course Mth 163 Ӈn~FԚЫ}assignment #016
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10:34:07 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 --> Because the variables for a and b will be set at a different power for and then multiplied. Example: 2^3 * 3^2 = 72 while (2*3)^(3*2) = 6^6 = 6*6*6*6*6*6 = 46656 confidence assessment: 3
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10:34:23 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 --> self critique assessment: 3
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10:35:05 Why is the follow erroneous: a^(-n) = - a^n
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RESPONSE --> because when the number is to the negative power you get a fraction. confidence assessment: 3
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10:36:38 Why is the following erroneous: a^n + a^m = a^(n+m)
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RESPONSE --> different powers gives a totally different anwer that could be low compared to a higher power that gives you a totally different answer. confidence assessment: 3
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10:36:54 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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10:37:09 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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10:37:28 Why is the following erroneous: a^n * a^m = a^(n*m).
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RESPONSE --> Same as the first question. confidence assessment: 3
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10:37:38 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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10:38:42 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 --> ok self critique assessment: 3
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10:40:23 give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 400 ( 1.07 ) ^ t
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RESPONSE --> (1, 428) and (2, 183184) confidence assessment: 3
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10:41:17 STUDENT 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.
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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. ok self critique assessment: 3
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10:41:56 STUDENT 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.
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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 assessment: 3
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10:43:41 problem 3. y = f(x) = 5 (1.27^x).
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RESPONSE --> y = 5(1.27^x) (1, 6.35) (2, 8.0645) confidence assessment: 3
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10:44:06 ** f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 **
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RESPONSE --> f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 self critique assessment: 3
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10:44:39 ** f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 1.27. **
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RESPONSE --> f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 1.27 self critique assessment: 3
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10:44:53 ** STUDENT RESPONSE: My points were 4.5 and 5.5, and the y ratio was again 1.27 **
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RESPONSE --> ok self critique assessment: 3
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10:45:12 ** 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. **
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RESPONSE --> 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 self critique assessment: 3
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10:45:38 ** 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. **
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RESPONSE --> 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 assessment: 3
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10:46:00 ** 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. **
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RESPONSE --> 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 assessment: 3
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10:46:20 ** 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. **
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RESPONSE --> What is the y = A b^x form of this function? ** 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 assessment: 3
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10:46:39 ** CORRECT STUDENT RESPONSE: this is the b value in the form y = A b^x. **
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RESPONSE --> this is the b value in the form y = A b^x. self critique assessment: 3
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10:46:59 ** 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 --> 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 assessment: 3
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10:47:20 ** 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 --> 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 assessment: 3
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10:47:36 ** 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 --> The growth factor is .85, which is 1 + growth rate. It follows that the growth rate is .85 - 1 = .15 self critique assessment: 3
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10:47:55 ** 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 --> 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 assessment: 3
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10:48:12 ** 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 --> 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 assessment: 3
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10:48:31 ** 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 --> 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 assessment: 3
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10:48:56 ** 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 --> 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. self critique assessment: 3
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10:49:23 ** 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 --> ok self critique assessment: 3
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10:49:43 ** 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 --> 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 assessment: 3
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10:50:02 ** 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 --> 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 assessment: 3
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10:50:19 ** 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 --> 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 assessment: 3
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10:50:59 ** 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 --> 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 self critique assessment: 3