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course MTH 279
1/30 12:20 pm
110131 pre-class cal1 qa10`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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dP/dt is equivalent to P' and that's of the form y' and thus P relates to y and p(t) is thus -k since p(t) can be a constant. So we can put dP/dt into the first-order linear homogeneous equation form.
dP/dt-kP=0 or P'-kP=0
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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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p(t)=-k we have e^(-∫ p(t) dt) so for this: P= e^(kt+c)= C*e^(kt) where C= constant is a general solution to the equation.
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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(t)= 1000*e^(0.06t)
P(0)= initial money invested.
P(5)= money after 5 years, assuming t is in years.
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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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Here we have a system of equations, well two anyway.
P(2)= 8000= C*e^(2k)
P(6)= 1100= C*e^(6k)
Solve one for C:
C= 1100/e^(6k)
Then sub into the other eq:
8000=1100/e^(6k)*e^(2k)= 1100*e^(-6k)*e^(2k)= 1100*e^(-4k) ---> k= (ln 800/1100)/-4 = 0.0796 --> C= 682.3
Thus k=0.0796 and P(0)= 682.3
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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 2000, and for t ranging from 0 to 12. Use an increment of 4 for t and an increment of 500 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 <= 5000 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 guess the graph below is like the one we're asked to draw but the P scale is way off so I drew my own.
In-Out
A: (0,1500)-(4.75, 2000): Top
B: (0,1250)-(7.75, 2000): Top
C:(0,250)-(12, 500): Right
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@& Good. I've corrected this problem. Should have had P running from 0 to 2, instead of 0 to 2000, with the units each representing 1000.
We'll talk more about that Wednesday.*@
Sketch a curve which passes through the point (0, 1000). At what point does this curve exit the box?
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(11.5, 2000)
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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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(0, 975)
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Describe the solution curve which passes through the origin.
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This curve is zero and is a straight line with 0 slope and this makes sense since P(0)=0 ---> P= 0*e^(0.06t)
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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.
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dP/dt= P' then P' relates to y' and then P relates to y and we can rearrange this to be:
P'-kP=m which is just like y'-ky=m
Now it's easy to see how p(t)= -k and g(t) = m since functions of t can be constants.
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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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We have an integrating factor of e^(∫ p(t) dt) so our rearranged equation becomes:
e^(-kt)[P'-kP=m] --> e^(-kt)*P-ke^(-kt)*P= m*e^(-kt)
The right side is the product rule derivative of P*e^(kt) so:
∫ e^(-kt)*P-ke^(-kt)*P= P*e^(-kt)+C= ∫ m*e^(-kt)
P(t) = e^(kt)*∫ m*e^(-kt) - C*e^(kt)
That is the general solution.
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`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)
If p(t) is the percent concentration of salt in the second bottle, as a function of clock time, then at what rate is salt leaving the bottle at clock time t?
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This is just what I thought at first run through after having read through the section in the book but still trying to reason it out on my own.
Here r(t)= rate of flow in and if c= concentration of inflowing solution then r*c= rate of inflowing salt where r= cm^3/s and c= kg/cm^3.
V here is constant and the concentration is salt per unit volume or: cr/V which is in kg/cm^3
The flow of salt out is the concentration of the new solution times the out flow rate or C*R
The volume of solution in must be equal the volume out or :
r(t)*t= R(t)*t so r(t)= R(t) Then:
p(t)= p_0 + (crt - CR)/V t then the derivative wrt t is:
dp/dt=cr-CR or dp/dt= R(c-C) then R =(c-C)^-1 dp/dt
Well I just realized that p(t) should be C so I'll definitely have to revise this. I think that p(t)= crt-Rt where R is outflow of salt so dp/dt= cr-R or R=cr-dp/dt might be closer.
#$&*... if second bottle isn't initially full and sealed ... etc.
@& You're on the right track. We'll work through this on Wednesday.*@
*#&!
@& Very good. Check my note(s).*@