asst 3 query

asst 3 query

Looks good. The early part of the assignment isn't here--might be in another file or something--but if you're doing OK here then you must have been OK earlier.

For the record, I didn't do an administrative withdrawal so you won't have to worry about that.

If you can do an assignment a day you can get done in less than a month, but unless your saturation point is much higher than most, you're going to experience information overaload. Most people can't compress this course into that short a time frame, even if they work 24 hours a day with time off just to eat and sleep. The brain needs time to adapt and process information.

If you can pull it off, great. If you can do half that much I'll give you extra time to finish.

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17:49:17 At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?

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RESPONSE --> y=200(1.1^x) Y=[200(1.1^20)] / 2 y = 672.7499944 672.7499944=200(1.1^x) x = 12.72745909 wow THAT DERIVE PROGRAM WORKS PRETTY GOOD.

Once you get used to it, it does a good job of most things.

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17:50:20 ** The t = 20 value is $200 * 1.1^20 = $1340, approx.

Half the t = 20 value is therefore $1340/2 = $670 approx..

By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx..

For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45).ÿ At 12.75=674.20 so it would probably be about12.72.ÿ

This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr.

This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **

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RESPONSE --> yesssssss.

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17:58:42 query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%

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RESPONSE --> y=200(1.1^x)

y=200(1.2^x)

y=200(1.3^x)

y=200(1.4^x) ooohhh. purdy.

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18:00:27 ** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double.

for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double.

Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double.

The final 4-year amount increases by more and more with each 10% increase in interest rate.

The doubling time decreases, but by less and less with each 10% increase in interest rate. **

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RESPONSE --> ok

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18:28:43 query #11. equation for doubling time

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RESPONSE -->

y^2 = 200 [ 1.1 ^ (2x) ].............?????????

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18:37:32 ** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore

P0 * (1+r)^t = 2 P0.

Note that this simplifies to

(1 + r)^ t = 2,

and that this result depends only on the interest rate, not on the initial amount P0. **

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RESPONSE --> ok.

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18:58:22 Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.

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RESPONSE --> 5000(2)= 5000(1.08^x) 2= (1.08^x)

This equation does give you the doubling time, but it's the doubling time starting at t = 0, where the principle is $5000.

Turns out that doubling time doesn't depend on when you start, but that's the point of this problem--to show that if you start at, say, t = 2 it still takes the same time to double.

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19:07:13 **dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get

1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2].

This can be written as

1.08^2 * 1.08^doublingtime = 2 * 1.08^2.

Dividing both sides by 1.08^2 we obtain

1.08^doublingtime = 2.

We can then use trial and error to find the doubling time that works. We get something like 9 years. **

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RESPONSE --> kinda confused but ok

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19:28:14 Desribe how on your graph how you obtained an estimate of the doubling time.

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RESPONSE --> ok. so 2= (1.08^x) so you look where y=2 is on the graphed line, and see what x is.

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19:29:16 **In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis.

The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **

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RESPONSE --> ok

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‹˜éz¡…ó¿’ɂɉ׊•û{˜å¡z assignment #004

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Calculus I 03-31-2005