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course Mth 279
2/15
q_a_03prior to class 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.
****
dP/dt = k * P and y ' + p(t) y = 0 are the same equation with the exception of p(t) = -k and y ' = dP/dt and y = P which make it true.
Really the only difference between the two are different letters for variables with the exception of p(t) = -k which makes these forms equal.
The forms are used for different purposes.
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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 = k * P
dP/P = k * dt
int (dP/P) = int (k * dt)
ln(P) = int (k * dt) + c
P = e^(int (k * dt) + c)
P = C e^(int (k * dt))
or
dP/dt = k * P
dP/dt - k * P = 0
which is in the form y ' + p(t) y = 0
and y = C e^(-int (p(t) dt))
P = C e^(int (k * dt))
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`q002. If k = .06 and P(0) = $1000, then what is the function P(t)?
****
p(t) = -k
If k = .06 then p(t) = -.06
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What is the meaning of P(0)?
****
P(t)
P(0) means that when t equals zero, P(t) = $1000
by the expression P(0) = $1000
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What is the meaning of P(5)?
****
A version of the function P(t), and means when t = 5.
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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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dP/dt = k * P
p(t) = -k
y ' + p(t) y = 0
u(t) = e^(int (p(t)dt))
y 'e^(int (p(t)dt)) + p(t) ye^(int (p(t)dt)) = 0
(ye^(int (p(t)dt)))' = 0
int ((ye^(int (p(t)dt)))' ) = 0
ye^(int (p(t)dt)) = C
y = C e^(-int (p(t)dt))
P(2) = $800
P(6) = $1100
no idea. ?
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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.
****
dP/dt = k P
dP/dt - k P = 0
P = C e^(int (k * dt))
P = C e^(int (0.06 * dt))
P = C e^(0.06t)
P = e^(0.06t)
p(0) = 1, p(12) = 2.05
P = 8e^(0.06t)
p(0) = 8, p(12) = 16.44
P = 3e^(0.06t)
p(0) = 3, p(12) = 6.16
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Sketch a curve which passes through the point (0, 4). At what point does this curve exit the box?
****
P = 4e^(0.06t)
p(0) = 4, p(12) = 8.22
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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?
****
P = 5.7 e^(0.06t)
p(0) = 5.7, p(12) = 11.7
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Describe the solution curve which passes through the origin.
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P = e^(0.06t) -0.5
This would pass through the origin but is not a solution curve for this problem.
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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'.
****
for all the solutions the intercept is the constant c
not sure about 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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This equation is therefore first-order linear and nonhomogeneous. 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= mAe^kt/(1+Ae^kt)
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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).
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?
****
No idea, couldn't find any notes or material on this.
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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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Write this as a differential equation and solve the equation.
****
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Find the particular solution for which r = 7 cm^3 / sec.
****
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Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#$&*
course Mth 279
2/15
q_a_03prior to class 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.
****
dP/dt = k * P and y ' + p(t) y = 0 are the same equation with the exception of p(t) = -k and y ' = dP/dt and y = P which make it true.
Really the only difference between the two are different letters for variables with the exception of p(t) = -k which makes these forms equal.
The forms are used for different purposes.
#$&*
Apply the techniques for solving a homogeneous first-order linear equation to the equation dP/dt = k * P.
****
dP/dt = k * P
dP/P = k * dt
int (dP/P) = int (k * dt)
ln(P) = int (k * dt) + c
P = e^(int (k * dt) + c)
P = C e^(int (k * dt))
or
dP/dt = k * P
dP/dt - k * P = 0
which is in the form y ' + p(t) y = 0
and y = C e^(-int (p(t) dt))
P = C e^(int (k * dt))
@&
Doing the integration you get
P = C e^(k t).
*@
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`q002. If k = .06 and P(0) = $1000, then what is the function P(t)?
****
p(t) = -k
If k = .06 then p(t) = -.06
#$&*
What is the meaning of P(0)?
****
P(t)
P(0) means that when t equals zero, P(t) = $1000
by the expression P(0) = $1000
#$&*
What is the meaning of P(5)?
****
A version of the function P(t), and means when t = 5.
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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)?
****
dP/dt = k * P
p(t) = -k
y ' + p(t) y = 0
u(t) = e^(int (p(t)dt))
y 'e^(int (p(t)dt)) + p(t) ye^(int (p(t)dt)) = 0
(ye^(int (p(t)dt)))' = 0
int ((ye^(int (p(t)dt)))' ) = 0
ye^(int (p(t)dt)) = C
y = C e^(-int (p(t)dt))
P(2) = $800
P(6) = $1100
no idea. ?
@&
The general solution to the equation is
P = C e^(k t).
P(2) = 800 implies that
800 = C e^(2 k)
and P(6) = 1100 implies
1100 = C e^(6 k).
So you have two equations with unknowns C and k.
Divide one equation by the other to eliminate C, solve for k, then plug that value into one of the equations to find C.
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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.
****
dP/dt = k P
dP/dt - k P = 0
P = C e^(int (k * dt))
P = C e^(int (0.06 * dt))
P = C e^(0.06t)
P = e^(0.06t)
p(0) = 1, p(12) = 2.05
P = 8e^(0.06t)
p(0) = 8, p(12) = 16.44
P = 3e^(0.06t)
p(0) = 3, p(12) = 6.16
@&
You haven't used the direction field to approximate your solution of this equation.
Of course this equation doesn't require a direction field, but the point here is to learn to use the direction field so you can use it when you have no other options.
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Sketch a curve which passes through the point (0, 4). At what point does this curve exit the box?
****
P = 4e^(0.06t)
p(0) = 4, p(12) = 8.22
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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?
****
P = 5.7 e^(0.06t)
p(0) = 5.7, p(12) = 11.7
#$&*
Describe the solution curve which passes through the origin.
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P = e^(0.06t) -0.5
This would pass through the origin but is not a solution curve for this problem.
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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'.
****
for all the solutions the intercept is the constant c
not sure about the box
@&
You need to solve this problem by sketching the direction field and approximating, without using the solution of the equation.
*@
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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.
#$&*
This equation is therefore first-order linear and nonhomogeneous. We have learned the technique for solving such equations.
What is the general solution to our equation dP/dt = k P + m?
****
p= mAe^kt/(1+Ae^kt)
#$&*
`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?
****
No idea, couldn't find any notes or material on this.
#$&*
What therefore is the net rate at which the amount of salt in the second bottle is changing?
****
#$&*
Write this as a differential equation and solve the equation.
****
#$&*
Find the particular solution for which r = 7 cm^3 / sec.
****
#$&*"
@&
You're given plenty of information to solve this problem.
7 cm^3 / second of a 10% solution is entering the second bottle. At what rate is salt therefore entering?
If q is the amount of salt in the second bottle at some later time, then how much salt is there in that bottle per cm^3? (The volume of the bottle is V, the amount of salt is q).
7 cm^3 of that solution is leaving every second. So in terms of the 7 cm^3 / second, and the values of q and V, at what rate is salt leaving the bottle?
OK, now q is continually changing so q is a function of t.
The rate at which the amount of salt in the bottle is changing is equal to the rate at which it is entering, minus the rate at which it is leaving.
What is this rate, in terms of q, V and that 7 cm^3 / second rate of flow?
q(t) is the amount of salt in the bottle. The rate at which the amount of salt is changing is therefore q ' (t). So the rate you wrote to answer the preceding is equal to q ' (t).
So set q ' (t) equal to that expression. What you get is a first-order differential equation in q.
Solve that equation.
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Good overall, but I'm going to ask you to submit revisions of problem `q004, and the last problem. I've given you notes on both.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@