course Mth 271 004. `query 4*********************************************
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Given Solution: `a You have to find the average rate of change between clock times t and t + `dt: ave rate of change = [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt = [ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt = [ 2 a t `dt + a `dt^2 + b `dt ] / `dt = 2 a t + b t + a `dt. Now if `dt shrinks to a very small value the ave rate of change approaches y ' = 2 a t + b. ** Self-critique Rating: While I didn’t approach the problem the same way the solution suggests, I feel that I did a good job explaining how to obtain the desired expression. ********************************************* Question: `q explain how we know that the depth function for rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for some constant c, and explain the significance of the constant c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Above I proved that the functions y = at^2 +bt + c is the same as y’ = 2at + b. The later equation represents the linear rate of change and can also be represented by y’ = mt + b. Because y = 2at + b and y = mt + b are the same, it is understood that their variables are the same quantities. The variable b is the identical in both equations, the coefficient 2a and m are both variables of t and so 2a = m. If 2a = m, then a = 1/2m. This means the depth function must be y = 1/2mt^2 + bt+ c
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Given Solution: `a Student Solution: If y = a t^2 + b t + c we have y ' (t) = 2 a t + b, which is equivalent to the given function y ' (t)=mt+b . Since 2at+b=mt+b for all possible values of t the parameter b is the same in both equations, which means that the coefficients 2a and m must be equal also. So if 2a=m then a=m/2. The depth function must therefore be y(t)=(1/2)mt^2+bt+c. c is not specified by this analysis, so at this point c is regarded as an arbitrary constant. c depends only on when we start our clock and the position from which the depth is being measured. ** Self-critique Rating: I feel very confident about my solution to the problem. ********************************************* Question: `q Explain why, given only the rate-of-depth-change function y' and a time interval, we can determine only the change in depth and not the actual depth at any time, whereas if we know the depth function y we can determine the rate-of-depth-change function y' for any time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the rate-of-depth-change function we can find the average rate of change over a given period of time. I read your explanation and it makes sense to me, but I wouldn’t have been able to come to this conclusion on my own.
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Given Solution: `a Given the rate function y' we can find an approximate average rate over a given time interval by averaging initial and final rates. Unless the rate function is linear this estimate will not be equal to the average rate (there are rare exceptions for some functions over specific intervals, but even for these functions the statement holds over almost all intervals). Multiplying the average rate by the time interval we obtain the change in depth, but unless we know the original depth we have nothing to which to add the change in depth. So if all we know is the rate function, have no way to find the actual depth at any clock time. ** Self-critique Rating: I had to read the given solution to understand why this would be true. ********************************************* Question: `q In terms of the depth model, explain the processes of differentiation and integration. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We used differentiation to find the rate-of-depth-change function from the equation y = at^2 + bt + c. Integration would be taking the rate-of-depth-change function to find the original equation.
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Given Solution: `a Rate of depth change can be found from depth data. This is equivalent to differentiation. Given rate-of-change information it is possible to find depth changes but not actual depth. This is equivalent to integration. To find actual depths from rate of depth change would require knowledge of at least one actual depth at a known clock time. ** Self-critique Rating: I feel fairly confident about this problem. ********************************************* Question: `q $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The growth rate is 10%, or .10. Growth factor is 1.1. P(t) = P0(1 + r)^t 400 = 200(1.1)^t 2 = 1.1^t ln2 = tln1.1 ln2/ln1.1 = t t = 7.27
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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 Rating: I feel very confident about this problem. ********************************************* 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: Using the equation above, 200(1.1^t), to find the principle at t = 20, I found the principle to equal 1345.5. Half of this is 672.75. To find the value of t at principle 672.75, I used the equation: 200(1.1^t) = 672.75 200(1.1^t) = 672.75 1.1^t = 672.75/200 1.1^t = 3.3638 t ln 1.1 = ln 3.3638 t = ln 3.3638/ln 1.1 t = 12.73 I didn’t understand what you meant by the principle doubling in the interval between 12.7 years and 20 years. Could you explain this to me?
