#$&* course Mth163 July 2,2013 22:47
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Given Solution: `a** The growth rate is .10 and the growth factor is 1.10 so the amount function is $200 * 1.10^t. This graph passes through the y axis at (0, 200), increases at an increasing rate and is asymptotic to the negative x axis. For t=0, 1, 2 you get p(t) values $200, $220 and $242--you can already see that the rate is increasing since the increases are $20 and $22. Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power functions (give that a quick review if you don't--it will definitely pay off in this course). $400 is double the initial $200. We need to find how long it takes to achieve this. Using trial and error we find that $200 * 1.10^tDoub = $400 if tDoub is a little more than 7. So doubling takes a little more than 7 years. The actual time, accurate to 5 significant figures, is 7.2725 years. If you are using trial and error it will take several tries to attain this accuracy; 7.3 years is a reasonable approximation for trail and error. To reach $300 from the original $200, using amount function $200 * 1.10^t, takes a little over 4.25 years (4.2541 years to 5 significant figures); again this can be found by trial and error. The amount function reaches $600 in a little over 11.5 years (11.527 years), so to get from $300 to $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& “Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power “- I thought this was pre-cal…am I doing the wrong assignment??? I see that I can use decimals in my powers to get a more precise answer &&& ********************************************* Question: `qAt what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& $200 * 1.1^20=$1345.5 $1340/2=$670 $200 * 1.1^12=670.67 To double $200 * 1.1^20=$1345.5 We say the t value of $1345 doubled &&&& confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The t = 20 value is $200 * 1.1^20 = $1340, approx. Half the t = 20 value is therefore $1340/2 = $670 approx.. By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx.. For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45). At 12.75=674.20 so it would probably be about12.72. This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qquery #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& $1(1.1^4)=$1.46 $1(1.2^4)=$2.07 $1(1.3^4)=$2.86 $1(1.4^4)=$3.84 My graph is increasing at an increasing rate &&& confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double. for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double. Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& I found that the rate doubled at $1(1.2^4)=$2.07 from initial principle It doubled at 20% on the fourth year Did you want us to find the doubling amount for every percentage??? &&&&& ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `qquery #11. What is the equation for doubling time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& P(2+DoubTime)=10,000 Reduced 1.8^(DoubTime)=-1.24
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Given Solution: `a** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The Resulting equation is therefore P0 * (1+r)^t = 2 P0. Note that this simplifies to (1 + r)^ t = 2, and that this result depends only on the interest rate, not on the initial amount P0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&& &&&& P(2+DoubTime)=10,000 Reduced 1.8^(DoubTime)=-1.24 &&&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The principal function is • P(t) = $5000 * 1.08^t. When t = 2 the principal is • P(2) = $5000 * 1.08^2. We want to find out how long it takes, starting at t = 2, for the principal to double. Doubling the principal $5000 * 1.08^2 results in principal 2 * ($5000 * 1.08^2). This principal will occur at some later time t = 2 + doubling time. At this time the principal will be P(2 + doubling time) = $5000 * 1.08^(2 + doubling time). Thus • P(2 + doubling time) = 2 * ($5000 * 1.08^2). To find the doubling time, we solve this equation for doubling time. The solution is as follows: Dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get • 1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2]. This can be written as • 1.08^2 * 1.08^doublingtime = 2 * 1.08^2. Dividing both sides by 1.08^2 we obtain • 1.08^doublingtime = 2. We can then use trial and error to find the doubling time that works. We get something like 9 years. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& 1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2]. To cancel we multiply both sides by 1.08 ^ 2 And that’s where I went wrong &&&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q Describe how you used your graph to obtain an estimate of the doubling time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& I doubled the initial time and found the y axis &&&& confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis. The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& You would find your doubled amount on the y axis then find the point on the graph that would lead you to the corresponding x axis &&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q#17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& Q(t)=(1+-.11)=.89 &&&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or Q(t)=550(.89)^t ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I was one step behind Q(t)=550(.89)^t The quantity is equal to initial dosage times growth rate raised to the t power (hours in this case) &&&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qHow much antibiotic is present at 3:00 p.m.? