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course Phy 279
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q_a_07________________________________________
110214
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Question: `q001. Solve the equation
dP/dt = .02 P ( 1 - P / 100 )
for P(0) = 10.
What is the limiting value of P, as t increases?
The solution is given in the notes for 110214, but you should work this out yourself without reference to the notes.
Determine how long it would take this function to reach 50% of its limiting value, and how long it would then take to get halfway from that value to its limiting value.
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Your solution:
dP/dt = .02 P ( 1 - P / 100 )
dP/dt = .0002P(100 - P)
dP/ P(100 - P) = .0002dt
Int(1/ P(100 - P) dP) , using partial fractions
1/P(100-P) = A/P + B/100-P = (A(100-P)+ BP)/P(100-P)
100A - AP + BP = 100A + P(B - A) = 1
No P in the numerator so P(B - A) = 0
100A must equal 1, so A = .01, going back to P(B - A) having A = .01 so
B = .01 so
Int(dP/(P(100 - P)) = Int((-.01/P) + (.01/100 - P) dP)
Int( (-.01/P) dP)n = -.01ln|P|
Int((.01/100 - P) dP)n = .01ln|100 - P|
Int(1/ P(100 - P) dP) = -.01ln|P| + .01ln|100 - P| = .01(ln|100 - P| - ln|P|) = .01ln|P/(100-P)|
.01ln|P/(100-P)| = .0002t
ln|P/(100-P)| = .02t + C
P/(100-P) = e^(.02t + C) = Ae^(.02t)
1 = Ae^(.02t)/(P/100 - P) = (100Ae^(.02t)) - Ae^(.02t))/P
APe^(.02t))/P = 100Ae^(.02t)
P(Ae^(.02t)) = 100Ae^(.02t)
P = 100Ae^(.02t)/( Ae^(.02t))
For P(0) = 10
10 = 100A/A
A = .1
P(t) = P = 10e^(.02t)/( .1e^(.02t))
????This doesn’t look right all the other problems like this I’ve answered so I more than likely messed something up????????
I found it, at this point I messed up
1 = Ae^(.02t)/(P/100 - P) = (100Ae^(.02t)) - Ae^(.02t))/P
Really is
P = Ae^(.02t)*(100 - P) = Ae^(.02t) - APe^(.02t)
P + APe^(.02t) = P(1 + Ae^(.002t))
P = Ae^(.02t)/ (1 + Ae^(.02t))
For P(0) = 10
10 = A/ (1 + A)
@& 10 = A / (1 + A) means that
10 ( 1 + A) = A,
which is easily solved to get A = 1/9.*@
????I don’t know where I’m going wrong?????????
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Question: `q002. Solve the equation
dP/dt = .02 P ( 1 - P / 100 ) + 3
for P(0) = 10
and explore how the solution differs from the solution to the preceding problem.
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Your solution:
Well everything is the same up till…
dP/dt = .02 P ( 1 - P / 100 ) + 3
dP/dt = .0002P(100 - P) + 3
dP/ P(100 - P) = 3.0002 dt
@& Again a = b + c doesn't give you a / b = c.*@
Int(1/ P(100 - P) dP) , using partial fractions
1/P(100-P) = A/P + B/100-P = (A(100-P)+ BP)/P(100-P)
100A - AP + BP = 100A + P(B - A) = 1
No P in the numerator so P(B - A) = 0
100A must equal 1, so A = .01, going back to P(B - A) having A = .01 so
B = .01 so
Int(dP/(P(100 - P)) = Int((-.01/P) + (.01/100 - P) dP)
Int( (-.01/P) dP)n = -.01ln|P|
Int((.01/100 - P) dP)n = .01ln|100 - P|
Int(1/ P(100 - P) dP) = -.01ln|P| + .01ln|100 - P| = .01(ln|100 - P| - ln|P|) = .01ln|P/(100-P)|
.01ln|P/(100-P)| = 3.0002t
ln|P/(100-P)| = 3.02t + C
P/(100-P) = e^(3.02t + C) = Ae^(3.02t)
P = Ae^(3.02t)*(100-P) = 100Ae^(3.02t) - P Ae^(3.02t)
P + P Ae^(3.02t) = Ae^(3.02t)
P(1 + Ae^(3.02t)) = Ae^(3.02t)
P = Ae^(3.02t)/( (1 + Ae^(3.02t)))
Fpr P(0) = 10
10 = A/(1+A)
10 + 10A = A
10A = -10
A = -1
P(t) = -e^(3.02t)/( (1 - e^(3.02t)))
????My solutions aren’t making sense for the situation we are observing. I believe I’m on the right track but something is not right. But by adding 3 wouldn’t that increase the exponent of e increasing the rate at which the population approaches the living capacity???????
