hw asst  4

course mth 272

I just wanted to remind you that I have submitted my proctor info to you and just want to make sure everything is ok with that. I plan on taking the test #1 on wed or thurs of this week at JTCC.

I've contacted your proctor and emailed you an acknolwedgement; thanks for reminding me.

????????????H?assignment #004004. `query 4

Applied Calculus II

06-22-2008

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08:40:29

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

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RESPONSE -->

dy/dt=ky

y=e^kt

5=e^k(3)

ln5 = 3k

(1/3)ln5 = k

k= .5364793041

y = 3e^.536479t

confidence assessment: 1

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08:48:13

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 -->

where did .5 come from?

The problem gives two points, (3, .5) and (4, 5).

You substitute the form y = C e^(k t) into this form and you get two simultaneous equations, which you solve for the two unknown parameters C and k.

you use the model and plug in your x and y into it.

.5 = C e^(3*k)

5 = Ce^(4k)

then divide the 2nd by the first

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

10 = e^k

k=2.3 (ln10=2.3)

to find C you plug in 2.3 to .5 = C e^(2.3*3)

solving for C you get .0005

y = .0005 e^(2.3 t)

self critique assessment: 2

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08:57:18

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

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RESPONSE -->

5.2=Ce^0

5.2=C

find k

y= 5.2e^kt

18=5.2e^k0

3.46=e^0k

1.24=k

y= 5.2e^1.24t

confidence assessment: 1

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09:03:32

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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RESPONSE -->

ok, I see that y = A e^(5.2 t) is the general function which satisfies dy/dt = 5.2 y

because 5.2=k

plug in

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

A = 18

y =18e^(5.2t)

self critique assessment: 2

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09:10:25

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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RESPONSE -->

use A=Pe^rt

for 10 yr amt

A=750e^.105(10)

A = 2143.24

for 25 yr amt

A = 750e^.105(25)

A=10353.43

confidence assessment: 1

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09:12:47

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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RESPONSE -->

3

self critique assessment: 3

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09:19:33

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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RESPONSE -->

i think you set up

5=Ce^300k

and solve for C, but you dont know k, so im confused

Again you have two pairs of values so you can get two equations for the two parameters C and k.

confidence assessment: 0

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09:28:29

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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RESPONSE -->

5 = C e^(300 k) 4 = C e^(400 k)

divide the first into the 2nd

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

5/4 = e^(-100 k)

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

then you substitue this into the first equation

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

C=9.8

plug in p= 9.8e^(-.0022 t)

self critique assessment: 2

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Good work. See my notes and let me know if you have questions. &#