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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 we get 674.20 so it would probably be about 12.72 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 Rating: I feel fairly confident about this problem. ********************************************* Question: `q Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the first four years, I obtained the following values: t = 1 t = 2 t = 3 t = 4 % principle % principle % principle % principle 10 220 10 242 10 266.2 10 292.82 20 240 20 288 20 345.6 20 414.72 30 260 30 338 30 439.4 30 571.22 40 280 40 391 40 548.8 40 768.32 After reading your given solution, I’m confused about what you were looking for in an answer.
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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 Rating: I’m confused about your given solution. The question asked to sketch a graph of principle vs. rate for the first four years with given rates, but in the solution you state something about the amount of time it will take to double.
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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 Rating: I feel confident about this problem. ********************************************* 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: 5000(1.08^(2 + doublingTime)) = 10000 1.08^(2 + doublingTime) = 2 1.08^2 + 1.8^doublingTime = 2 1.08^doublingTime = 2 doublingTime(ln1.08) = ln2 doublingTime = 9.0064
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Given Solution: 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 Rating: I feel very confident about this problem. ********************************************* Question: `q Desribe how on your graph how you obtained an estimate of the doubling time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To sketch a graph of doublingTime, I have to plot the initial principle on the y axis. Then plot the doubled amount, also on the y axis; from here I draw a line straight down to the x axis, which represents time.
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Given Solution: `aIn 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 Rating: I’m very confident about this problem. ********************************************* 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: Using the solution given above for time (P0(1 + r) ^ t), where r is the growth rate, and I solved for t. The growth rate of .11 is negative because it is decreasing. Q(t) = 550(1 - .11)^t Q(t) = 550(.89)^t
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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 Rating: I feel very confident about this problem. ********************************************* Question: `qHow much antibiotic is present at 3:00 p.m.? Your solution: I used the solution I found for the previous problem and plugged the value for t in. t is 5 because from 10:00 am to 3:00 pm is 5 hours. Q(t) = 550(.89)^5 Q(t) = 307.12 mg 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 = 307.123mg in the blood ** Self-critique Rating: I feel extremely confident about this problem. ********************************************* Question: `qDescribe your graph and explain how it was used to estimate half-life. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the half life, I did the same as the graph mentioned above. I plotted half of the principle amount on the graph and moved down to the x axis to find the time.
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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. ** ********************************************* Question: `qWhat is the equation to find the half-life? What is the most simplified form of this equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Q(t) = 1/2Q0
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Given Solution: `a Q(doublingTime) = 1/2 Q(0)or 550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have .89^doublingTime = .5. We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). ** Self-critique Rating: I feel fairly confident about this problem. ********************************************* 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: t = -24, Q(t) = .101 t = -30, Q(t) = .057 t = -47, Q(t) = .0113 t = -55, Q(t) = .00528
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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 Rating: I feel confident about this problem. ********************************************* Question: `qexplain why the negative t axis is a horizontal asymptote for this function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The value of Q(t) approaches 0 but will never reach it.
