course Mth 173 09-28-2008`??????w??????
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21:10:03 `qNote that there are 11 questions in this assignment. `q001. Suppose that you expect for the next 6 years to receive a steady stream of extra income, at the rate of $20,000 per year. This income is expected to 'flow in' at a constant rate, day by day. Suppose furthermore that you expect that your money will grow at a constant annual rate of 8%. Assuming you do not use any of the money or interest, the question we want to answer is how much you would therefore expect to have at the end of 6 years. This problem is complicated by the fact that the money that goes in, say, today will earn interest over a longer period than the money you earn tomorrow. In fact, the money you receive in one minute will earn interest for a different length of time than the money you receive in a different minute. To begin to answer the problem you could start out by saying that while some of the money will earn interest for 6 years, some will go in at the very end of the 6-year period and will therefore not earn any interest at all, so the average time period for earning interest will be 3 years. You could then calculate the interest on the full amount for 3 years, and get a fairly good idea of the final amount. As we will see, you can't get a precise estimate this way, but it gives you a reasonable starting estimate. What would be the simple interest on the total 6-year amount of money flow for 3 years, and what would be the final amount?
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RESPONSE --> y = k (x+2)(x-1)(x-5). At x = 0 the function has value 2, 2 = k (0+2)(0-1)(0-5) => 2 = 10 k. k = .2 the function is y = .2 ( x+2)(x-1)(x-5) confidence assessment: 3
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21:12:14 The total money flow would be $20,000 per year for 6 years, or $120,000. 8% simple interest for 3 years would be 3 * 8% = 24%, and 24% of $120,000 is $28,800, so your first estimate would be $28,800. You therefore expect to end up with something not too far from $148,800.
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RESPONSE --> I was looking at the wrong question...so The total money flow would be $20,000 per year for 6 years, or $120,000. 8% simple interest for 3 years would be 3 * 8% = 24%, and 24% of $120,000 is $28,800, so your first estimate would be $28,800. $148,800. I get it self critique assessment: 3
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21:14:00 `q002. Money usually doesn't earn simple interest. Interest is almost always compounded. If the interest on $120,000 was compounded annually at 8% for 3 years, what would be the final amount?
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RESPONSE --> $120,000*1.08 = 139,968 confidence assessment: 3
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21:14:09 There is a more efficient way to calculate this, but we'll see that shortly. 8% of $120,000 is $9600, so after 1 year we will have $120,000 + $9600 = $129,600. Note that this result could have been obtained by multiplying $120,000 by 1.08. We will use this strategy for the next two years. After another year we will have $129,600 * 1.08 = $139,968. If you wish you can take 8% of $129,600 and add it to $129,600; you will still get $139,968. After a third year you will have $139,968 * 1.08 = $151,165. Note that this beats the $148,800 you calculated with simple interest. This is because each year the interest is applied to a greater amount than before; previously the interest was just applied to the starting amount.
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RESPONSE --> ok self critique assessment: 3
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21:14:51 `q003. Money can earn interest that is compounded annually, quarterly, weekly, daily, hourly, or whatever. For given interest rate the maximum interest will be obtained if the money is compounded continuously. When initial principle P0 is compounded continuously at rate r for t interest periods (e.g., at rate 8% = .08 for t years), the amount at the end of the time is P = P0 * e^(r t). If the $120,000 was compounded continuously at 8% annual interest for 3 years, what would be the amount at the end of that period?
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RESPONSE --> We would have P0 = $120,000 r = .08 t = 3. P = $120,000 * e^(.08 * 3) = $120,000 * e^(.24) = $152,549. confidence assessment: 3
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21:14:56 We would have P0 = $120,000, r = .08 and t = 3. So the amount would be P = $120,000 * e^(.08 * 3) = $120,000 * e^(.24) = $152,549. This beats the annual compounding by over $1,000.
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RESPONSE --> ok self critique assessment: 3
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21:16:32 `q004. This keeps getting better and better. We end up with more and more money. Now let's see how much money we would really end up with, if the money started compounding continuously as soon as it 'flowed' into the account. As a first step, we note that 6 years is 72 months. About how much would the money flowing into the account during the 6th month be worth at the end of the 72 months?
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RESPONSE --> monthly flow is $20,000 / 12 = $1,666.67. The $1,666,67 that flows in during the 6th month will have about 72 - 5.5 = 66.5 months to grow. 66.5 months is 66.5/12 = 5.54 years $1,666,67 would grow at the continuous rate of 8 % for 5.54 years. $1,666,67 * e^(.08 * 5.54) = $2596.14. confidence assessment: 3
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21:16:38 Flowing into the account at a constant rate of $20,000 per year, we see that the monthly flow is $20,000 / 12 = $1,666.67. The $1,666,67 that flows in during the 6th month will have about 72 - 5.5 = 66.5 months to grow. [ Note that we use the 5.5 instead of 6 because the midpoint of the 6th month is 5.5 months after the beginning of the 72-month period (the first month starts at month 0 and ends up at month 1, so its midway point is month .5; the nth month starts at the end of month n-1 and ends at the end of month n so its midway point is (n - 1) + .5). ] 66.5 months is 66.5/12 = 5.54 years, approximately. So the $1,666,67 would grow at the continuous rate of 8 percent for 5.54 years. This would result in a principle of $1,666,67 * e^(.08 * 5.54) = $2596.14.
