course Mth 173 ??}???`?J?s??????assignment #005
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07:18:32 `qNote that there are 9 questions in this assignment. `q001. We see that the water depth vs. clock time system likely behaves in a much more predictable detailed manner than the stock market. So we will focus for a while on this system. An accurate graph of the water depth vs. clock time will be a smooth curve. Does this curve suggest a constantly changing rate of depth change or a constant rate of depth change? What is in about the curve at a point that tells you the rate of depth change?
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RESPONSE --> Constant change in rate Slope of line shows rate confidence assessment: 2
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07:18:44 The steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change.
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RESPONSE --> self critique assessment: 3
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07:20:38 `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 40 and t = 90?
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RESPONSE --> t=10 y(t)=71 t=40 y(t)=26 t=90 y(t)=-9 confidence assessment: 2
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07:21:40 At t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm. At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm.
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RESPONSE --> oops, wrong middle parameter self critique assessment: 2
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07:24:14 `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals?
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RESPONSE --> (54-71)/(20-10)=-1.7 (-9-54)/(90-20)=-0.9 confidence assessment: 2
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07:25:55 From 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s. From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 40 sec and t = 90 sec, so the change in clock time is a9 0 sec - 40 sec = 50 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 50 sec = -1.26 cm/s.
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RESPONSE --> Did I miss a time 40? self critique assessment: 2
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07:34:42 `q004. What is the average rate at which the depth changes between t = 10 and t = 11, and what is the average rate at which the depth changes between t = 10 and t = 10.1?
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RESPONSE --> t=11 y(t)=69.21 71-69.21=1.79 m=-1.79 t=10.1 y(t)=70.8201 71-70.8201=0.1799 m=0.1799/-0.1=-1.799 confidence assessment: 2
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07:35:06 At t=10 the depth is y(10) = .01(10^2) - 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=11 the depth is y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21, representing a depth of 69.21 cm. The average rate of depth change between t=10 and t = 11 is therefore change in depth / change in clock time = ( 69.21 - 71) cm / [ (11 - 10) sec ] = -1.79 cm/s. At t=10.1 the depth is y(10.1) = .01(10.1^2) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201, representing a depth of 70.8201 cm. The average rate of depth change between t=10 and t = 10.1 is therefore change in depth / change in clock time = ( 70.8201 - 71) cm / [ (10.1 - 10) sec ] = -1.799 cm/s. We see that for the interval from t = 10 sec to t = 20 sec, then t = 10 s to t = 11 s, then from t = 10 s to t = 10.1 s the progression of average rates is -1.7 cm/s, -1.79 cm/s, -1.799 cm/s. It is important to note that rounding off could have hidden this progression. For example if the 70.8201 cm had been rounded off to 70.82 cm, the last result would have been -1.8 cm and the interpretation of the progression would change. When dealing with small differences it is important not around off too soon.
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RESPONSE --> ok self critique assessment: 3
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07:40:09 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
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RESPONSE --> using the tan function I come up with -3.078 confidence assessment: 0
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07:41:23 The progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8.
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RESPONSE --> I don't understand why my answer did not work but I understand this point. self critique assessment: 2
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07:51:17 `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times?
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RESPONSE --> y(t1)=.01(t1)^2-2(t1)+90 y((t1)+'dt)=.01((t1)+'dt)^2-2((t1)+'dt)+90 confidence assessment: 2
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07:51:36 At clock time t = t1 the depth is y(t1) = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + `dt the depth is y(t1 + `dt) = .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90.
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RESPONSE --> ok self critique assessment: 3
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08:00:24 `q007. What is the change in depth between these clock times?
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RESPONSE --> .01(t1)^2-2(t1)+90-(.01((t1)+'dt)^2-2((t1)+'dt)+90) confidence assessment: 2
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08:02:13 The change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2.
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RESPONSE --> Didn't simplify but follow process self critique assessment: 2
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08:14:38 `q008. What is the average rate at which depth changes between these clock time?