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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. ** Self-critique Rating: I feel fairly confident about this problem. ********************************************* Question: `q#22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = Ab^x y = 12(e^-.5x) e^-.5 = .606 = b
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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 Rating: I feel very confident about this problem. ********************************************* Question: `qwhat is b for the function y = .007 ( e^(.71 x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = Ab^x y = .007(e^.71x) e^.71 = 2.033 = b
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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 Rating: I feel very confident about this problem. ********************************************* Question: `qwhat is b for the function y = -13 ( e^(3.9 x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = Ab^x y = -13(e^3.9x) e^3.9 = 49.40 = b
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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 Rating: I feel very confident about this problem. ********************************************* Question: `qList these functions, each in the form y = A b^x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 12(e^-.5x) = 12(.606^x) y = .007(e^.71x) = .007(2.033^x) y = -13(e^3.9x) = -13(49.4^x)
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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 Rating: I feel very confident about this problem. ********************************************* Question: `q0.4.44 (was 0.4.40 find all real zeros of x^2+5x+6 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 + 5x + 6 (x + 3)(x + 2) x + 3 = 0 x = -3 x + 2 = 0 x = -2
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Given Solution: `a We can factor this equation to get (x+3)(x+2)=0. (x+3)(x+2) is zero if x+3 = 0 or if x + 2 = 0. We solve these equations to get x = -3 and x = -2, which are our two solutions to the equation. COMMON ERROR AND COMMENT: Solutions are x = 2 and x = 3. INSTRUCTOR COMMENT: This error generally comes after factoring the equation into (x+2)(x+3) = 0, which is satisfied for x = -2 and for x = -3. The error could easily be spotted by substituting x = 2 or x = 3 into this equation; we can see quickly that neither gives us the correct solution. It is very important to get into the habit of checking solutions. ** Self-critique Rating: I feel extremely confident about this problem. ********************************************* Question: `qExplain how these zeros would appear on the graph of this function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (-2, 0) and (-3, 0)
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Given Solution: `a We've found the x values where y = 0. The graph will therefore go through the x axis at x = -2 and x = -3. ** Self-critique Rating: I feel very confident about this problem. ********************************************* Question: `q 0.4.50 (was 0.4.46 x^4-625=0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^4 – 625 = 0 x^4 = 625 take the fourth root x = +-5
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Given Solution: `a Common solution: x^4 - 625 = 0. Add 625 to both sides: x^4 = 625. Take fourth root of both sides, recalling that the fourth power is `blind' to the sign of the number: x = +-625^(1/4) = +-[ (625)^(1/2) ] ^(1/2) = +- 25^(1/2) = +-5. This is a good and appropriate solution. It's also important to understand how to use factoring, which applies to a broader range of equations, so be sure you understand the following: We factor the equation to get (x^2 + 25) ( x^2 - 25) = 0, then factor the second factor to get (x^2 + 25)(x - 5)(x + 5) = 0. Our solutions are therefore x^2 + 25 = 0, x - 5 = 0 and x + 5 = 0. The first has no solution and the solution to the second two are x = 5 and x = -5. The solutions to the equation are x = 5 and x = -5. ** Self-critique Rating: I feel very confident about this problem ********************************************* Question: `q0.4.70 (was 0.4.66 P = -200x^2 + 2000x -3800. Find the x interval for which profit is >1000 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -200x^2 + 2000x – 3800 > 1000 -200x^2 + 2000x – 4800 = 0 Divide by -200 x^2 – 10x + 24 = 0 (x -6)(x – 4) x = 6, x = 4
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Given Solution: `a You have to solve the inequality 1000<-200x^2+2000x-3800, which rearranges to 0<-200x^2+2000x-4800. This could be solved using the quadratic formula, but it's easier if we simplify it first, and it turns out that it's pretty easy to factor when we do: -200x^2+2000x-4800 = 0 divided on both sides by -200 gives x^2 - 10 x + 24 = 0, which factors into (x-6)(x-4)=0 and has solutions x=4 and x=6. So -200x^2+2000x-4800 changes signs at x = 4 and x = 6, and only at these values (gotta go thru 0 to change signs). At x = 0 the expression is negative. Therefore from x = -infinity to 4 the expression is negative, from x=4 to x=6 it is positive and from x=6 to infinity it's negative. Thus your answer would be the interval (4,6). COMMON ERROR: x = 4 and x = 6. INSTRUCTOR COMMENT: You need to use these values, which are the zeros of the function, to split the domain of the function into the intervals 4 < x < 6, (-infinity, 4) and (6, infinity). Each interval is tested to see whether the inequality holds over that interval. COMMON ERROR: Solution 4 > x > 6. INSTRUCTOR COMMENT: Look carefully at your inequalities. 4 > x > 6 means that 4 > x AND x > 6. There is no value of x that is both less than 4 and greater than 6. There are of course a lot of x values that are greater than 4 and less than 6. The inequality should have been written 4 < x < 6, which is equivalent to the interval (4, 6). ** Self-critique Rating: I feel very confident about this problem.