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course M 279
2-2 1Partial assignment
For P = 2000, then what should be the slope of the direction field, according to the equation dP/dt = .06 P?
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dP/dt = 120
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Do the slopes here appear to match the slope you just calculated?
The slope you should have obtained was over 100. The slopes on this direction field all appear to be less than 1. We need to either reconcile this, or discard the direction field of the figure.
Let's first check out a solution to the problem. As you can easily verify, dP/dt = k P gives us general solution P(t) = C e^(k t). To get the solution passing through the P axis at 2000, we apply the initial condition P(0) = 2000. Our particular solution for this initial condition is P(t) = 2000 e^(k t). Our direction field is for k = .06; for P(0) = 2000 our function is therefore P(t) = 2000 e^(.06 t). A graph of this function is consistent with the direction field.
Verify this. Following the trend of the direction field, estimate the likely value of P(3). What is your estimate?
2,500
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What do you get if you evaluate P(3)? Is this consistent with your estimate?
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approx. 2,394, yes is consistent
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According to the direction field, does the solution curve through (0, 2000) exit the box to the right or through the top?
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right
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Check this out by evaluating P(12). What is your conclusion?
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Correct. It is easy for me to visualize that if you start at 2,500 and follow lines you hit top right corner.
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Reasonable answers to these questions will tend to reinforce our confidence in the direction field.
Consider the direction field line through (8, 4000). Based on the scale of your graph, estimate the rise of that segment. It should be obvious that the run of the segment is 2. What therefore is the slope represented by that segment?
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approx. 400/2 which equals 200 or .2 if we scale down.
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The slope appears to be less than 1. Explain why that segment actually represents a slope in the hundreds.
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slope = rise/run = 400/2 = 200
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Construct the slope field for the equation dP/dt = .06 P, for the interval 10 000 <= P <= 20 000, 0 <= t <= 12. Scale your slope segments appropriately.
Based on your slope field, at what value of t do you estimate the value of P(t) will double, starting with P = 10 000 at t = 0?
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P = 333,333.3333 approx.
dP/dt = .06P
2 = .06P
P = 2/.06 = 33.3333, where P is on scale of 10,000
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`q005 (from 110131). 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.
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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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`q006 (from 110131). 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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Express your preceding result as a differential equation, find the general solution of the equation, and find the particular solution for which r = 7 cm^3 / sec.
Question `q001 (non-italicized), for the 2/02/11 class, was given previously. Subsequent questions are all non-italicized and are for the 2/02/11 class only.
`q002. A quantity y is directly proportional to a quantity x if there exists a constant k such that y = k x.
The rate, with respect to time, at which the temperature of an object left to cool in a room at temperature T_room is directly proportional to the difference between the temperature of the object, which we will represent by T(t), and the temperature of the room.
Letting T ' (t) stand for the rate at which the temperature T(t) changes with respect to clock time, write down the proportionality.
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T ' (t) = -k(T(t) - T_room)
Because as temp of object approaches temp of room rate of temp change is less and less. When object reaches temp of room unless another factor is involved temp change stops. k is neg because room is cool.
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You should have written down a differential equation. That equation is linear. Find its general solution.
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T ' (t) = -k(T(t) - T_room)
T ' (t)/ (T(t) - T_room) = -k
Int(T ' (t)/ (T(t) - T_room)) = Int(-k T`(t) )
ln|(T(t) - T_room)| = -kt +c
T(t) - T_room = e^(-kt +c) = e^(-kt)*e^(c) = A e^(-kt)
T(t) = T_room + A e^(-kt)
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If the object has an initial temperature of 50 Celsius, and its temperature initially changes at -5 Celsius per minute when placed in a room at temperature 20 Celsius, then what is the specific function that models its temperature as a function of time?
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50 = 20 + A e^(-k*0)
30 = A
Knowing temp changes at -5C/min, then at t=1 T(t) = 45
45 = 20 + 30e^(-k*1)
25 = 30e^(-k)
ln|25/30| = -k
k = approx. .1823
So we have
T(t) = 20 + 30e^(-.1823t)
@& The rate does not remain -5 for the first minute. The rate begins to decrease immediately, so the temperature at t = 1 is greater than 45.
All we know is the instantaneous rate. Fortunately that is the derivative:
T ' (t) = -30 k e^(-k t),
so
T ' (0) = -30 k e^(-k * 0)
and with T ' (0) = -5
-5 = -30 k.
Thus k = 1/6 and our solution is
T (t) = 30 e^(- t / 6) + 20.
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`q003. Consider the equation dy/dt = -.01 y^2.
Explain why we cannot solve this equation using the techniques we have been applying to first-order linear equations.
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Its not a linear equation because of the y^2
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We will see why shortly, but for now accept that this equation can be rearranged into the form
dy / y^2 = -.01 dt
and integrate both sides to get a general solution.
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Int( 1/y^2 dy)
Int(y^-2 dy) = -1/y
Int(-.01 dt)
-.01t
dy / y^2 = -.01 dt
-1/y = -.01t +c
-1 = (-.01t + c)*y
y = 1/(.01t + c)
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`q004. Consider the equation dy/dt = -.01 t y^2. Algebraically rearrange this equation so that dy is a factor of one side, with all other y factors on that side, and dt, along with all other t factors, is on the other.
Integration both sides of the equation to get a general solution.
1/y = -.01 t + c
@& The problem should have been dy/dt = .01 t y
The integral of the right-hand side .01 t is 0.005 t^2.*@
y = 1 / (-.01 t + c)
Find the particular solution for which y = 25 when t = 0.
Show that the function f(t, y) = -.01 t y^2 is defined and continuous for all values of y and t.
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25 = 1/c
c= 1/25
@& Any number t can be multiplied by the square of any other number y, so the function is defined.
Continuity is obvious but would be tedious to prove, so we'll skip that for now.
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Show that there exist values of t for which your solution is neither defined nor continuous.
@& The denominator of the function
y = 1 / (-.005 t^2 + .04)
is -.005 t^2 + .04, which is zero when t = sqrt(8) = 2 sqrt(2).
This makes the function undefined, and therefore discontinuous, at t = 2 sqrt(2).
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Whenever (-.01 t + c) = 0 fn is neither defined nor continuous
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@& Good work, but once more be sure to see my notes.*@