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RESPONSE --> .01('dt)^2+.02('dt)(t1)-2('dt) confidence assessment: 1
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08:17:50 The average rate is ave rate = change in depth / change in clock time = ( .02 t1 `dt + - 2 `dt + .01 `dt^2 ) / `dt = .02 t1 - 2 + .01 `dt. Note that as `dt shrinks to 0 this expression approaches .02 t1 - 2.
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RESPONSE --> Screwed up on a sign when I reduced Forgot to divide by run self critique assessment: 2
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08:19:08 `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results?
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RESPONSE --> The value of the equation is 1.8 and that was the original hypothesis as to what the slope was at t=10. confidence assessment: 3
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08:19:22 At t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10.
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RESPONSE --> ok self critique assessment: 3
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??}???`?J?s?????? assignment #005 005. 07-10-2007
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07:18:32 `qNote that there are 9 questions in this assignment. `q001. We see that the water depth vs. clock time system likely behaves in a much more predictable detailed manner than the stock market. So we will focus for a while on this system. An accurate graph of the water depth vs. clock time will be a smooth curve. Does this curve suggest a constantly changing rate of depth change or a constant rate of depth change? What is in about the curve at a point that tells you the rate of depth change?
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RESPONSE --> Constant change in rate Slope of line shows rate confidence assessment: 2
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07:18:44 The steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change.
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RESPONSE --> self critique assessment: 3
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07:20:38 `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 40 and t = 90?
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RESPONSE --> t=10 y(t)=71 t=40 y(t)=26 t=90 y(t)=-9 confidence assessment: 2
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07:21:40 At t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm. At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm.
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RESPONSE --> oops, wrong middle parameter self critique assessment: 2
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07:24:14 `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals?
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RESPONSE --> (54-71)/(20-10)=-1.7 (-9-54)/(90-20)=-0.9 confidence assessment: 2
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07:25:55 From 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s. From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 40 sec and t = 90 sec, so the change in clock time is a9 0 sec - 40 sec = 50 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 50 sec = -1.26 cm/s.
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RESPONSE --> Did I miss a time 40? self critique assessment: 2
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07:34:42 `q004. What is the average rate at which the depth changes between t = 10 and t = 11, and what is the average rate at which the depth changes between t = 10 and t = 10.1?
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RESPONSE --> t=11 y(t)=69.21 71-69.21=1.79 m=-1.79 t=10.1 y(t)=70.8201 71-70.8201=0.1799 m=0.1799/-0.1=-1.799 confidence assessment: 2
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07:35:06 At t=10 the depth is y(10) = .01(10^2) - 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=11 the depth is y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21, representing a depth of 69.21 cm. The average rate of depth change between t=10 and t = 11 is therefore change in depth / change in clock time = ( 69.21 - 71) cm / [ (11 - 10) sec ] = -1.79 cm/s. At t=10.1 the depth is y(10.1) = .01(10.1^2) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201, representing a depth of 70.8201 cm. The average rate of depth change between t=10 and t = 10.1 is therefore change in depth / change in clock time = ( 70.8201 - 71) cm / [ (10.1 - 10) sec ] = -1.799 cm/s. We see that for the interval from t = 10 sec to t = 20 sec, then t = 10 s to t = 11 s, then from t = 10 s to t = 10.1 s the progression of average rates is -1.7 cm/s, -1.79 cm/s, -1.799 cm/s. It is important to note that rounding off could have hidden this progression. For example if the 70.8201 cm had been rounded off to 70.82 cm, the last result would have been -1.8 cm and the interpretation of the progression would change. When dealing with small differences it is important not around off too soon.
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RESPONSE --> ok self critique assessment: 3
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07:40:09 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
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RESPONSE --> using the tan function I come up with -3.078 confidence assessment: 0
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07:41:23 The progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8.
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RESPONSE --> I don't understand why my answer did not work but I understand this point. self critique assessment: 2
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07:51:17 `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times?
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RESPONSE --> y(t1)=.01(t1)^2-2(t1)+90 y((t1)+'dt)=.01((t1)+'dt)^2-2((t1)+'dt)+90 confidence assessment: 2
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07:51:36 At clock time t = t1 the depth is y(t1) = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + `dt the depth is y(t1 + `dt) = .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90.
