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course Mth 279
2/22
q_a_04prior 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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according to the equation dP/dt = .06 P
Slope is dP/dt for a graph of P and t
Therefore if dP/dt = .06 P and P = 2000,
Then dP/dt = .06 P = .06 (2000) = 120
dP/dt = 120
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Do the slopes here appear to match the slope you just calculated?
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The slopes on the graph appear to have been drawn by hand, without grid paper,
its hard to tell what the slope of the lines on the graph above are.
Although, using a graphing calculator I can plug in y = 120x
setting the window accuratly, with the same dimmensions as our graph,
I get a line similar to that of those at 2000.
The reason why I think this worked is the formula for slopes being y=mx+b
where m is the slope, and since I pluged in 120 they so appear to match.
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The graph is accompanied by a scale.
Any of the segments corresponding to P = 2000 has a run of 2 seconds. It is possible on the vertical scale of the graph to estimate the rise of a segment.
That should be the basis for an answer to this question.
It is certainly useful to scale the graphing calculator and match the result to the graph, but that would provide a confirmation of your calculation as opposed to being a justifcation for your result.
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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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by following the path of the direction feild for the function P(t) = 2000 e^(.06 t)
I would estimate that P(3) = 2300
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What do you get if you evaluate P(3)? Is this consistent with your estimate?
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P(t) = 2000 e^(.06 t)
P(3) = P(3) = 2000 e^(.06 (3)) = 2000 e^(0.18) = 2394
This is fairly accurate with my estimation showing that the direction feild works for this solution.
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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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According to the direction feild it appears to exit through the right side, but its close.
I could further varify this by using a graphing calculator set to the same window settings.
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Check this out by evaluating P(12). What is your conclusion?
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P(t) = 2000 e^(.06 t)
P(t) = 2000 e^(.06 (12)) = 2000 e^(0.72) = 4109
This number shows that it would exit the box through the right side.
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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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By looking at the graph it appears that the rise of the line segment is about 400 or 500.
And with the run being 2, the slope would be rise/run = 500/2 = 250
or 400/2 = 200
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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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Scaling of the graph can account for this, because its not square. If we were to make each of the sides equal meansurements,
then the graphed slopes would look highly more accurate.
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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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After constructing a graph and direction feild of this, I can use the function P(t) = 10 000 e^(0.06t),
and estimate that P(11.5) would get a double value of 20 000 from the previous P(0) = 10 000.
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The point is not to use the function P(t) = 10 000 e^(.06 t) as a basis for your estimate. The point is to use your constructed direction field as the basis.
In this simple example you can solve the equation and the direction field would in fact be pointless in this situaiton. However if the equation was, say
P(t) = .06 sqrt(t^3 - t) / | 2+ cos(t) |
you would not be able to solve the equation. You would, however, be able to construct a direction field and estimate solution curves.
The reason for using an easily-solved equation here is to allow you to compare your estimates to the actual solution, not to replace the process of estimation.
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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)
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You should have written down a differential equation. That equation is linear. Find its general solution.
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The differential equation T ' (t) = k * (T(t) - T_room) is in the form of dP/dt = k*P
and a general solution would be T(t) -T_room = C * e^(kt)
which would be T(t) = C * e^(kt) + T_room
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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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using these 2 functions we can plug in all the information and get a time model.
T ' (t) = k * (T(t) - T_room) and T(t) = C * e^(kt) + T_room
first using, T(t) = C * e^(kt) + T_room
T(0) = C * e^(k*0) + 20 = 50
T(t) = C + 20 = 50
C = 30
then using the original equation T ' (t) = k * (T(t) - T_room)
T ' (0) = k * (T(0) - 20) = -5
T ' (0) = k * (30 - 20) = -5
T ' (0) = k * (10) = -5
k = -0.5
This information gives us the funtion
T(t) = 30 e^(-0.5t) + 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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We cannot solve this equation using the techniques we have been applying to first-order linear equations,
because this equation is first-order, but not linear.
The y^2 makes it not linear, and therefore harder to solve.
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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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dy / y^2 = -.01* dt
First I reconize this form to be easier for me to see how to solve.
y^(-2) dy = -.01* dt
int (y^(-2) dy) = int (-0.01* dt)
-y^(-1) = -0.01t + c
1/y = 0.01t - c
1 = (0.01t - c) * y
y = 1/(0.01t - 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.