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RESPONSE --> ok self critique assessment: 3
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21:18:02 `q005. How much will the money flowing into the account during the 66th month be worth at the end of the 72 months?
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RESPONSE --> $1,666,67 would flow into the account during this month. The midpoint of this month is 65.5 months after the start 6.5 / 12 = .54 years to grow. the money that flows in during the 66th month will grow to $1,666,67 * e^(.08 * .54) = $1737 confidence assessment: 3
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21:18:08 Again $1,666,67 would flow into the account during this month. The midpoint of this month is 65.5 months after the start of the 72-month period, so the money will have an average of about 6.5 months, or 6.5 / 12 = .54 years, approx., to grow. Thus the money that flows in during the 66th month will grow to $1,666,67 * e^(.08 * .54) = $1737, approx..
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RESPONSE --> ok self critique assessment: 3
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21:18:30 `q006. We could do a month-by-month calculation and add up all the results to get a pretty accurate approximation of the amount at the end of the 72 months. We would end up with $152,511, not quite as much as our last estimate for the entire 72 months. Of course this approximation still isn't completely accurate, because the money that comes in that the beginning of a month earns interest for longer than the money that comes in at the end of the month. We could chop up the 72 months into over 2000 days and calculate the value of the money that comes in each day. But even that wouldn't be completely accurate. We now develop a model that will be completely accurate. We first imagine a short time span near some point in the 72 months, and calculate the value of the income flow during that time span. We are going to use symbols because our calculation asked to apply to any time span at anytime during the 72 months. We will use t for the time since beginning of the 6-year period, to the short time span we are considering; in previous examples t was the midpoint of the specified month: t = 5.5 months in the first calculation and 66.5 months in the second. We will use `dt (remember that this stands for the symbolic expression delta-t) for the duration of the time interval; in the previous examples `dt was 1 month. For a time interval of length `dt, how much money flows? Assume `dt is in years. If this money flow occurs at t years from the beginning of the 6-year period, then how long as it have to grow?
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RESPONSE --> The amount in 1 year is $20,000 so the amount in `dt years is $20,000 * `dt. Money that flows in near the time t years from the beginning of the 6-year period will have (6-t) years to grow confidence assessment: 3
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21:18:43 The amount in 1 year is $20,000 so the amount in `dt years is $20,000 * `dt. Money that flows in near the time t years from the beginning of the 6-year period will have (6-t) years to grow.
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RESPONSE --> right self critique assessment: 3
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21:19:35 `q007. How much will the $20,000 `dt amount received at t years from the start grow to in the remaining (6-t) years?
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RESPONSE --> Amount P0 grows to P0 e^( r t) in t years so in 6-t years amount $20,000 `dt = $20,000 `dt e^(.08 (6-t) ) confidence assessment: 3
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21:19:44 Amount P0 will grow to P0 e^( r t) in t years, so in 6-t years amount $20,000 `dt will grow to $20,000 `dt e^(.08 (6-t) ).
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RESPONSE --> ok self critique assessment: 3
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21:20:22 `q008. So for a contribution at t years from the beginning of the 6-year period, at what rate would we say that money is being contributed to the final t = 6 year value? Let y stand for the final value of the money, and `dy for the contribution from the interval `dt.
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RESPONSE --> `dy = $20,000 `dt * e^(.08 (6-t) ), the rate is dy / `dt = [ $20,000 `dt * e^(.08 (6-t) ) ] / `dt = $20,000 e^(.08 ( 6-t) confidence assessment: 3
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21:20:33 The money that flows in during time interval `dt will grow to `dy = $20,000 `dt * e^(.08 (6-t) ), so the rate is about `dy / `dt = [ $20,000 `dt * e^(.08 (6-t) ) ] / `dt = $20,000 e^(.08 ( 6-t) ). As `dt shrinks to 0, this expression approaches the exact rate y ' = dy / dt at which money is being contributed to the final value.
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RESPONSE --> As `dt shrinks to 0, this expression approaches the exact rate y ' = dy / dt at which money is being contributed to the final value self critique assessment: 3
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21:21:26 `q009. As we just saw the rate at which money accumulates is y' = dy / dt = 20,000 e^(.08(6-t)). How do we calculate the total quantity accumulated given the rate function and the time interval over which it accumulates?
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RESPONSE --> The rate function is the derivative of the quantity function. An antiderivative of a rate-of-change function is a change-in-quantity function. change in a quantity from the rate function can be found by finding the change in this antiderivative confidence assessment: 3
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21:21:33 The rate function is the derivative of the quantity function. An antiderivative of a rate-of-change function is a change-in-quantity function. We therefore calculate the total change in a quantity from the rate function by finding the change in this antiderivative.
