Querry 04

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course Mth 272

6/24/13 around 10:45also did not go through

004. `query 4

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Question: `q4.6.1 (previously 4.6.06 (was 4.5.06)) y = C e^(kt) thru (3,.5) and (4,5)

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Your solution:

.5 = C e^(3K)

5= C e^(5k)

Divide equations

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

10 = e^k

K = ln10 or 2.3

5 = C e^(4*2.3)

5 = C e^(9.2)

C= 5/ e^(9.2)

C= .0005

Y= .0005e^(2.3t)

confidence rating #$&*: 2

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Given Solution:

`a 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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Self-critique (if necessary):

ok

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Self-critique Rating:

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Question: `q 4.6.2 (previously 4.6.10 (was 4.5.10)) solve dy/dt = 5.2 y if y=18 when t=0

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Your solution:

Dy/dt = 5.2y divide y

Dy /y = 5.2dt integrate both sides

Ln y = 5.2t + c

Y = e^(5.2t + c) take constant out of the power

Y = e^C e^(5.2t) e^C>0 = X

Y = X e^(5.2t)

Plug in numbers

18 = X e^0

18 = x

The function is y = 18e^(5.2t)

confidence rating #$&*: 2

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Given Solution:

`a 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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Self-critique (if necessary):

ok

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Self-critique Rating:

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Question: `q4.6.5 (previously 4.6.25 (was 4.5.25)) Init investment $1000, rate 12%.

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Your solution:

A = A0 e^(rt)

2000 = 1000 e^(.12t)

2 = e^(.12t)

Ln2 = .12t

T = ln2 / .12 = 5.77 years

For 10 years

A = 1000e^(.12*10) = 3320

For 25 years

A = 1000e^(.12*25) = 20085

confidence rating #$&*: 3

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Given Solution:

`a

Rate = .12 and initial amount is $1000 so we have

amt = $1000 e^(.12 t).

The equation for the doubling time is

1000 e^(.105 t) = 2 * 1000.

Dividing both sides by 1000 we get

e^(.12 t) = 2. Taking the natural log of both sides

.12t = ln(2) so that

t = ln(2) / .12 = 5.8 yrs approx.

after 10 years we have

• amt = 1000e^(.12(10)) = $3 320

after 25 yrs we have

• amt = 1000 e^(.12(25)) = $20 087

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q 4.6.8 (previously 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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Your solution:

5 = C e^(k300)

4 = C e^(k400)

5/4 = (C e^(k300)) / (C e^(K400))

5/4 = e^(k-100) get k by itself and take the ln of both sides

K = (-100)/ ln5/4

K = -448

5 = C e^(-448*300)

C = 5 / e^(-448*300)

confidence rating #$&*: 1

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Given Solution:

`a 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:

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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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Self-critique (if necessary):

#### messed up my algebra while solving for K.

Self-critique Rating 3

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