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course Mth 279
2-19 2
q_a_04________________________________________
prior to class 110202
`q001. The direction field shown in the figure below applies to the equation dP/dt = k P, with k = .06.
For P = 2000, 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?
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yes
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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?
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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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approx. 400/2 which equals 200 or .2 if we scale down.
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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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@& Good.
Compare to the following, which also has links to a few graphs.
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.
An approximate curve, somewhat below the curve starting at (0, 2000), is depicted below
(link to picture is http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/differential_equations/q_a_041.gif ):
Another approximate curve, starting at (2000, 0), has been added. According to this curve it appears P(3) is about 2300 (link to picture at http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/differential_equations/q_a_042.gif ).
Plugging into the function we find that the solution is about P(3) = 2400.
The curve appears to come pretty close to the upper right-hand corner of the box.
P(12) = 2000 * e^(.06 * 12) = 4100, approx..
The upper right-hand corner is at (12, 5000), so the curve does exit to the right.
Projection lines from the ends of the segment indicate a rise of perhaps 300 and a run of 2, giving the 'slope line' a slope of 300/2 = 150 on the scale of the graph (link at http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/differential_equations/q_a_043.gif )
The vertical axis has a scale which is hundreds of times that of the horizontal axis. The slope represented by a segment must be determined by the rise and run, on the scale of the graph.
The figure below sketches a thin blue solution curve through (0, 1000-). That curve appears reach value 20000 around t = 10, so the graph below estimates a doubling in about 10 years. (link at http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/differential_equations/q_a_044.jpg )
If we work it out, the doubling time turns out to be about 20 years.
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`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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@& Good. The solution below includes the solutions for the initial conditions:
The difference between the temperatures is T - T_room.
The proportionality is therefore
T ' (t) = k ( T - T_room)
The equation can be written as
T ' - k T = T_room
This equation is first-order linear nonhomogenous.
The integrating factor e^(- k t) puts the equation into the form
(T e^(-k t) ) ' = T_room e^(-k t) + C
with solution
T = T_room + C e^(k t)
If T(0) = 50 Celsius and T_room = 20 Celsius then
T(0) = 20 Celsius + C e^(k * 0) = 50 Celsius, so
20 Celsius + C * 1 = 50 Celsius and
C = 30 Celsius.
Thus
T = 20 Celsius + 30 Celsius * e^(k t).
Thus T ' = k * 30 Celsius * e^(k t), and the condition
T ' (0) = -5 Celsius / minute
gives us
k * 30 Celsius * e^(k * 0) = -5 Celsius / minute
so that
k = -5 Celsius / minute / (30 Celsius) = -.167 / minute.
Our function is therefore
T (t) = 20 Celsius + 30 Celsius * e^(-.167 / minute * t).
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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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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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The term y^2 is not linear. The equation is of the form y ' + .01 y^2 = 0, which is y ' + p(t) y^2 = 0 with p(t) = .01.
This is not of the first-order linear form y ' + p(t) y = g(t).
Integrating, we get
ln | y | = 0.01 t + c,
so that
| y | = e^(.01 t + c) = e^c * e^(.01 t).
e^c is positive for any value of c, and for the appropriate value of c the expression e^c can equal any positive number, so we will denote our solution as
| y | = A e^(.01 t), A > 0.
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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.
Integrate both sides of the equation to get a general solution.
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dy/.01y^2 = t dt
.01*(-1/2y) = t^2/2 + C
you wouldn't get the 2 y. You've been watching me too long.
1/y = -200t^2/2 + C = -100t^2 + C
y = -.01/(t^2 + C)
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Find the particular solution for which y = 25 when t = 0.
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For
-.01/(0^2 + C) = -.01/C = 25
C = -2500
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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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continuous because .01 t y^2 is continuous on entire ty plane as is the partial derivative of .01 t y^2 which is .02ty therefore the conditions are meet open rec. R defined by -infin
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Show that there exist values of t for which your solution is neither defined nor continuous.