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& .89^4(550)=345.08mg &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 3:00 p.m. is 5 hours after the initial time so at that time there will be Q(5) = 550 mg * .89^5 = 307.123 mg in the blood ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I put 4 hours instead of 5, which I know is wrong it should have been 5 this makes sense Q(5) = 550 mg * .89^5 = 307.123 mg I get the same amount &&&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qDescribe your graph and explain how it was used to estimate half-life. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& My graph is decreasing at an increasing rate (it is also neg) You use your initial coordinates and your doubled coordinates and your half-life is somewhere between there &&& confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point. The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down. The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& ------------------------------------------------ Self-critique rating:1 ********************************************* Question: `qWhat is the equation to find the half-life? What is its most simplified form? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& [.89^dt(550)]/550=275/550 .89^dt=1/2 &&& confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Q(halvingTime) = 1/2 Q(0)or 550 mg * .89^halvingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have .89^halvingTime = .5. We can use trial and error to find an approximate value for halvingTIme (later we use logarithms to get the accurate solution). ** STUDENT QUESTION ??? Mr. Smith, please explain how you came up with this equation, because I want to understand it and I have not seen this in our notes. Thank-you.??? INSTRUCTOR RESPONSE Your notes include a similar treatment of doubling time. The halving time is the time required for a quantity to fall to half its value. If a function is exponential its halving time will be uniform, meaning that the time to fall to half its value is the same no matter at what instant we choose to begin. If the function is Q(t), then for example Q(0) is its value at t = 0. • If the value of Q then falls to half this value, its new value will be 1/2 Q(0). • If we refer to the time required to fall as halvingTime, then after this time has passed the value will clock time will be 0 + halvingTime = halvingTIme. • At this instant the value of our function will be Q(halvingTime), and this will be 1/2 Q(0). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q#19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0. For what values of t did Q(t) lie between .005 Q0 and .01 Q0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& Q(t)=Q(0)(1.1^t) Q(t)=Q(0)(1.1^.05) Q(t)=Q(0)(1.1^.1)
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Given Solution: `a** Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0. Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0. Solving Q(t) = .05 Q0 we rewrite this as Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get 1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -31.4 approx. Solving Q(t) = .1 Q0 we rewrite this as Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get 1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -24.2 approx. (The solution for .005 Q0 is about -55.6, for .01 is about -48.3 For this solution any value between about t = -48.3 and t = -55.6 will work). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& ????Any value between about t = -24.2 and t = -31.4 Where did this come from ????
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Given Solution: `a** The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. Another way to see this: The function 1.1^t has a constant doubling time (it doubles at an approximate interval of 7.2). So if you trace it back in the negative direction its value halves at regular intervals, and therefore approaches 0.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& As our graph reaches infinity at the y axis is also does so at the x axis? ???Another way to see this: The function 1.1^t has a constant doubling time (it doubles at an approximate interval of 7.2). So if you trace it back in the negative direction its value halves at regular intervals, and therefore approaches 0.** when you say track it back are you saying this is not the end result??? &&&
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Given Solution: `a** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qwhat is b for the function y = .007 ( e^(.71 x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& y = .007 ( e^(.71 x) ) . y=.007(2.03)^x .b=2.03 &&&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx. So this function is of the form y = A b^x for b = 2.041 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qwhat is b for the function y = -13 ( e^(3.9 x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& y = -13 ( e^(3.9 x) ) .y=13(49.4024) .b=49.4 &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx. So this function is of the form y = A b^x for b = 49.4 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qList these functions, each in the form y = A b^x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& y = .007 *2.03^x. y = -13* 49.4^x. &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& I forgot to include the y=12(.607)^x &&&&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: What is the equation for the doubling time of the function y = 32 * 1.3^t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& 64=32(1.3)^t Divide 32 by both sides 2=(1.3^t) &&&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: Explain why the positive t axis is a horizontal asymptote for the function y = 400 * 2^(-t). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& your function is negatively increasing as it started off very close to x=0 but never actually touched x axis that’s why it is called as asymptote Can you explain the question a little more???