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course Mth 279
2-19 5
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.
dP/dt = y’
k*P = p(t) y so that
dP/dt = k * P is equal to y ' = - p(t) y
when equation is in the form y’ = p(t) y then for p(t) = k then to set equation equal to zero
y’ - p(t) y = 0 or y’ + p(t) y = 0 where p(t) is neg.
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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, rearranging
dP/dt - k*P = 0, integration constant ie e^(Int(k dt)) = e^(kt + C). Allowing e^(kt) to sub in for P
P’ = ke^(kt) so that we see that
P’ - k*P = 0 ke^(kt) - ke^(kt) = 0, which checks out.
P(t) = e^(kt + C) = Ae^(kt)
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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) = A e^(.06 t) (see the preceding solution).
P(0) = A e^(.06 * 0) = A, so since P(0) = $1000, we have A = $1000.
For this condition, then,
P(t) = $1000 * e^(.06 t)
and we have
P(5) = $1000 * e^(.06 * 5) = $1000 * e^(.30), which when evaluated yields the result #1350, approximately.
COMMON ERROR: Students will often integrate k and get k^2 / 2. However this would be the integral with respect to k, not with respect to t. The integral is with respect to t, the same variable with respect to which we differentiate to get y '.
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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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Our solution is
P = A e^(k t).
Our two conditions tell us that
$800 = A e^(k * 2)
and
$1000 = A e^(k * 6).
We solve these equations simultaneously to get the values of A and k.
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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.
(0, 8) exits thru top
(0, 1) exits thru right
approx. (0, 6) exits thru corner in right
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Sketch a curve which passes through the point (0, 4). At what point does this curve exit the box?
right
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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?
approx. (0, 6)
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Describe the solution curve which passes through the origin.
Has a constant slope = 0, runs along t axis.
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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'.
Intercept with P axis would be the init investment into the account, while the coord at which the curve leaves the box will state that whatever P is when box is exited is your total amount in the account at whatever the t value is.
Ex.) for the coord. (6, 12) this means at the 6 yr mark your Principle is worth $12,000
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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) = - m and g(t) = m.
For dP/dt = k P + m, this can be written as P’ = kP + m, where k is a fn of P and m is a fn of t. Which is saying the same thing as y ' = p(t) y + g(t) except 2nd variable is in terms of P instead of y. After that it’s easy to see p(t) = -k because it must be moved to left side while g(t) stays put giving us a +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?
dP/dt = k P + m
P’ - kP = m, integral constant = e^(-k dt) = e^(-kt) mult thru
e^(-kt)P’ - ke^(-kt)P = e^(-kt)m
(e^(-kt)*P)’ = e^(-kt)P’ - ke^(-kt)P, therefore
(e^(-kt)*P)’ = e^(-kt)m, taking integrals of both sides
e^(-kt)*P = Int(e^(-kt)m) = (e^(-kt)m/-k) + C, dividing thru by e^(-kt)
P = (m/-k) + Ce^(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?
(q(t)/V)*r, this would be outflow
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What therefore is the net rate at which the amount of salt in the second bottle is changing?
inflow - outflow
.1r - ((q(t)/V)*r)
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Write this as a differential equation and solve the equation.
dQ/dt = .1r - ((q(t)/V)*r)
Q’ + (r/V)Q = .1r, integration constant is e^((r/V)t) we get
Qe^((r/V)t) = Int( .1re^((r/V)t) = .1r((V/r)e^((r/V)t) + C = .1Ve^((r/V)t) + C, dividing thru by e^((r/V)t)
Q(t) = .1V + Ce^((r/V)t)
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There should be a - sign on the exponent:
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).
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Find the particular solution for which r = 7 cm^3 / sec.
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Self-critique (if necessary):
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Self-critique rating:
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&#Good work. See my notes and let me know if you have questions. &#