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RESPONSE --> ok self critique assessment: 3
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08:00:24 `q007. What is the change in depth between these clock times?
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RESPONSE --> .01(t1)^2-2(t1)+90-(.01((t1)+'dt)^2-2((t1)+'dt)+90) confidence assessment: 2
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08:02:13 The change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2.
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RESPONSE --> Didn't simplify but follow process self critique assessment: 2
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08:14:38 `q008. What is the average rate at which depth changes between these clock time?
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RESPONSE --> .01('dt)^2+.02('dt)(t1)-2('dt) confidence assessment: 1
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08:17:50 The average rate is ave rate = change in depth / change in clock time = ( .02 t1 `dt + - 2 `dt + .01 `dt^2 ) / `dt = .02 t1 - 2 + .01 `dt. Note that as `dt shrinks to 0 this expression approaches .02 t1 - 2.
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RESPONSE --> Screwed up on a sign when I reduced Forgot to divide by run self critique assessment: 2
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08:19:08 `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results?
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RESPONSE --> The value of the equation is 1.8 and that was the original hypothesis as to what the slope was at t=10. confidence assessment: 3
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08:19:22 At t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10.
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RESPONSE --> ok self critique assessment: 3
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Section 1.1 1. The population of the city is 35 million in 1962 (12=1950). 2. What would be the thickness of the egg with a concentration 200 ppm 3. a) a car is worth $6,000 after five years b) it should be decreasing c) y is age of car, x is value of car 4. (0,0) and (1,1) M=(1-0)/(1-0)=1 Y=X+b 0=0+b 0=b Y=x 6. (-2,1) and (2,3) M=(3-1)/(2-(-2))=2/4=1/2 Y=1/2 x + b 3=1/2*2+b b=2 Y=1/2 x + 2 10. a) V b) IV c) I d) VI e) II f) III 12. 1=(-1/5)2+b 7/5 = b Y=(1/5)x + 7/5 17. Domain 1-6 Range 1-5 22. S=h^2 24. N=1/l^2 25. a) II b) I c) III 27. 31. a) 9/5 b) F=(9/5)C+32 c) F=(9/5)20+32=68 d) x=(9/5)x+32 -(4/5)x=32 X=-40 32. a) The temperature of the object is 10 degrees C after 30 minutes b) Point A represents the temperature of the object when it is placed outside while point b represents the time at which it the temperature reaches zero. 35. a) q=b-(2/5)p 100=b-(2/5)550 B=320 Q=320-(2/5)p P=b-(5/2)q 550=b-(5/2)100 B=800 P=800-(5/2)q . Radioactive decay of plutonium Plutonium is very dangerous stuff. It's great for making nuclear bombs. Also, the amount required to almost certainly give you lung cancer, should you inhale it, is most appropriately measured in millionths (maybe billionths) of a gram. Another thing about plutonium: if you have a sample of the stuff, its radioactivity doesn't decrease all that rapidly. In fact, it takes something like 10,000 years before it loses half of its radioactivity. Suppose that plutonium loses 7% of its radioactivity in a millennium. This actually means that 7% of the plutonium will change into something else in a millennium. We will therefore have only 93% as much plutonium at the end of the millennium as at the beginning. Before reading further, suppose we have a sample of 1 gram of plutonium: determine the amount of plutonium left after 1 millennium, then after 2, 3, ..., 10 millennia. We can approach this problem as a growth rate problem. The only difference between this and compound interest is that the amount of stuff we have is decreasing rather than increasing. We take care of this by simply saying that the rate is negative. In the present example, where we lose 7% of our stuff every millennium, the rate is -7%, or -.07. The