#$&*
course Mth 279
10/6 10
q_a_03________________________________________
prior 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.
****
When you rearrange the equation dP/dt = k * P you get dP/dt - k*P = 0 which gives it the form of y ' + p(t) y = 0. Also when you rearrange the equation y = k*y you get y -k*y = 0 and from the theorem you use the expression to integrate that is in front of the y.
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Apply the techniques for solving a homogeneous first-order linear equation to the equation dP/dt = k * P.
****
dP/dt - k*p =0 so taking the negative integral of - k we get kt then we put that to the exponent. So our p(t) = e^(kt) r
dP/dt = ke^(kt) so plugging these expressions into the equation we get ke^(kt)- ke^(kt) = 0.
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`q002. If k = .06 and P(0) = $1000, then what is the function P(t)? P(t) = 1000e^(-.06t)
What is the meaning of P(0)? The value of the principle when the time is zero.
What is the meaning of P(5)? The value of the principle when the time is at 5 seconds.
`q003. Under different conditions, suppose that the equation dP/dt = k * P holds, and we know that P(2) = $800 and P(6) = $1100.
What are the values of k and P(0)? 800 = e^(-2k) to find k take the natural log of both sides
k= -3.34 and P(0) = 3.625 ???I am confused here. I think I did something wrong.
@&
The general solution would be P = A e^(k t).
P(2) = 800 and P(6) = 1100 then yield equations
800 = A e^(2 k)
1100 = A e^(6 k),
which constitute two simultaneous equations in the two unknowns A and k.
To solve I suggest you start by dividing the second equation by the first, which eliminates A.
*@
`q004. Assuming k = .06, sketch the direction field of the equation dP/dt = k P for P ranging from 0 to 12, and for t ranging from 0 to 12. Use an increment of 4 for t and an increment of .5 for P.
Based on your sketch, plot a variety of solution curves.
Each curve will leave the 'box' defined by 0 <= t <= 12, 0 <= P <= 12 at some point. Any solution curve must leave the box by the top and some by the right side. Be sure you have included curves with both properties.
All your curves can be extended to the left until they intersect the y axis. If necessary, extend your curves accordingly.
For three different curves, at least one of which exits the box to the right and at least one of which exits from the top, indicate the coordinates of the point at which the curve enters, and the point at which it exits.
****
I am not exactly sure how to do this.
#$&*
@&
You appear to have also left the direction field questions blank on an earlier assignment.
Do you have the DVD's?
*@
Sketch a curve which passes through the point (0, 4). At what point does this curve exit the box?
****
I am not sure.
#$&*
Sketch the curve which exits the box through the top right-hand corner. At what point does this curve enter the box from the left?
****
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Describe the solution curve which passes through the origin.
****
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If P is principle in thousands of dollars, then what is the interpretation of each of the solution curves you have sketched? In particular be sure you have stated the meaning of the intercept of the graph with the P axis, and of the point at which the curve leaves the 'box'.
****
I need further explanation on this.
#$&*
`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) = - m 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?
****
Write the equation as y - k*y = m
Begin by multiplying both sides by e^(-kt)
This gives us ye^(-kt) = me^(-kt)
@&
If you multiply both sides by e^(-kt) the left-hand side does not become y e^(-kt). The left hand side will have two terms, which constitute (y e^(-kt) ) ' .
*@
Now take the derivative with respect to t of both sides.
This gives us ye^(-kt) = -1/k * me^(-kt) + c
Now solve for y which gives us y =( -1/k)*(me^-kt + c)/(e^(-kt))
When simplified we get y = (1/k)*(1+ ce^(kt))
@&
You need to rework from my note. I don't think you'll have difficulty doing so.
*@
#$&*
`q006. I can mix a 10% salt solution into pure water by siphoning the salt solution from one bottle, into a second bottle which is initially full of pure water. The second bottle was initially full, so it will overflow.
We will make the simplifying assumption that the salt water flowing into the second bottle is instantly distributed equally throughout that bottle.
Assume that the rate of flow of salt solution is r (e.g., r might be, say, 7 cm^3 / second). The second bottle is assumed to have volume V (e.g., the second bottle might have a volume of 500 cm^3, corresponding to half a liter).
It should be clear that the salt solution in the second bottle will become increasingly concentrated, resulting in an increasing amount of salt in that bottle.
At what rate is salt entering the second bottle from the first?
If q(t) represents the amount of salt in the second bottle, as a function of clock time, then at clock time t, at what rate is salt leaving the bottle in the overflow?
****
Rate salt entering : 10% r = .10*r
Rate salt leaving: q(t)/V *r
#$&*
What therefore is the net rate at which the amount of salt in the second bottle is changing?
****
Net Rate = .10r - q(t)/V *r
#$&*
Write this as a differential equation and solve the equation.
****
q + r/V * q = .10r
#$&*
Find the particular solution for which r = 7 cm^3 / sec.
****
Multiply everything by e^(-r^2/2v)
You get r/v * q e^(-r^2/2v) = .10r e^(-r^2/2v)
@&
When you multiply by the integrating factor, you get two terms on the left-hand side, the first of which will contain q ' as a factor.
