course mth 163 —¡ÑÙâÆX³—¥Í½ùê“œ‰v|«æ…“^yõassignment #016
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20:32:42 behavior and source of exponential functions problem 1, perversions of laws of exponents Why is the following erroneous: a^n * b^m = (ab) ^ (n*m)
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RESPONSE --> the exponents should be added and not multiplied confidence assessment: 3
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20:32:55 STUDENT 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) **
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RESPONSE --> ok self critique assessment: 3
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20:33:35 Why is the follow erroneous: a^(-n) = - a^n
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RESPONSE --> 2^-2 is (1/4) and not -2^2 confidence assessment: 3
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20:33:46 STUDENT 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 **
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RESPONSE --> ok self critique assessment: 3
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20:36:21 Why is the following erroneous: a^n + a^m = a^(n+m)
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RESPONSE --> 2^1 + 2^2 = 6 and not 2^3 confidence assessment: 3
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20:36:27 STUDENT 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.**
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RESPONSE --> ok self critique assessment: 3
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20:37:16 STUDENT RESPONSE: Why is the following erroneous: a^0 = 0
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RESPONSE --> 2^0 = 1 not zero self critique assessment: 3
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20:37:29 4^0 is not equal to 0 INSTRUCTOR COMMENT: ** a^(-n) * a^n = 0 but neither a^(-n) nor a^n need equal zero **
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RESPONSE --> ok self critique assessment: 3
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20:39:26 Why is the following erroneous: a^n * a^m = a^(n*m).
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RESPONSE --> 2^2 * 2^3 = 32 and not 2^6 confidence assessment: 3
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20:39:32 STUDENT 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).
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RESPONSE --> ok self critique assessment: 3
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20:41:21 problem 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) )
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RESPONSE --> didn't understand 2-5 on this assignment because I never understood the ratio. I believe another example of this would help. confidence assessment: 0
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20:43:04 STUDENT 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
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RESPONSE --> do not understand where this ratio comes from self critique assessment: 2
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20:48:53 problem 6 P(n+1) = (1+r) P(n), with r = .1 and P(0) = $1000. What are P(1), P(2), ..., P(5)?
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RESPONSE --> P(1)= $1100, P(2)= $1210, P(3)= $1331, P(4)=$1464.10, P(5)=$1610.51 confidence assessment: 3
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20:49:20 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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20:50:19 problem 8. Q(n+1) = .85 Q(n), Q(0) = 400. What are Q(n) for n = 1, 2, 3 and 4 /
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RESPONSE --> 1= 340, 2= 289, 3=245.65, 4=208.8025 confidence assessment: 3
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20:50:34 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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20:52:29 What is the growth rate for this equation?
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RESPONSE --> .85 confidence assessment: 1
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20:53:58 ** The growth factor is .85, which is 1 + growth rate. It follows that the growth rate is .85 - 1 = .15 **
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RESPONSE --> Understand that the rate is subtracted one because it is decreasing and if increasing it would have been plus one self critique assessment: 2
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20:55:21 problem 9. interest rate 12%, initial principle $2000. What is your difference equation?
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RESPONSE --> P(n+1) = (1+.12) P(n) confidence assessment: 3
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20:55:32 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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20:58:37 How did you use your difference equation to find the principle after 1, 2, 3 and 4 years and what did you get?
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RESPONSE --> plugged in the 2000 for the orginial and then got the next P(1)= 2240 and then use that to get P(2)= 2508.80 and use that P(3)= 2809.86 use that P(4)= 3147.04 confidence assessment: 3
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20:58:53 ** 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 **
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RESPONSE --> ok self critique assessment: 3
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21:04:26 problem 11. Texcess(t) = 50 (.97 ^ t). What is your estimate of the time required to fall to 1/8 of the original value?
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RESPONSE --> 41.6 confidence assessment: 2
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21:06:24 ** 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. **
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RESPONSE --> 6.25 in to y and solve out for t self critique assessment: 2
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21:07:46 What are your ratios of temperature excess to average rate, and are they nearly constant?
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RESPONSE --> -.44, -.55592, -.85208, -1.28 and they are a constant confidence assessment: 2
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21:09:03 ** If the function is 50 (.97^t) then at t = 0, 17, 34, 51, 68 we have function values 50, 29.79130219, 17.75043372, 10.57617070, 6.301557949 and average rates of change -1.18756929, -0.7082863803, -0.4220154719, -0.2514478090, -0.1498191533. Trapezoids have 'average altitudes' 39.89565109, 23.77086796, 14.16330221, 8.438864327. 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. **
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RESPONSE --> copied down a wrong answer and understand the answer given here self critique assessment: 2
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21:09:32 What are your estimates of the times required to fall to half of the three values?
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RESPONSE --> 23.349, 34, and 41.6 confidence assessment: 2
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21:10:56 ** 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. **
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RESPONSE --> understand 50/2 and then half of that on down self critique assessment: 2
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21:11:29 Give the original and the simplified equation to determine the time required for Texcess to fall to half its original value.
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RESPONSE --> confidence assessment: 0
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21:12:10 ** 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. **
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RESPONSE --> understand to plug in half of 50 which is 25to the y and then simplify by dividing both by 50 self critique assessment: 2
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21:14:59 problem 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?
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RESPONSE --> temp(t)=50(.97^25) confidence assessment: 2
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21:16:24 ** Using 75 for room temperature and realizing that temperature is room temperature + temperature excess we obtain the function Temp(t)=50(.97^t)+75.**
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RESPONSE --> add the temp and not place it in the t because 75 is room temp which is excess which would be added self critique assessment: 2
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21:17:16 Identify the values of A, b and c in the generalized form y = A b^x + c.
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RESPONSE --> A= 50, b=.97 and c= 75 confidence assessment: 3
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21:17:23 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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21:19:59 problem 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?
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RESPONSE --> 14mg/hr confidence assessment: 3
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21:20:07 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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21:20:58 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> how equations can still be formed in simple ways to solve things that seem to be a guess and check before self critique assessment: 2
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