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For
y = -.01/(t^2 - 2500), whenever t = +-50 we have
y = -.01/(-50^2 - 2500) = -.01/(2500 - 2500) = -.01/0, which is undefined and would
be the same result we would have gotten for
t=50
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We get
dy/y^2 = .01 t dt.
Integrating both sides we get
-1 / y = .005 t^2 + c, so that
y = -1 / (.005 t^2 + c)
y(0) = 1/(-0.005 * 0^2 + c).
y(0) = 25
Thus 1 / c = 25 and c = 1/25 = .04.
Any value of t and any value of y give a real-number value for .01 t y^2.
The y and t derivatives are respectively .02 y t and .01 y^2, both continuous functions.
The function
y = 1 / (-.005 t^2 + .04)
is undefined when the denominator is zero, i.e., when
-.005 t^2 + .04 = 0. Solving for t we get
t = +- sqrt(-.04 / (-.005)) = +- sqrt(8), about +- 2.818.
The equation is of the form y ' + f(t, y) = 0, with f(t, y) = .01 t y^2, a function which is defined and continuous for all y and all t. If the theorem held for all continuous functions f(t, y), then it would apply to the case given here.
The solution shown above is not defined for t = +- sqrt(8), so clearly the theorem does not apply to this example.
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While a linear equation of the form y ' + p(t) y = g(t) always has a solution which is defined and continuous, on any interval for which p(t) and g(t) are defined and continuous, this example shows that the same is not necessarily so for an equation of the form y ' + f(t, y) = 0. Explain this statement.
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For the form y ' + p(t) y = g(t) we are dealing with first order linear equations where y ' + f(t, y) = 0 are of the form first order nonlinear equations.
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`q005. The rate, with respect to time, at which the velocity of a spherical object moving through water changes is directly proportional to the square of its velocity.
Write this as a differential equation.
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dv/dt = k*v^2
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Solve the equation.
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dv/dt = k*v^2
dv/v^2 = k dt
Int(dv/v^2) = Int(k dt)
Int(1/v^2 dv) = Int(v^-2 dv) = -1/v
Int(k dt) = kt + c
-1/v = kt
-1 = v*(kt +c)
v = -1/(kt+c)
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@& The equation would be
dv/dt = k v^2.
The equation can be separated to yield
dv/v^2 = k dt
with solution
-1/v = k t + c
so that
v(t) = 1 / (k t + c).
v(0) = 1 / (k * 0 + c) = 1 / c, and
v(0) = 30 cm/s so
1/c = 30 cm/s and
c = 1/(30 cm/s) = .0333... s / cm.
v ' (t) = -k / (k t + c)^2 = -k / (k t + .0333... s/cm) so
v ' (0) = -k / (.0333... s/cm)^2 = 1000 cm/s^2.
Thus
k = -1000 cm/s^2 * (.0333... s / cm)^2 = 1.11 ... cm^-1.
The velocity function is therefore
v(t) = 1 / (1.11... cm^-1 * t + .0333... s/cm).
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If the object is initially moving at 30 cm/s and changing velocity at the rate of 1000 cm/s^2, then what function describes its velocity?
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For dv/dt = k*v^2, where
dv/dt = 1000 cm/s^2
v^2 = (30)^2 = 900
We solve for k,
1000 = k*900
k = 10/9 = approx. 1.1111
dv/dt = 1.1111*v^2 and
v = -1/(1.1111t + c)
?????Im not sure this is correct??????
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`q006. Recall implicit differentiation. If you need a review of implicit differentiation see the q_a_ document at http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa16.htm . If you need a review of the chain rule, seehttp://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa12.htm and http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa13.htm .
If y is a function of t, then what is the derivative with respect to t of the function H(t, y) = t cos y + 1/y?
