^x, approx. This is in the form y = A b^x for A = 3 and b = 1.23. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhat does the value of 2 ^ .3 have to do with this situation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It is the b value in the y form that we just saw = A b^x confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** CORRECT STUDENT RESPONSE: this is the b value in the form y = A b^x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qproblem 6 P(n+1) = (1+r) P(n), with r = .1 and P(0) = $1000. What are P(1), P(2), ..., P(5)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we start with n = 0 we get P(0 + 1) = (1 + .1) * P(0), or P(1) = 1.1 * P(0). Since we know that P(0) = $1000 we have P(1) = 1.1 * $1000 = $1100. If n = 1 we get P(1 + 1) = (1 + .1) * P(1), or P(2) = 1.1 * P(1). Since we know P(1) = $1000 we have P(2) = 1.1 * $1100 = $1210. If n = 2 we get P(2 + 1) = (1 + .1) * P(2), or P(3) = 1.1 * P(2). Since we know P(2) = $1000 we have P(3) = 1.1 * $1210 = $1331. If n = 3 we get P(3 + 1) = (1 + .1) * P(3), or P(4) = 1.1 * P(3). Since we know P(3) = $1000 we have P(4) = 1.1 * $1331 = $1464/1. If n = 4 we get P(4 + 1) = (1 + .1) * P(4), or P(5) = 1.1 * P(4). Since we know P(4) = $1000 we have P(5) = 1.1 * $1464.1 = $1610.51. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If n = 0 we get P(0 + 1) = (1 + .1) * P(0), or P(1) = 1.1 * P(0). Since P(0) = $1000 we have P(1) = 1.1 * $1000 = $1100. If n = 1 we get P(1 + 1) = (1 + .1) * P(1), or P(2) = 1.1 * P(1). Since P(1) = $1000 we have P(2) = 1.1 * $1100 = $1210. If n = 2 we get P(2 + 1) = (1 + .1) * P(2), or P(3) = 1.1 * P(2). Since P(2) = $1000 we have P(3) = 1.1 * $1210 = $1331. If n = 3 we get P(3 + 1) = (1 + .1) * P(3), or P(4) = 1.1 * P(3). Since P(3) = $1000 we have P(4) = 1.1 * $1331 = $1464/1. If n = 4 we get P(4 + 1) = (1 + .1) * P(4), or P(5) = 1.1 * P(4). Since P(4) = $1000 we have P(5) = 1.1 * $1464.1 = $1610.51. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qproblem 8. Q(n+1) = .85 Q(n), Q(0) = 400. What are Q(n) for n = 1, 2, 3 and 4 ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For n = 0 Q(0 + 1) = .85 Q(0) therefore Q(1) = .85 * 400 = 340. For n = 1 Q(1 + 1) = .85 Q(1) therefore Q(2) = .85 * 340 = 289. For n = 2 Q(2 + 1) = .85 Q(2) therefore Q(3) = .85 * 400 = 245.65. For n = 3 Q(3 + 1) = .85 Q(3) therefore Q(4) = .85 * 400 = 208.803. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** For n = 0 we have Q(0 + 1) = .85 Q(0) so Q(1) = .85 * 400 = 340. For n = 1 we have Q(1 + 1) = .85 Q(1) so Q(2) = .85 * 340 = 289. For n = 2 we have Q(2 + 1) = .85 Q(2) so Q(3) = .85 * 400 = 245.65. For n = 3 we have Q(3 + 1) = .85 Q(3) so Q(4) = .85 * 400 = 208.803. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhat is the growth rate for this equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Growth rate = .85 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The growth factor is .85, which is 1 + growth rate. It follows that the growth rate is .85 - 1 = -.15 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qproblem 9. interest rate 12%, initial principle $2000. What is your difference equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The growth rate is 12% = .12 The growth factor is 1 + .12 Difference equation = P(n+1)=(1+.12)P(n), P(0)=2000 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The growth rate is 12% = .12 The growth factor is therefore 1 + .12 and the difference equation is P(n+1)=(1+.12)P(n), P(0)=2000. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qHow did you use your difference equation to find the principle after 1, 2, 3 and 4 years and what did you get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used the equation P(0+1)=(1+.12)2000 So we get: P1=2240 P2=2508.8 P3=2809.856 P4=3147.03872 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT RESPONSE P(0+1)=(1+.12)2000 and so on up to P(4) was found. P1=2240 P2=2508.8 P3=2809.856 P4=3147.03872 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qproblem 11. Texcess(t) = 50 (.97 ^ t). What is your estimate of the time required to fall to 1/8 of the original value? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At t = 0 Texcess(0) = 50 * .97^0 = 50. 1/8 of this value = 1/8 * 50 = 6.25. We then plug this in finding 50 * .97^t = 6.25. Then you divide both sides by 50 and get .97^t = 6.25 / 50 which is then 97^t = .125. Then you can use trial and error to find t: Try t = 10: .97^10 = .74 which is too high Try t = 100: .97^100 = .04 which is too low. So try a number between 10 and 100 but closer to 100 since that was closer to the number we wanted Try 65: .97^65 = .138 which is still too high so maybe try something a little more like 68 Try 68: .97^68 = .126 which gives us the right number closest to it confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The original value takes place at t = 0 and is Texcess(0) = 50 * .97^0 = 50. 1/8 of the original value is therefore 1/8 * 50 = 6.25. You have to find t such that 50 * .97^t = 6.25. Dividing both sides by 50 you get .97^t = 6.25 / 50 or .97^t = .125. Use trial and error to find t: Try t = 10: .97^10 = .74 approx. That's too high. Try t = 100: .97^100 = .04 approx. That's too low. So try a number between 10 and 100, probably closer to 100. Try 70: .97^70 = .118. Lucky guess. That's close to .125 but a little low. {Try 65: .97^65 = .138. Too high. Try a number between 65 and 70, closer to 70 but not too much closer. Try 68: .97^68 = .126. That's good to the nearest whole number. The process could be continued and refined to get more accurate values of t. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhat are your ratios of temperature excess to average rate, and are they nearly constant? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The ratios are 39.9 / (-1.19), 23.8 / (-.708), 14.2 / (-.422), 8.44 / (-.251 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If the function is 50 (.97^t) then at t = 0, 17, 34, 51, 68 we have function values temp excesses: 50, 29.79130219, 17.75043372, 10.57617070, 6.301557949 and average rates of change ave rates: -1.18756929, -0.7082863803, -0.4220154719, -0.2514478090, -0.1498191533. On the corresponding trapezoidal graph of temperature excess vs. clock time we have four trapezoids, and their slopes correspond to the average rates of change. The 'altitudes' of the trapezoids correspond to the temperature excesses. Each trapezoid has a single slope, but two 'altitudes', corresponding to the fact that each interval has two temperature excesses but only a single average rate. To calculate the desired ratio for an interval, we need a single value of the temperature excess to compare with the single rate. Rather than using either of the two temp excesses or 'altitudes', it's more appropriate to use their average. The four trapezoids have 'average altitudes' 39.89565109, 23.77086796, 14.16330221, 8.438864327, each corresponding to the average of the initial and final 'tempearture excesses' on the associated interval. Ratio of temp excess to ave rate, using 'average altitudes' as temp excess, are therefore 39.9 / (-1.19), 23.8 / (-.708), 14.2 / (-.422), 8.44 / (-.251). These quantities vary slightly but all are close to the same value around 33. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qWhat are your estimates of the times required to fall to half of the three values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the temperature falls to 50/2 = 25 at t = 22.75. The temperature falls to 25/2 = 12.5 at t = 45.51 The temperature falls to 12.5/2 = 6.25 at t = 68.26. The time interval = 22.75 so the half-life is constant The function = Texcess(t) = 50 (.97 ^ t) We plug in different values to get t values like 25/2 = 12.5 maybe 25/3 = 8.3333 We keep trying until t = 45.51 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT RESPONSE: The temperature falls to 50/2 = 25 at t = 22.75. The temperature falls to 25/2 = 12.5 at t = 45.51 The temperature falls to 12.5/2 = 6.25 at t = 68.26. The time interval required for each subsequent fall is very close to 22.75, demonstrating that the half-life is constant. ** STUDENT QUESTION I dont understand how to get the t values. INSTRUCTOR RESPONSE You evaluate the function Texcess(t) = 50 (.97 ^ t) at various values of t until you find the result you're looking for. For example the temperature falls to 25/2 = 12.5 at some value of t. You can plug in t = 10 but the result will be greater than 12.5. You could plug in higher values of t until you find a result that's lower than 12.5. Then you can keep trying numbers in between until you find a t value that gets you reasonably close to T = 12.5. That occurs around t = 45.51. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qGive the original and the simplified equation to determine the time required for Texcess to fall to half its original value. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The equation = 25 = 50 * .97^t. To simplify this equation we have to divide both sides by 50 to get .97^t = 1/2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Texcess has an original value at t = 0, which gives us Texcess(0) = 50 * .97^0 = 50. Half its original value is therefore 25. So our equation is 25 = 50 * .97^t. This equation is simplified by dividing both sides by 50 to get .97^t = 1/2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qproblem 12. Texcess(t) = 50(.97^t), room temperature {{ 25 if you used Celsius and 75 if you used Farenheit in your observations What function Temp(t) gives temperature as a function of time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We use 75 for room temperature and we know that room temperature + temperature excess will gives us the function so we get: Temp(t)=50(.97^t)+75 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Using 75 for room temperature and realizing that temperature is room temperature + temperature excess we obtain the function Temp(t)=50(.97^t)+75.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qIdentify the values of A, b and c in the generalized form y = A b^x + c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A = 50, b = .97 and c = 75 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Since the function is Temp(t) = 50(.97^t)+75 , the y = A b^x + c has y = Temp(t), A = 50, b = .97 and c = 75. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qproblem 14. Antiobiotic removal, 40 mg/hour when there are 200 milligrams present At what rate would antibiotic be removed when there are 70 milligrams present? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rate of removal is directly proportional to the quantity present so we have y = k x y = the rate of removal x = the amount present. We know that y = 40 when x = 200 and plugging these values in we get 40 = k * 200 so that k = 40/200 = .2. Therefore y = .2 x. And if x = 70 we get y = .2 * 70 = 14 So we know that when there are 70 mg present the rate of removal is 14 mg/hr. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If the rate of removal is directly proportional the quantity present then we have y = k x where y is the rate of removal and x the amount present. Since y = 40 when x = 200 we have 40 = k * 200 so that k = 40/200 = .2. Thus y = .2 x. If x = 70 then we have y = .2 * 70 = 14. When there are 70 mg present the rate of removal is 14 mg/hr. ** Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Explain why valid or not valid. if invalid give valid form for both sides: (n^p)^q = n^(p+q) 2^(.3 * x) = (2^.3) * x 2^(.3 + x) = 2^.3 * 2^X a^(b^c) = (a^b)^c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (n^p)^q = n^(p+q) is incorrect because if you substitute numbers in for n, p, & q you will see that you do not arrive at the same number on both sides. You would need to multiply p & q together to get the same number instead of adding them 2^(.3 * x) = (2^.3) * x this is incorrect because of the parentheses. If you use PEMDAS as a guide and do parentheses first you will arrive at different numbers on both sides