#$&* course mth 174 5/18 2:20 pm assignment # 10 ......!!!!!!!!...................................
.............................................
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 $2000 then the equation becomes 2000 e^(.05 T) + 20,000 ( e^(.05 T) - 1 ) = 10,000, or 22,000 e^(.05 T) = 30,000 so e^(.05 T) = 30/22, .05 T = ln(30/22) T = ln(30/22) / .05 = 6.2.
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: problem 8.7.21 (4th edition 8.7.20 3d edition 8.6.20 formerly page 414 #12) death density function f(t) = c t e^-(kt). What is c in terms of k? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are trying to solve F9t)=c t e^(-kt) in terms of c. we begin by factoring out the constant c getting C(f(t) = t e^(-kt) We set this up as c * the integral from 0 to infinity of t e^(-kt) We let u = t u’=1 v’=e^(-kt) v=-1/(ke^(-kt)) So we have c[t*(-1/(ke^-(kt)) - integral -1/(ke^(-kt)) C(-t/(ke^(-kt)) - 1/(k^2 e^(-kt)) Evaluating at 0 we get c(0/ke^0 - 1/k^2e^0) Getting c(-1/k^2) we know that all probability function should equal 1 so we set this equal to 1 and get c(-1/k^2)=1 -c/k^2 = 1 -c=k^2 C=-k^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. Note: The questions below are 'followup questions'. You need not insert answers to these questions, but you may do so. If you do, please precede your answer with **** and add #$&* in the line immediately after your insertion. You will be able to recognize similar 'followup questions' on many questions in subsequent assignments.
.................................................
What is the cumulative death distribution function?
.................................................
**********The cumulative death distribution function is the integral of the density function which is cte^(-kt) but we substitute in -k^2 for c and get integral k^2 (te^(-kt) and we evaluate this integral from 0 to t and use x as our variable of integration so we have The integral from 0 to t of -k^2 x e^(-kx) We pull the k^2 to the outside since it represents a constant and we solve by letting u = x u’=1 v’=e^(-ktx v=-1/(ke^(-kx)) So we have k^2[x*(-1/(ke^-(kx)) - integral -1/(ke^(-kx)) -k^2(-x/(ke^(-kx)) - 1/(k^2 e^(-kx)) #$&* What integral did you use to obtain the cumulative death distribution function and why? *******I used the integral of the density function to obtain the cumulative death distribution function except I used the found value of c which was -k^2. I did this because it is the same method we used in density problems to find the second moment???? #$&*
.................................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### I am not sure how to solve using derive. I have a TI-84 calculator and I am not familiar with using derive? Solving approximately using Derive, with trial interval starting at 0, we get k = .4045.?????????????????? ------------------------------------------------ Self-critique Rating:3
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have y=4 and want to know if it is a probability function or a cumulative function. We know that in order to be a cumulative distribution function, the function must be increasing and must begin at zero, this function does neither so we know it must be a probability function and therefore we also know it must equal 1. Now we want to find the value of c The integral from 0 to c of 4 The antiderivative is 4y and we evaluate this from 0 to c getting 4c -0 or 4c We know that this must equal 1 since it is a probability distribution so 4c=1 c = 1/4 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: In order to be a probability function the x values must be between a and b in this case they are between 0 and ? so since there isn’t a defined end point it can’t be a probability function and must therefore be a cumulative function. We know cumulative function start at 0 and end at 1 and this function starts at 0 and ends at 3c so therefore 3c must equal 1 so 3c=1 c = 1/3 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ###Is my understanding and explanation of why this isn’t a probability distribution function correct or am I misunderstanding??? ------------------------------------------------ Self-critique Rating:3
.............................................
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.
......!!!!!!!!...................................
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): ok ------------------------------------------------ Self-critique Rating: ********************************************* 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.
......!!!!!!!!...................................
01:37:23 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We want to construct a graph of the probability distribution for a swinging pendulum observed at a large number of random times. If a pendulum is swinging and we are randomly observing it, we would be more likely to observe the pendulum at times when it is moving slower which would be at either endpoint. The pendulum would move fastest when it is near the center so we would have less observations at the center. So the graph would have more area underneath the 2 endpoints and less underneath the center so the graph would look much like the arc of the pendulum and would be in the shape of a upward turned U. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ###Ok I think, my description is similar?????