resulting growth factor is thus (1 + r) = (1 + -.07) = .93. Note that this growth factor is less than 1. If you multiply by a number less that 1, the value decreases, which is what happens to the activity of a plutonium sample. Thus, according to this model: After one millenium we have .93 (1 gram) = .93 grams of plutonium. After two millenia we have .93 ^2 (1 gram) = .865 grams of plutonium. After three millenia we have .93 ^3 (1 gram) = .804 grams of plutonium. After four millenia we have .93 ^4 (1 gram). After five millenia we have .93 ^5 (1 gram). After ten millenia we have .93 ^10 (1 gram) = .484 grams of plutonium. The quantity Q(t) of plutoniuim after t years is Q(t) = 1 gram (1 + r) ^ t = 1 gram (.93 ^ t). Exercise 15 15. If the amount of plutonium decreases by 7% per millennium, then how much of a 16-gram sample will remain after 10, 20, 30, 40 and 50 millenia? 16*(1-.07)^10=7.74 16*(1-.07)^20=3.75 16*(1-.07)^30=1.81 16*(1-.07)^40=0.88 16*(1-.07)^50=.42 Sketch a graph of the amount of plutonium vs. the number of millennia, for t = 0 to 50 millennia. 15.5. If we start with a quantity Q0 of plutonium, and if the amount of plutonium decreases by 7% per millennium, then what expression represents the quantity Q(t) of plutonium after t years (after t years, not after t millenia)? Qt=Q0*(1-.00007)^t Radioactive decay: initial quantity Q0, rate r If we start with quantity Q0 of a radioactive substance, and if the amount changes at rate r (per period), then the growth rate r is negative, the growth factor (1 + r) is less than 1, and the amount after t periods is Q(t) = Q0 (1 + r) ^ t. This expression is identical in form to the expression P(t) = P0 (1+r) ^ t for the principle in a compound interest problem. The only difference is that for radioactive decay, r is negative and the growth factor (1+r) is less than 1, both corresponding to a decrease over time in the amount of the substance. Half-life As implied above the half-life is the time required for a quantity to decrease to half its original amount. As another example, consider an antibiotic in the body. It is gradually filtered out by the kidneys, so that the amount in the bloodstream decreases with time. It turns out that the amount filtered out in a given time tends to be directly proportional to the amount present. Thus the amount in the bloodstream is well approximated by an exponential function of time. Suppose that an individual receives a dose of an antibiotic which results in an initial quantity of 200 milligrams in the bloodstream. Suppose also that 6% of the quantity initially in the bloodstream has been filtered out after 1 hour. The growth rate associated with this situation is -6%, so r = -.06. The growth factor is therefore (1 + r) = 1 + -.06 = .94, and the amount remaining after t hours is Q(t) = 200 mg (.94) ^ t. The half-life is the time `dt required for Q(t) to drop by half. Starting from t = 0, when Q(t) = 200 mg, the half-life will be the time `dt at which Q(t) reaches 100 mg. So we have Q(`dt) = 200 mg (.94) ^ `dt = 100 mg. The resulting equation 200 mg (.94) ^ `dt = 100 mg can be rearranged (dividing by 200 mg) to give .94 ^ `dt = 1/2. This equation will later be solved for `dt using logarithms. For now we can solve it by trial and error, using different values of `dt until we find one that makes the expression .94 ^ `dt sufficiently close to 1/2. We can use a graph of .94 ^ t vs. t to approximate the desired value of `dt. We see that .94 ^ t has initial value 0 when t = 0, and falls to .5, half of its initial value, in a time interval of a little over 11 units. We thus see that the half-life of this antibiotic is about 11 hours. Exercises 16-17 16. The quantity of a certain radioactive element decreases by 15% per day. The initial amount present is 30 grams. What function Q(t) gives the quantity Q as a function of time t? Q(t)=30*(1-.15)^t Sketch a graph of Q(t) vs. t for 0 <= t <= 10 days. Use this graph to estimate the half-life of the element. A Touch more then 4 yrs. Write the equation you would solve to find the half-life of this element. Simplify the equation as much as possible. ?