Test 3

I'm glad you're looking at Set 7. If you get a problem like this right on Test 3, naturally I would count it.

Note that in the q_a_ for Calculus I, to which I believe I referred you earlier, asst #10 is specifically about future value, building the concept step by step. I don't know if that treatment would be necessary for you, but I'm mentioning it in case you find it helpful.

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20:29:36 Query problem 8.6.8 (8.4.8 in 3d edition) $1000/yr continuous deposit at 5%

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RESPONSE --> ok

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20:48:13 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 --> I know that I am supposed to use the formula for the future value of a streaming income to figure this problem out and I know that the future value must be 10,0000 in both instances. The difference between the two integrals would be the number P(t). In the first it would be int(1000e^-.05(M-t)), t, 0, M) Then set the integral equal to 10,000 and solve for M. I am not exactly sure how to do this. in the second, it would be: int(2000e^-.05(M-t)), t, 0, M). this integral would have to be set equal to 10,000 and solved for M. I am not exactly sure how to do that though.

You have a very good outline of how to proceed.

** 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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20:48:27 What integral did you use to solve the first problem, and what integral did use to solve the second?

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RESPONSE --> int(1000e^-.05(M-t)), t, 0, M) int(2000e^-.05(M-t)), t, 0, M)

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20:48:33 What did you get when you integrated?

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RESPONSE --> not sure yet

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20:53:17 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 believe you would use the future value integral for a streaming income. Especially since the question says, obtain the amount after a certain number (T) years. int(P(t)e^r(M-T)dt, t, 0, M)

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20:58:35 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 --> ok

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20:59:34 Explain how the previous expression is built into a Riemann sum.

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RESPONSE --> sum(P(t)delta(t)e^r(M-t) is the rieman sum for this

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21:00:04 Explain how the Riemann sum give you the integral you used in solving this problem.

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RESPONSE --> you take the limit of the sum as the subdivisions move towards zero (get smaller and smaller). this gives you the integral

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��d�ǎ��际��ր��������Y�Ԯ assignment #010 ��g������m���k�۟ͯE�wұҚ�R Physics II 04-19-2006

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23:34:03 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 --> ok

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23:38:36 query 8.7.20 (8.6.20 ed editin) death density function f(t) = c t e^-(kt)

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RESPONSE --> ok

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23:49:49 what is c in terms of k?

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RESPONSE --> To find C in terms of K, you would solve for C then plug it back into the equation i believe. I am not 100% sure that this is what this question is looking for however. For death density: c=f(t)/te^-kt*delta(t) therefore f(t)=(f(t)/te^-kt*delta(t))*te-kt*delta(t) I need a little help here because i could not find an example in the book that talked about this particular problem. I also did not see anything like this in the class notes. Perhaps you can help me here.

This is an instance of a probability density function, and is solved by the standard techniques. Once more you have most of the solution, so you should be in good shape to understand the following.

Ironically you used 100% in your note, in what is actually a very similar context. The one key idea missing from your solution is that the entire probability density function gives you 100%, or 1.00:

** 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 . **

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You are doing well here; in each case I believe you have the right idease and are missing only one detail. Let me know if you have questions.