course Mth 272
6/21/104:10pm
Assignment/ 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:
y=Ce^kt
.5 =Ce^3k / .5Ce^4k
10=e^k
ln10= k*lne
ln10/lne
k=2.3026
.5 = C e^(2.3 * 3)
.5 = C e^(6.9)
C = .5 / e^(6.9) = .0005
y =.0005 e^(2.3 t).
confidence rating #$&*3
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Given Solution:
`a Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations
.5 = C e^(3*k)and
5 = Ce^(4k) .
Dividing the second equation by the first we get
5 / .5 = C e^(4k) / [ C e^(3k) ] or
10 = e^k so
k = 2.3, approx. (i.e., k = ln(10) )
Thus .5 = C e^(2.3 * 3)
.5 = C e^(6.9)
C = .5 / e^(6.9) = .0005, approx.
The model is thus close to y =.0005 e^(2.3 t). **
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Self-critique (if necessary):OK
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Self-critique rating #$&*3
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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
dive both sides by y
(1/y)dy = 5.2dt
integrate both sides
ln | y | = 5.2t +C
y = e^(5.2 t + c)
e^c e^(5.2 t)
18: A e^0. e^0 = 1
A: =18
y = 18 e^(5.2 t)
confidence rating #$&*3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Given Solution:
`a The details of the process:
dy/dt = 5.2y. Divide both sides by y to get
dy/y = 5.2 dt. This is the same as
(1/y)dy = 5.2dt. Integrate the left side with respect to y and the right with respect to t:
ln | y | = 5.2t +C. Therefore
e^(ln y) = e^(5.2 t + c) so
y = e^(5.2 t + c). This is the general function which satisfies dy/dt = 5.2 y.
Now e^(a+b) = e^a * e^b so
y = e^c e^(5.2 t). e^c can be any positive number so we say e^c = A, A > 0.
y = A e^(5.2 t). This is the general function which satisfies dy/dt = 5.2 y.
When t=0, y = 18 so
18 = A e^0. e^0 is 1 so
A = 18. The function is therefore
y = 18 e^(5.2 t). **
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Self-critique (if necessary):OK
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Self-critique rating #$&*3
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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 = $1000 e^(.12 t).
2*1000 = $1000e^.12t
e^(.12 t) = 2
.12t = ln2
ln2/.12 = t
t = 5.8 yrs
10 yrs: 1000e^(.12(10)) = $3 320
25yrs:1000 e^(.12(25)) = $20 087
confidence rating #$&*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=$4, x = 400
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Your solution:
5 = C e^(300 k) / 4 = C e^(400 k)
5/4 = e^(300 k) / e^(400 k)
k = ln(5/4) / (-100)
k=-.0022
Solve for C
5 = C e^(300 k)
C = 5 / e^(300 k) = 5 / [ e^(300 ln(5/4) / -100 ) ] = 5 / [ e^(-3 ln(5/4) ]
C = 9.8
p = 9.8 e^(-.0022 t)
confidence rating #$&*3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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) s
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 #$&*3"
&#Very good work. Let me know if you have questions. &#
#$&*