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For H(t, y) = t cos y + 1/y,
(tcos(y) + 1/y) = tcos(y) + (1/y)
tcos(y) = t*cos(y) + t*cos(y)
= cos(y) - t*sin(y)*y
(1/y) = (-1/y^(-2))y
H(t, y) = cos(y) - t*sin(y)*y - (-1/y^(-2))y
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`q007. Verify that the derivative of the function H(t, y) = t^2 sqrt(y) is 2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y '.
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(t^2 sqrt(y)) ' = (t^2)*`sqrt(y) + t^2 (sqrt(y))
= 2t`sqrt(y) + t^2*1/(2*sqrt(y)), so
(t^2 sqrt(y)) ' = 2t`sqrt(y) + t^2*1/(2*sqrt(y))*y
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Verify that the equation
2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0
is of the form
( H(t, y) ) ' = 0,
where H(t, y) = t^2 sqrt(y) and ' indicates the derivative with respect to t.
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We have 2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0, which is of the form
n(y)(dy/dt) + m(t) = 0
Where
n(y)(dy/dt) = t^2 / (2 sqrt(y) ) * y '
m(t) = 2 t sqrt(y)
Also h(t, y) = y, so (H(t, y)) is really just derivative of H(t, y) or h(t, y)
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Verify that the equation
( H(t, y) ) ' = 0
is equivalent to the equation
H(t, y) = c,
where c is any constant number.
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For n(y)(dy/dt) + m(t) = 0, we can solve initial value problems by reversing the chain rule.
n(y)(dy/dt) + m(t) = 0, we can let N(y) is any antideriv. of n(y)
d/dy(N(y)) = n(y), if y is differentiable fn of t then chain rule leads to
d/dt(N(y)) = (dN/dy)*(dy/dt)
=n(y)(dy/dt), this shows n(y)(dy/dt) can be written as dN/dt Note M(t) is antider. of m(t) then it to can be written as dM/dt
Now we have equation
n(y)(dy/dt) + m(t) =(d/dt) N(t) +(d/dt) M(t) = d/dt [N(y) + M(t)]
this equation reduces to, d/dt [N(y) + M(t)] = 0
term N(y)+M(t) is a fn of t(involving unknown soln y=y(t)] so,
N(y) + M(t) = C, where C is an arbitrary constant, where we can always determine C by imposing the initial cond y(t_0) = y_0. Finding
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C = N(y_0) + M(t_0)
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For the given function H(t, y) = t^2 sqrt(y), solve the equation
H(t, y) = c
for y as a function of t.
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H(t, y) = 2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0
y = -(2 t sqrt(y))/(t^2/(2`sqrt(y))) = -(4t`sqrt(y)^2)/t^2) = -4t^-1y = -4y/t
So we have y = -4y/t
y/4y + 1/t = 0, taking integrals we find
Int(y/4y ) = (1/4)Int(1/y dy) = (ln|y|)/4
Int(1/t dt ) = ln|t|
(1/4)ln|y| + ln|t| = C
Mult through by e^1
(1/4)y + t = e^(C), where C absorbs the e
Sol.ve for y
(1/4)y + t = A
y = 4(A - t)
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Show that this function y(t) satisfies the equation
2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0.
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y = -(2 t sqrt(y))/(t^2/(2`sqrt(y))) = -(4t`sqrt(y)^2)/t^2) = -4t^-1y = -4y/t
Int( 1/y dy) = ln|y|
Int(-4/t dt) = -4Int( 1/t dt ) = -4(ln|t| + C)
ln|y| = -4(ln|t| + C), mult through by e^1
y = -4(t + e^C) = -4(t +A)
y = 4(A - t)
???????????I messed up in my algebra somewhere but cant see where. Maybe because A is arbitrary it absorbs something I missed but for the two equations my constant A has opposite signs?????
Im pretty sure I understand the point you want us to see though.
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@& Check the solution in Class Notes for 110207, for comparison on that last question.
I've also included solutions for most of these questions, mostly for comparison, occasionally for some correction.*@