Query 4

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course Mth 272

June 3, 20128:15pm

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^(k3) and 5 = C e^(k4)

.5=C*ln(3k) and 5=C*ln(4k)

confidence rating #$&*: 0

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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): I got stuck and wasn’t sure what to do. I figured I should isolate C, but wasn’t sure how to go about solving the rest of the problem. I figured I had done it wrong by the time I got to that step.

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Right. You can't just replace the exponential expressions with logarithmic expressions.

Do you understand the given solution?

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Self-critique Rating: 1

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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(18)

18d/0d=93.6

Multiply each side by 18 to isolate the variable d

1684.6=d

confidence rating #$&*:

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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.

@&

d is not a variable. dy/dt indicates the derivative of y with respect to t.

The technique of solving this problem is covered in the text and in the Class Notes.

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Self-critique Rating: 2

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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:

Initial investment=1000e^.12

This is the formula I used to solve similar problems with the given information; however, the problem isn’t asking me over what period of time I am solving for.

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): A time period was missing in order to solve this problem.

@&

The full problem is stated in your text.

It is expected that you will have solved the problems in the text before completing the Query.

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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:

This information was from the same section as the first number that I need to review.

@&

Once more, having solved this problem as assigned you will have seen the full statement and either done the necessary steps, or posed questions.

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confidence rating #$&*: 0

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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:

}

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): OK.

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Self-critique Rating: 2

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@&

Check my notes.

Be sure you are working the entire set of assigned problems in your text.

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