Integrate both sides of the equation to get a general solution.
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dy/dt = .01 t y^2
dy/y^2 = .01 t dt
int (dy/y^2) = int (0.01 t dt)
int (y^(-2)dy) = 0.01 * int (t dt)
-y^(-1) = 0.01 (1/2)t^2 + c
1/y = -0.005 t^2 - c
y = -1/(0.005 t^2 + c)
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Find the particular solution for which y = 25 when t = 0.
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y = -1/(0.005 t^2 + c)
y = 25, t = 0.
25 = -1/(0.005 (0)^2 + c)
25 = -1/c
c = -1/25
y = -1/(0.005 t^2 - (1/25))
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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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The solution to this equation is undefined when 0.005t^2 - (1/25) = 0
because we cannot divide by zero.
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Show that there exist values of t for which your solution is neither defined nor continuous.
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0.005t^2 - (1/25) = 0
0.005t^2 = (1/25)
t^2 = 8
Therefore the funtion is undefined when t = + or - sqrt(8)
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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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The examples showed that the general solution to the form y ' + f(t, y) = 0 would always be a form of y = -1/(k t^2/2 + c)
If the bottom of the equation were to equal zero then the function would be undefined, because you cannot divide by zero.
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The point is made for this example specifically because the f(t, y) function is defined and continuous for all values of t, while the solution to y ' + f(t, y) = 0 i snot.
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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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a(t) = k * v(t)^2
since a(t) = dv/dt which is acceration.
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Solve the equation.
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a(t) = k * v(t)^2
dv/dt = k * v^2
dv/v^2 = k dt
v^(-2) dv = k dt
int (v^(-2) dv) = int (k dt)
-v^(-1) = kt + c
1/v = -kt - c
v = 1/(-kt - c) or v = -1/(kt + c)
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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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first we need these equations to solve this:
a(t) = k * v(t)^2 and v(t) = -1/(kt + c)
and v(0) = 30,
a(0) = 1000
v(t) = -1/(kt + c)
v(0) = -1/(k(0) + c) = 30
v(0) = -1/(c) = 30
c = -1/30
a(t) = k * v(t)^2
a(0) = k * v(0)^2
1000 = k * 30^2
1000 = k * 900
k = 10/9
This information gives us the equation
v(t) = -1/((10/9)t - (1/30))
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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, see http://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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H(t, y) = t cos y + 1/y
= t cos(f(t)) + 1/f(t)
d/dt (t cos(f(t)) + 1/f(t)) = (-f '(t) / f(t)^2) - t * f '(t)sin(f(t)) + cos(f(t))
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Good. That shows specifically how the y ' terms in the following arise.
Since f(t) and f ' (t) stand for y and y ' respectively, we would then write this as
-t sin(y) y ' + cos(y) - (1/y^2) y '
where the y ' arises by the chain rule whenever we take a derivative of 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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This is clearly the derivative form:
d/dt ( f(t) * g(t) ) = f '(t)g(t) + f(t)g'(t)
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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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If H(t, y) = t^2 sqrt(y) and,
df/dt = 2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y '
Then these appear to be in that form.
and if one equals zero than so does the other because they are equal.
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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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These are equivilant because the derivative of a constant equals zero.
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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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c = t^2 sqrt(y(t))
sqrt(y(t)) = (c/t^2)
y(t) = (c/t^2)^2
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This could then be written as
c / y^4, c > 0.
The c of this equation is the square of the c of the preceding, which makes it positive. (That c would also have been positive anyway because the square root has to be positive).
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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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2 t sqrt(y(t)) + t^2 / (2 sqrt(y(t)) ) * y '(t)= 0
2 t sqrt((c/t^2)^2) + t^2 / (2 sqrt((c/t^2)^2) ) * y '(t) = 0
y'(t) = -c^2 (1/3)t
2 t sqrt((c/t^2)^2) + t^2 / (2 sqrt((c/t^2)^2) ) * (-c^2 (1/3)t) = 0
kinda confused now
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You are using y = c^2 / t^4, to put it in simplified form.
So y ' = -4 c^2 / t^3, and sqrt(y) = c / t^2.
If you plug these expressions into
2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y '
you get
2 t c / t^2 + t^2 / (2 c/t^2) * (-4 c^2 / t^3)
I believe this simplifies to zero.
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&#Good responses. See my notes and let me know if you have questions. &#