#$&* course Mth 174 assignment # 10......!!!!!!!!...................................
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Given Solution: In a short time interval `dt at t years after the start the amount deposited will be 1000 * `dt. Suppose that T stands for the time required to grow to $10,000. Then the $1000 * `dt will be able to grow for T - t years at a 5% continuous rate. It will therefore grow to $1000 * `dt * e^(.05 ( T - t) ). Adding up all the contributions and taking the limit as `dt -> 0 we get the integral of 1000 e^(.05 ( T - t) ) with respect to t, integrated from t = 0 to t = T. An antiderivative is -1000 / .05 e^(.05 T - t)); evaluating at the limits and subtracting we get 20,000 ( e^(.05 T) - e^0) = 20,000 (e^(.05 T) - 1) Setting this equal to 10,000 we get e^(.05 T) - 1 = .5, or e^(.05 T) = 1.5. Taking ln of both sides gives .05 T = ln(1.5). Thus T = ln(1.5) / .05 = 8.1093. It takes 8.1093 years for the principle to reach $10,000 with initial principle $1000. If initial principle is $2000 then the equation becomes 2000 e^(.05 T) + 20,000 ( e^(.05 T) - 1 ) = 10,000, or 22,000 e^(.05 T) = 30,000 so e^(.05 T) = 30/22, .05 T = ln(30/22) T = ln(30/22) / .05 = 6.2.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: problem 8.7.21 (4th edition 8.7.20 3d edition 8.6.20 formerly page 414 #12) death density function f(t) = c t e^-(kt). What is c in terms of k? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From t=0 to infinity, the integral of c t e^(-kt) is 1. An antiderivative is F(t) = -c (t e^(-kt) / k+ e^(-kt) / k^2) = -c e^(-kt) (kt +1) / k^2. The limit as t approaches infinity for F(t)= 0. Also, F(0) = c(-1/ k^2) = -c/ k^2. Therefore, the integral from zero to infinity is c / k^2. The integral of has to be 1, so c / k^2 = 1. C= k^2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The integral of any probability density function should be 1, which is equivalent for the present problem to saying that every sick individual will eventually die. Thus the integral of c t e^(-kt), from t=0 to infinity, is 1. This is the relationship you solve for c. An antiderivative of c t e^(-kt) is F(t) = - c (t e^(-kt) / k + e^(-kt) / k^2) = -c e^(-kt) ( k t + 1) / k^2. lim{t -> infinity)(F(t)) = 0. F(0) = c(- 1/k^2) = -c/k^2. So the integral from 0 to infinity is c / k^2. This integral must be 1. So c / k^2 = 1 and c = k^2 . ** ** See also the previous note, where we see that in order for this function to be a probability distribution, the constant c must take the value k^2. We now have the information that 40% die within 5 years, so that the integral of f(t) from 0 to 5 is .4. This integral is in terms of c and k and yields an equation relating c and k. Combining this information with our previously found relationship between c and k you can find both c and k. We have for the proportion dying in the first 5 years: integral ( k^2 t e^-(kt) dt, t from 0 to 5) = .4. Using the antiderivative F(t) = -c e^(-kt) ( k t + 1) / k^2 = - k^2 e^(-kt) ( k t + 1) / k^2 = -e^(-kt) ( kt + 1) we get F(5) - F(0) = .4 1 - e^(-5 k) ( 5 k + 1) = .4 e^(-5 k) ( 5 k + 1) = .6. This equation presents a problem because it can't be solved exactly. If we graph the left-hand side as a function of k we see that there are positive and negative solutions. We are interested only in positive solutions because otherwise the limit of the original antiderivative at infinity won't be 0 and the integral will be divergent. Solving approximately using Derive, with trial interval starting at 0, we get k = .4045. the cumulative function is just the integral of the density function--the integral from 0 to t of f(x), where x is our 'dummy' integration variable. We have P(t) = cumulative distribution function = integral ( k^2 x e^-(kx) dx, x from 0 to t). Using the same antiderivative function as before this integral is P(t) = F(t) - F(0) = -e^(-kt) ( kt + 1) - (- e^(-k*0) ( k*0 + 1) ) = 1 - e^(-kt) ( kt + 1). Note that for k = .4045 this function is P(t) = 1 - e^(-.4045 t) ( -.4045 t + 1). You can check that for this function, P(5) = .4 (40% die within 5 years) and lim{t -> infinity}(P(t)) = 1. ** STUDENT QUESTION Can you walk me thru how you obtained your antiderivative??????????????????????? INSTRUCTOR REPSONSE To get an antiderivative of t e^(-k t) you use integration by parts, with u = t and dv = e^(-k t) dt. This gives you du = dt and v = -1/k e^(-kt). Thus u v - int(v du) = -t / k e^(-k t) - int(-1/k e^(-k t) dt) = -t / k e^(-k t) - 1 / k^2 e^(-k t). This algebraically rearranges to the given forms. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: **** query problem 8.7.4 (3d edition 8.6.4) fn y = 4 for 0 < x < c. **** Is the function a probability distribution function or a cumulative probability function? What is the value of c?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is a probability distribution function, and the integral of the function has to be 1. The integral of y=4 from x=0 to x=c is equal to 4c. Therefore, 4c=1 and c= ¼. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** A cumulative distribution function must be increasing, and it starts at 0. This is a prob distribution function. The integral of the function must be 1. The integral of y = 4 from x = 0 to x = c is 4 c. So 4c = 1 and c = ¼. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: **** query problem 8.7.9 (4th edition 8.7.8 3d edition 8.6.8) Function linear from (0,0) - (2,c) - (4, 3c) - (?, 3c) **** Is the function a probability distribution function or a cumulative probability function? What is the value of c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This function cannot be a probability distribution function because it would have infinite area. Therefore, it is a cumulative probability function. These functions start at 0 and end up with the value 1. This function, however, starts at 0 and ends with the value 3c. Therefore, 3c=1 and c= 1/3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This function would have infinite area and can’t be a probability distribution function. It’s a cumulative probability function. Cumulative probability functions start at 0 and end up with value 1. This one starts at 0 and ends up with value 3c. So 3c = 1 and c = 1/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: **** query problem 8.7.13 (4th edition 8.7.12 3d edition 8.6.12). F(x) cum distr fn heights of trees in meters **** What is the meaning of F(7) = .6? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because of the cumulative distribution function, we know that for any x the probability that the quantity ruled by the function will be less than or equal to x. It is the cumulative probability of all occurrences up to the value. F(7) = 0.6 means that 0.6 of the trees are up to 7 meters in height and can equal 7 meters, as well. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The cumulative distribution function tells you for any x the probability that the quantity being governed by the function will be less than or equal to x. It is the cumulative probability of all occurrences up to that value. Thus F(7) = .6 means that .6 of the trees are up to and including 7 meters in height. ** **** Which is greater, F(6) or F(7)? Justify in terms of trees.
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20:35:10 ** There are more trees up to and including 7 meters tall than trees up to and including 6 meters tall, since the first characterization includes the second. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: problem page 415 #18 probability distribution function for the position of a pendulum bob This question has been omitted from recent editions of the text. You should, however, answer it and self-critique your answer. If the position of the bob of a swinging pendulum is observed at a large number of random times, we can use our observations to construct a graph of the probability distribution for the position of the pendulum. Describe the expected characteristics of this graph.
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01:37:23 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The pendulum bob is least likely to be found where it is most the quickest, which is near the center, and most likely to be found where it is moving the slowest, which is near its ends. The velocity consistently increases as we get closer to the center, and consistently decreases as we move farther from the center. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The pendulum bob is most likely to be found where it is moving most slowly, and least likely where it is moving most quickly. It's slowest near its ends and quickest near the center. So the probability density would be low in the middle and high near the ends. Velocity increases smoothly as we approach the center, and decreases smoothly as we move away from the center, so the probability would decrease smoothly from x = -a to x = 0, then increase smoothly as we move from x = 0 to x = a. ** the curve is a parabola opening upward &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I should add that the probability would decrease smoothly from x = -a to x = 0, then increase smoothly as we move from x = 0 to x = a. Also, the curve is a parabola that opens upward. ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!