@& You changed the equation in about the third step.
You're doing just about everything right, but after that step you're solving a different equation.
Considering that, of course, it's a good thing your solutions aren't making sense.*@
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Question: `q003. If an object of mass m experiences a drag force equal to k * v when it is moving at velocity v, and if the only other force acting on the object is a constant force F_applied, then what differential equation governs its velocity as a function of time? What is the solution of this equation?
What is the limiting value of this solution as t approaches infinity? Can you explain why this makes sense?
Can you make sense of the following statement?
'The rate at which velocity changes is proportional to the difference between the velocity and the limiting velocity, so the solution is expected to be an exponential function of this difference'.
Also, the statement could use specific wording. Can you provide it?
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Your solution:
F_net = -k*v + F_applied
mv’ = F_applied - k*v
????In class you divided mv’ by F_applied - k*v. Why cant you just subtract F_applied - k*v to get it on the other side? is it because when you would try and take the integral on right side you wouldn’t be able to take the integral of 0?????????
mv’/( F_applied - k*v) = 1
Int(mdv/( F_applied - k*v)) = Int( dt)
Int(m/ ( F_applied - k*v) dv) = m*Int(1/ ( F_applied - k*v) dv) , using partial fractions
1/k( F_applied/k -v) = A/k + B/( F_applied/k - v) = (A(F_applied/k - v))+ Bk)/ k( F_applied/k -v)
A(F_applied/k ) - Av + Bk = 1
????Having a problem completing the Partial Fractions process with in this form??????
@& You could get the form
m v '' + k v = F_applied.
That's a second-order nonhomogeneous equation with constant coefficients, and could be solved using methods we have just developed in relation to Chapter 4. Let v = e^r and solve the homogeneous equation, then use variation of parameters.
However at this point you need to go with the fact that the equation is separable.
You can't factor that denominator any further. Partial fractions requires that you factor into linear and quadratic factors, and it's already linear.
Integrate the left-hand side by substitution. Let u = F_applied - k v so that du = - k dv.*@
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Question: `q004. If an object of mass m experiences a drag force equal to k * v^2 when it is moving at velocity v, and if the only other force acting on the object is a constant force F_applied, then what differential equation governs its velocity as a function of time?
What is the solution of this equation?
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Your solution:
mv’ = F_a - kv^2
dv/(F_a - kv^2) = dt
Int(1/(F_a-kv^2) dv), using partial fractions
Let (F_a - kv^2) = F_a(1 - (k/F_a)v^2)
1/(1 - (k/F_a)v)(1 + (k/F_a)v) = A/(1 - (k/F_a)v) + B/(1 + (k/F_a)v)
=A(1 + (k/F_a)v) + B(1 - (k/F_a)v)/( 1 + (k/F_a)v) (1 + (k/F_a)v)
A(1 + (k/F_a)v) + B(1 - (k/F_a)v) = A+B + (k/F_a)(A-B), (k/F_a) = 0
A = .5 and B = .5
Int(1/(1-(k/F_a)v^2) dv) = Int(.5/(1+(k/F_A)v) + .5/(1 - (k/F_a)v) dv) = .5ln|F_A - kv| + .5ln|F_a + kv|
@& When you integrate .5 / (1 - (k/F_a) v) you get -.5 ln | 1 - (k/F_a) v |.