*@
Take the integral of both sides with respect to r
You get -qr^2/v^2 e^(-r^2/2v) = (-.10r^2/ v * e^(-r^2/2v) + .10 e^(-r^2/2v) + c) / e^(-r^2/2v)
Solve for q, plug in givens and get
q = (23675 + 250000c) / 51.46
#$&*
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#$&*
course Mth 279
10/6 10
q_a_03________________________________________
prior 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.
****
When you rearrange the equation dP/dt = k * P you get dP/dt k*P = 0 which gives it the form of y ' + p(t) y = 0. Also when you rearrange the equation y = k*y you get y k*y = 0 and from the theorem you use the expression to integrate that is in front of the y.
#$&*
Apply the techniques for solving a homogeneous first-order linear equation to the equation dP/dt = k * P.
****
dP/dt k*p =0 so taking the negative integral of - k we get kt then we put that to the exponent. So our p(t) = e^(kt) r
dP/dt = ke^(kt) so plugging these expressions into the equation we get ke^(kt)- ke^(kt) = 0.
#$&*
`q002. If k = .06 and P(0) = $1000, then what is the function P(t)? P(t) = 1000e^(-.06t)
What is the meaning of P(0)? The value of the principle when the time is zero.
What is the meaning of P(5)? The value of the principle when the time is at 5 seconds.
`q003. Under different conditions, suppose that the equation dP/dt = k * P holds, and we know that P(2) = $800 and P(6) = $1100.
What are the values of k and P(0)? 800 = e^(-2k) to find k take the natural log of both sides
k= -3.34 and P(0) = 3.625 ???I am confused here. I think I did something wrong.
@&
The general solution would be P = A e^(k t).
P(2) = 800 and P(6) = 1100 then yield equations
800 = A e^(2 k)
1100 = A e^(6 k),
which constitute two simultaneous equations in the two unknowns A and k.
To solve I suggest you start by dividing the second equation by the first, which eliminates A.
*@
`q004. Assuming k = .06, sketch the direction field of the equation dP/dt = k P for P ranging from 0 to 12, and for t ranging from 0 to 12. Use an increment of 4 for t and an increment of .5 for P.
Based on your sketch, plot a variety of solution curves.
Each curve will leave the 'box' defined by 0 <= t <= 12, 0 <= P <= 12 at some point. Any solution curve must leave the box by the top and some by the right side. Be sure you have included curves with both properties.
All your curves can be extended to the left until they intersect the y axis. If necessary, extend your curves accordingly.
For three different curves, at least one of which exits the box to the right and at least one of which exits from the top, indicate the coordinates of the point at which the curve enters, and the point at which it exits.
****
I am not exactly sure how to do this.
@&
You appear to have also left the direction field questions blank on an earlier assignment.
Do you have the DVD's?
*@
#$&*
Sketch a curve which passes through the point (0, 4). At what point does this curve exit the box?
****
I am not sure.
#$&*
Sketch the curve which exits the box through the top right-hand corner. At what point does this curve enter the box from the left?
****
#$&*
Describe the solution curve which passes through the origin.
****
#$&*
If P is principle in thousands of dollars, then what is the interpretation of each of the solution curves you have sketched? In particular be sure you have stated the meaning of the intercept of the graph with the P axis, and of the point at which the curve leaves the 'box'.
****
I need further explanation on this.
#$&*
`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) = - m 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.
#$&*
This equation is therefore first-order linear and nonhomogeneous. We have learned the technique for solving such equations.
What is the general solution to our equation dP/dt = k P + m?
****
Write the equation as y k*y = m
Begin by multiplying both sides by e^(-kt)
This gives us ye^(-kt) = me^(-kt)
@&
If you multiply both sides by e^(-kt) the left-hand side does not become y e^(-kt). The left hand side will have two terms, which constitute (y e^(-kt) ) ' .************
*@
Now take the derivative with respect to t of both sides.
This gives us ye^(-kt) = -1/k * me^(-kt) + c
Now solve for y which gives us y =( -1/k)*(me^-kt + c)/(e^(-kt))
When simplified we get y = (1/k)*(1+ ce^(kt))
@&
You need to rework from my note. I don't think you'll have difficulty doing so.
*@
#$&*
`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?
****
Rate salt entering : 10% r = .10*r
Rate salt leaving: q(t)/V *r
#$&*
What therefore is the net rate at which the amount of salt in the second bottle is changing?
****
Net Rate = .10r q(t)/V *r
#$&*
Write this as a differential equation and solve the equation.
****
q + r/V * q = .10r
#$&*
Find the particular solution for which r = 7 cm^3 / sec.
****
Multiply everything by e^(-r^2/2v)
You get r/v * q e^(-r^2/2v) = .10r e^(-r^2/2v)
@&
When you multiply by the integrating factor, you get two terms on the left-hand side, the first of which will contain q ' as a factor.
*@
Take the integral of both sides with respect to r
You get qr^2/v^2 e^(-r^2/2v) = (-.10r^2/ v * e^(-r^2/2v) + .10 e^(-r^2/2v) + c) / e^(-r^2/2v)
Solve for q, plug in givens and get
q = (23675 + 250000c) / 51.46
#$&*
"
@&
You are missing a step in reasoning out solutions to nonhomogeneous equations.
Let me know if you have the DVD's, which include explanations of direction fields.
*@