course Mth 271 I could not get the Derive worksheet done because the ti web site discontinued the download for it. I went to the learning lab and could not find it in there, either. I find this first assignment easy, other than the few mistakes that I made because of carelessness, I hope that it continues to be this easy. {́zoX߷ѱassignment #001
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11:45:46 `qNote that there are four questions in this assignment. `q001. If your stocks are worth $5000 in mid-March, $5300 in mid-July and $5500 in mid-December, during which period, March-July or July-December was your money growing faster?
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RESPONSE --> The change between March-July is $5000 - $5300= -$300 and the change in July-December is $5300 - $5500= -$200. So the money is growing faster in the period of March-July. confidence assessment: 2
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11:48:15 The first period (4 months) was shorter than the second (5 months), and the value changed by more during the shorter first period (increase of $300) than during the longer second perios (increase of $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.
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RESPONSE --> I forgot to count the difference in months. This could have cause an error in a different problem but I have gotten this problem right. self critique assessment: 2
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11:52:48 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> The change in the first period was $300 and the change in the second period was $200. So 300/200 = 1.5 dollars/month. confidence assessment: 2
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11:57:22 From mid-March thru mid-July is 4 months. A change of $300 in four months gives an average rate of change of $300 / (4 months) = $75/month. From mid-July through mid-December is 5 months, during which the value changes by $200, giving an average rate of $200 / (5 months) = $40 / month. Thus the rate was greater during the first period than during the second.
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RESPONSE --> I misunderstood the question. I do understand that for each period a rate must be given rather than combining the two periods. So the change for the first period $300 must be divided by the change in onths which is 4 and the change for the second period $200 must be divided by the change in months which is 5. $300 / 4 months = $75 / month for the first period. $200 / 5 months = $40 / month for the second period. self critique assessment: 2
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12:11:22 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?
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RESPONSE --> The change in t = 10 sec and t = 40 sec with y = 80 cm and y = 40 cm is: ( 40 - 80 ) / ( 40 - 10 ) = -40 / 30 = -1.33 cm/sec The change in t = 40 sec and t = 90 sec with y = 40 cm and y = 20 cm is: ( 40 - 20 ) / ( 40 - 90 ) = 20 / -50 = -0.4 cm/sec So the depth is changing more quickly between t = 40 sec and t = 90 sec. confidence assessment: 2
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12:15:00 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.
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RESPONSE --> I did my calculations right but I put that the depth is changing more quickly in the second time interval because I got confused with the negatives. I am used to thinking that -.4 is greater than -1.33. But in this problem the negative doesn't work the same way because you are subtracting cm/sec and the more that is subtracted in one sec. is the quicker the depth is changing. self critique assessment: 2
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12:18:24 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?
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RESPONSE --> The two questions are both dealing with the rate of change in a certain amount of time. The mathematical reasoning is the same in both cases because the same formula is being used. You are calculating the change in something verses the change in time for both problems. A simpler way of saying this is that the rate of change is being calculated for both problems only different units of measure are being used. confidence assessment: 2
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12:19:54 In each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt.
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RESPONSE --> self critique assessment: 3
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fƨϵݞz assignment #001 001. `Query 1 Applied Calculus I 09-07-2007
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15:13:16 Section 0.1.26 solve x/2-x/3>5
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RESPONSE --> First you have to solve for x: ( x/2 ) * 2 - x/3 > 5 * 2
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15:17:10 It's easiest to avoid denominators where possible. So the preferred first step is to multiply both sides of the original equation by the common denominator 6: 6(x/2) - 6(x/3) = 6 * 5, which gives you3x - 2x = 6 * 5 which gives you x > 6 * 5 which simplifies to x > 30. The interval associated with this solution is 30 < x < infinity, or (30, infinity). To graph you would make an arrow starting at x = 30 and pointing to the right, indicating by an open circle that x = 30 isn't included.**
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RESPONSE --> I did not remember that you could get rid of the denominator by multiplying both sides of the equation by the common denominator. When I worked this problem on paper I had some trouble remembering what it was that I ws suppose to do to keep the x and get rid of the denominator. I understand now that I could have done this to get the right answer. self critique assessment: 2
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15:22:18 Section 0.1.28 solve 2x^2+1<9x-3
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RESPONSE --> First the equation needs to be put together to form a quadratic: 2x^2 - 9x + 4 < 0 then it must be factored properly: ( 2x - 1) ( x - 4) from this information we know that there are two zeros one at -1/2 and the other at 4 this gives -1/2 < x < 4 when you test the points on the outside of these two numbers the equation does not fit and when you test the numbers between these two points the equation is right so the graph would have a point at -1/2 and 4 and the x could fall anywhere in between these two points. confidence assessment: 2
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15:25:42 The given inequality rearranges to give the quadratic 2x^2 - 9 x + 4 < 0. The left-hand side has zeros at x = .5 and x = 4, as we see by factoring [ we get (2x-1)(x-4) = 0 which is true if 2x-1 = 0 or x - 4 = 0, i.e., x = 1/2 or x = 4. ] The left-hand side is a continuous function of x (in fact a quadratic function with a parabola for a graph), and can change sign only by passing thru 0. So on each interval x < 1/2, 1/2 < x < 4, 4 < x the function must have the same sign. Testing an arbitrary point in each interval tells us that only on the middle interval is the function negative, so only on this interval is the inequality true. Note that we can also reason this out from the fact that large negative or positive x the left-hand side is greater than the right because of the higher power. Both intervals contain large positive and large negative x, so the inequality isn't true on either of these intervals. In any case the correct interval is 1/2 < x < 4. ALTERNATE BUT EQUIVALENT EXPLANATION: The way to solve this is to rearrange the equation to get 2 x^2 - 9 x + 4< 0. The expression 2 x^2 - 9 x + 4 is equal to 0 when x = 1/2 or x = 4. These zeros can be found either by factoring the expression to get ( 2x - 1) ( x - 4), which is zero when x = 1/2 or 4, or by substituting into the quadratic formula. You should be able to factor basic quadratics or use the quadratic formula when factoring fails. The function can only be zero at x = 1/2 or x = 4, so the function can only change from positive to negative or negative to positive at these x values. This fact partitions the x axis into the intervals (-infinity, 1/2), (1/2, 4) and (4, infinity). Over each of these intervals the quadratic expression can't change its sign. If x = 0 the quadratic expression 2 x^2 - 9 x + 4 is equal to 4. Therefore the expression is positive on the interval (-infinity, 1/2). The expression changes sign at x = 1/2 and is therefore negative on the interval (1/2, 4). It changes sign again at 4 so is positive on the interval (4, infinity). The solution to the equation is therefore the interval (1/2, 4), or in inequality form 1/2 < x < 4. **
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RESPONSE --> I forgot to change my sign on ( 2x - 1 ) when I simplified it into a point. I had the right idea going for the problem but I don't understand how I didn't catch this when I entered my test points and got the right answer. self critique assessment: 2
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