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course Mth 279
February 15 around 5pm.
q_a_03Class 110131
`q001. If your money grows at some rate, and is compounded continuously, then the rate of change of your principle is a multiple of your principle. In the form of a differential equation, this says that
dP/dt = k * P
for some constant k.
This equation is of the same type as y ' = k * y, which can be arranged to the form y ' + p(t) y = 0, with p(t) = -k (a constant function). This equation is first-order linear and homogeneous. Explain in detail the connection between the given equation dP/dt = k * P and the form y ' + p(t) y = 0, and how we conclude that p(t) = -k.
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This is a first order linear homogeneous.
P’ - kP = 0
General solution: P=e^(-kt+c)
=e^(c)*e^(-kt)
=Ae^(-kt) with A>0
We conclude that p(t)=-k because if we look at P’ - kP = 0, you see that -k is in front of P, which is p(t).
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Apply the techniques for solving a homogeneous first-order linear equation to the equation dP/dt = k * P.
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dP/dt=kP
dP/P=kdt
Int (dP/P) = Int (k dt)
=ln |P| = kt+C
=e^(ln|P|) = e^(kt)+C
P=e^(kt+c)
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@& Good. Note that e^(k t + c) = e^c * e^(k t), that e^c is always a positive number and can take any positive value, so that we can replace e^c with A, as long as we specify A > 0.
Thus the solution is
P = A e^(k t).
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`q002. If k = .06 and P(0) = $1000, then what is the function P(t)?
What is the meaning of P(0)?
What is the meaning of P(5)?
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P=Ae^(-kt)=Ae^(0.06t)
P(0)=$1000
A=$1000
P(t)=$1000e^(0.06t)
P(0)=Ae^(0.06(0))
=A
P(5)=Ae^(0.06(5))
=1.35A
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@& Good.
Since A = $1000, your solution for P(5) would be
P(5) = $1000 e^(.06 * 5) = $1350.
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`q003. Under different conditions, suppose that the equation dP/dt = k * P holds, and we know that P(2) = $800 and P(6) = $1100.
What are the values of k and P(0)?
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[Ae^(6k)/Ae^(2k)]=(1100/800)=11/8
e^(4k)=11/8 so 4k=ln|11/8|
k=1/4 ln|11/8|
=0.0796
Ae^(2k)=$800
A=800/e^(2*0.0796)
=$700
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`q004. Assuming k = .06, sketch the direction field of the equation dP/dt = k P for P ranging from 0 to 12, and for t ranging from 0 to 12. Use an increment of 4 for t and an increment of .5 for P.
Based on your sketch, plot a variety of solution curves.
Each curve will leave the 'box' defined by 0 <= t <= 12, 0 <= P <= 12 at some point. Any solution curve must leave the box by the top and some by the right side. Be sure you have included curves with both properties.
All your curves can be extended to the left until they intersect the y axis. If necessary, extend your curves accordingly.
For three different curves, at least one of which exits the box to the right and at least one of which exits from the top, indicate the coordinates of the point at which the curve enters, and the point at which it exits.
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I’m going to submit the next 4 questions (‘q004) when I finish working on this problem.
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Sketch a curve which passes through the point (0, 4). At what point does this curve exit the box?
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Sketch the curve which exits the box through the top right-hand corner. At what point does this curve enter the box from the left?
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Describe the solution curve which passes through the origin.
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If P is principle in thousands of dollars, then what is the interpretation of each of the solution curves you have sketched? In particular be sure you have stated the meaning of the intercept of the graph with the P axis, and of the point at which the curve leaves the 'box'.
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`q005. The equation dP/dt = k P + m is of the same form as y ' + p(t) y = g(t), with p(t) = -k and g(t) = m
Explain in detail why the equation is of this general form, specifically explaining how we get p(t) = - k and g(t) = m.
The equation can be rearranged to get
P ' - k P = m,
which is of the form y ' + p(t) y = m, with p(t) = -k and g(t) = m.