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RESPONSE --> ok self critique assessment: 3
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21:21:53 `q010. If the money that flows into the account t years along the 6-year cycle adds to the final value of the money at the previously mentioned rate y' = dy / dt = 20,000 e^(.08(6-t)), then what function describes the final value of the money accumulated through time t?
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RESPONSE --> The function must be an antiderivative of the rate function y ' = $20,000 e^(.08 ( 6-t) confidence assessment: 3
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21:22:02 The function must be an antiderivative of the rate function y ' = $20,000 e^(.08 ( 6-t) ). The y ' function can be expressed as y ' = $20,000 e^(.08 * 6) e^(-.08 t) = $20,000 * 1.616 e^(-.08 t) = $32,321 e^(-.08 t). The general antiderivative of e^(-.08 t) is -1/.08 e^(-.08 t) + c = -12.5 e^(-.08 t) + c. The general antiderivative of y ' = $32,321 e^(-.08 t) is y = $32,321 * -12.5 e^(-.08 t) = -404,160 e^(-.08 t) + c. The value of the constant c could be determine if we knew, for example, the amount of money present at t = 0. However as we will see in the next step, the value of c doesn't affect the change in the quantity we are considering in this problem.
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RESPONSE --> ok I understand self critique assessment: 3
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21:22:33 `q011. How much money accumulates during the 6 years?
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RESPONSE --> y represents the money accumulated through t years. We subtract the money assumulated at 0 years from the money accumulated at 6 years to get the amount accumulated during the time interval from t = 0 to t = 6 years. The result is -404,160 e^(-.08 * 6 ) + c - [ -404,160 e^(-.08 * 0 ) + c ] = $154,072 confidence assessment: 3
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21:22:39 y represents the money accumulated through t years. We subtract the money assumulated at 0 years from the money accumulated at 6 years to get the amount accumulated during the time interval from t = 0 to t = 6 years. The result is -404,160 e^(-.08 * 6 ) + c - [ -404,160 e^(-.08 * 0 ) + c ] = $154,072.
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RESPONSE --> ok self critique assessment: 3
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21:22:47 **** redo the last few steps -- we need to be able to make the step from f(t) `dt to the integral, though the steps taken here can illuminate the Fundamental Theorem
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RESPONSE --> ok self critique assessment: 3
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?v???J??H???[{ assignment #010 ??K???????? Calculus I 09-28-2008
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21:25:45 query problem 1.6.9 (was 1.10.16)cubic polynomial representing graph. What cubic polynomial did you use to represent the graph?
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RESPONSE --> zeros at x = -2, 1, 5, so y = k (x+2)(x-1)(x-5). At x = 0 the function has value 2, so 2 = k (0+2)(0-1)(0-5), or 2 = 10 k. y = .2 ( x+2)(x-1)(x-5).
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21:25:51 *&*& The given function has zeros at x = -2, 1, 5, so y = k (x+2)(x-1)(x-5). At x = 0 the function has value 2, so 2 = k (0+2)(0-1)(0-5), or 2 = 10 k. Thus k = .2 and the function is y = .2 ( x+2)(x-1)(x-5). **
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RESPONSE --> ok
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21:26:15 Query problem 1.6.15 (formerly 1.4.19) s = .01 w^.25 h^.75what is the surface area of a 65 kg person 160 cm tall?
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RESPONSE --> sa = .01 *65^.25 *160^.75 = 1.277m^2
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21:26:19 ** Substituting we get s = .01 *65^.25 *160^.75 = 1.277meters^2 **
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RESPONSE --> ok
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21:27:11 What is the weight of a person 180 cm tall whose surface area is 1.5 m^2?
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RESPONSE --> 1.5 = .01 w^.25*180^.75 . 1.5/180^.75 .01w^.25. 3.05237 = w^.25 Taking the fourth power of both sides: w = 3.052^4 = 86.806
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21:27:16 ** Substituting the values we get 1.5 = .01 w^.25*180^.75 . Dividing both sides by 180: 1.5/180^.75 .01w^.25. Dividing both sides by .01: 3.05237 = w^.25 Taking the fourth power of both sides: w = 3.052^4 = 86.806 **
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RESPONSE --> ok
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21:28:07 For 70 kg persons what is h as a function of s?
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RESPONSE --> s = .01 *70^.25 h^.75 s = .02893 h^.75 s/.02893 = h^.75. take 1.333 power of both sides (s/.02893)^1.333 = h h = 110.7s^1.333 = 110.7 s^(4/3
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21:28:12 ** Substituting 70 for the weight we get s = .01 *70^.25 h^.75 s = .02893 h^.75 s/.02893 = h^.75. Taking the 1/.75 = 1.333... power of both sides: (s/.02893)^1.333 = h h = 110.7s^1.333... = 110.7 s^(4/3). **
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RESPONSE --> ok
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