Assignment Four

course Mth 272

Confused still

??????????assignment #003003. `query 3

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Applied Calculus II

07-10-2007

?T?????????w????l?assignment #004

004. `query 4

Applied Calculus II

07-10-2007

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15:03:28

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RESPONSE -->

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15:07:10

4.6.06 (was 4.5.06) y = C e^(kt) thru (3,.5) and (4,5)

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I got 2.303 = 4k/3k

I basically solved for C using the first set of #'s. Once I got C I pluged that in along with the second set of #'s. From there I was a little confused and try to simplify as best I could.

confidence assessment: 1

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15:12:53

Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations

.5 = C e^(3*k)and

5 = Ce^(4k) .

Dividing the second equation by the first we get

5 / .5 = C e^(4k) / [ C e^(3k) ] or

10 = e^k so

k = 2.3, approx. (i.e., k = ln(10) )

Thus .5 = C e^(2.3 * 3)

.5 = C e^(6.9)

C = .5 / e^(6.9) = .0005, approx.

The model is thus close to y =.0005 e^(2.3 t). **

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RESPONSE -->

I dont get why you divide the second equation by the first. Is that some kind of rule?

As long as none of the quantities is zero, it's a valid step, involving one of the most basic rules of algebra. However it's not one that readily occurs to most students until they have seen it.

Also when you do that how does C e^(4k) / [ C e^(3k) ] get canceled out so that you get 10 = e^k? Basically I dont get this question

This involves very basic algebra and the laws of exponents, with which you should be familiar. A summary of the steps:

C e^(4k) / [ C e^(3k) ] = (C / C) * ( e^(4k) / e^(3k) ). Since C / C = 1 and 1 * ( e^(4k) / e^(3k) ) this simplifies to just
( e^(4k) / e^(3k) ). Among the laws of exponents: a^m / (a^n) = a^(m - n). So
( e^(4k) / e^(3k) ) = e^(4 k - 3 k) = e^k.

self critique assessment: 1

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15:21:10

4.6.10 (was 4.5.10) solve dy/dt = 5.2 y if y=18 when t=0

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i havnt a clue how to do this.

confidence assessment: 0

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15:28:52

The details of the process:

dy/dt = 5.2y. Divide both sides by y to get

dy/y = 5.2 dt. This is the same as

(1/y)dy = 5.2dt. Integrate the left side with respect to y and the right with respect to t:

ln | y | = 5.2t +C. Therefore

e^(ln y) = e^(5.2 t + c) so

y = e^(5.2 t + c). This is the general function which satisfies dy/dt = 5.2 y.

Now e^(a+b) = e^a * e^b so

y = e^c e^(5.2 t). e^c can be any positive number so we say e^c = A, A > 0.

y = A e^(5.2 t). This is the general function which satisfies dy/dt = 5.2 y.

When t=0, y = 18 so

18 = A e^0. e^0 is 1 so

A = 18. The function is therefore

y = 18 e^(5.2 t). **

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I get the process up until you have the intergrate the left side with repect to y and the right with respect to t.

self critique assessment: 0

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15:30:47

4.6.25 (was 4.5.25) init investment $750, rate 10.5%, find doubling time, 10-yr amt, 25-yr amt) New problem is init investment $1000, rate 12%.

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confidence assessment: 0

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15:32:04

When rate = .105 we have

amt = 1000 e^(.105 t) and the equation for the doubling time is

750 e^(.105 t) = 2 * 750. Dividing both sides by 750 we get

e^(.105 t) = 2. Taking the natural log of both sides

.105t = ln(2) so that

t = ln(2) / .105 = 6.9 yrs approx.

after 10 years

amt = 750e^.105(10) = $2,143.24

after 25 yrs

amt = 7500 e^.105(25) = $10,353.43 *

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I am confused this isnt even the problem in the book

self critique assessment: 0

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15:40:57

4.6.44 (was 4.5.42) demand fn p = C e^(kx) if when p=$5, x = 300 and when p=$4, x = 400

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I first plugged in the numbers for the first set of numbers and solved for C. Plugged C in with the second set of #'s. Now I am looking to solve for k. the problem is I always end up with a K / K ...basically I am lost.

confidence assessment: 0

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15:48:30

You get 5 = C e^(300 k) and 4 = C e^(400 k).

If you divide the first equation by the second you get

5/4 = e^(300 k) / e^(400 k) so

5/4 = e^(-100 k) and

k = ln(5/4) / (-100) = -.0022 approx..

Then you can substitute into the first equation:

}

5 = C e^(300 k) so

C = 5 / e^(300 k) = 5 / [ e^(300 ln(5/4) / -100 ) ] = 5 / [ e^(-3 ln(5/4) ] .

This is easily evaluated on your calculator. You get C = 9.8, approx.

So the function is p = 9.8 e^(-.0022 t). **

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your answers dont even match up with the numbers and answers in the book ... this does not help at ALL.... I am confused

self critique assessment: 0

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You are having a lot of trouble with some first-semester calculus procedures and also with algebra.

Of course it doesn't help if the problem don't match those in the text. I'll check into that tomorrow and get back to you.

You should, however, review (as I've recommended), and also go back and review and practice the laws of exponents. There is a chapter in the text which includes a review of algebra and basis precalculus.

I'll work with you to clarify these ideas, and if necessary I can give you extra time to finish the course. You need to begin, however, with a fairly extensive review.