* 30 = 30 * (1-.15)^?t ?= .85^?t 17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What function Q(t) gives the quantity Q of the antibiotic as a function of the time t since 10:00 a.m.? Q(t) = 550 * (1 - .11)^t How much antibiotic will be present at 3:00 p.m.? Q(5) = 550 * (1 - .11)^5 = 550 * .89^5 = 307 mg Sketch a graph of Q(t) vs. t for 0 <= t <= 10 hours. Use this graph to estimate the half-life of the element. Almost 8 yrs. Write the equation you would solve to find the half-life of this element. Simplify the equation as much as possible. ?.896?t Suggested Exercise: The value of the half-life of plutonium given here might or might not be very accurate. Find out what the actual half-life of plutonium really is, and determine how long it would take to reduce an initial quantity to 1% of its original activity. ?= .93^?t, or 9,551 years .01 = .93^?t or 63,457 years Challenge Exercise for Calculus-bound students: Suppose that the activity of a sample of plutonium is initially 5 decays per second. Using the model above, determine the activity after 5, 10, 15, and 20 millenia. Then find the approximate average number of decays per second from 0-5 millenia, 5-10 millenia, 10-15 millenia and 15-20 millenia. Use these averages to determine the total approximate number of decays during each of these time intervals, and the total for the entire 20 millenia. Speculate on how many decays there will be from the initial time to eternity. Q(5)= 5dps * (1-.07)^ 5 or 3.48dps Q(10) = 5dps * .93^10 or 2.42dps Q(15) = 5dps * .93^15 or 1.68dps Q(20) = 5dps * .93^20 or 1.17dps 0-5 (5+3.48)/2 = 4.24dps; total decays 6.78 * 1011 5-10 (3.48+2.42)/2=2.95dps; total decays 4.72 * 1011 10-15 (2.42+1.68)/2=2.02dps; total decays 3.23 * 1011 15-20 1.68+1.17)/2=1.43dps; total decays 2.29 * 1011 Total over 20 millenia 1.70 * 10 12 Over infinity, there would have to approach 100% of total # of sec/5 millenia=60*60*24*365.25(or so)*1000*5 = 1.6 * 1011 Asymptotes It should be clear that the quantity of a radioactive element decreases in such a way as to eventually approach 0. This is clear from the fact that after one half-life, 1/2 of the original quantity is present, while after two half-lives 1/2 of this quantity, or 1/4 of the original quantity, is present; after three half-lives 1/2 of this, or 1/8 of the original quantity, is present; and so on. After the fourth through tenth half-lives the quantities will be 1/16, 1/32, 1/64, 1/128, 1/256, 1/512, and 1/1024 of the original amount. From the form Q(t) = Q0 (1+r) ^ t, we can see that if r is negative 1+r is less than l, so that higher and higher powers of (1 + r) will get smaller and smaller. For example if r = -.03, so that 1+r = .97, the t = 1, 2 and 3 powers will be .97^1 = .97, .97^2 = .94 and .97^3 = .91. Every time we multiply by a number less than 1, our result decreases. Furthermore the result decreases toward 0: it doesn't level off at something like .4, or .1, or even .00001, but rather continues decreasing to a value as close to 0 as we might choose. This means that as t gets large, Q(t) will approach 0. This means that the graph will approach the x axis as an asymptote. Exercises 17-20 17. For the function Q(t) = Q0 (.9 ^ t), if t = 3, then Q(t) = Q0 (.9^3) = .729 Q0. Find a value of t for which Q(t) lies between .05 Q0 and .1 Q0. .075=.9^t T=24 Find values of t for which Q(t) lies within each of the following ranges: between .005 Q0 and .01 Q0 ,0075=.9^t T=46 between .0005 Q0 and .001 Q0. .00075=.9^t T=68 Sketch a reasonably accurate graph of Q(t) vs. t, with the range of t sufficient to allow Q(t) to fall below .01 Q0. In terms of this exercise explain why the positive x axis is a horizontal asymptote for this function. As we multiply each successive result by .9, we will decrease the value an cause it to approach zero 18. For each of the following functions find a value of t such that Q(t) lies between .05 Q0 and .1 Q0: Q(t) = Q0 (.8 ^ t) .075=.8^t T=11 Q(t) = Q0 (.7 ^ t) .075=.7^t T=7 Q(t) = Q0 (.6 ^ t) .075=.6^t T=5 Q(t) = Q0 (.5 ^ t). .075=.5^t T=3