course Mth 174 dDzɶ۽ۇw{]ވassignment #010
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21:50:28 Query problem 8.6.8 (8.4.8 in 3d edition) $1000/yr continuous deposit at 5%
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RESPONSE --> If I understood this problem correctly, what I need to do here is find a future value of $10,000 with a continuous yearly deposit of $1000 into an account bearing 5% interest. Okay, here goes: a) I can find future value by taking the present value of a continuous income account and multiplying it by e^(rM): Future = e^(rM) * int(P(t)e^(-rt)dt, t, 0, M) = 10,000 Since $1000 is the yearly continuous income and r= 0.05, the function is P(t) = 1000 and the integral becomes: int( 1000e^(-0.05t)dt, t, 0, M) Now I can take that and try and solve it for M by finding the antiderivative: int = -20,000e^(-0.05t) from 0 to M = -20,000e^(-0.05M) + 20,000 Now we can multiply this by e^(0.05M) to find the future value: 10,000 = (-20,000e^(-0.05M) + 20,000)e^(0.05M) 10,000 / e^(0.05M) = -20,000e^(-0.05M) + 20,000 10,000e^(-0.05M) = -20,000e^(-0.05M) + 20,000 30,000e^(-0.05M) = 20,000 e^(-0.05M) = 2/3 -0.05M = ln(2/3) M = ln(2/3) / -0.05 = 8.1093 years So it will take just a bit more than 8 years to have $10,000 in the account. This matches with the integral over that period: int( 1000e^(0.05(8.1093 - t)) dt, t, 0, 8.1093) b) Since it starts with a $2000 balance, I think all I need to do is add the future balance of the $2000 to the future balance of the continual $1000 deposit, and then go through the same process that I used on the prior answer: 10,000 = (-20,000e^(-0.05M) + 20,000)e^(0.05M) + 2000e^(0.05M) 10,000e^(-0.05M) = -20,000e^(-0.05M) + 20,000 + 2000 30,000e^(-0.05M) = 22,000 e^(-0.05M) = 22/30 -0.05M = ln(22/30) M = ln(22/30) / -0.05 = 6.203099 years. Whew!
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21:51:17 how long does it take the balance to reach $10000, and how long would take if the account initially had $2000?
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RESPONSE --> According to my calculations, it will take 8.1093 years for the balance to reach $10,000 with no initial deposit. With an initial deposit of $2000, it will take 6.2031 years.
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21:52:47 What integral did you use to solve the first problem, and what integral did use to solve the second?
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RESPONSE --> First problem: e^(rM) * int( 1000e^(-0.05t) dt, t, 0, M) Second problem: e^(rM) * int( 1000e^(-0.05t) dt, t, 0, M) + 2000e^(rM)
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21:54:35 What did you get when you integrated?
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RESPONSE --> For the first problem: 9999.996757 Second problem: 10000.00215
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21:56:28 Explain how you would obtain the expression for the amount after T years that results from the money deposited during the time interval `dt near clock time t.
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RESPONSE --> I think that the t variable in the Present balance equation would turn into (M - t), or the total time minus the point at which the interval is calculated.
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21:58:49 The amount deposited in the time interval `dt of the previous question is $1000 * `dt and it grows for T - t years. Use your answer consistent with this information?
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RESPONSE --> Yes. I used a slightly different flavor of the future value equation, namely: e^(rM) * Present value which is equal to: int( P(t)e^(r (M - t) ) dt, t, 0, M) I think using the alternate form made it easier to calculate the value of M.
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22:03:22 Explain how the previous expression is built into a Riemann sum.
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RESPONSE --> The growth factor is the rate of growth at a point t, and since it is continuous it uses logarithmic growth. Therefore, the equation is actually: 1000 e^(0.05 (T - t) ) 'dt Summing this over the entire period creates the Riemann sum: sum( 1000 e^(0.05 (T - t) ) 'dt ) This is like finding the area of a strange shape where the wierd side is given to us by the formula 1000 e^(0.05 (T - t) ) 'dt.
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22:04:47 Explain how the Riemann sum give you the integral you used in solving this problem.
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RESPONSE --> To get to the integral here, all we need to do is take the limit of the Riemann sum as t->0.
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22:50:41 query 8.7.20 (8.6.20 ed editin) death density function f(t) = c t e^-(kt)
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RESPONSE --> Ah, the weird one. (Yes, I did misspell ""weird"" previously.) a) The first part is to find c in terms of k. I take this to mean I solve for c and leave k on the other side of the equation. However, I am at a loss as to how to figure this out. Since this is a density function, the only value that I know it will have is: int( f(t) 'dt, t, -infinity, infinity) = 1 But do I need to find the antiderivative of this and evaluate it, or am I missing something simpler?
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22:50:46 what is c in terms of k?
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RESPONSE --> I am not sure.
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22:50:52 If 40% die within 5 years what are c and k?
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RESPONSE --> Again, not sure.
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22:50:57 What is the cumulative death distribution function?
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RESPONSE --> Not sure.
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23:19:21 If you have not already done so, explain why the fact that the total area under a probability distribution curve is 1 allows you to determine c in terms of k.
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RESPONSE --> Hmm. I guess my previous guess was right. Now, to figure it out... Well, I think I got most of the way there. I used this relationship: 1 = int( cte^(-kt) 'dt, t, -infinity, infinity) 1 = -(c / k) e^(-kt) between -infinity and infinity 1 = - (c / k) e^(-k * infinity) + (c/k)e^(-k * -infinity) (-k * infinity turns the first function into 0) 1 = 0 + (c/k)e^(-k * -infinity) 1 = (c/k)e^(-k * -infinity) c = k e^(-k * infinity) (Sign change on infinity) c = k e^(-k * infinity) But doesn't this mean that c = 0? Hmm. I think I am still lost.
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23:20:54 What integral did you use to obtain the cumulative death distribution function and why?
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RESPONSE --> The integral of the function: c t e^(-kt) 'dt should be the cumulative death distribution function. The death density function is actually a rate and is therefore the derivative of the cumulative death distribution function.
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23:21:25 query problem page 415 #18 probability distribution function for the position of a pendulum bob
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RESPONSE --> There is no problem #18 on p. 415.
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23:21:36 describe your density function in detail -- give its domain, the x coordinates of its maxima and minima, increasing and decreasing behavior and concavity.
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RESPONSE --> I don't think I have enough info to work on this.
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23:21:44 Where is the bob most likely to be found and where is at least likely to be found, and are your answers consistent with your description of the density function?
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RESPONSE --> ...
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23:22:56 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Again, interesting stuff, but I preferred the previous section dealing with volumes and such.
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