ASSN10

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03:36:40

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

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03:46:19

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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Using the formula (1000t)(1.05^t) = B$

I got 7.08 years to reach $10000

and assuming the acount had $2000 in it already it would take it would take 5.98 years.

** 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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03:49:02

What integral did you use to solve the first problem, and what integral did use to solve the second?

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for the first I use (1000t)(1.05)^t= $B

for the second I used (2000+(1000t))(1.05)^t=$B

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03:58:14

What did you get when you integrated?

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by integrating on the interval I found when the acounts would reach the 10,000 mark.

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04:01:29

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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assuming that you are still adding the $1000 per year then you would plug in t to the equation and this would give you the amount with the interest it has gained.

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04:04:56

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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I Think that my equation is correct, Pretty sure. If it is not the right one then it isn't accounting for the interest that takes place during the year before the 1000 is added. In which case it would have to be figured out by how many years it is (1000*1.07^t)+(1000*1.07^t)... with the number of ""segments"" corresponding directly to 'dt.

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04:04:56

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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04:05:58

Explain how the previous expression is built into a Riemann sum.

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Because instead of using the indeninate integral and having an equation for it, you use the Riemann sum to find the definate integral on the interval.

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04:06:26

Explain how the Riemann sum give you the integral you used in solving this problem.

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It gives the amount of money the account earns.

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04:21:10

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

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04:23:59

what is c in terms of k?

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k = ln(ct) / t

** 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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** 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 so the probability would decrease smoothly from x = -a to x = 0, then increase smoothly again from x = 0 to x = a. **

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Once more, be sure you understand the process of partitioning, considering a typical interval with arbitrary sample point, Riemann sum, then integral.

See my notes and let me know if you have questions.