So your integral becomes
.5 ln | (1 + (k/F_a) v) / (1 - (k/F_a) v) |
This changes things, and you should work it out accordingly.*@
=.5ln|F_a^2 - (kv)^2|
.5ln|F_a^2 - (kv)^2| = F_a*t + c
ln|F_a^2 - (kv)^2| = 2F_a*t + C
F_a^2 - (kv)^2 = Ae^(2F_a*t)
-(kv)^2 = Ae^(2F_a*t) - F_a^2
-k^2v^2 = Ae^(2F_a*t) - F_a^2
v^2 = Ae^(2F_a*t) - F_a^2 + k^2
v(t) = `sqrt(Ae^(2F_a*t) - F_a^2 + k^2)
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Question:
`q005. If velocity v is a function v(t) of clock time t, then acceleration is dv/dt.
If velocity is a function v(x) of its position x, then acceleration, being dv/dt, would be expressed (using the chain rule) as
dv/dt = dv/dx * dx/dt.
In terms of motion, what is the meaning of dx/dt?
What therefore is another way to write dx/dt?
How can we therefore express dv/dt in terms of just the two variables v and t?
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Your solution:
dx/dt is the change of position wrt the change in time.
This could also be considered the velocity .
dx/dt = v and dv/dt = v’(t).
????I went back and saw my mistakes but didn’t want to correct them so that you might comment on what I’m seeing incorrectly.???
@& You're pretty much got it, but I'm not sure you put it together:
By the chain rule dv/dt = dv/dx * dx/dt, so
dv/dt = v * dv/dx.*@
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Question: `q006. If an object of constant mass m is subject to a net force proportional to the inverse square of its position r, then we have
F_net = k / r^2.
By Newton's Second Law this is equal to m a, where a is the second derivative with respect to time of the position function. Thus
m r '' = k / r^2.
Is this equation linear? Is it first-order? Can you see how to solve it?
Now r ' is just the velocity v of the object, so r '' is just dv/dt. So our equation could be written
m dv/dt = k / r^2.
This looks good, because now appears to be a first-order equation, though still nonlinear. However it's not as simple as it appears. There are three variables, v, r and t, involved in this equation.
We can get around this. v ' means dv/dt. If we regard v as a function of r, then we have
dv/dt = dv/dr dr/dt
and since v is the derivative dr/dt of the position function, we have
dv/dt = v dv/dr.
Replacing dv/dt with v dv/dr we get the equation
m v dv/dr = k / r^2.
Now we have only the variables r and v, and we recognize the equation as a separable first-order equation.
Solve this equation for m = 1, k = -5 and initial condition v(r_0) = v_0, where r_0 = 100 and v_0 = 0.
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Your solution:
Not completely understanding what’s going on here.
@& By Newton's Second Law this is equal to m a, where a is the second derivative with respect to time of the position function. Thus
m r '' = k / r^2.
Is this equation linear? Is it first-order? Can you see how to solve it?
Now r ' is just the velocity v of the object, so r '' is just dv/dt. So our equation could be written
m dv/dt = k / r^2.
This looks good, because now appears to be a first-order equation, though still nonlinear. However it's not as simple as it appears. There are three variables, v, r and t, involved in this equation.
We can get around this. v ' means dv/dt. If we regard v as a function of r, then we have
dv/dt = dv/dr dr/dt
and since v is the derivative dr/dt of the position function, we have
dv/dt = v dv/dr.
Replacing dv/dt with v dv/dr we get the equation
m v dv/dr = k / r^2.
Now we have only the variables r and v, and we recognize the equation as a separable first-order equation.
Solve this equation for k = -5, m = 1 and initial condition v(r_0) = v_0, where r_0 = 100 and v_0 = 0.
Our equation becomes
m v dv = k / r^2 dr.
Integrating we get
m v^2 / 2 = - k / r + c.
(Physics students will note that the left-hand side is the KE of the object, and might recognize the right-hand side as a measure of potential energy).
So
v(r) = -(2 k / m) / r + c.
If k = -5 our initial condition is
0 = (10 / 1) / 100 + c
so that c = -1 / 10.
Our function is therefore
v = -10 / r - 1/10.
*@
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03-06-2011
03-06-2011