#$&*
course M 279
2-2 12
`q001. If your money grows at some rate, and is compounded continuously, then the rate of change of your principle is a multiple of your principle. In the form of a differential equation, this says thatdP/dt = k * P
for some constant k.
This equation is of the same type as y ' = k * y, which can be arranged to the form y ' + p(t) y = 0, with p(t) = -k (a constant function). This equation is first-order linear and homogeneous. Explain in detail the connection between the given equation dP/dt = k * P and the form y ' + p(t) y = 0, and how we conclude that p(t) = -k.
****
Connection is …
dP/dt = y '
k * P = p(t) y
where dP/dt = k * P can be arranged dP/dt - k * P = 0
In equation when k * P is moved to left side k*P becomes neg, so we have -k
#$&*
Apply the techniques for solving a homogeneous first-order linear equation to the equation dP/dt = k * P.
****
Solve e^(Int(k dk)) = e^(k^2/2)
For P = e^(k^2/2)
dP/dt = ke^(k^2/2), when plugged into either dP/dt = k * P or dP/dt - k * P = 0
k e^(k^2/2) = k e^(k^2/2), which is true or….
ke^(k^2/2) - k e^(k^2/2) = 0, which is true as well.
#$&*
`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)?
`q003. Under different conditions, suppose that the equation dP/dt = k * P holds, and we know that P(2) = $800 and P(6) = $1100.
What are the values of k and P(0)?
`q004. Assuming k = .06, sketch the direction field of the equation dP/dt = k P for P ranging from 0 to 2000, and for t ranging from 0 to 12. Use an increment of 4 for t and an increment of 500 for P.
Based on your sketch, plot a variety of solution curves.
Each curve will leave the 'box' defined by 0 <= t <= 12, 0 <= P <= 5000 at some point. Any solution curve must leave the box by the top and some by the right side. Be sure you have included curves with both properties.
All your curves can be extended to the left until they intersect the y axis. If necessary, extend your curves accordingly.
For three different curves, at least one of which exits the box to the right and at least one of which exits from the top, indicate the coordinates of the point at which the curve enters, and the point at which it exits.
****
!!!!!!!!!!!!!!!With P ranging from 0 to 12, not 0 to 2000!!!!!!!!!!!!!!!!
To right
(0, 4) enters, (12, 7) exits
Thru top
(0, 10) enters, (6, 12) exits
Top right corner
(0, 7.5) enters, (12, 12) exits
#$&*
Sketch a curve which passes through the point (0, 1000). At what point does this curve exit the box?
****
In class you said not to use this scale.
#$&*
Sketch the curve which exits the box through the top right-hand corner. At what point does this curve enter the box from the left?
****
Approx (0, 7.5)
#$&*
Describe the solution curve which passes through the origin.
****
Runs along x axis due to constant slope of 0.
#$&*
`q005. The equation dP/dt = k P + m is of the same form as y ' + p(t) y = g(t), with p(t) = -k and g(t) = m
Explain in detail why the equation is of this general form, specifically explaining how we get p(t) = - m and g(t) = m.
****
?????I think there is a misprint, instead of p(t) = - m maybe p(t) = - k????????
@& The equation can be rearranged to get
P ' - k P = m,
which is of the form y ' + p(t) y = m, with p(t) = -k and g(t) = m.
*@
If so…
This problem is same as one of the previous ones..
where p(t) = -k, because to get into the right form kP must be moved to left side, making kP neg or giving us –k
where g(t) = m because it stays on right side so m stays positive.
Everything else has same explanation as before.
#$&*
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?
****
We have rearranged equation
dP/dt - k P = m
First we find
e^(Int(-k)) = e^(-k^2/2)
Mult. dP/dt - k P = m by e^(-k^2/2)
@& The integral is with respect to t, the same variable with respect to which we differentiate to get y '.*@
e^(-k^2/2)*dP/dt - e^(-k^2/2)*k P = e^(-k^2/2)*m
We check to make sure (P* e^(-k^2/2))’ = left hand side
(P* e^(-k^2/2))’ = P’* e^(-k^2/2) + P*(e^(-k^2/2))’
P’* e^(-k^2/2) = dP/dt* e^(-k^2/2)
P*(e^(-k^2/2))’ = P*(e^(-k^2/2)*-k)) = - e^(-k^2/2)*k P
(P* e^(-k^2/2))’ = e^(-k^2/2)*dP/dt - e^(-k^2/2)*k P, works
Thus
(P* e^(-k^2/2))’ = e^(-k^2/2)*m and then
Int((P* e^(-k^2/2))’ dt) = Int(e^(-k^2/2)*m dt)
P* e^(-k^2/2) = Int(e^(-k^2/2)*m dt)
Note k = p(t), that’s why derivative is taken wrt t!!!!!
@& You integration of k was, as pointed out, in error.
Otherwise you followed the steps of the process well.*@
@& 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).
*@
#$&*
`q006. I can mix a 10% salt solution into pure water by siphoning the salt solution from one bottle, into a second bottle which is initially full of pure water. The second bottle was initially full, so it will overflow.
We will make the simplifying assumption that the salt water flowing into the second bottle is instantly distributed equally throughout that bottle.
Assume that the rate of flow of salt solution is r (e.g., r might be, say, 7 cm^3 / second). The second bottle is assumed to have volume V (e.g., the second bottle might have a volume of 500 cm^3, corresponding to half a liter)
If p(t) is the percent concentration of salt in the second bottle, as a function of clock time, then at what rate is salt leaving the bottle at clock time t?
****
r
Is both in flow and out flow
p(t)
percent concentration of salt in the second bottle
V
second bottle assumed volume
dp/dt
rate salt is leaving the bottle at clock time t
Because p(t) is concentration of salt and equals a/V where a(amount of salt present) is divided by V(total volume of bottle),so a = p(t)*V. So p(t)*V gives us concentration of salt per unit of volume. Where dp/dt is rate of salt leaving 2nd bottle, then…
dp/dt = (p(t)*V)*r
So rate of change of salt leaving the bottle wrt time is concentration of salt per unit of volume in 2nd bottle times the rate at which that volume is exiting the bottle.
#$&*
@& 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).
*@
@& Good work, but not without a couple of errors, so be sure to see my notes.*@