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dP/dt=kP+m
dP/dt=P’=y’
k=-p(t)
P=y
m=g(t)
P’ -kP=m
OR multiply both sides by e^(kt)
=ye^(kt) = (1/k)*m*e^(kt) + c
=y= (1/k)*((me^(kt) + c)/(e^(kt)))
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This equation is therefore first-order linear and non-homogeneous. We have learned the technique for solving such equations.
What is the general solution to our equation dP/dt = k P + m?
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P’ - kP = m
p(t)=-k so P(t)=(-k^2/2) so mu(t)=e^(-k^2/2)
e^(-k^2/2)P’ - e^(-k^2/2)kP = e^(-k^2/2)*m
e^(-k^2/2)P=Int (e^(-k^2/2)*m)
Solve for P.
Would this be the general solution? Or would P’ - kP = m?
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@& You appear to be integrating k with respect to k.
As was done previously, to avoid confusion we will use y(t) instead of P(t) in the process of solving the problem.
The equation will therefore be expressed for the moment as
y ' - k y = m.
This is a nonhomogenous linear first-order equation.
To solve such an equation we multiply by the integrating factor e^integral(p(t) dt) = e^(-k t).
Our equation becomes
e^(- k t) y ' - k e^(-k t) y = m e^(- k t),
which you can verify is the same as
(y e^(-k t) ) ' = m e^(-k t).
Integrating both sides we have
y e^(-k t) = integral ( m e^-(k t) ) so that
y e^(-k t) = -m / k * e^(-k t) + c. Solving for y (divide both sides by e^(-k t)):
y = -m / k + c / e^(- k t) or
y = -m / k + c e^(k t).
Returning to P(t) for our population function, our solution is
P(t) = -m / k + c e^(k t).*@
`q006. I can mix a 10% salt solution into pure water by siphoning the salt solution from one bottle, into a second bottle which is initially full of pure water. The second bottle was initially full, so it will overflow.
We will make the simplifying assumption that the salt water flowing into the second bottle is instantly distributed equally throughout that bottle.
Assume that the rate of flow of salt solution is r (e.g., r might be, say, 7 cm^3 / second). The second bottle is assumed to have volume V (e.g., the second bottle might have a volume of 500 cm^3, corresponding to half a liter).
It should be clear that the salt solution in the second bottle will become increasingly concentrated, resulting in an increasing amount of salt in that bottle.
At what rate is salt entering the second bottle from the first?
If q(t) represents the amount of salt in the second bottle, as a function of clock time, then at clock time t, at what rate is salt leaving the bottle in the overflow?
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P(t)=q(t)/V
q’ + (r/v)q = 0.10r
In: (10%)*r=0.10r
Out: (q(t)/V)*r
Rate at which salt ENTERS is dq/dt.
Rate is: dq/dt=0.10r-((q(t)/V)*r) first order linear non-homogeneous
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What therefore is the net rate at which the amount of salt in the second bottle is changing?
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Net: In-Out=0.10r-((q(t)/V)*r)
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Write this as a differential equation and solve the equation.
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q’ + (r/v)q = 0.10r (solve for “q”)
I would have to find the derivative of (r/v). WRT what?! I would then do mu(t)=e^(that deriv) and multiply both sides by this. Integrate and solve for q. Right?!
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@& The rate of change is with respect to clock time t, so the integral is with respect to t.
The rate at which amount in second bottle changes is dq/dt, so the equation is
dq/dt = +.10 r - q(t) / V * r.
Putting this into the form
q ' + r / V * q = .10 r
we see that it is first-order linear nonhomogeneous.
The integrating factor is e^(r / V * t) and the equation becomes
(q e^(r / V * t) ) ' = .10 r * e^(r / V * t).
Integrating we get
q e^(r / V * t) = .10 V / r * e^(r / V * t) + c
so that
q = .10 V / r + c e^(-r / V * t).*@
Find the particular solution for which r = 7 cm^3 / sec.
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I would have to solve the equation first before I could plug in 7cm^3/sec.
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@& See if you can find the particular solution.*@
@& You're doing well. See my notes, which should get you over a couple of issues, and if you aren't completely sure of your solutions resubmit all or part of the document, marking new insertions in the usual manner.*@